NEET

CHEMISTRY
MODULE - 7

Optical Isomerism

Haloalkanes and Haloarenes

Alcohols, Phenols & Ethers

Aldehydes, Ketones, & Carboxylic Acids

Amines

Biomolecules
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 INDEX
S. No. Chapter Name Page No.

1. Optical Isomerism

Theory and Exercise 1.1 to 1.2 1-12

Exercise 2 13-17
Exercise 3 18-19
Exercise 4 (Previous Year’s Questions) 20-20

Answer Key 21
2. Haloalkanes and Haloarenes

Theory and Exercise 1.1 to 1.3 23-40

Exercise 2 41-49
Exercise 3 50-53

Exercise 4 (Previous Year’s Questions) 54-56

Answer Key 57
3. Alcohols, Phenols & Ethers

Theory and Exercise 1.1 to 1.3 59-83

Exercise 2 84-91

Exercise 3 92-96
Exercise 4 (Previous Year’s Questions) 97-100

Answer Key 101

4. Aldehydes, Ketones, and Carboxylic Acids

Theory and Exercise 1.1 to 1.4 103-135

Exercise 2 136-148
Exercise 3 149-154

Exercise 4 (Previous Year’s Questions) 155-160

Answer Key 161

5. Amines

Theory and Exercise 1.1 to 1.3 163-180

Exercise 2 181-189
Exercise 3 190-194

Exercise 4 (Previous Year’s Questions) 195-198

Answer Key 199

6. Biomolecules

Theory and Exercise 1.1 to 1.3 201-222

Exercise 2 223-228
Exercise 3 229-234
Exercise 4 (Previous Year’s Questions) 235-237
Answer Key 238
Optical Isomerism CHEMISTRY

Chapter
OPTICAL
1 ISOMERISM

Introduction
Chapter Summary
Definition:
• Introduction
Compound having same molecular formula, same structural
• Symmetry elements formula but different stereo chemical formula with different
• Representation of optical behaviour towards plane polarised light (PPL) are called optical
isomers isomers and the phenomenon is known as optical isomerism.
• Configuration of optical • Behaviour of optical isomers can be predicted by polarimeter
isomers experiment.
• Calculation of number of Polarimeter experiment:
optical isomers
(A)

(B)

Polarimeter
(light (ordinary Nicol PPL Nicol prism
tube
source) light) prism (Anlyser)
(Polarizer) (filled with (C)
solution of
compound
under
observation)
Observation (A): Compound is optically active, rotating the
plane of PPL in clockwise direction. It is known as
dextrorotatory and is represented by d/+

Observation (B): Compound is optically active, rotating the

plane of PPL in anti clockwise direction. It is known as
laevorotatory and is represented by l/–.

Observation (C): This observation indicates that

(i) Compound is optically inactive and not an optical isomer.
Or

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(ii) Compound is optical isomer but is optically inactive, it is known as meso isomer.
Or
(iii) Polarimeter tube was filled with equimolecular mixture of d and l mirror images molecules, it is known
as racemic mixture and is represented by dl/±.
Note. The angle of rotation of plane of polarised light produced by solution depends on
(a) The concentration of solution (gm/ml) (C)
(b) The length of polarimeter tube (dm) (l)
(c) The temperature at the time of experiment
(d) The wave length of light.

Condition for optical Activity:

• For a molecule to be optically active, it must be chiral or asymmetrical
• A molecule is chiral or asymmetric, if it is nonsuperimposable on its mirror image
• For a molecule to be non superimposable on its mirror image, symmetry elements [plane of symmetry
(pos) and centre of symmetry (Cos)] must be absent.
Note. Chirality of a molecule does not depend on presence or absence of chiral -C in it. Chiral C is Sp3 C which
is bonded to four different atoms or groups (units). It is also called asymmetric carbon.
COOH
Ex.
H C Sp3 C with different units
H3C NH2

Symmetry elements
(1) Plane of symmetry (pos): An imaginary plane passing through a molecules which divides the molecule
in two such equal halves that each half of the molecule is mirror image of the other half

Me Me
H H
Ex. C C Pos
H H H H
Pos Pos
Pos
F Cl
Pos
F
H , C
Br H

H H
(2) Centre of symmetry (COS): An imaginary point present in a molecule in such a way that every straight
line passing through the point has similar units at its opposite ends present at the same distance from
that point.
If even a single line passing through the point has different units at its opposite ends, COS is considered
to be absent.
(different units)
H Cl H3C Cl
C=C C=C
Cl H H CH3
(COS present) (COS absent)

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(3) Axis of symmetry (AOS): If the new image obtained on rotating a molecule about an axis passing
through the molecule by an angle θ is similar to the previous image, the molecule has Cn axis of
symmetry where (n = 360/θ)
every molecule at least has C1 AOS, therefore this symmetry element does not decide optical activity
of molecule. For a molecule to be optically active both COS & POS must be absent

Representation of optical isomers

(1) Perspective formula (3–D)
(2) Wedge dash formula (3–D)
(3) Fischer-projection formula (2–D)

(1) Perspective formula (3–D): The molecule is visualised in such a way that only chiral-c is maintained in
plane and units on chiral-c are outside and inside.
Ex. CH3–CH(OH)–COOH HOOC–CH(OH)–CH(OH)–COOH
Lactic acid Tartaric acid
COOH COOH
H OH
C C
H OH C
CH3 H OH
COOH

(2) Wedge dash formula (3–D): The molecule is visualised in such a way that out of four units on chiral
carbon any two units along with chiral-c are maintained in a plane and among rest of the units, one is
out side and other is inside the plane.
H
HO COOH
COOH
Ex. OH H
H3C HOOC
H OH

(3) Fischer-projection formula (2–D) : It is a combination of horizontal and vertical lines. The molecule is
visualised in such a way that only chiral c is maintained in a plane and units out side the plane are on
horizontal line and inside the plane are on vertical line
COOH COOH
H OH
H OH H OH
COOH
CH3

Note. Units on horizontal line are outside the plane or towards the observer in actual 3–D view and units on
vertical line are inside the plane or away from the observer in actual 3–D view.

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Conversion of one fisher projection formula into another:
An optical isomer with one chiral-c can be represented by twelve different fisher projection formula,
so if we have one fisher projection formula then the rest of the eleven fisher projection formula can be
made with the help of this fisher projection formula, It is as follows

Technique–I : If a Fischer projection formula is rotated by 180° in the plane (not by lifting from the
plane) the new fisher projection formula represents same isomer
COOH 180° CH3

H OH HO H (same isomer)
In the plane

CH3 COOH

Technique-II : If in a fisher projection any one unit is kept fixed and other three are shifted clockwise
or anti clock wise, the new fisher projection formula represents the same isomer
COOH COOH
fixed

H OH H3C H (same isomer)

CH3 OH

Note: A compound which show optical isomerism essentially has d & l isomers (theoretically) but may or may
not have meso isomer
To have meso isomer following condition are required.
(1) Compound should have more than one chiral-c
(2) Compound should be symmetrical (in respect of IUPAC numbering)
* *
Ex. (i) HOOC − C(OH) − (H/ OH)− COOH
• Symmetrical compound
• Number of chiral-c is two
• Possible isomers (all three types d, l & meso)
* * *
(ii) H3 C − CH(Cl) − CH(Cl) − CH(Cl) − CH3
• Symmetrical compound
• Number of chiral-c is three
• Possible isomers (all three types d, l & meso)
• The centre most c is pseudo chiral -c
Pseudo chiral-c: If chirality of a carbon depends on configuration of units attached with it, is called
pseudo chiral-c.
* *
H3 C − CH(Cl) − CH(Cl) − CH(Cl) − CH3
Configuration S achiral S
R achiral R
R chiral S
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Meso isomer : An optically inactive isomer having more than one chiral-c with symmetry element (Pos
or cos) is called meso isomer
COOH
H * OH
Ex. Plane of symmetry (POS)
H * OH
COOH
• It is always achiral molecule since it is superimposable on its mirror image
• Mirror image of a meso isomer always represents identical molecule (Homomer)
• Optical inactivity of a meso isomer is due to internal compensation / Intramolecular neutralization
/molecular symmetry.

Racemic mixture : An equimolar mixture of d & l mirror image molecules is called racemic mixture.
• It is a resolvable mixture and separation of d & l isomers from the mixture is called optical resolution.
• Conversion of d or l form into a racemic mix is known as racemization.
• Optical inactivity of racemic mixture is due to external compensation/Intermolecular neutralization/
optical impurity.

Enantiomers : d & l mirror image compounds are called enantiomers of each other
COOH COOH
H OH HO H
HO H H OH
COOH COOH
• Enantiomers are always chiral molecules since these are non-super imposable on mirror images.
• Enantiomers have similar physical and chemical properties but different biological properties.
• Enantiomers rotate the plane of light in opposite direction by same angle.

Diastereomers : stereoisomers which are not a mirror image of each other are called diastereomers
COOH COOH
H OH HO H
HO H HO H
COOH COOH
Both optically active

• Diastereomers maybe Both optically inactive

One optically active & one optically inactive

• Physical properties of diastereomers like BP, MP solubility are different.
• Diastereomeric mixtures can be separated by using fractional distillation method.

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Exercise 1.1
1. Which of the following can show optical COOH
isomerism :
6. Meso-tartaric acid H OH is optically
(1) 1-Chlorobutane H OH
(2) t-Butyl chloride
COOH
(3) sec-Butyl chloride
(4) Isobutyl chloride inactive due to the presence of-
(1) Molecular symmetry
2. Which of the following compound can show (2) Molecular asymmetry
optical isomerism: (3) External compensation
(4) Two asymmetric carbon atoms
Me Me Cl D
(1) (2) H
H H 7. Which is optically active molecule-
H (1) C6H5—C—OH (2) CH3—CH—C2H5
C Br
(3) (4) All O OH
C Br
H (3) C6H5—CH—OH (4) C6H5—CH—CH3
H CH3
3. Which of the following statement is true for
the given compound:
OH 8. Correct statement about the compound
Me
H (i), (ii) and (iii) is -
H COOCH3 COOH
Me
HO (i) H OH (ii) H OH
(1) Optically active H H
OH OH
(2) Has plane of symmetry COOH COOCH3
(3) Has centre of symmetry COOH
(4) 1 & 3 both H OH
(iii)
HO H
4. Which of the following contain chiral carbon
COOCH3
atom:
(1) (i) & (ii) are identical
(1) CH3–CH–CH2–CH2–CH3
(2) (i) & (ii) are diastereomers
OH
(3) (i) & (iii) are enantiomers
(2) CH3–CH–CH–CH3
(4) (i) & (ii) are enantiomers
Br CH3
(3) CH3–CH2–CH–CH3
9. If optical rotation produced by the
Br
(4) All the above compound [A] is + 65°, then produced by the
compound [B] is –
5. Which of the following compound is a meso CH3 CH3
compound: H OH HO H
H Cl Br Cl HO H H OH
(1) (2)
CH3 CH3
Cl H H H [A] [B]
Cl Cl H Cl (1) + 65° (2) –65°
(3) (4)
(3) Zero (4) Unpredictable
H H Br H
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10. The molecules below are: 14. Which of the following is a chiral molecule
F H H CH3 CH3

H F (1) (2)
F H CH3
F CH3
(1) Constitutional isomers
(2) Enantiomers CH3 CH3
(3) Diastereomers
(4) Identical (3) (4)
CH3 H3 C
11. Which one of the following shows optical
activity 15. How many chiral center present in
H
tetracycline –
(1) HO—C—COOH CH3
H CH3
H OH H N
H OH
(2) CH3—C—COOH
C—NH2
Cl O OH
O
CH3
(1) 6 (2) 4
(3) CH3—C—COOH (3) 8 (4) 5
OH
CH3 16. Among the following which one can have a
(4) CH3—C—COOH meso form
(1) CH3CH(OH)CH(Cl)C2H5
Cl (2) CH3CH(OH)CH(OH)CH3
(3) C2H5CH(OH)CH(OH)CH3
12. Which one of the following may exist in pair (4) HOCH2CH(Cl)CH3
of enantiomers?
CH3 17. Which of the following compounds are meso
(1) CH3—CH—COOH forms?
(2) CH2 = CHCH2CH2CH3 CH3
CH3 CH3
NH2
H OH H Cl
(3) CH3—CH—CH3
H OH Cl H
NH2
CH3 CH3 CH3
(4) CH3—CH2—CH—CH3 1 2
3

(1) 1 Only (2) 1 and 3

13. Which of the following compounds may not (3) 1 and 2 (4) 2 and 3
exist as enantiomers?
18. A compound whose molecule is super-
(1) CH3CH(OH)CO2H
imposable on its mirror image despite
(2) CH3CH2CH(CH3)CH2OH
containing chiral carbon atoms is called:
(3) C6H5CH2CH3 (1) Threo isomer (2) Meso compound
(4) C6H5CHClCH3 (3) Enantiomer (4) No special name

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19. The pair of enantiomers among the following 20. Which of the following statements is
compound is: correct?
CH3 CH3 (1) The presence of chiral carbon is essential
condition for enantiomerism.
Ph H H Br (2) Functional isomerism is a kind of
stereoisomerism.
Br Ph
(3) The compounds containing one chiral
(I) (II)
carbon are always chiral.
CH3 H
(4) All statements are wrong.
Ph Br Br CH3

H Ph
(III) (IV)
(1) I and IV (2) II and IV
(3) II and III (4) I and I

Configuration of optical isomers

(1) Absolute configuration (R/S configuration)
(2) Relative configuration (D/L configuration)
(1) Absolute configuration :
R ⇒ Rectus ⇒ Right/clockwise
S ⇒ Sinister ⇒ left/Anticlockwise
R/S configuration in wedge dash formula
Rules :
(i) Assign priority order to units attached with chiral-c according to CIP sequence rule.
(ii) Neglect the least priority unit and check the arrangement of other three in their decreasing priority
order if it is
Clock wise ⇒ R-Configuration
Anticlockwise ⇒ S-Configuration
(iii) In assigning configuration, the least priority unit should be at the maximum distance
(i.e. on dash)
If it is not then apply change to shift it on dash

Note. Change that can be applied is, keep any one unit fixed and swift other three in clockwise or
anticlockwise direction.
2
COOH
4

Ex. (1) 3 H
H3C OH
1

as per CIP sequence rule priority order of units is

–OH>–COOH>–CH3>–H

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Neglect the least priority unit i.e.–H and check arrangement of other three in decreasing priority order
that is
–OH>–COOH>–CH3
as it is anticlockwise so configuration is 'S'
2

C2H5 fixed 2
C2H5
1
H2N 4 H 3
H Change CH3
(2) H3C H2N 1
3

(Clockwise: R–configuration)
H 3

Br COOH
1 'S'
2
2 (2s, 3s)
(3) H 'S'
H3C
3 1 Cl

R/S configuration in fisher projection formula :

Rules are similar to wedge-dash
Note. As the least priority unit should be at maximum distance from observer (i.e.dash) so it should be on
vertical line in fisher projection formula since units on vertical represent units away from observer
(i.e.dash)
Ex. (i) (ii)
2
CH=O 2
COOH
2

Fixed COOH
3 1 4 1 1 3
Change
H3C OH H Cl Cl CH3

H CH3 H
3

(S-configuration) (R-configuration)

(2) Relative configuration (D/L configuration)

(i) Glyceraldehyde is considered as a reference molecule to assign D/L configuration to other compounds
(ii) In Fischer projection formula of glyceraldehyde in which longest c-chain should be on vertical line and
most oxidised 'C' should be on top of vertical line
In this case 'OH' group is on
R.H.S. → D – Configuration
L.H.S. → L – Configuration
CH=O CH=O

H OH HO H

CH2–OH CH2–OH
(D-Glyceraldehyde) (L-Glyceraldehyde)

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CHEMISTRY Optical Isomerism
(iii) The system is used mainly for carbohydrates, Amino acids and comparable molecules.
Ex. Carbohydrates
CH=O
H OH
HO H
(Right)
H OH
H OH

CH2OH Last chiral-C (according to IUPAC)

(D-glucose)

Note. If more than one chiral c are having –OH group then D/L configuration is assigned on the basis of
left/Right position of the–OH group present on last chiral –C (according to IUPAC)
Ex. Amino acid
COOH COOH

α α
H NH2 H2N H

G(≠H) G(≠H)
(D-Amino acid) (L-Amino acid)
Note. In amino acid D/L configuration is assigned on the basic of left/Right position of-NH2 group on alpha–C

Calculation of number of optical isomers

Type of compound d+l Meso isomer Total isomers
(1) Unsymmetrical 2n Zero 2n
(2) Symmetrical
n=even no. 2n-1 2n/2–1 2n–1+2n/2–1

n=odd no 2n–1–2(n–1)/2 2(n-1)/2 2n–1

n= number of chiral-c or stereo sites where optical isomerism is possible
* *
Ex. (1) CH3 − CH ( Cl ) − CH (Br ) − CH3
Unsymmetrical compound
n=2
(d+l) = 2n = 22 = 4
Note. Unsymmetrical compound does not has meso isomer
* *
Ex. (2) CH3 − CH ( Cl ) − CH ( Cl ) − CH3
Symmetrical compound
n=2 (even number)
(d+l) = 2n-1= 22–1 = 2
Meso = 2n/2–1 =22/2–1 = 2º = 1

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Exercise 1.2
1. How many stereoisomers are possible for 6. The configuration of the given compound is:
the given compound: Br
C
H3C H
(1) 2 (2) 4 Cl
(3) 3 (4) 8 (1) E (2) R
(3) S (4) Z
2. Absolute configuration of the given
compound is:
CHO 7. The total number of stereoisomers possible
for 2,3-Dichlorobutane:
HO H (1) 2 (2) 3
(3) 4 (4) 5
CH2OH
(1) L (2) D 8. Correct configuration of the following is:
(3) S (4) 1 & 3 both CH3
H OH
3. How many stereoisomers does this
molecules have CH3 OH
CH3CH = CHCH2CHBrCH3 H
(1) 8 (2) 2 (1) 2S, 3S (2) 2S, 3R
(3) 4 (4) 6
(3) 2R, 3S (4) 2R, 3R
4. The R-isomers among the following are
CHO H 9. The priority of groups –OH, –COOH, –CHO,
–OCH3 attached to a chiral carbon is in
H OH D OH order -
(1) – OH > – COOH > – CHO > – OCH3
CH2OH CH3 (2) – OCH3 > – OH > – CHO > – COOH
(i) (ii) (3) – OCH3 > –OH > – COOH > – CHO
CH3 COOH (4) None

H OH H3C NH2
10. No. of optically active isomers possible in
CH3–CH–CH–CH–CH2OH are -
CH2CH3 H | | |
(iii) (iv) OH OH OH
(1) (i) and (ii) (2) (i) and (iii) (1) 2 (2) 4
(3) (ii) and (iii) (4) (iii) and (iv) (3) 6 (4) 8

5. The absolute configurations of the C2 and C3 11. The correct absolute configuration assigned
atoms in the given molecule is for compound (I) and (II) respectively is :
1
CH3 COOH COOH
Cl H
2
(I)H NH2 (II)H NH2
3
H OH
CH3 CH2SH
4
CH3
(1) R, R (2) R, S
(1) 2S, 3S (2) 2R, 3S
(3) S, S (4) S, R
(3) 2S, 3R (4) 2R, 3R

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CHEMISTRY Optical Isomerism
12. Which of the following has D configuration 13. Total number of stereoisomer of the given
CHO compound is :
H OH CH3–CH=CH–CH–CH=CH–C2H5
(1) |
H OH CH3
CH2OH (1) 2 (2) 4
CHO (3) 6 (4) 8
H OH
(2)
HO H 14. is named as :-
HO H
CH2OH
(1) (2R, 3E)-pent-3-en-2-ol
CHO (2) (2 S, 3 E-pent-3-en-2-ol
HO H
(3) (3) (2E, 3R)-pent-2-en-3-ol
HO H (4) (2 E, 3 S-pent-2-en-3-ol
CH2OH
CHO 15. How many compounds are optically active
H OH in the isomers of C4H9Br
(4)
H CH2OH (1) 1 (2) 2
OH (3) 3 (4) 4

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Exercise 2
1. The following pair of compounds is best 4. In which of the following compound, posses
described as : plane of symmetry as well as centre of
symmetry?

Cl Cl

(1) Identical (1) (2)

(2) Diastereomers Cl
(3) Enantiomers
(4) Constitutional isomers CH3 CH3
(3) (4)
2. Which of the following structures is not
meso-2,3-butanediol? CH3 CH3

CH3
H OH 5. Which of the following is optically inactive
(1)
H CH3 H CH3
OH
(1) HO CH3 (2) HO CH3
OH H3C OH HO CH3
H CH3
(2) H H
H CH3
CH3 OH
OH
(3) HO H (4) H3C H
CH3
HO H H3C OH
HO H
(3) CH3 H
H CH3
OH
H 6. Which is not a pair of geometrical isomers ?
OH CH3 Cl Me Br Me
(4)
H OH (1) C=C and C=C
Br H Cl H
CH3
Ph OH Ph
3. Which of the following is a meso compound? (2) C=N and C=N
CH3 CH2—CH3 Me Me OH
H Br H OH Me
(1) (2) Ph H Ph
H Br H OH (3) C=C and C=C
CH3 CH2—CH3 H Me Me Me
H H Br
(3) (4) All of these (4) and Br
Br Br
OH OH

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CHEMISTRY Optical Isomerism
7. Which of the following is optically active: 12. Number of optically active isomers of
OH tartaric acid is –
OH CH(OH)COOH
|
CH3 CH(OH)COOH
(1) (2) (1) 2 (2) 3 (3) 4 (4) 5
OH
OH CH3 13. Which molecule is achiral?
H Br Br
(3) (4)
H Br Cl H Cl H
H Br Cl Br Br Cl
8. Which of the following are not enantiomeric
pair. Cl H H
C2H5 C2H5 I II III
CH3 Cl
(1) I (2) II
H Cl H CH3 (3) III (4) All of these
(A) (B)
14. Which pair of structures represents the same
C2H5 H C2H5 CH3 compound?
H CH3 CH3 CH3
Cl CH3 Cl H OH H OH H OH
(C) (D) H OH HO H HO H
(1) A and C (2) A and D H OH H OH HO H
(3) B and C (4) C and D CH3 CH3 CH3
I II III
9. The absolute configuration of the given CH3 CH3
compound is- HO H H OH
CH3 HO H H OH
H Cl H OH HO H
CH3 CH3
Cl H IV V
C2H5 (1) I and II (2) II and III
(1) 2S, 3R (2) 2S, 3S (3) III and IV (4) III and V
(3) 2R, 3S (4) 2R, 3R
15. The two compounds shown below are:
10. Which of the following is a meso-compound Br F
COOH CH3
H Cl H
OH H OH
(1) (2) F Br Cl
H
H OH HO H
CH2OH CH3
(1) Enantiomers
CH2OH (2) Diastereomers
COOH
(3) Constitutional isomers
HO H H Cl
(3) (4) (4) Identical
HO H H Cl
CH3 CH2OH 16. The following compound can exhibit
CH3 H
11. Number of asymmetric C atoms are present C=C H
in- CH3 C
(i) 1,2-Dimethyl cyclohexane CH COOH
3
(ii) 3-Methyl cyclopentene (1) Tautomerism
(iii) 3-Methyl cyclohexene (2) Optical isomerism
(1) Two, one, one (2) One, one, one (3) Geometrical isomerism
(3) Two, none, two (4) Two, none, one (4) Geometrical and optical isomerism both
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17. Out of the following the alkene that exhibits 22. The compound which has maximum number
optical isomerism is of chiral carbon is :-
(1) 2-Methyl-2-pentene
OCH3
(2) 3-Methyl-2-pentene
(3) 4-Methyl-1-pentene (1) Br NH2 (2)
(4) 3-Methyl-1-pentene
Cl
18. Which of the following compounds can have OH
superimposable mirror image? (3) (4)
H Cl
(1) H (2) H H
Br
Br Br Br 23. Which of the following compound has plane
Br Br of symmetry
(3) (4) all of these
H H H3C CH3 Cl Cl
(1) (2)
19. Predict the number of stereoisomers in
CH2OH.(CHOH)4.CHO H3C CH3
(1) 16 (2) 8
Cl Br
(3) 4 (4) 2
(3) (4) C H
20. Meso stereoisomer is possible in which of
I Cl
the following compounds?
Cl
(I) 2,4-Dibromopentane
(II) 2,3-Dibromopentane
(III) 3-Bromo-2-pentanol
24. Which of the following statement is true :-
(IV) cis-1,3-Dimethylcyclohexane
(V) trans-1,3-Dimethylcyclohexane (1) Compound having chiral- C essentially
(1) I (2) I and III shows optical isomerism
(3) I and IV (4) None of these (2) Compound having one chiral- C can not
21. Which of the following pairs of compounds have meso isomer
are enantiomers (3) Chiral molecule may or may not have
CH3 CH3 chiral-C
HO H HO H
(1) and (4) All
H OH HO H
CH3 CH3
CH3 CH3 25. Which of following compound is chiral :-
H OH HO H CH3
(2) and H Cl
HO H H OH (1) (2)
CH3 CH3 H Cl
CH3 CH3 CH3
H OH HO H CH3
(3) and
HO H HO H
CH3 CH3 Cl
CH3 CH3 (3) (4) Cl
Cl
H OH H OH
(4) and
HO H H OH CH3
CH3 CH3
Sarvam Career Institute 15
CHEMISTRY Optical Isomerism
26. Which of the following cyclopentane 31. The molecules represented by the following
derivative is optically inactive two structures are
OH H CH3 H
OH H
H OH HO CH3
(1) (2)
H OH H OH Br H H CO2H
CH3 CH3 H CH3 CO2H Br
(3) (4) (1) Identical (2) Enantiomers
H H HO H (3) Diastereomers (4) Epimers
27. Which of the following is optically active? 32. Which is optically active?
CH–CH3
(1) H3C CH3 (2) HO CH3
CH3
(1) (2) Cl

COOH
H OH
(3) (4) All of these (3) (4)
H OH
COOH
28. Which of the following compounds can have
meso isomer? 33. The necessary and sufficient condition for a
molecule to be optically active is that
(1) (2) (1) It must contain asymmetric carbon
atoms
(2) It must be symmetric
(3) It must be identical with its mirror image
(3) (4) All of these (4) It must be non-identical with its mirror
Br image
29. Which of the following compound will not
show optical isomerism ? COOH COOH
CH3 H OH H3C OH
34. and are-
CH3
HO CH3 HO H
(1) (2)
CH3
COOH COOH
(1) Enantiomers
CH3
CH3 (2) Position isomers
(3) Geometrical isomers
(3) (4) All of these (4) Homomers

CH3 35. The molecules shown are:

CH3 H
30. Which of the following molecule has
nonsuperimposable mirror images? Cl H Cl CH3
CH3 C2H5 H Cl Cl H

(1) Cl Me (2) D CH2–CH3 CH3 CH3

(1) Enantiomers
H H (2) Diastereomers
CH3 Et
(3) Constitutional
(3) Br H (4) H CH2–CH3 (4) Two different conformations of the same
molecules
D Cl
16 Sarvam Career Institute
Optical Isomerism CHEMISTRY

36. Which of the following are diastereomers? 38. Which of the following is not chiral :
Me Me (1) 2–Butanol
H Br Br H (2) 2,3–Dibromo pentane
(I). (II). (3) 3–Bromo pentane
H Br H Br
(4) 2–Hydroxy propanoic acid
COOH COOH
(I) (II)
39. A pair of diastereomer are
COOH COOH (1) A pair of such stereoisomers which are
H Br Br H not mirror image of each other
(III). (IV).
H Br H Br (2) A pair of non-superimposable mirror
Me Me image of each other
(III) (IV) (3) Always optically active
(4) Both (1) and (3)
(1) (I) and (III) (2) (II) and (IV)
(3) (I) and (II) (4) None 40. The number of chiral molecules and achiral
37. Which compound would exhibit optical molecules are respectively among given
isomers: molecules
COOH O2N H OH H
H
(1) H3C H3C ;
; H3C
NO2 HOOC OH Br CH3
Br Br
CH3
Br
(2) C OH
; CH3CH2CH2CH2Br
HO COOH CH3 ;
H OH
HOOC H
C (1) 3, 3
(3) (2) 4, 2
C
(3) 2, 4
H COOH
(4) None of the above
O
C
(4)

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CHEMISTRY Optical Isomerism
Exercise 3
1. Assertion: Diastereomers are not mirror 4. Assertion: An organic compound is optically
image of each other. active only when its mirror image is non-
super imposable irrespective of the fact
Reason: Diastereomers may be optically whether it contains a chiral carbon atom or
active. not.
(1) If both Assertion & Reason are True & the Reason: Organic compounds which do not
Reason is a correct explanation of the contain chiral carbon atoms can not be
Assertion optically active.
(2) If both Assertion & Reason are True but (1) If both Assertion & Reason are True & the
Reason is not a correct explanation of Reason is a correct explanation of the
Assertion
the Assertion
(2) If both Assertion & Reason are True but
(3) If Assertion is True but the Reason is
Reason is not a correct explanation of
False the Assertion
(4) If both Assertion & Reason are False (3) If Assertion is True but the Reason is
False
2. Assertion: Racemic mixture is optically (4) If both Assertion & Reason are False
inactive.
Reason: Racemic mixture is mixture of meso 5. Assertion: Meso tartaric acid is optically
structures. inactive.
(1) If both Assertion & Reason are True & the Reason: Its optically inactivity is due to
internal compensation.
Reason is a correct explanation of the
(1) If both Assertion & Reason are True & the
Assertion
Reason is a correct explanation of the
(2) If both Assertion & Reason are True but
Assertion
Reason is not a correct explanation of (2) If both Assertion & Reason are True but
the Assertion Reason is not a correct explanation of
(3) If Assertion is True but the Reason is the Assertion
False (3) If Assertion is True but the Reason is
(4) If both Assertion & Reason are False False
(4) If both Assertion & Reason are False
3. Assertion: Chiral objects are not
superimpose on their mirror image. 6. Assertion : Diastereo isomers have different
physical properties
Reason: Chiral objects have plane of
Reason: Diastereo isomers are always
symmetry.
optically active
(1) If both Assertion & Reason are True & the (1) Both Assertion and Reason are correct
Reason is a correct explanation of the and Reason is correct explanation for the
Assertion Assertion
(2) If both Assertion & Reason are True but (2) Both Assertion and Reason are correct
Reason is not a correct explanation of but Reason is not correct explanation for
the Assertion Assertion
(3) If Assertion is True but the Reason is (3) Assertion is correct but Reason is
False incorrect
(4) Assertion is incorrect but Reason is
(4) If both Assertion & Reason are False
correct

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Optical Isomerism CHEMISTRY

7. Assertion: CH2=C=CH2 exhibits optical 9. Statement-I: Molecules that are not

isomerism. superimposable on their mirror images are
Reason: Its mirror image is chiral.
nonsuperimposable. Statement-II: All chiral molecules have
(1) Both Assertion and Reason are correct chiral centres.
and Reason is correct explanation for the (1) Statement-I is correct but Statement-II
Assertion is incorrect.
(2) Both Assertion and Reason are correct (2) Both statements-I and II are correct.
but Reason is not correct explanation for (3) Both Statement-I and II are incorrect
Assertion (4) Statement-I is incorrect but Statement-II
(3) Assertion is correct but Reason is is correct.
incorrect
(4) both Assertion and Reason are incorrect 10. Statement-I: A racemic mixture show zero
optical rotation.
8. Given below are two statements: Statement-II: Diastereomers mixture can be
Statement-I : Compound which are Non separated by fractional distillation method
superimposable on their mirror images are (1) Statement-I is correct but Statement-II
called chiral & property is knows as chirality. is incorrect.
Statement-II : All compounds which are (2) Both statements-I and II are correct.
superimposable mirror images of each other (3) Both Statement-I and II are incorrect
are called meso compounds. (4) Statement-I is incorrect but Statement-II
In the light of the above statements, choose is correct.
the most appropriate answer from the
options given below:
(1) Statement-I is correct but Statement-II
is incorrect.
(2) Both statements-I and II are correct.
(3) Both Statement-I and II are incorrect
(4) Statement-I is incorrect but Statement-II
is correct.

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CHEMISTRY Optical Isomerism

Exercise 4 (Previous Year's Questions)

1. Two possible stereo-structures of 3. Total number of possible isomers (both
CH3CHOH.COOH, which are optically structural as well as stereoisomers) of
active, are called : [AIPMT PRE-2015] cyclic ethers of molecular formula C4H8O is:
(1) Enantiomers [NEET- 2025]
(2) Mesomers (1) 10 (2) 11
(3) Diastereomers (3) 6 (4) 8
(4) Atropisomers

2. Which of the following biphenyls is

optically active ? [NEET-I 2016]
O 2N

(1)

Ι
Br Br

(2)

Ι Ι
Ι

(3)

Ι
CH3

(4)

CH3

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Optical Isomerism CHEMISTRY

ANSWER KEYS
Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 1 1 4 3 1 2 4 2 1 2 4 3 2 4 2 2 2 3 3

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 2 3 3 1 4 2 2 1 3 4 2 1 4 1 2

Exercise 2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 1 4 4 3 4 2 1 2 4 1 1 3 4 1 2 4 4 1 3
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 2 3 1 4 1 3 4 2 2 3 1 4 4 4 1 3 1 3 1 1

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10
Ans. 2 3 3 3 1 3 4 1 1 2

Exercise 4 (Previous Year's Questions)

Que . 1 2 3
Ans. 1 2 1

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 Notes

Sarvam Career Institute

CHEMISTRY Haloalkanes and Haloarenes

Chapter
HALOALKANES
2 AND
HALOARENES
Chapter Summary Introduction
Substitution of hydrogen atom (s) of an aliphatic or aromatic
• Introduction hydrocarbon by halogen atom (s) results in formation of
• Methods of preparation of halogen derivatives.
haloalkanes
• These are organic compound in which halogen is directly
attached with carbon atom.
• Preparation of Haloarenes • These are called as haloalkanes (halogen bonded to alkyl
• Physical properties group) and haloarene (halogen bonded to aryl group)
• Chemical Reaction of
Ex. R – X (Haloalkanes)
X (Haloarene)
haloalkanes
• Reaction of haloarenes • In these compounds hybridisation of carbon is Sp3 (alkyl
halides) and Sp2 (Aryl halides)
• Polyhalogen compounds
Classification:
(A) On the basis of number of halogen atoms:
(i) Mono halides : consist of single halogen atom
Ex. CH3–Cl, CH3–CH2–CH2–Br
(ii) Dihalides: consist of two halogen atoms are furthers
categorised mainly as
• Gem dihalides (Two halogens on same carbon):
X
Ex. R–CH
X
• Vicinal dihalides (Two halogens on adjacent carbon):
Ex. R–CH–CH2
X X
(iii) Polyhalides: consist of three, four …… etc halogen atoms
• Trihalides: consist of three halogen atoms
Ex. CHX3 (Haloform)
• Tetra halides: Consist of four halogen atoms
Ex. CCl4
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 CHEMISTRY Haloalkanes and Haloarenes
(B) On the basis of type of C to which halogen is bonded:
(I) Compounds containing Sp3 C–X Bond (x = F, Cl, Br, I)
(a) Alkyl halides or haloalkanes (R–X):
• General formula : CnH2n+1X
• Depending on nature of carbon to with halogen is attached, these are categorised as
1°
Primary (1°) halide R − CH2 − X
R 2°
Secondary (2°) halide CH–X
R
R 3°
Tertiary (3°) halide R CH–X
R

(b) Allylic halides: Halogen atom bonded to Sp3C adjacent to carbon – carbon double bond (C = C)
X
(Allylic carbon)
CH2–X
(Allylic carbon)

(c) Benzylic halides: halogen atom bonded to Sp3 C atom attached to benzene ring.

C
X
(Benzylic carbon)
(II) Compounds containing Sp2 C–X Bond

(a) Vinylic halides: Halogen atom bonded to Sp2C atom of carbon-carbon double bond.
(Vinylic carbon)
X
X
(Vinylic carbon)

(b) Aryl halides: halogen atom bonded to Sp2 C atom of benzene ring
CH3
X
X
Nature of C–X bond: it’s a polar bond, since halogen is more electronegative than carbon so C atom
bears partial positive and halogen atom bears partial negative charge
δ⊕ δΘ
C X

Order of C–X bond length : C– I > C – Br > C– Cl > C–F

Bond strength : C– I < C – Br < C– Cl < C–F
Dipole moment of different CH3–X : CH3Cl > CH3F > CH3Br > CH3I

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Haloalkanes and Haloarenes CHEMISTRY
Methods of preparation of haloalkanes
(1) By halogenation of Alkanes:
Uv light or
R − H + X 2 heat
→R − X + HX
(detailed mechanism is given in hydrocarbon)

(2) By hydrohalogenation of Alkenes:

C=C + H–X → C–C

H X
(detailed mechanism is given in hydrocarbon)

(3) By halogenation of alkenes:

Br
CCl4
C=C + Br2 C–C
Br
(detailed mechanism is given in hydrocarbon)

(4) From alcohols:

(a) Reaction with conc. HX
R − OH + H − X ( Conc.) 
→R − X + H2O

mechanism:
Θ
••
••
⊕ -H2O ⊕ X
R–O–H + H–X R–O–H R R–X
••
H
• Formation of carbocation is rate determining step.
• Reactivity of different HX : HI > HBr > HCl
Note: In reaction of alcohol with conc. HCl, either alcohol should be highly reactive (reference 3° alcohol) or
the reaction is carried out in presence of anhydrous ZnCl2.
Function of anhydrous ZnCl2 :
(i) Absorbs water formed in r.d.s.
⊕ Θ
(ii) Improves ionisation of HCl : HCl + ZnCl2  H+ ZnCl3
Note: Mixture of conc. HCl and anhydrous ZnCl2 is known as Lucas reagent. It is used to differentiate 1°, 2° &
3° alcohols.
ZnCl (anhyd)
R − OH + H − Cl(conc.) 
2
→ R − Cl(Turbidity) + H2O
R–OH time taken to appear turbidity
3° instant/immediate (2 -3 sec.)
2° 5-7 min.
1° No turbidity at room temp.
(on heating turbidity appear after 30 min.)

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 CHEMISTRY Haloalkanes and Haloarenes
(b) Reaction with PCl5
Cl
R
O + PCl3 R–Cl + POCl3 + HCl
H
Cl
Note: In reaction with PCl5, POCl3 is formed as by product.

(c) Reaction with PCl3

3 R − O − H + P Cl3 → 3 R − Cl + H3PO3
Note: In reaction with PCl3, H3PO3 is formed as by product.
For reaction with PBr3 [P+Br2] and PI3 [P+I2] is used in reaction.

(d) Reaction with SOCl2 (Thionyl chloride)

R Cl
O+S=O R–Cl + SO2 + HCl
H
Cl
Note: In reaction with SOCl2, SO2 is formed as by product.

Note: In preparation of Haloalkanes from alcohols, Thionyl chloride is preferred because in this reaction alkyl
halide is formed along with gases SO2 and HCl. The two gases products are escapable, hence the
reaction gives pure alkyl halide.
The method used to prepare alkyl halide from alcohol is not applicable for preparation of aryl halides
from phenol because the carbon-oxygen bond in phenol has a partial double bond character and is
difficult to break being stronger than a single bond.

(a) By halogen exchange method:

(a) Finkelstein Reaction:
R − X + NaI →
Acetone
R − I + NaX
(x = Cl,Br) (PPt .)

NaCl or NaBr formed is precipitated in dry acetone. It facilitates the reaction in forward direction.

(b) Swarts Reaction:

R − X + AgF 
water
→R − F + Agx
(x = Cl,Br) (PPt .)

Metallic fluoride such as Hg2F2, CoF2 or SbF3 can also be used in the reaction.

Preparation of Haloarenes
(1) From Aromatic hydrocarbon (Arene):
CH3 CH3 CH3
x
Fe
+ x2 dark +

x
• The reaction involves electrophilic substitution mechanism
• Reaction with iodine is reversible in nature and requires the presence of an oxidising agent (like
HNO3/HIO3) to oxidise HI formed during iodination.
• Fluoro compounds are not prepared by this method.

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Haloalkanes and Haloarenes CHEMISTRY
(2) Form diazonium salt:
(a) N2Cl Ι
KΙ

(b) Sandmeyer's reaction:

N2Cl x
Cu2X2

(c) Gattermann reaction:

N2Cl x
H–X
Cu powder

(d) Balz Schiemann reaction

N2Cl F
HBF4
∆

Exercise 1.1
1. In Finkelstein reaction, which reagent is CH2OH
used :
(1) NaI + C2H5OH +HCl ∆ (A) NaI
(B)
4. Acetone
(2) NaCl + Acetone
(3) NaBr + CH3COCH3 OH
(4) NaI + CH3COCH3
In the above reaction product B is:

2. Which of the following leads to the CH2Ι CH2Ι

formation of an alkyl halide :
RedP+Br2
(1) C2H5OH  → (1) (2)
SOCl2
(2) C2H5OH 
→
KBr+Conc.H2 SO4
(3) C2H5OH  → OH Ι

(4) All the above CH2Cl CH2Cl

3. The best reagent for converting ethanol to

(3) (4)
chloroethane is –
(1) PCl3 (2) PCl5
OH Ι
(3) SOCl2 (4) HCl + ZnCl2

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 CHEMISTRY Haloalkanes and Haloarenes
5. CH3–CH2–CH–CH3 (i) NaBr,H2SO4 8. In which reaction alkyl halide is not
[x]
OH
(ii) Mg, dry ether
Product obtained?
(iii) CH3OH (1) R − OH + NaBr →
Product [X] formed in the above reaction is (2) R − OH + H − Br →
(1) CH3–CH2–CH–CH3 (3) R − OH + SOCl2 →
OCH3 (4) R − OH + PCl5 →
(2) CH3–CH=CH–CH3
(3) CH3–CH2–CH2–CH3 9. In which of the following reaction alkyl
(4) CH3–CH2–CH–CH3 halide will be obtained majorly :
Br (1) CH3 − CH2 − CH2 − OH 
HCl
→
6. The major product formed in the following NaCl
CH3—CH2—CH—OH
(2)
reaction is :
CH3 HBr
CH3
CH3–CH=CH–CH (3) CH3—CH2—CH—OH KI
CH3 Conc.H2SO4
CH3
(1) CH3–CH2–CH2–C CH3
CH3 KI
Br (4) CH3—CH2—CH—OH
Conc.H3PO4
CH3
(2) Br–CH2–CH2–CH2–CH CH3
CH3
CH3 10. Which is correct reaction
(3) CH3–CH–CH2–CH
CH3
Br (1) R − X + NaI 
Dry Acetone
→R − I + NaX
CH3 (X=Cl, Br) (Finkelstein reaction)
(4) CH3–CH2–CH–CH CH
3 (2) CH3 − Br + AgF → CH3 − F + AgBr
Br
(Swarts reaction)
7. Which of the following reaction occur most H2 SO4
(3) R − OH + NaBr  →
rapidly?
(1) ( CH3 )2 CHOH 
HBr
→ R − Br + NaHSO4 + H2O
(4) All
(2) ( CH3 )2 CHOH →
HI

(3) ( CH3 )3 C − OH 
HBr
→
(4) ( CH3 )3 C − OH →
HI

Physical properties
• Alkyl halides are colourless when pure, Bromides and Iodides develop colour when exposed to light.
• Due to stronger intermolecular forces of attraction, boiling points of halogen derivatives are higher
than those of hydrocarbons of comparable molecular mass.
• For the same alkyl groups the boiling points of alkyl halides decrease in the order :
R-I > R-Br > R-Cl > R-F, this is because with the increase in mass, the magnitude of vander waal forces
increases.
• The boiling points of isomeric haloalkanes decreases with increase in branching.
CH3
Ex. CH3–CH2–CH2–CH2–Br > CH3CH2CHCH3 > H3C–C–CH3
Br Br
• In isomeric dichlorobenzene
Boiling point : Ortho > para > meta
Melting point : Para > ortho > meta

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Haloalkanes and Haloarenes CHEMISTRY
• Haloalkanes are very slightly soluble in water. In order to dissolve haloalkanes in water energy is
required to overcome the attraction between haloalkane molecules and break
H-bonds between water molecules, but less energy is released when new attractions are set up
between haloalkanes and water molecules as a result of this, solubility of haloalkanes in water in low.

Chemical Reaction of haloalkanes

Reactions of haloalkanes can be categorized as:
(i) Nucleophilic Substitution reactions
(ii) Elimination reactions
(iii) Reaction with metals.
1. Nucleophilic substitution reaction:
In this reaction nucleophile attacks on organic substrate (haloalkane) to replace halogen called leaving
group which departs as halide ion.
Θ δ⊕ δΘ Θ
•• ••
Nu + —C–X —C–Nu + X
(Reagent) (Organic substrate)
Mechanism: The reaction has been found to proceed by two different mechanism which are
(a) Unimolecular nucleophilic substitution (SN1)
(b) Bimolecular nucleophilic substitution (SN2)
Point of SN1 SN2
comparison
1 Mechanism Step-I: δ⊕ δΘ Slow ⊕ Θ
•• R R R
R–L R+L •• δΘ δΘ Θ
Nu + C Nu C X C+X
Θ
⊕ •• H X H H Nu H
Step-II: R + Nu Fast R – Nu H H
(T.S)
2. Number of steps Complete in two steps Complete in one step
3. Rate equation r = K [R–L] Θ
••
r = K [R–CH2–X] [Nu]
4. Carbocation Possible since carbocation is Not possible since no intermediate is
rearrangement formed as intermediate formed
5. Solvent Generally carried out in polar Can be carried out in polar protic and
protic solvents like water, polar aprotic solvents
alcohol, acetic acid etc.
6. Nucleophile Rate of reaction does not Rate of reaction increases with the
depend on strength and increase in strength or concentration
concentration of nucleophile of nucleophile
7. Reactivity of R–X : 3° > 2° > 1° > CH3 X R–X : CH3X > 1° > 2° > 3°
halides : R–Ι > R–Br > R–Cl > R-F : R–Ι > R–Br > R–Cl > R-F
8. Stereochemistry Since nucleophile attack on Since nucleophile attacks on organic
of the reaction. carbocation (Planar attacking substrate at 180° opposite to the
site) so both retention and leaving group so only inversion in
inversion in configuration is configuration is observed.
observed
9. Optical activity Racemic mixture that is Single optically active compound with
of product optically inactive product is inversion of configuration is formed
formed. as product.
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 CHEMISTRY Haloalkanes and Haloarenes
2. Elimination reaction: In this reaction base attacks on organic substrate that is haloalkane to remove
hydrogen atom from β-carbon and halogen atom is also eliminated from α-carbon atom. as a result, an
alkene is formed as product, since β-Hydrogen atom is involved in elimination. The reaction is often
called β-elimination.
H Θ •• Θ
β α
–C—C– B (Base)
••
C=C + B–H + X
X
Mechanism: The reaction has been found to proceed with two Mechanisms which are:
(a) Unimolecular elimination (E1)
(b) Bimolecular elimination (E2)
Point of E1 E2
comparison
1 Mechanism Step-I: δΘ
B
H H Θ H
H Θ
•• Θ
••
•• B
–C—C–
Slow
–C—C– + X –C—C– –C C– –C = C– + B–H+ X
⊕
X X XδΘ
Step-II: (T.S.)
H Θ
••
B
–C—C– C=C + B–H
⊕ Fast
2. Number of Completes in two steps Completes in one step
steps
3. Rate equation H H Θ
••
r = K –C—C– r = K –C—C– [B]
X X
4. Carbocation Possible since carbocation is Not Possible since no intermediate is formed,
rearrangement formed as intermediate transition state is only formed in reaction.
5. Temperature High temperature is a High temperature is favorable condition to get
favorable condition to get product with good yield.
product with good yield.
6. Reactivity R–X : 3° > 2° > 1° R–X : 3° > 2° > 1°
: R–Ι > R–Br > R–Cl > R-F : R–Ι > R–Br > R–Cl > R-F
7. Major Product If more than one types of In general saytzeff's product is major but
alkenes are expected, more depends on some factors as
alkylated alkene (Saytzeff's (i) If attacking base is bulky like (t-butoxide
product) is formed as major ion), Hofmann's product is formed as major
product and less alkylated product
alkene (Hofmann's product) (ii) If leaving group is a bad leaving group (like-
is formed as minor product. F among all halogens) major product depends
on carbanion stability on two different
attacking sites of β-carbon.

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Haloalkanes and Haloarenes CHEMISTRY
Θ
•• H H Θ
••
α
Ex. B B CH3–CH2–CH2–CH=CH2
CH3–CH2–CH = CH–CH3 CH3–CH2–CH–CH–CH2
β β
Saytzeff's product X (Hofmann's product)
(Alexander Zaitsev product) (less alkylated alkene)
(more alkylated alkene)

Elimination versus nucleophilic substitution:

It has been observed that in the same reaction mixture there is a competition between elimination
reaction (E1 and E2) and nucleophilic substitution (SN1 and SN2).
It is because elimination involves attack of base on organic substrate and nucleophilic substitution
reaction involves attack of nucleophile on organic substrate and as we know all nucleophiles behave
as base and vice versa, so two parallel reaction are observed in the same reaction mixture. Now there
are some factors which decide that whether elimination reaction or nucleophilic substitution reaction
will dominate. These factors are.

(1) Attacking reagent: If attacking reagent is a weak nucleophile but good base, elimination reaction
dominates.
If attacking reagent is a good nucleophile as well as good base, elimination reaction dominates (on the
basis of acid-base reaction)

(2) Temperature: High temperature is favorable condition for elimination reaction. If a reaction is
elimination at a specific temperature and if temperature is decreased then reaction shifts from
elimination to substitution

(3) Solvent: Solvent also decides the reaction whether elimination or nucleophilic substitution reaction
will dominate.

(4) Organic substrate: degree of halides (1°/2°/3°) also decides main reaction.

Note: Similar to elimination and nucleophilic substitution there is competition between SN1 and SN2 reaction
also.

(1) SN1 reaction is Preferred when attacking nucleophile is weak like H2O, R–OH.
Θ Θ Θ
(2) SN2 reaction is Preferred when attacking nucleophile is good like OH , R − S , X, NH3 .etc.
(3) organic substrate also decide like
CH3–X mainly undergoes SN2 reaction.
(CH3)3 C–X mainly undergoes SN1 reaction.

Examples of nucleophilic substitution and elimination reaction.

Θ Θ
•• ••
SN: R–X + Nu R–Nu + X
(Organic (Reagent) (Product)
substrate)
H
Θ Θ
••
Elimination: –C—C– + B•• C=C + B–H + X
X (Reagent) (Product)
(Organic
substrate)

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 CHEMISTRY Haloalkanes and Haloarenes

Reagent Nature of reagent Product Remarks

H2O Θ
•• R-OH Mainly follows SN1 mechanism
H2O attacks as Nu
R'–OH Θ
•• R–OR' Mainly follows SN1 mechanism
R'OH attacks as Nu
KOH(aq.) Θ Θ
•• R–OH Mainly follows SN2 mechanism
OH attacks as Nu
KOH(alco.) Θ
C=C Mainly follows E2 mechanism
OH attacks as base
NaNH2 Θ
C=C Mainly follows E2 mechanism
NH2 attacks as base
NH3 Θ
•• Mixture of R–NH2 + Mainly follows SN2 mechanism
NH3 attacks as Nu ⊕Θ at high temperature or in 3° R–X
R2NH + R3N + R4 NX
reaction shifts to E2 mechanism.
R'ONa Θ Θ
•• R–OR' • Mainly follows SN2 mechanism
R' O attacks as Nu
• Reaction is known as
Williamson's synthesis
• In 2° & 3° R–X Reaction shifts to
E2 mechanism
NaCN/KCN Θ R–CN + R–NC Reaction follows SN2 mechanism
CN attacks as
(ionic) Θ
(Major) (minor)
••
ambident Nu
AgCN Θ
•• R–NC + R–CN Reaction follows SN1 mechanism
Ag CN attacks as Nu
(Covalent) (Major) (minor)
NaNO2/KNO2 Θ R-ONO + R–NO2 Reaction follows SN2 mechanism
O− NO attacks as
(Ionic) Θ
(Major) (Minor)
••
ambident Nu
AgNO2 Θ
•• R–NO2 + R-ONO Reaction follows SN1 mechanism
Ag-O–No attacks as Nu
(Major) (Minor)
R'–SNa Θ Θ
•• R-S-R' Reaction follows SN2 mechanism
R'— S attacks as Nu
LiAIH4 Θ Θ
•• R-H • Reaction follows SN2
H attacks as Nu
mechanism
• In 2° & 3° R–X E2 dominates

3. Reaction with metals:

(a) Reaction with Mg:
δΘ δ⊕ δΘ
R–X + Mg dry ether R–Mg X (Organometallic compounds)
• R–Mgx is known as grignard's reagent.
• C–Mg bond is polar covalent bond, C bears negative charge.
• Mgx bond is ionic.
• R–Mgx reacts with the compound having active hydrogen (the hydrogen bonded to O/N/X/S/Csp) to give
hydrocarbon.

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Haloalkanes and Haloarenes CHEMISTRY
Ex. R'–Mgx + H–OH
R'–H
H–OR
H–NH–R
H–C≡C–R
• To avoid acid-base reaction, grignard's reagent is prepared in absolute dry condition.

(b) Reaction with Na [Wurtz reaction]

2R–X + Na dry ether R–R + NaX

Exercise 1.2
1. Which of the following will give fastest SN2 5. Identify Z in the sequence,
− +
C2H5O Na
reaction ? CH3CH2 − =
CH CH2 
HBr
H2O2
→ Y  →Z
(X )

(1) MeO (1) CH3–CH–CH2–O–CH2–CH3

Br
CH3
(2) O2N (2) CH3–CH2–CH–O–CH2–CH3
Br
CH3
(3) H3C (3) CH3–(CH2)3–O–CH2–CH3
Br (4) CH3–(CH2)4–O–CH3
(4) 6. 2,2-Dichloropropane on hydrolysis yields:
Br
(1) Acetone
2. Rate of SN2 will be negligible in : (2) 2,2-Propane diol
(3) Isopropyl alcohol
Cl Cl
(4) Acetaldehyde
(1) (2) 7. A Halogen compound ‘A’ on hydrolysis with
dilute alkali followed by acidification gives
Cl acetic acid. The compound A is :
(1) ClCH2CH2Cl (2) CH3CHCl2
(3) (4) All (3) ClCH2CHCl2 (4) CH3CCl3

8. Which statement is correct about the

3. The order of decreasing SN1 reactivities of following reaction?
the halides is : Me
Θ
I. CH3CH2CH2Cl Me Br+MeO Product
II. CH2=CHCHClCH3 Me
III. CH3CH2CHClCH3 Me
(1) Major product CH2 is formed
(1) I > II > III (2) II > I > III Me
(3) II > III > I (4) III > II > I by elimination reaction
Me
4. The decreasing nucleophilic order of the Me OMe
(2) Major product is
following is : Me
i. FΘ ii. ClΘ iii. BrΘ iv. IΘ formed by substitution reaction
(1) i > ii > iii > iv (2) iv > iii > ii > i (3) Both (1) and (2) are formed in equal
proportion
(3) ii > i > iii > iv (4) ii > i > iv > iii
(4) None
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 CHEMISTRY Haloalkanes and Haloarenes
9. Which statement is correct about the given NaCN
(A)
reaction?
Me
Θ 13. Cl
Me O + Me–Br
Me AgCN
(B)
Me
(1) Product A and B will be respectively :
CH2 (I) is formed by
Me + –
(1) C≡N, N≡C
elimination reaction
+ –
Me N≡C, C≡N
(2)
(2) Product Me OMe (II) is formed
Me (3) C≡N, C≡N
by SN2 reaction + – + –
(3) Product (II) is formed by SN1 reaction (4) N≡C, N≡C
(4) Both (1) and (2) are correct
14. Tertiary butyl halide on boiling with water
10. Which one is a nucleophilic substitution
gives tertiary butyl alcohol. The reaction
reaction among the following :-
H+ follows –
(1) CH3–CH=CH2+H2O→CH3–CH–CH3
(1) E2 mechanism (2) SN1 mechanism
OH (3) SN2 mechanism (4) E1 mechanism
(i)R′MgX
(2) RCHO R–CH–R′
(ii)H2O/H+ 15. Which of the following statements is
OH correct about the following reactions?
CH3 Cl 
Aq.NaOH
(C) ←
Alc.
NaOH
 Me → (B)
(3) CH3–CH2–CH–CH2Br + NH3 → (A)
CH3 (1) (B) is obtained by elimination reaction
CH3–CH2–CH–CH2NH2 (2) (C) is obtained by substitution reaction
(4) CH3CHO + HCN → CH3CH(OH)CN (3) The molecular formula of (B) is C3H6
11. The given reaction is called as: and that of (C) is C3H8O.
C2H5ONa +C2H5Br → C2H5–O–C2H5 + NaBr (4) (B) is an isomer of ethyl methyl ether, while
(1) Frankland reaction (C) is dehydrated compound of (B)
(2) Wurtz reaction
(3) Williamson's synthesis 16. Identify the product X and Y in the given
(4) Cannizzaro reaction reaction,
Dry ether D2O
12. Which of the following reactions give CH3—CH—CH3+Mg X Y
substitution product? Br
Br
(1)X = CH3—CH—CH2Mg, Y = CH3CH2CH2OH
KOH
(1) H 2O Br
Br (2) X = CH3—CH—CH3, Y = CH3—CH—CH3
Θ ⊕
t −BuO K D
(2)  ∆
→ MgBr
Br (3) X = CH3—CH—CH3, Y = CH3—CH—CH3

NaNH2
MgBr OD
(3) ∆
→
(4) X = CH3—CH—CH2Mg, Y = CH3—CH—CH3
(4) alc.KOH
 → Br OH
∆
Br
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Haloalkanes and Haloarenes CHEMISTRY
CH3 19. Given reaction mainly proceeds through –
17. Cl
NBS KCN
A DMF B; CH3ONa in
CH3OH/∆

Product B is :- (1) E1 (2) SN1

Br CN (3) SN2 (4) E2

CH3
(1) (2)
H Br
20. →
NaOH,DMSO
Product is
Br Br H D
CH3
(3) (4) CH3 CH3
H OH HO H
CN (1) (2)
H D H D
CH3 CH3 CH3
Θ
H Na⊕SCH2CH3
18. Prodcuct CH3 CH3
Br H Br H D
(3) (4)
Product is :- H OH H OH
CH3 CH3 CH3
(1) H
SCH2CH3
CH3 CH3

(2) H
SCH2CH3
CH3

(3) SCH2CH3
H
CH3 CH3

(4) S–CH2CH3
H

Reactions of haloarenes
Reactions of haloarenes can be categorised as:
(i) Nucleophilic substitution
(ii) electrophile substitution
(iii) Reaction with metals

(1) Nucleophilic substitution: Aryl halides are extremely less reactive towards nucleophilic substitution
reaction due to:

(i) Resonance effect: due to resonance C–X bond acquires a partial double bond character so bond
cleavage in haloarenes is difficult than in haloalkanes.
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 CHEMISTRY Haloalkanes and Haloarenes
(ii) Hybridisation of carbon: In haloarenes Sp hybridised carbon with a greater s-character is more
2

electronegative and can hold the electron pair of C–X bond more tightly then sp3 hybridised carbon in
haloalkanes with less s-character.

(iii) Instability of phenyl carbocation:

The phenyl cation formed as a result of self-ionisation will not be stabilised by resonance.
Ex. Alkaline hydrolysis of aryl halides
Since aryl halides are extremely less reactive towards nucleophilic substitution reaction, so reaction is
carried out in the presence of high temperature and pressure condition.
Cl OH
(i) NaOH(aq), 623K, 300 atm
(ii) H⊕

Mechanism:
Cl HO Cl HO Cl HO Cl OH
Θ Θ
OH •• Θ
•• Step II
••
Step I Θ
RDS –Cl
••
Θ

HO Cl

Reactivity of ∝ eΘ withdrawing 1
∝ Θ
aryl halides for group(EWG) e releasing
SN reaction on ring group(EDG)
on ring

Cl Cl Cl Cl
Reactivity order: O2N NO2 NO2

NO2 NO2 NO2

2. Electrophilic substitution reaction:

(i) Halogenation:
Cl Cl Cl
Cl
+ Cl2 anhyd. AlCl3 +

Cl
(major) (minor)
(ii) Nitration:
Cl Cl Cl
NO2
Conc.(HNO3 +H 2SO4)
+

NO2
(major) (minor)

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Haloalkanes and Haloarenes CHEMISTRY
(iii) Sulphonation:
Cl Cl Cl
SO3H
Conc H2SO4 +

SO3H
(major) (minor)
(iv) Friedel craft reaction:
a) Alkylation:
Cl Cl Cl
CH3
Anhyd.AlCl3 +
+ CH3–Cl

CH3
(major) (minor)
b) Acylation:
Cl Cl Cl O
O C
Anhyd.AlCl3 CH3
+ CH3–C–Cl +

C
O CH3
(major) (minor)
Note: Although halogen (– X) is deactivating group but it is ortho para directing.
3. Reaction with metals:
A) Wurtz fittig reaction:
X R
dry ether
+ Na + R–X ∆
+ R–R + + Nax

b) fittig reaction:
X
2 dry ether
+ Na + Nax
∆
Diphenyl

Polyhalogen compounds
(a) Dichloromethane (Methylene chloride, CH2Cl2)
(i) Dichloromethane is widely used as a solvent as a paint remover, as a propellant in aerosols, and as a
process solvent in the manufacture of drugs. It is also used as a metal cleaning and finishing solvent.
(ii) Methylene chloride harms the human central nervous system(CNS).
(iii) In humans, direct skin contact with methylene chloride causes intense burning and mild redness of the
skin. Direct contact with the eyes can burn the cornea.

(b) Trichloromethane (Chloroform, CHCl3)

(i) Chemically, chloroform is employed as a solvent for fats, alkaloids, iodine and other substances. The
major use of chloroform today is in the production of the freon refrigerant R-22.
(ii) As might be expected from its use as an anaesthetic, inhaling chloroform vapours depresses the central
nervous system.
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 CHEMISTRY Haloalkanes and Haloarenes
(iii)Chronic chloroform exposure may cause damage to the liver (where chloroform is metabolised to
phosgene) and to the kidneys, and some people develop sores when the skin is immersed in chloroform.
(iv) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl
chloride, also known as phosgene. It is therefore stored in closed dark coloured bottles completely
filled so that air is kept out.
light
2CHCl3 + O2  → 2COCl2 + 2HCl
Phosgene

(c) Triiodomethane (Iodoform, CHI3)

Iodoform was used earlier as an antiseptic but the antiseptic properties are due to the liberation of free
iodine and not due to iodoform itself.

(d) Tetrachloromethane (Carbon tetrachloride, CCl4)

(i) Carbontetrachloride is produced in large quantities for use in the manufacture of refrigerants and
propellants for aerosol cans.
(ii) It is also used as feedstock in the synthesis of chlorofluorocarbons and other chemicals,
pharmaceutical manufacturing, and general solvent use.
(iii) It was also widely used as a cleaning fluid, both in industry, as a degreasing agent, and in the home,
as a spot remover and as fire extinguisher.
(iv) When carbon tetrachloride is released into the air, it rises to the atmosphere and depletes the ozone
layer.

(e) Freons
(i) The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are
extremely stable, unreactive, non-toxic, non-corrosive and easily liquefiable gases.
(ii) Freon 12 (CCl2F2) is one of the most common freons in industrial use. It is manufactured from
tetrachloromethane by Swarts reaction.
(iii) These are usually produced for aerosol propellants, refrigeration and air conditioning purposes.
(iv) In stratosphere, freon is able to initiate radical chain reactions that can upset the natural ozone
balance.

(f) p,p’-Dichlorodiphenyltrichloroethane(DDT)
(i) DDT, the first chlorinated organic insecticides, was originally prepared in 1873,
(ii) The use of DDT increased enormously on a worldwide basis after World War II, primarily because of its
effectiveness against the mosquito that spreads malaria and lice that carry typhus.
(iii) The chemical stability of DDT and its fat solubility compounded the problem. DDT is not metabolised
very rapidly by animals; instead, it is deposited and stored in the fatty tissues.
Cl

H Cl Cl
Cl3C–HC=O Conc.
Cl Cl
(Chloral) H2SO4
H Cl Cl H
(Chlorobenzene) DDT

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Haloalkanes and Haloarenes CHEMISTRY
Exercise 1.3
1. Identify the correct order of reactivity for 4. What is final product of the following
the following pairs towards the respective reaction :
mechanism
Cl
Br Br
(A) SN2: > Mg H2O

Et 2O
→ A → B
Br
(B) SN1: Br
> MgCl
(1) (2)
(C) Nucleophilic substitution:
Br OH
Br
< (3) (4) None of these

NO2 5. Which of the following is incorrect ?

Choose the correct answer from the options Na
(1) CH3–CH2–Cl Dry CH3–CH2–CH2–CH3
given below : Ether
(1) (A), (B) only (2) (A), (B), (C)
Wurtz reaction
(3) (A), (C) only (4) (B) Only
Mg
2. The product of the following reaction is : (2) Cl dry MgCl
ether
Cl
O2N NO2 Grignard reaction
+ H2O Warm
→ ?
Na
NO2 (3) Cl dry
ether
Cl
O2N NO2 Cl Fittig reaction
(1) (2) NaI
(4) Cl dry I
OH acetone
OH OH Swarts reaction
O2N NO2 O2N NO2
(3) (4)
6. Which of the following compound will have
NO2 NO2
the lowest rate towards nucleophilic
3. The structure of the major product formed aromatic substitution on treatment with
in the following reaction is : Θ
OH
Cl
NaCN Cl Cl
DMF
Ι (1) (2)
CN NO2
CN
(1) (2)NC NO2

CN Cl Cl
NO2 NO2
(3) Cl (4) CN
(3) (4)
CN Ι NO2

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 CHEMISTRY Haloalkanes and Haloarenes
7. The end product (Q) in the following 10. Which reaction is least feasible?
sequence of reactions is Cl OH
Cl2/FeCl3 Na/ether
P Q NaOH
(1) High temp

Cl CH3
Cl OH
(1) (2)
NaOH
OH (2) ∆
(3) (4)
NO2 NO2
Cl OH
Cl OMe OMe
NaOH
Cl2
(3)
8. X + Y
Anhyd. FeCl3 (Major) (Minor)
Cl OH
Find the correct statement - NO2 NO2
(1) Boiling point of X is greater than that of Y (4) NaOH
(2) Melting point of X is more than that of Y
(3) X is ortho dichlorobenzene and Y is para NO2 NO2
dichlorobenzene
(4) All of these

9. Consider following reaction:

Cl OH
O 2N NO2 O 2N NO2
Warm H2O Reaction (A)

NO2 NO2
Cl Cl

conc.H2SO4 Reaction (B)

SO3H
Find correct statement :-
(1) Both reaction are example of
substitution reaction
(2) Reaction A is electrophilic substitution
reaction
(3) Reaction B is nucleophilic substitution
reaction
(4) All are correct

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Haloalkanes and Haloarenes CHEMISTRY
Exercise 2
CH3 5. The major product for the following reaction
1. Br 
NaOC2H5
→ Major Product will be : and mechanism involved is respectively :
NaOCH3
CH3 H3C CH3OH
CH3
(1) OC2H5 (2) OC2H5
H3C
Cl
OC2H5
CH3 H3C
(1) ; SN2 mechanism
(3) (4)
H3C OCH3

2. The reaction : (2) H3C ; E2 mechanism

H2O
O 2N Cl+ NaOH Heat H3C

(3) ; E1 mechanism
O 2N OH + NaCl
H3C
is an example of : CH3
(1) Nucleophilic substitution proceeding
through addition elimination mechanism (4) H C ; SN1 mechanism
(2) Nucleophilic substitution proceeding 3
H3CO CH3
through elimination addition mechanism
(3) Electrophilic substitution proceeding
through addition elimination mechanism 6. The order of decreasing nucleophilicities of
(4) Nucleophilic substitution proceeding the following species is :
through SN2 mechanism (1) CH3SΘ>CH3OΘ>CH3COOΘ>CH3OH
(2) CH3COOΘ>CH3SΘ>CH3OΘ>CH3OH
3. Which of the following substrate can give (3) CH3OH>CH3SΘ>CH3COOΘ>CH3OΘ
exactly same products during SN1 and SN2 (4) CH3OΘ> CH3SΘ >CH3COOΘ> CH3OH
mechanism :

(1) (2) 7. Which will give white ppt. with AgNO3 :

Br
Cl (1) Cl (2) Cl
Cl
(3) CH2Cl (4) Both (1) & (3)
(3) (4) None of these

8. Consider the following sequence of reaction :

4. The following benzyl halides are subjected to +
KCN H3 O
hydrolysis under SN1 reaction conditions: C2H5Cl →
DMSO
X ∆
→Y
(P) CH2Cl The products (X) and (Y) are, respectively:
(1) C2H5CN and C2H5CH2NH2
(Q) CH3 CH2Cl (2) C2H5CN and C2H5CONH2
(3) C2H5NC and C2H5NHCH3
(R) CH3O CH2Cl
(4) C2H5CN and C2H5COOH
(S) O2N CH2Cl
9. During reaction of alcohol with KI, which
The order of reactivity is : acid can be used?
(1) S > R > Q > P (2) R > Q > S > P (1) Conc. H2SO4 (2) Conc. H3PO4
(3) R > Q > P > S (4) Q > R > P > S (3) Both (4) None of these
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 CHEMISTRY Haloalkanes and Haloarenes
10. Mechanism involved and product obtained 14. Identify major product for the following
when 3-iodo-3-ethylpentane is treated with reactions
sodium methoxide in methanol is : Ι
(1) Ether, SN1 (2) Alkene, E1 O2N OMe
Aqueous/Acetone
(3) Ether, E1 (4) Alkene, E2
OH
Me
H C + NaN3
DMF
Product (1) O2N OMe
11. Br
Et
The correct representation of the product is OH
Me Me (2) O2N OMe
(1) H N3 (2) N3 H
Et Et OH
N3 N3 (3) O2N OMe
(3) Et H (4) Me Et
Me H OH
(4) O2N OMe
12. The increasing order of the boiling points of
the major products A, B and C of the
following reactions will be: 15. In which of the given pair, the first
(C6H5COO)2 compound is more reactive than second
(a) + HBr A
towards SN2 reaction.
(b) + HBr B (A)
Cl CH2Cl
(c) + HBr C
Br
(1) C < A < B (2) B < C < A (B)
Br
(3) A < B < C (4) A < C < B
(C) CH3CH2CHCH2Br CH3CHCH2CH2Br
| |
13. For the given reaction : CH3 CH3
CH = CHBr 1. NaNH2 Cl Cl
2. Red hot iron tube, 873 K A
CH3 (Major Product) (D)
What is ‘A’
(1) CH=CH–NH2 (1) B, C & D Only (2) B & D Only
(3) B & C only (4) A, B, C & D
CH3
Cl2 /hν aq.KOH
16. (A)  → (B)  → (C)  →
[O]

(2) CH3CHO, Identify A, B & C -

H3C CH3 (1) Ethylalcohol, Ethyl chloride & Ethane
(3) (2) Ethane, Ethylchloride &
CH3 – CH2 – OH
CH3 (3) Propane, Propylchloride &
(4) CH3CH2CH2NH2 CH3 – CH2 – CH2 – OH
(4) All the above

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Haloalkanes and Haloarenes CHEMISTRY
17. For reaction CH3Br + OH¯ → CH3OH + Br¯ 22. The major product ‘Y’ in the following
The rate of reaction is given by the reaction is:
expression:
(1) Rate = k[CH3Br] F EtONa HBr
X Y
Heat
(2) Rate = k[OH¯]
(3) Rate = k[CH3Br][OH¯]
(4) Rate = k[CH3Br]° [OH¯]° (1) (2) HO
Br
18. From left to right, correct statements are: Br
CH3 CH3 CH3 (3) (4) Br
CH3–C–Cl CH3–C–Br CH3–C–I
H H H
(A) Rate of SN1 mechanism increases in 23. Which of the following reactivity order

polar protic solvent towards SN2 reaction is incorrect.

(B) Rate of SN2 mechanism increases in (1) <
Cl Br
DMSO < I
(C) Rate of E2 mechanism increases
(2) C6H5C(CH3)(C6H5)Br < C6H5CH(C6H5)Br
(D) Rate of E1 mechanism increases
< C6H5CH(CH3)Br < C6H5CH2Br
(1) Only A & D (2) Only B, C & D
(3) CH3CH2CH2CH2Br > (CH3)2CHCH2Br
(3) Only A, B & D (4) A, B, C & D
> CH3CH2CH(Br)CH3 > (CH3)3CBr
EtOH–H2O
19. nBuBr + KCN 
→ Product is (4) C6H5Cl > CH3–Cl > CH3–CH2–CH2–Cl
CN
(1) (2)
CN 24. In the following reaction sequence, the
correct structure(s) of X is
(3) (4)
Me N3
1) Nal, Me2CO
X
20. Which of the following is most easily 2) NaN3, HCONMe2

hydrolyzed enantiomerically pure

CH2–Cl
Cl Me Br
CH (1)
(1) (2)
Me Br
Cl
Cl O 2N NO2 (2)
(3) (4)
NO2
(3)
21. 2-Bromopentane is heated with potassium Me Br
ethoxide in ethanol. The major product is :
(1) trans-2-pentene (2) 2-Ethoxypentane
(4)
(3) 1-Pentene (4) cis-2-pentene Me Br

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 CHEMISTRY Haloalkanes and Haloarenes
25. Consider the following reaction sequence 29. Arrange the following alkyl halides in
CH3 decreasing order of the rate of β-elimination
CH3–C– CH=CH2 (i) HCl
 →? reaction with alcoholic KOH.
(ii) CH3CH2OH/KOH

CH3 (I) CH3–CH–CH2Br (II) CH3–CH2–Br

Final major product is CH3
CH3 CH3 (III) CH3–CH2–CH2–Br
(1) CH3–C– CH–CH3 (2) CH3–C– CH–CH3 (1) I > II > III (2) III > II > I
CH3 OH OH CH3 (3) II > III > I (4) I > III > II
CH3
(3) CH3–C– CH=CH2 (4) CH3–C = C–CH3 30. Arrange the following compounds in
decreasing order of rate of SN1 reaction
CH3 CH3 CH3
(A) Cl (B) Cl
26. CH3–CH2–C≡CH → (P) → HBr
(2 mole)
(i)Alc. KOH
(ii) NaNH2
H2
(Q) 
Pd/BaSO4
→ (R) (C) (D)
Cl Cl
Final product ‘R’ is
CH3 H (1) C > B > D > A (2) C > D > A > B
(1) CH3–CH2–CH2–CH3 (2) C=C
CH3
(3) D > A > C > B (4) D > C > A > B
H
H H H3C– H2C H
(3) C=C (4) C=C
H3C CH3 H
31. Which of the following alkyl halides will
H
undergo SN1 reaction most rapidly?
27. How many optically active compounds are
(1) (CH3)3 C–F (2) (CH3)3 C–Cl
formed on monochlorination of given
compound (3) (CH3)3 C–Br (4) (CH3)3 C–I
Me
32. The negative part of the addendum (the
Me molecule to be added) adds on to the carbon
Me
atom of the double bond containing the least
(1) 2 (2) 4
(3) 6 (4) 8 number of hydrogen atoms. This rule is
known as
28. Rate determining transition state can be (1) Saytzeff's rule (2) Peroxide rule
represented for the following reaction by
(3) Markovnikov's rule (4) Hoffmann rule.
CH3–Br + CH3ONa →
H 33. Which of the following molecules has
δ⊕ δΘ
(1) H C------Br highest dipole moment?
H (1) CH3Cl (2) CH2Cl2
H δΘ (3) CHCl3 (4) CCl4
Br
(2) H C δΘ 34. Arrange the following compounds in
OCH3
H decreasing order of their boiling points.
H (i) CH3Br
δ⊕ δ⊕
(3) H C ----- Br -----Na (ii) CH3CH2Br
H (iii) CH3CH2CH2Br
H H (iv) CH3CH2CH2CH2Br
δΘ δΘ (1) (i) > (ii) > (iii) > (iv) (2) (iv) > (iii) > (ii) > (i)
(4) H3CO-----C-----Br
(3) (i) > (iii) > (ii) > (iv) (4) (iii) > (iv) > (i) > (ii)
H

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Haloalkanes and Haloarenes CHEMISTRY
35. In the given reaction: 38. Which of the following statements regarding
Br the SN1 reaction shown by alkyl halide is not
CH3–C≡CNa
Et2O/∆ correct?
the products are: (1) The added nucleophile plays no kinetic
C≡C–CH3
role in SN1 reaction.
(1) and
(2) The SN1 reaction involves the inversion
20% 80%
of configuration of the optically active
C≡C–CH3
(2) substrate.
+
20% 80% (3) The SN1 reaction on the chiral carbon
C≡C–CH3 ends up with racemisation of the
(3) product.
100% (4) The more stable the carbocation
intermediate the faster the SN1 reaction.
(4)
100% 39. Which of the following products as shown by
36. In the reaction given below, the dehydrohalogenation of alkyl halides
CH3CH2 CH2CH3 with sodium ethoxide in ethanol is correctly
Θ
CH3 C–Cl+OH OH–C CH3+ClΘ marked as major product?
H H
Br C2H5ONa
which of the following statements is correct? (i) CH3+ CH2
CH3 C2H5OH
(1) The reaction proceeds via SN2 mechanism
(Minor) (Major)
hence inversion of configuration takes
place. Br
(2) The reaction proceeds via SN1 mechanism (ii) CH3–C–CH2CH3 CC22HH55ONa
OH
hence inversion of configuration takes
CH3
place.
CH2=C–CH2CH3+CH3–C=CHCH3
(3) The reaction proceeds via SN2 mechanism
hence their is no change in the CH3 CH3
(Minor) (Major)
configuration.
(4) The reaction proceeds via SN1 mechanism CH3 CH3
C2H5ONa
hence there is no change in the (iii) CH3— C — C—CH2CH3 C2H5OH
configuration. CH3 Br
37. Arrange the following compound in order of CH3 CH3
their reactivity towards SN2 reaction. CH3— C — C=CH—CH3+
(i) CH3(CH2)3CH2Br CH3 CH3 CH2
(ii) (CH3)2CHCH2CH2Br (Major)
CH3— C — C—CH2— CH3
CH3
CH3
(iii) CH3CH2–CH–CH2Br (Minor)
CH3
(1) Only (i) and (ii)
(iV) CH3–C–CH2Br
(2) Only (i) and (iii)
CH3
(3) Only (ii) and (iii)
(1) (i) > (ii) > (iii) > (iv) (2) (ii) > (iii) > (iv) > (i)
(4) only (ii)
(3) (iii) > (i) > (ii) > (iv) (4) (iv) > (ii) > (i) > (iii)

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 CHEMISTRY Haloalkanes and Haloarenes
44. The relative reactivity of following halides
NBS Na/ether
40. (X) ; towards SN2 reaction follows the order :-
Peroxide
O
X is Cl ; Cl ;
Cl ; Cl
(1)
P Q R S

(1) Q>S>R>P (2) P>S>R>Q

(2) (3) S>R>Q>P (4) P>R>S>Q

45. Rate of SN1 reaction is:-

(3) Br CH2–Br

(P) (Q)
CH3
CH2–CH2–Br CH
(4) None of these Br
(R) (S)
41. Which of the following compounds will give (1) S>Q>R>P (2) S>R>P>Q
racemic mixture on nucleophilic
– (3) P>Q>R>S (4) S>R>Q>P
substitution by OH ion?
Br 46.
(I) CH3—CH—Br (II) CH3—C—CH3 Me
Θ
C2H5 C2H5 OH
H
Me HO
(I)
(III) CH3—CH—CH2Br CH2CH3
H Cl
C2H5 Me Me
Θ
(1) (I) (2) (I), (II) & (III) CH2CH3 OH
H OH+HO H
(3) (II) & (III) (4) (I) & (III) (II)
CH2CH3 CH2CH3
42. Correct order for reaction with alcoholic Reaction I and II respectively are:
KOH is (1) Both SN1 (2) Both SN2
Cl
(3) (I) SN1, (II) SN2 (4) (I) SN2, (II) SN1
Br Br Br
(i) (ii) (iii) (iv)
47. Consider the reaction:
(1) i > ii > iii > iv (2) i > iii > ii > iv
CH3 − CH2 − CH2 − Br + NaCN
(3) iv > ii > iii > I (4) i > iv > ii > iii
→ CH3CH2CH2 − C ≡ N + NaBr
43. The major product of the following reaction The correct statement is :-
is
(1) The reaction will be fastest in water
Br
KOH alc.(excess) (2) The reaction will be fastest in N,N-
∆
Ph dimethylformamide (DMF)
Br (3) Transition state of SN2 is tetrahedral and
(1) (2) sp3 hybridized
Ph Ph (4) If conc. of alkyl bromide is tripled and

conc. of CN is reduced to half rate of SN2
(3) (4)
Ph Ph increased by 2 times

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Haloalkanes and Haloarenes CHEMISTRY
48. Which of the following compound will give 53. A SN2 reaction at an asymmetric carbon of a
curdy precipitate with AgNO3 solution: compound always give:
Cl (1) An enantiomer of the substrate
(2) A product with opposite rotation
(3) A mixture of diastereomer
(1) (2) Cl (4) A single stereoisomers

Cl 54. DDT is manufactured by the acid catalysed

Cl condensation of:-
(1) Chloroform and acetone
(3) (4) (2) Chloroform and nitric acid
(3) Chlorobenzene and chloral
(4) Chlorobenzene and chloroform
55. Which of the following compound is used as
49. Which of the following reaction is correct:
refrigerant?
(1) CH3CH2 − Br + KCN → CH3CH2 − NC
(1) Urea (2) Acrolein
(2) CH3CH2 − Br + AgCN → CH3CH2 − CN
(3) CCl2F2 (4) CH2—CH2
O
(3) CH3CH2—Br + KNO2→CH3—CH2—N OH OH
O
56. Which of the following is added to chloroform
O
in a small quantity before is bottled for sale?
(4) CH3—CH2—Br + AgNO2→CH3—CH2—N
O (1) CH3OH (2) C2H5Cl
(3) C2H5OH (4) AgNO3
50. Which of the following compound is most Cl
reactive toward SN2 displacement?
(1) CH3CH2CH2CH2 − Cl concHNO3 (i)NaOH/443K
A (ii) H⊕ B
57. concH2SO4
(2) ( CH3 )3 C − Cl
(3) CH3—CH—CH—CH3 B as a major product is-
Cl OH
CH3 Cl OH NO2
(4)CH3—CH—CH2—Cl
(1) (2)
CH3
NO2
51. Which of the following statement is incorrect:-
OH Cl
(1) Boiling point of alkylhalide is R—I > R—Br
> R—Cl > R—F
(2) Alkyl halide are soluble in organic solvent (3) (4)
and less soluble in water
(3) Freons are used for aerosol propellant, NO2 OH
refrigeration and air conditioning purpose.
58. Arrange the following in decreasing order of
(4) o– and m-dichlorobenzene has higher
their leaving tendency
melting point than those of p-isomer
O
52. Which of the following is most reactive BrΘ H3C–C–OΘ CH3CH2OΘ Ph–OΘ
towards nucleophilic substitution reaction?
(I) (II) (III) (IV)
(1) CH2 = CH–Cl
(2) C6H5 –Cl (1) I > II > III > IV
(2) I > III > II > IV
(3) CH3 – CH = CH – Cl
(3) I > II > IV > III
(4) Cl – CH2 – CH = CH2
(4) I > IV > III > II

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 CHEMISTRY Haloalkanes and Haloarenes
59. Which is correct order 63. Identify major product for the following
Cl Cl Cl reaction.
Cl Cl
(1) > > Cl
NaOMe/MeOH
Cl
O 2N NO2
Cl
Major product is?
(Order of boiling point) OMe
Cl Cl Cl (1) MeO
Cl
O 2N NO2
(2) > >
Cl
Cl
MeO
Cl (2)
(Order of melting point) O 2N NH2
Br > > Br Cl
(3)
Br MeO
(3)
(Order of boiling point)
O 2N NO2
(4) All
OMe
(4) Cl
60. The order of reactivity of the following alkyl
halides for SN1 and SN2reaction is - O 2N NO2
(1) RF > RCl > R − Br > R − Ι
64. Which is correct order of reactivity in SN2
(2) R − F > R − Br > R − Cl > R − Ι
reaction of following compound?
(3) R − Cl > R − Br > R − F > R − Ι
(4) R − Ι > R − Br > R − Cl > R − F (1) Cl < CH2–Cl

(2) Cl < I
61. Among the bromides given below, the order
of reactivity of SN1 reaction is – (3) Ph–CH–CH3 < Ph–CH2–Br
O Br
(I) (II) (III) (4) All
O

Br Br Br 65. Which of the following gives cyanide when

(1) III > I > II (2) III > II > I reacts with KCN
(A) Chloro benzene (B) Allyl bromide
(3) II > III > I (4) II > I > III
(C) Vinyl chloride (D) Benzyl bromide
(1) A, B and C only
62. The correct order of rate of following (2) A and B only
compounds towards SN2 reaction (3) B and D only
(4) A and C only
Br Br > Br
66. The decreasing basic strength order of the
Br following is :
(A) (B) (C) (D)
i. FΘ ii. ClΘ iii. BrΘ iv. IΘ
(1) A > B > C > D (2) D > C > B > A (1) i > ii > iii > iv (2) iv > iii > ii > i
(3) A > C > B > D (4) C > D > A > B (3) ii > i > iii > iv (4) ii > i > iv > iii
48 Sarvam Career Institute
Haloalkanes and Haloarenes CHEMISTRY
67. The decreasing nucleophilic order of the 71. In dehydrohalogenations the base (alcoholic
following compounds is: KOH) abstracts :
Θ Θ Θ
(1) The halide ion.
i. CH3 ii. NH2 iii. OH iv. FΘ
(2) The proton present on the carbon next to
(1) i > ii > iii > iv (2) iv > iii > ii > i
the carbon to which the halogen is
(3) ii > i > iii > iv (4) ii > i > iv > iii
attached.
68. SN1 & SN2 are not favourable in
(3) The proton present on the carbon to
(A) H2C = CH–Cl (B) Ph–CH2–Cl
which the halogen is attached.
(C) Ph–Cl (D) H2C=CH–CH2–Cl
(4) The proton as well as halide ion
(1) A, B & C Only (2) A & C Only
(3) B & D Only (4) A, B, C & D
72. In the elimination reactions, the reactivity
69. For the given reaction order of halogens in alkyl halides is:
R1 R1 (1) R-I > R-Br > R-Cl (2) R-Cl > R-Br > R-I
HOH (3) R-Br > R-Cl > R-I (4) R-I > R-Cl > R-Br
R–C–X R – C – OH
R2 R2 73. The elimination of HX from an alkyl halide
Which substrate will give maximum forms an alkene. The order of the elimination
racemisation?
reaction is :
CH3
(1) 3° halide > 2° halides > 1° halides
(1) C6H5 – C – Br (2) 1° halide > 2° halides > 3° halides
C2H5 (3) 1° halide = 2° halides > 3° halides
CH3 (4) 2° halide > 1° halides > 2° halides
(2) CH2 = CH – C – Br
Br
C2H5 Alc.KOH
74. +
Br
(Y%) (X%)
(3) C6H5–C OCH3
X and Y respectively are:
(1) 81, 19 (2) 50, 50
(3) 19, 81 (4) 100, 0
CH3
Br 75. Consider the following reaction sequence.
(4) C6H5–C NO2
CH3MgBr
CH3C ≡ CH  CH3CH2Br
→I  →II
The final product (II) formed is :
(1) CH3C ≡ CCH3
NH3 (2) CH3C ≡ CCH2CH3
⊕
(3) CH3C ≡ CMgBr
70. C4H9Br 
alc.KOH
∆
→(B)  (i)O3
(ii) Zn/H2O
→ CH3–CHO CH3
(A)
(4) CH3–C=CHCH2CH3
(A) and (B) in the above reaction sequence
are:
(1) sec-Butyl bromide, α-Butylene
(2) tert-Butyl bromide, Isobutylene
(3) sec-Butyl bromide , β-Butylene
(4) n-Butyl bromide, α-Butylene

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 CHEMISTRY Haloalkanes and Haloarenes

Exercise 3
1. Assertion: The boiling points of alkyl 4. Assertion: Aryl halides are highly reactive
halides decrease in the order : towards nucleophilic substitution
RI > RBr > RCl > RF reactions.
Reason: The boiling points of alkyl chlorides, Reason: In case of haloarenes, halogen
bromides and iodides are considerably higher atoms is attached to sp hybridised carbon
than that of the hydrocarbon of comparable atom.
molecular mass. (1) If both assertion and reason are true
(1) If both assertion and reason are true and reason is the correct explanation of
and reason is the correct explanation of assertion.
(2) If both assertion and reason are true but
assertion.
reason is not the correct explanation of
(2) If both assertion and reason are true but
assertion.
reason is not the correct explanation of
(3) If assertion is true but reason is false.
assertion.
(4) If both assertion and reason are false.
(3) If assertion is true but reason is false.
(4) If both assertion and reason are false.
5. Assertion: Electrophilic substitution
2. Assertion: Aryl halides cannot be prepared reactions in haloarenes occur slowly and
require more drastic conditions as
by replacement of hydroxyl group of phenol
compared to those in benzene.
by halogen atom.
Reason: Halogens are ortho and para-
Reason: Phenols react with halogen acids
directors.
violently.
(1) If both assertion and reason are true
(1) If both assertion and reason are true
and reason is the correct explanation of
and reason is the correct explanation of
assertion.
assertion. (2) If both assertion and reason are true but
(2) If both assertion and reason are true but reason is not the correct explanation of
reason is not the correct explanation of assertion.
assertion. (3) If assertion is true but reason is false.
(3) If assertion is true but reason is false. (4) If both assertion and reason are false.
(4) If both assertion and reason are false.
6. Assertion: Presence of a nitro group at
3. Assertion: The boiling point of the ortho or para position increases the
compounds increases in the order: reactivity of haloarenes towards
Isopropylchloride < 1-Chloropropane <1- nucleophilic substitution.
Chlorobutane. Reason: Nitro group, being an electron
Reason: Boiling point depends upon the withdrawing group decreases the electron
molecular mass and surface area. density over the benzene ring.
(1) If both assertion and reason are true (1) If both assertion and reason are true
and reason is the correct explanation of and reason is the correct explanation of
assertion. assertion.
(2) If both assertion and reason are true but (2) If both assertion and reason are true but
reason is not the correct explanation of reason is not the correct explanation of
assertion. assertion.
(3) If assertion is true but reason is false. (3) If assertion is true but reason is false.
(4) If both assertion and reason are false. (4) If both assertion and reason are false.

50 Sarvam Career Institute

Haloalkanes and Haloarenes CHEMISTRY
7. Assertion: It is difficult to replace chlorine 10. Given below are two statements; one is
by –OH in chlorobenzene in comparison to labelled as Assertion (A) and the other is
that in chloroethane. labelled as Reason(R).
Reason: Chlorine-carbon (C–CI) bond in Assertion: Alkyl chlorides are best
chlorobenzene has a partial double bond
prepared from alcohols by reaction of
character due to resonance.
SOCl2.
(1) If both assertion and reason are true
Reason: The other two products are easily
and reason is the correct explanation of
escapable. So reaction gives pure Alkyl
assertion.
chlorides. In the light of the above
(2) If both assertion and reason are true but
reason is not the correct explanation of statements, choose the most appropriate
assertion. answer from the options given below :
(3) If assertion is true but reason is false. (1) Both (A) and (R) are correct and (R) is
(4) If both assertion and reason are false. the correct explanation of (A).
(2) Both (A) and (R) are correct but (R) is
8. Assertion: Rate of alkaline hydrolysis of not the correct explanation of (A).
methyl chloride to methanol is higher in (3) (A) is correct but (R) is not correct.
DMF than in water (4) (A) is not correct but (R) is correct.
Reason: alkaline hydrolysis of methyl
chloride follows second order kinetics. 11. Given below are two statements; one is
(1) If both assertion and reason are true
labelled as Assertion (A) and the other is
and reason is the correct explanation of
labelled as Reason(R).
assertion.
Assertion: Boiling points of chlorides,
(2) If both assertion and reason are true but
bromides and iodides are considerably
reason is not the correct explanation of
higher than those of the parent
assertion.
(3) If assertion is true but reason is false. hydrocarbons.
(4) If both assertion and reason are false. Reason: Due to greater polarity as well as
higher molecular mass as compared to
9. Assertion: Alkyl iodide can be prepared by parent hydrocarbon, the intermolecular
treating alkyl chloride/bromide with Nal in forces of attraction are stronger in halogen
acetone. derivatives.
Reason: formed NaCl/NaBr is precipitated In the light of the above statements, choose
in dry acetone. the most appropriate answer from the
(1) If both assertion and reason are true options given below :
and reason is the correct explanation of (1) Both (A) and (R) are correct and (R) is
assertion.
the correct explanation of (A).
(2) If both assertion and reason are true but
(2) Both (A) and (R) are correct but (R) is
reason is not the correct explanation of
not the correct explanation of (A).
assertion.
(3) (A) is correct but (R) is not correct.
(3) If assertion is true but reason is false.
(4) (A) is not correct but (R) is correct.
(4) If both assertion and reason are false.

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 CHEMISTRY Haloalkanes and Haloarenes
12. Given below are two statements; one is 14. Given below are two statements:
labelled as Assertion (A) and the other is Statement-I: 2-Bromo pentane on reaction
labelled as Reason(R). with alcoholic KOH gives 1-pentene as a
Assertion: Chlorobenzene does not react major product.
with NaOH whereas ethyl chloride reacts. Statement-II: In dehydrohalogenation
Reason: The partial double bond between
reactions the preferred product is always
carbon and chlorine in chlorobenzene
that alkene which has lesser number of
causes less reactivity towards Nucleophilic
alkyl groups attached to the doubly bonded
substitution reactions.
c -atoms.
In the light of the above statements, choose
In the light of the above statements, choose
the most appropriate answer from the
options given below : the most appropriate answer from the

(1) Both (A) and (R) are correct and (R) is options given below:
the correct explanation of (A). (1) Both statements-I and II are correct.
(2) Both (A) and (R) are correct but (R) is (2) Statement-I is correct but Statement-II
not the correct explanation of (A). is incorrect.
(3) (A) is correct but (R) is not correct. (3) Both Statement-I and II are incorrect
(4) (A) is not correct but (R) is correct. (4) Statement-I is incorrect but Statement-

13. Given below are two statements; one is II is correct.

labelled as Assertion (A) and the other is
labelled as Reason(R). 15. Given below are two statements:
Assertion : Alkyl iodides are often prepared Statement-I: NaCl or NaBr formed is
by the reaction of alkyl chlorides/bromides precipitated in dry acetone in Finkelstein
with NaI in dry acetone. reaction.
Reason : The synthesis of alkyl fluorides is Statement-II: It facilitates the reverse
best accomplished by heating an alkyl
reaction according to Le-chatliers
chloride/ bromide in presence of metallic
principle.
fluoride such as AgF, Hg2F2, CoF2 or SbF3.
In the light of the above statements, choose
In the light of the above statements, choose
the most appropriate answer from the
the most appropriate answer from the
options given below:
options given below :
(1) Both (A) and (R) are correct and (R) is (1) Statement-I is correct but Statement-II

the correct explanation of (A). is incorrect.

(2) Both (A) and (R) are correct but (R) is (2) Both statements-I and II are correct.
not the correct explanation of (A). (3) Both Statement-I and II are incorrect
(3) (A) is correct but (R) is not correct. (4) Statement-I is incorrect but Statement-
(4) (A) is not correct but (R) is correct. II is correct.

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Haloalkanes and Haloarenes CHEMISTRY
16. Match the reactions given in column I with 18. Match the following
the type of reaction mentioned in column II Column-I Column-II
CH3 –CH2 –CH2Cl + KOH(Aq.) →
n-Propyl
and mark the appropriate choice. (a) (p)
bromide
Column I Column II (b) CH3 –CH = CH2 + HBr → (q) Propene
A CH3–CH–CH–CH3 + C2H5OH i β-Elimination Peroxide Isopropyl
(c) = CH2 + HBr 
CH3 –CH → (r)
Br CH3 bromide
OC2H5
CH3 – CH2 – CH2 – Cl + KOH(Alco.) → n-Propyl
(d) (s)
CH3CH2–C–CH3 alcohol
CH3 (1) (a-q), (b-r), (c-p), (d-s)
(2) (a-s), (b-r), (c-p), (d-q)
B CH3CH2Br NaI ii SN1 Reaction (3) (a-s), (b-p), (c-r), (d-q)
Acetone
(4) (a-q), (b-p), (c-r), (d-s)
CH3CH2OH
19. Match the structures of compounds given in
C CH3CH = CH2 + HBr iii SN2 Reaction column I with the classes of compounds
given in column II.
Peroxide
→ Column-I Column-II
CH3CH2CH2Br (i) CH3–CH–CH3 (a) Aryl halide
X
D CH3CH2Br + alc.KOH → iv Kharasch
(ii) CH2= CH–CH2–X (b) Alkyl halide
CH2 = CH2 effect (iii) X (c) Vinyl halide
(1) A → (iv), B → (i), C → (ii), D → (iii)
(2) A → (ii), B → (iii), C → (iv), D → (i)
(iv) CH2 = CH –X (d) Allyl halide
(3) A → (i), B → (ii), C → (iv), D → (iii) (1) (i–b), (ii–c), (iii–a), (iv–d)
(2) (i–d), (ii–b), (iii–a), (iv–c)
(4) A → (iii), B → (i), C → (ii), D → (iv)
(3) (i–b), (ii–d), (iii–a), (iv–c)
(4) (i–c), (ii–b), (iii–a), (iv–d)
17. Match the column I with column II and mark
20. Match the column-I and column-II and
the appropriate choice. select the correct answer :
Column I Column II Column-I Column-II
(a) Cl (p) E1
A Carbon (i) Paint remover
aq.KOH
tetrachloride
B Methylene (ii) Refrigerators and (b) Cl (q) SN1
alc.KOH
chloride air conditioners
C DDT (iii) Fire-extinguisher
(c) Cl (r) SN2
D Freons (iv) Non-biodegradable H2O

insecticide
(1) A → (ii), B → (iii), C → (i), D → (iv) (d) OH (s) E2
⊕
H /∆
(2) A → (iv), B → (iii), C → (ii), D → (i)
(3) A → (i), B → (ii), C → (iii), D → (iv) (1) (a–s); (b–r); (c–p); (d–q)
(4) A → (iii), B → (i), C → (iv), D → (ii) (2) (a–r); (b–s); (c–q); (d–p)
(3) (a–r); (b–s); (c–p); (d–q)
(4) (a–q); (b–p); (c–r); (d–s)
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 CHEMISTRY Haloalkanes and Haloarenes

Exercise 4 (Previous Year's Questions)

1. Which of the following compounds will 5. Correct order of reactivity towards
undergo racemisation when hydrolysed by elimination reaction [AIIMS-2015]
solution of KOH [AIPMT-2014] (i) C6H5CH2CH2–Br
CH2Cl (ii) CH3CH2CH2–Br
(1) (iii) C6H5CH2CH–CH3
Br
(2) CH3CH2CH2Cl
(iv) H3C–CH–CH3
CH3
Br
(3) H3C–CH–CH2Cl
(1) i > ii > iii > iv
CH3 (2) ii > i > iii > iv
C (3) iii > iv > i > ii
(4) H Cl
C2H5 (4) iii > i > iv > ii

2. Correct order of nucleophilicity :- 6. In the reaction

(1)NaNH2 /liq.NH3
[AIIMS-2014] H – C ≡ CH 
(2)CH CH Br
→X
3 2
– –
(1) CH3O > CH3 > NH2 > CH3COO–
–
(1)NaNH2 /liq.NH3

→Y(2)CH3CH2Br
(2) CH3– > NH2– > CH3O– > CH3COO–
X and Y are : [NEET-I-2016]
(3) NH2– > CH3– > CH3O– > CH3COO–
(1) X = 1-Butyne ; Y = 3-Hexyne
(4) CH3– > CH3O– > NH2– > CH3COO–
(2) X = 2-Butyne ; Y = 3-hexyne
(3) X = 2-Butyne ; Y = 2-Hexyne
3. In an SN1 reaction on chiral centres, there
(4) X = 1-Butyne ; Y = 2-Hexyne
is: [Re-AIPMT-2015]
(1) 100 % retention
7. For the following reactions :-
(2) 100 % inversion
(a) CH3CH2CH2Br + KOH →
(3) 100 % racemization
CH3CH = CH2 + KBr + H2O
(4) inversion more than retention leading to CH3 H3C CH3
(b) H3C
partial racemization +KOH +KBr
Br OH
4. Which of the following reaction(s) can be Br
used for the preparation of alkyl halides? (c) +Br2
Br
[Re-AIPMT-2015]
Which of the following statement is
anh.ZnCl2
(I) CH3CH2OH + HCl  ∆
→ correct? [NEET-I-2016]
(II) CH3CH2OH + HCI → (1) (a) and (b) are elimination reaction and
(III) (CH3)3COH + HCl → (c) is addition reaction
(IV) (CH3)2CHOH + HCl  anh.ZnCl2
→ (2) (a) is elimination, (b) is substitution and
(1) (IV) only (c) is addition reaction
(2) (III) and (IV) only (3) (a) is elimination, (b) and (c) are
(3) (I), (III) and (IV) only substitution reactions
(4) (a) is substitution, (b) and (c) are
(4) (I) and (II) only
addition reaction

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Haloalkanes and Haloarenes CHEMISTRY
8. Consider the reaction 12. The incorrect statement regarding chirality
CH3CH2CH2Br+NaCN→CH3CH2CH2CN+NaBr is : [NEET(UG)-2022]
(1) SN1 reaction yields 1 : 1 mixture of both
[NEET- II-2016] enantiomers.
This reaction will be the fastest in (2) The product obtained by SN2 reaction of
(1) N, N'-dimethylformamide (DMF) haloalkane having chirality at the
reactive site shows inversion of
(2) Water configuration.
(3) Ethanol (3) Enantiomers are superimposable
mirror images on each other.
(4) Methanol
(4) A racemic mixture shows zero optical
rotation.
9. Arrange the following compounds in order
13. Which of the following is suitable to
of reactivity towards SN1 reaction.
synthesize chlorobenzene?
[AIIMS-2016] [NEET(UG)-2022]
(a) CH3–CH2–CH2–CH3 (1) Benzene, Cl2, anhydrous FeCl3
(2) Phenol, NaNO2, HCl, CuCl
Br
(3) ,HCl
(b) H3C–CH–CH2–CH3
Br NH2
(4) , HCl, Heating
(c) OHC–CH–CH2–CH3
Br
14. The given compound
(1) a > b > c (2) b > a > c
CH=CH–CH–CH2 CH3
(3) c > b > a (4) a > c > b
X

10. The hydrolysis reaction that takes place at

Is an example of_____ [NEET(UG)-2023]
the slowest rate, among the following is
(1) Benzylic halide (2) Aryl halide
[NEET-2019(ODISHA)] (3) Allylic halide (4) Vinylic halide
aq. NaOH Θ⊕
(1) Cl ONa
15. Consider the following reaction and
CH3 CH3 identify the product (P). [NEET(UG)-2023]
(2) H3C–CH2–Cl →
aq.NaOH
H3C–CH2–OH CH3–CH–CH–CH3
HBr
 → Product (P)
(3)H2C=CH-CH2Cl → H2C=CH–CH2OH
aq.NaOH CH3 OH
3-Methylbutan-2-ol
aq. NaOH
(4) CH2Cl CH2OH Br
(1) CH3–C–CH2–CH3
CH3
11. The correct sequence of bond enthalpy of
(2) CH3 CH=CH–CH3
‘C–X’ bond is: [NEET(UG)-2021]
(3) CH3–CH–CH–CH3
(1) CH3−F < CH3−Cl < CH3−Br < CH3−I
(2) CH3−F > CH3−Cl > CH3−Br > CH3−I CH3 Br
CH3
(3) CH3−F < CH3−Cl > CH3−Br > CH3−I
(4) CH3–C–CH2 Br
(4) CH3−Cl > CH3−F > CH3−Br > CH3−I
CH3

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 CHEMISTRY Haloalkanes and Haloarenes
16. Identify the product in the following 19. Given below are two statements : one is
reaction [NEET(UG)-2023] labelled as Assertion (A) and the other is
+ – labelled as Reason (R). [NEET-2025]
N2Cl (i) Cu2Br2/HBr
Assertion (A): I undergoes SN2
(ii) Mg/dry ether
Product
(iii) H2O reaction faster than
Cl
OH
Reason (R): Iodine is a better leaving group
(1) (2) because of its large size.
In the light of the above statements, choose
MgBr OH the correct answer from the options given
below :
(3) (4) (1) A is true but R is false
Br (2) A is false but R is true
(3) Both A and R are true and R is the
17. Which amongst the following reactions of correct explanation of A
alkyl halides produces isonitrile as a (4) Both A and R are true but R is not the
major product ? [NEET(Manipur)-2023] correct explanation of A
(A) R-X+ HCN →
(B) R - X + AgCN →
(C) R-X+ KCN →
H2O
(D) R-X + NaCN →
C2H5 OH

Choose the most appropriate answer

from the options given below:
(1) (D) only
(2) (C) and (D) only
(3) (B) only
(4) (A) and (B) only

18. Identify ‘X’ in the following reaction.

[NEET(Manipur)-2023]
dry D2O
Br Cl+Mg ether Intermediate X
[1.0mol] [1.0mol]

(1) Cl D

(2) DO OD

(3) D D

(4) D Br

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Haloalkanes and Haloarenes CHEMISTRY
ANSWER KEYS

Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10
Ans. 4 4 3 1 3 3 4 1 4 4

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 4 3 2 3 1 4 1 2 3 3 1 1 2 4 2 2 3 4 2

Exercise 1.3
Que . 1 2 3 4 5 6 7 8 9 10
Ans. 2 3 4 2 4 2 3 2 1 3

Exercise 2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 1 3 3 2 1 4 4 2 4 1 2 3 4 2 2 3 4 1 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 3 4 1 4 3 2 4 4 2 4 3 1 2 2 1 1 2 3 3
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 1 2 1 4 1 4 2 4 4 1 4 4 4 3 3 3 3 3 4 4
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. 1 1 3 4 3 1 1 2 3 3 2 1 1 1 2

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 2 3 1 4 2 1 1 1 1 1 1 1 2 3 1 2 4 2 3 2

Exercise 4 (Previous Year's Questions)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Ans. 4 2 4 3 4 1 2 1 2 1 2 3 1 3 1 2 3 1 3

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 Notes

Sarvam Career Institute

Alcohols, Phenols & Ethers CHEMISTRY

Chapter ALCOHOLS,
3 PHENOLS &
ETHERS
Chapter Summary Introduction
• Alcohols and phenols are formed when a hydrogen atom in a
• Introduction hydrocarbon, aliphatic and aromatic respectively is replaced by
• Preparation of alcohols –OH group
• An alcohol contain one or more hydroxyl (OH) groups directly
• Preparation of phenol
attached to carbon atom(s) of an aliphatic system while a
• Physical properties phenol contains –OH group(s) directly attached to carbon
• Chemical Reaction of atom(s) of an aromatic system (C6H5OH)
Alcohols
Classification of alcohol:-
• Chemical reaction of (a) Classification According to Number of —OH Groups:
phenol (i) Monohydric [one –OH] → CH3CH2—OH
• Preparation of ethers (ii) Dihydric [two –OH] → CH2—CH2

• Chemical reactions of OH OH
ether (iii) Trihydric [three –OH] → CH2—CH—CH2
OH OH OH
(iv) Polyhydric [n –OH] → CH2—CH CH2
OH OH OH
(b) Classification According to Nature of Carbon:
(i) p or 1° – alcohol → CH3CH2–OH
(ii) s or 2° – alcohol → (CH3)2CH–OH
(iii) t or 3° – alcohol → (CH3)3C–OH

• Allylic alcohols:- OH group is attached to a Sp3 hybridized

carbon atom adjacent to the carbon-carbon double bond.

H –C–
Ex. CH2=CH–CH2–OH CH2=CH–C–OH CH2=CH–C–OH
–C– –C–
(Primary) (Secondary) (Tertiary)

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CHEMISTRY Alcohols, Phenols & Ethers
• Benzylic alcohols:- OH group is attached to a Sp3 hybridized carbon atom next to an aromatic ring.

H –C–
CH2OH C–OH C–OH
Ex. –C– –C–

(Primary) (Secondary) (Tertiary)

Allylic & benzylic alcohols may be primary, secondary or tertiary

• Vinylic alcohols:- OH group bonded to a carbon–carbon double bond

Ex. CH2=CH–OH
Classification of phenol on the basis of number of –OH group
(i) monohydric (One–OH) (ii) Dihydric (two–OH) (iii) Trihydric(three–OH)
OH OH OH OH
CH3 OH OH
,

OH
Some important common name
OH Phenol OH O-cresol
CH3

OH m-cresol OH p-cresol

CH3
CH3
OH Catechol OH Resorcinol
OH

OH
OH Hydroquinone CH3OH Methyl alcohol
or quinol

OH
CH3CH2OH Ethyl alcohol CH3–CH2–CH2–OH n-propyl alcohol
CH3–CH–CH3 Iso propyl alcohol CH3–CH2–CH2–CH2–OH n-butyl alcohol
OH
CH3–CH–CH2–CH3 Sec. Butyl alcohol CH3–CH–CH2–OH Iso butyl alcohol
OH CH3

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Alcohols, Phenols & Ethers CHEMISTRY

CH3 tert-butyl alcohol CH2 – CH2 ethylene glycol

CH3–C–OH OH OH
CH3
CH2 – CH – CH2 Glycerol
OH OH OH

Comparison of Bond Angles in Phenols, Alcohols and Ethers:

•• ••
O •• 109° H O ••
H H H
H C H O•
•• •
H C C
108.5° 111.7° H
H H H
Bond angle increases with the increase in hindrance.

Preparation of alcohols
(1) From alkanes (By oxidation):
CH3 CH3
KMnO4
Ex. CH3–C–H CH3–C–OH
∆
CH3 CH3

(2) From alkenes

(a) By acid catalysed hydration:
H+/H2O
Ex. CH3–CH=CH2 CH3–CH–CH3 (Mechanism : Hydrocarbon chapter)
OH

(b) By hydroboration oxidation:

(i) B2H6
Ex. CH3–CH=CH2 Θ
CH3–CH2–CH2–OH (Mechanism: Hydrocarbon chapter)
(ii) H2O2/OH(Aq.)

(c) By oxymercuration Demercuration:

(i) Hg(OAC)2,H2O
Ex. CH3–CH=CH2 Θ CH3–CH–CH3 (Mechanism: Hydrocarbon chapter)
(ii) NaBH4,OH
OH
(3) From carbonyl compounds:
(a) By reduction of aldehyde and ketones:
Ni/Pd/Pt + H2
Ex. R–CHO R–CH2–OH (Primary alcohol)
or
NaBH4
or
LiAlH4
O OH
Ni/Pd/Pt + H2
R–C–R' R–CH–R' (Secondary alcohol)
or
NaBH4
or
LiAlH4

(b) By reduction of carboxylic acids and esters:

(i) LiAlH4
Ex. R–COOH R–CH2–OH
(ii) H2O

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CHEMISTRY Alcohols, Phenols & Ethers
However, LiAlH4 is an expensive reagent and therefore used for preparing special chemical only.
commercially, acids are reduced to alcohol by converting them to the esters followed by their reduction
using Hydrogen in the presence of catalyst (catalytic Hydrogenation)
O
R'–OH H2
R–COOH R–C–OR' Pd R–CH2–OH+R'–OH
H⊕

(4) From Alkyl halides:

Aq. KOH
Ex. CH3–CH2–Cl CH3–CH2–OH (Mechanism: Haloalkane Chapter)
S N2

(5) From Grignard reagents:

δΘ δ⊕ Θ ⊕
Ex. H2 O
O + R – Mg – X → C –OMg – X C –OH + Mg (OH)X
R R
Adduct
The overall reaction using different aldehyde and ketones are as follows:
H2 O
HCHO + RMgX → R–CH2OMgX R–CH2OH + Mg(OH)X

R' R'
H2 O
R–CHO + R'MgX → R–CH–OMgX R–CH –OH + Mg(OH)X
O R' R'
H2 O
R–C–R + R'MgX → R–C–OMgX R–C –OH + Mg(OH)X
R R
Note. The reaction of grignard reagents with methanal produces a primary alcohol, with other aldehyde
secondary alcohols and with ketones tertiary alcohols.

Preparation of phenol
Phenol also known as carbolic acid, was first isolated in the early nineteenth century from cool tar.
Nowadays, phenol is commercially produced synthetically. In the laboratory, phenols are prepared
from benzene derivatives by following methods.
(1) From Haloarenes
Θ⊕
Cl ONa OH
623 K H⊕
+ NaOH 300 atm

(2) From benzene sulphonic acid

SO3H OH
oleum (i) NaOH(Molten)
(H2S2O7) (ii) H⊕

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Alcohols, Phenols & Ethers CHEMISTRY
(3) From diazonium salts
⊕ Θ
NH2 N2Cl OH
NaNO2 + HCl H2 O
+ N2 + HCl
warm
(Aniline) (Benzene diazonium
Chloride)
(4) From cumene
CH3
H3C CH3 H3C
CH C–O–OH OH
O
O2 H⊕
H2 O
+ CH3 – C–CH3

Cumene Cumene Phenol Acetone

Hydroperoxide
Note. Acetone is by – product of this reaction which is obtained in large quantities by this method

Exercise 1.1
1. A tertiary alcohol is obtained when CH3MgI 5. 3-methyl-1-butene on oxymercuration-
reacts with:
demercuration yields ______ as the major
(1) CH3CHO (2) CH3CH2CHO
(3) CH3–CH–CHO (4) CH3COCH3 product.
CH3 (1) 3-methyl-2-butanol
(2) 2-methyl-2-butanol
2. BD3 Product A; A is:
Θ
H2O2/OH (3) 3-methyl-1-butanol
CH3 (4) 2-methyl-1-butanol
D
(1) OH (2)
OH
CH3 6. Which reagent converts propene to
OH
(3) OH (4)
D D 1-propanol?

Θ
(1) H2O, H2SO4
B2H6 H2O2 /OH
3. Alkene  → → 2 alcohol °
(2) B2H6, H2O2, OH¯
The alkene would be:
(1) CH3 – CH = CH2 (3) Hg(OAc)2, NaBH4/H2O
(2) CH3CH2 – CH=CH2 (4) Aq. KOH
CH3
(3) CH3–C=CH2
7. When CH2=CH–COOH is reduced with
(4) CH3–CH=CH–CH3
LiAlH4, the compound obtained will be:
4. Iso-butylene when subjected to (1) CH3–CH2 –CH2OH
hydroboration oxidation reaction yields
(1) Sec-butyl alcohol (2) CH3–CH2–CHO
(2) Tert-butyl alcohol (3) CH3–CH2–COOH
(3) Iso-butyl alcohol
(4) CH2=CH–CH2OH
(4) n-butyl alcohol
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CHEMISTRY Alcohols, Phenols & Ethers
8. In the given reaction: CH3 Br
CH3 AlC. KOH HBr
(i) Hg(OAc)2/CH3OH
C=CH–CH3  →[x] 11. (A ) (B )
CH3 (ii) NaBH 4 H 2O 2

[x] will be: Identify ‘A’ and ‘B’ respectively

OCH3 Br
(1) and
(1) CH3–C–CH2–CH3
CH3 Br
OCH3 (2) and
(2) CH3–CH–CH–CH3
Br
CH3
(3) and
OH
(3) CH3–C–CH2–CH3 Br
CH3 (4) and
O
(4) CH3–C–CH–CH3 12. Find out the major products from the
CH3 following reactions.

Hg(OAc)2, H2O BH3,THF

B A
NaBH4 H2O2/OH–
H3O + C6H5MgBr
9.  → X  → Y; Y is OH
(1) A= OH , B=
OH
OH OH
(1) O (2) (2) A= , B=
OH
Ph
OH OH
H (3) A= , B=
(3) (4)
Ph
(4) A= OH , B= OH

H2O/H+
X 13. The product formed in the following
Θ
(i) B2H6 (ii) H2O2/OH(aq.) reaction is
10. Y CH3
(i) (CH3COO)2Hg, H2O H3C (i) BH3–THF
Z Product
(ii) NaBH4 (iI) H2O2/OH–
CH3
Most acidic compound is CH3 OH
CH3
(1) H3C C (2) H3C
(1) X
HO CH3 CH3
(2) Y
(3) Z (3) H3C OH (4) H2C–CH2–C(CH3)2
CH3
(4) All are equally acidic OH OH

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Alcohols, Phenols & Ethers CHEMISTRY
14. Alcohol is not formed as product in
(i) A
NaNO2 +HCl(aq) 15. HCHO CH2–OH; A is :
(1) CH3 − CH2 − NH2  → (ii) H2O

(i) CH3 −MgBr

(2) CH3 − CHO (ii) H O
→ (1) MgX (2) CH3
2

O
(3) OH (4) OCH3
(3) H3C–C–NH2 LiAlH4
KOH(aq)
(4) CH3 − CH2 − Cl  →

Physical properties
(a) Boiling point:-
B.P ∝ molecular weight
1
B.P ∝ (when Mw is same)
Branching ( Sidechain )
Ex. Order of B.P
(i) CH3CH2–CH2–CH2–OH > CH3–CH2–CH2–OH > CH3–CH2–OH
CH3
(ii) CH3–CH2–CH2–CH2–OH > CH3–CH–CH2–OH > CH3–C–OH
CH3 CH3
(iii) CH3–CH2–CH2–OH < CH3–CH–CH2 < CH2–CH–CH2
OH OH OH OH OH
(Number of OH increase, H–bonding increases)
(iv) CH3–CH2–OH > CH3–O–CH3 > CH3–CH2–CH3
Note. Boiling point of alcohols and phenols are higher in comparison to other classes of compound, namely
Hydrocarbons, ether, haloalkanes and Haloarenes of comparable molecular masses due to inter
molecular H–bonding

(b) Solubility:- Solubility of alcohols and phenols in water is due to their ability to form Hydrogen bonds
with water molecules.
Solubility ∝ tendency to form H–bonds with water
1
Solubility ∝
Molecular weight of Alkyl / Aryl Part
When Mw is same then
Solubility ∝ Branching (side chain)
Ex. Order of solubility:
(i) CH3–OH > CH3–CH2–OH > CH3–CH2–CH2–OH > CH3–CH2–CH2–CH2–OH
(ii) CH3CH2CH2OH < CH2–OH < CH2–OH
CH–OH CH–OH
CH3 CH2–OH
CH3
(iii) CH3–CH2–CH2–CH2–OH < CH3–CH–CH2–OH < CH2–C–OH
CH3 CH3

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CHEMISTRY Alcohols, Phenols & Ethers
Chemical Reaction of Alcohols
Alcohols are versatile compounds. They react both as nucleophile and electrophile.

(i) Alcohols as nucleophiles:

The bond between O–H is broken when alcohol react as nucleophiles
H
•• ⊕ ⊕
R–OH
••
+ C— → R – O– C— → R – O – C – + H
⊕

(ii) Protonated alcohols as electrophiles

The bond between C–O is broken when they react as electrophiles. Protonated alcohols react in this
manner
⊕
R − CH2 − OH + H+ → R − CH2 − OH2
Θ ⊕
Br + CH2 – OH2 → Br–CH2–R + H2O
R
• Based on the cleavage of O–H and C–O bonds, the reaction of alcohols and phenols may be divided into
two groups

(A) Reaction involving cleavage of O–H bond

(1) Acidity of alcohols:
The acidic character of alcohol is due to the polar nature of O–H bond. An electron-releasing group (–
CH3, –C2H5) increases electron density on oxygen tending to decreases the polarity of O–H bond. This
decreases the acid strength.
CH3 CH3
Ex. CH3 CH2OH CH–OH CH3 C–OH (Acidic strength)
CH3 CH3
Primary alcohol Secondary alcohol Tertiary alcohol

(2) Reaction with metals:

Alcohol reacts with active metals such as sodium, potassium and Aluminium to yield corresponding
alkoxide and Hydrogen gas
Ex. 2R–O–H + 2Na → 2R–ONa + H2
sodium alkoxide
6CH3–CH2–OH + 2Al → 2(CH3–CH2O)3 Al + 3H2
Aluminium ethoxide

(3) Esterification:
• alcohol react with carboxylic acid to give ester
• conc. H2SO4 is used as catalyst and dehydrating agent
18 Conc. H2SO4 18
Ex. R—C—OH + H— O—R R–C–OR + H2O
O O
(Carboxylic acid) (alcohol) (ester)
Similarly

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O O O O
⊕
H
R–C–O–C–R + HO– R' R–C–OR' + R–C–OH
O O
Pyridine
R–C–Cl + HO– R' R–C–OR' + HCl
reaction of alcohol with acid chloride is carried out in the presence of a base (pyridine) so as to
neutralise HCl which is formed during the reaction. It shift the equilibrium to the right hand side.

(B) Reaction involving cleavage of C–O bond

(1) Reaction with hydrogen halides (H–X):
H-X
Ex. R–CH2–OH  → RCH2–X + H2O
Mechanism: H–X → H⊕ + XΘ
Θ
•• H⊕ ⊕ –H2O ⊕ X
R—CH2—OH R—CH2—O—H r.d.s.
R—CH2 R—CH2X
H (Carbocation) (Product)
(Unstable)
• Formation of carbocation is r.d.s step.
• Since carbocation is formed as intermediate hence, rearrangement can take place
• Reactively order of alcohol towards the reaction
Reactivity ∝ stability of carbocation
so reactivity order : 3° alcohol > 2° alcohol > 1° alcohol
• Reactivity order for H–X: HI > HBr > HCl
• Mechanism used:
3° alcohol 2° alcohol 1° alcohol
SN1 SN1 SN2
• Catalyst used in reactions:
conc. H–I (usually no. Catalyst is used)
conc. H–Br + H2SO4
conc. H–Cl + Anhyd. ZnCl2 → It is called Lucas Reagent (L.R.)
• Alcohol are soluble in Lucas reagent but their halides are immiscible and produce turbidity in solution
therefore this reaction is used to identify 1°, 2° and 3° alcohol by noticing time taken to appear turbidity
L.R.
3° alcohol or allylic alcohol or benzylic alcohol → Immediate turbidity
L.R.
2° alcohol → turbidity in 5–7 min.
L.R.
1° alcohol → No turbidity at room temperature (turbidity appear only on heating after 30 Min.)

(2) Reaction with phosphorus trihalides

3R—OH + PCl3 → 3RCl + H3PO3
R—OH + PCl5 → R—Cl + POCl3 + HCl

(3) Reaction with thionyl chloride (SOCl2)

Pyridine
R − OH + SOCl2  →R − Cl + SO2 ↑ + HCl
( gas )

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(4) Acidic dehydration of Alcohol.
Alcohols undergo dehydration (removal of a water molecule) to form alkenes on treating with a protic
acid i.e. Concentrated H2SO4 or H3P04, or catalysts such as anhydrous zinc chloride or alumina
H3C CH3 H⊕/∆
H C–C CH3–CH=CH–CH3 + CH3–CH2–CH=CH2
H H
OH (Major) (Minor)
Mechanism:
H
•• H⊕ Slow(RDS) ⊕ –H⊕
CH3–CH2–CH–O–H
••
CH3–CH2–CH–O–H H2O + CH3–CH2–CH–CH3 CH3–CH=CH–CH3
⊕
Major
CH3 CH3
(Carbocation) +
(Protonated alcohol)
CH3–CH2–CH=CH2
Miner
• Formation of carbocation is r.d.s step
• Rearrangement is possible
• Reaction follow E1 mechanism
• Ease of dehydration of alcohol ∝ Stability of carbocation formed in r.d.s. step
• Order of rate of dehydration of alcohol → 3° OH > 2° OH > 1° OH
• Sytzeff alkene is formed as major product

Ex. correct order of ease of dehydration of alcohol is-

CH3
(i) CH3–CH2–CH2–OH < CH3–CH–OH < CH3–C–OH
CH3 CH3
(1°alcohol) (2°alcohol) (3°alcohol)
OH OH OH

(ii) < <

(5) Oxidation of Alcohol.

(a) By mild oxidising agent
(I) By PCC (Pyridinium chloro chromate)
O
+ CrO3 + HCl → Cl–Cr=O
N N⊕ Θ
O
H
(Structure)

(II) By PDC (pyridinium dichromate)

2–
Cr2O7
N⊕ 2

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(III) By collin's reagent
O
N---Cr---N Chromium trioxide-Pyridine complex in
CH2Cl2 (DCM) solvent
O O
[O]
Ex. (i) CH3–CH2–CH2–OH CH3–CH2–CHO
(1°–OH) (Aldehyde)
OH O
[O]
(ii) CH3–CH–CH3 CH3–C–CH3
(2°–OH) (Ketone)
CH3
(ii) CH3–C–OH [O] No oxidation
CH3
(IV) By Cu/∆
Cu/573K
(i) CH3–CH2–CH2–OH CH3–CH2–CHO
(1°–OH) (Aldehyde) (Dehydrogenation)
OH O
Cu/573K
(ii) CH3–CH–CH3 CH3–C–CH3
(2°–OH) (Ketone)
CH3 CH3
(iii) CH3–C–OH Cu/573K
CH3–C=CH2 (Dehydration)
CH3 (Alkene)
(3°–OH)

Note. When the vapour of a primary or a secondary alcohol are passed over heated copper at 573K,
dehydrogenation takes place and an aldehyde or ketone is formed while tertiary alcohols undergo
dehydration.

(b) By strong oxidising agent

⊕
(I) K2Cr2O7/ H
K Cr O /H+
(i) CH3–CH2–CH2–OH 2 2 7 CH3–CH2–COOH + Cr+3
(orange)
(1°–OH) (Acid) (green)

OH O
(ii) CH3–CH–CH3 K2Cr2O7/H CH3–C–CH3 + Cr+3
+

(orange)
(2°–OH) (ketone) (green)
CH3
K2Cr2O7/H+
(iii) CH3–C–OH (orange) No oxidation (at normal condition)
CH3
(3°–OH)

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(II) By KMnO4(Acidic or alkaline or neutral)
KMnO4
(i) CH3–CH2–CH2–OH(purple) CH3–CH2–COOH + MnO2
(Brown)
OH O
KMnO4
(ii) CH3–CH–CH3 CH3–C–CH3 + MnO2
(purple)
(Brown)
CH3
(iii) CH3–C–OH KMnO4 No oxidation (At normal condition)
(purple)
CH3
(III) By Jones Reagent (CrO3 + Aq. H2SO4)→ in acetone solvent
CrO3–H2SO4
(i) CH3–CH2–CH2–OH CH3–CH2–COOH

OH O
CrO3–H2SO4
(ii) CH3–CH–CH3 CH3–C–CH3

CH3
(iii) CH3–C–OH CrO3–H2SO4 No oxidation (At normal condition)
CH3
(c) oxidation by HIO4:
Condition for Oxidation By HIO4
(i) At least 2 —OH or 2 C = O or 1 —OH and 1 C = O should be at vicinal carbons.
(ii) One HIO4 breaks one bond and adds one –OH to each carbon.
HIO4
Ex. CH2—CH2 CH2—OH + HO — CH2 –2H2O HCHO + HCHO
OH OH OH OH
HIO4 −3H2O
Ex. CH2—CH — CH2 HO—CH2 + HO—CH—OH + HO—CH2  → HCHO+HCOOH+HCHO
OH OH OH OH OH OH
CH3 OH CH3

Ex. CH3—CH—CH—C—CH3 2HIO4 CH3—CH—OH + CH—OH + HO — C — CH3

OH OH OH OH OH OH
–3H2O
O

CH3CHO + HCOOH + CH3 — C — CH3

O O O
1HIO4
Ex. CH3—C—CH—CH2—C—H CH3—C—OH + HO—CHCH2 — C —H

O OH OH
–H2O

CH3COOH + CHO—CH2—CHO
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Chemical reaction of phenol

(1) Acidity of phenols: Reaction of phenol with metals and sodium hydroxide indicates its acidic nature
OH ONa

(i) + Na → + ½ H2

(Phenol) (sodium phenoxide)

OH ONa

(ii) + NaOH → + H2O

(sodium phenoxide)

Note. The hydroxyl group of phenol is directly attached to sp2 hybridised carbon of benzene ring which acts
as an electron withdrawing group.
•• ••
•
•
•
••
O—H
••
⊕ O—H
••
⊕ O—H
⊕ O—H O—H
•

Θ Θ

Θ
• Due to resonance oxygen acquires positive charge thus facilitate the release of H+.
OH O—

+H+

• Phenoxide ion is also resonance stabilised

OΘ O O O OΘ
Θ Θ

Θ
• Both phenol and phenoxide ion are resonance stabilised but due to charge separation phenol is less
stable than phenoxide ion so it readily releases H+ and forms phenoxide ion.
• In substitute phenols, the presence of electron withdrawing groups such as–NO2,–CN, – CHO enhances
the acidic strength of phenol. This effect is more pronounced when such a group is present at ortho and
para position. it is due to the effective delocalization of negative charge.
OH OH OH OH
NO2 NO2 O 2N NO2

NO2 NO2
• Electron releasing group such as alkyl groups does not favour the formation of phenoxide ion resulting
in decrease in acidic strength.
OH OH OH OH
CH3 CH3 H3C CH3

CH3 CH3

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(2) Reaction with neutral Fecl3: Phenol gives violet colouration with FeCl3 solution (neutral) due to
formation of a complex.
PhOH + Fecl3 → H3 Fe ( OPh )6 
( violet color complex )
(3) Acetylation of phenol
O
OH O–C–CH3
O
Pyridine
Ex. + CH3–C–Cl + HCl

O
OH O–C–CH3
O O
Ex. H⊕
+ CH3–C–O–C–CH3 + CH3COOH

COOH COOH O
OH ⊕
O–C–CH3
Ex. + (CH3CO)2O H + CH3COOH

Salicylic acid (Aspirin)

Note. (i) Acetylation of salicylic acid produces aspirin
(ii) Aspirin has analgesic, anti inflammatory and antipyretic properties

(4) Electrophilic aromatic substitution reaction

(i) Nitration:
OH OH OH
NO2
dilute HNO3 +

NO2
(O–Nitrophenol) (P–Nitrophenol)

• ortho nitro phenol and para nitro phenol can

be separated by stem distillation
• ortho nitro phenol is steam volatile
OH
Conc. HNO3 O2N NO2
+ oxidized product

NO2
2,4,6 – Trinitrophenol → Poor yield obtained
(Picric acid)
Note. To get picric acid in good yield, first do sulphonation so that phenol becomes less activated then do
nitration
OH OH OH
SO3H Conc. HNO3 2 O N NO2
Conc. H2SO4
H2SO4

SO3H NO2
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(ii) Halogenation:
• Phenol reacts with bromine in CCl4 to form mixture of o–and p–bromo phenol.
OH OH OH

CHCl3 or CS2 or CCl4 Br

+ Br2 →
Low temp.
+

Br

• Phenol reacts with bromine water to form a white ppt. of 2,4,6 tribromophenol.
OH OH

H2O
Br Br
+ 3Br2 → + 3HBr

Br

(5) Kolbe's reaction:

OH ONa OH
COOH
NaOH (i) CO2
(ii) H⊕

(salicylic acid)
Mechanism:
⊕
Θ
Na Θ
Θ
OH O O O OH
O
Θ H O
Θ Θ C H ⊕ COOH
OH O=C=O OH
C O
(–H2O)
O
(Chelation)

(6) Reimer – Tiemann reaction:

Θ⊕ Θ ⊕
OH ONa ONa OH
CHCl2 CHO CHO
CHCl3 + aq.NaOH NaOH H⊕

Intermediate Salicylaldehyde

Mechanism:
Cl ⊕ Θ Cl
Cl C— H + NaOH→ Na C Cl NaCl + : CCl2
Cl + Cl dichloro carbene(DCC)
H2O (electrophile)

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CHEMISTRY Alcohols, Phenols & Ethers
Θ ⊕ Θ
OH ONa Cl O O Cl
:C H CH
NaOH Cl Θ Cl Cl
(–H2O) C
Cl

⊕
Na
Θ O H Θ OH O
O O O H H
Θ C–O–H C Θ C
C OH Cl OH Cl
H

OH
CHO
H⊕

Note. Abnormal Reimer- Tiemann rxn→

OH OH
COOH
+CCl4 (i)NaOH
+ (ii) H

(salicylic acid)

(7) Reaction with zn dust

OH

+zn + zno

(8) oxidation
OH O
Na2Cr2O7
H2SO4 + zno

O
(Benzoquinone)
• oxidation of phenol with chromic acid produces a conjugated diketone known as benzoquinone. In the
presence of air, phenols are slowly oxidised to dark colored mixtures containing quinones.

Some commercially important alcohols

1. Methanol : (Wood spirit):
( ZnO+Cr O )
CO + 2H2  2 3
200 −300 atm
→ CH3OH
573−673 K

It is highly poisonous in nature, causes blindness and even death. It is used in paints and varnishes.

2. Ethanol:
It is obtained commercially by fermentation of sugars. The sugar in molasses, sugarcane or fruits such
as grapes is converted to glucose and fructose

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Invertase
C12H22O11  → C6H12O6 + C6H12O6
glucose Fructose

Glucose and fructose undergoes fermentation in the presence of zymase enzyme, which is found in
yeast.
Zymase
C6H22O6 → C2H5OH + 2CO2
The action of zymase is inhibited once the percentage of alcohol reaches 14% under anaerobic
(absence of O2) condition. If air gets into fermentation mixture, the O2 oxidises ethanol to ethanoic acid,
which destroys the taste of alcoholic drinks.

Exercise 1.2
1. The –OH group of methyl alcohol cannot be OH
replaced by chlorine by the action of: CH3
Br2 (Excess) Br2
6. ‘A’ ← → ‘B’
(1) NaCl H2O CS2

(2) Hydrogen chloride ‘A’ & ‘B’ respectively are:

(3) Phosphorus trichloride OH
(4) Phosphorus pentachloride CH3
(1) Both are
2. Which of the following Reaction of alcohol
does not show cleavage of R-O linkage: Br
(1) ROH + PCl5 (2) ROH + SOCl2 OH
Br CH3
(3) ROH + HCl (4) ROH + Na
(2) Both are
3. Phenol reacts with bromine in CS2 at a low
Br
temperature, the product is:
OH OH
(1) m-Bromophenol Br CH3 CH3
(2) p-Bromophenol only (3) &
(3) o-and p-Bromophenol
(4) 2, 4, 6-Tribromophenol Br Br
OH OH
4. Phenol reacts with conc. HNO3 in the CH3 Br CH3

presence of conc. H2SO4 to give: (4) &

(1) Meta nitrophenol
Br Br
(2) Ortho nitrophenol
(3) Ortho and para nitrophenol
7. An alcohol gives turbidity with HCl/ZnCl2
(4) Picric acid
(Lucas reagent) only on heating. The
alcohol is :
5. Phenol can be converted into 1,3,5-
OH
tribromo benzene by using the sequence OH
(1) (2) Ph–CH–CH3
(1) Br2/H2O
(2) (i) Br2/H2O (ii) Zn-Dust OH CH2–Cl
(3) (i) Br2/CS2 (ii) Zn-Dust (3) (4)
(4) (i) Zn-Dust (ii) Br2/H2O OH

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CHEMISTRY Alcohols, Phenols & Ethers
8. The following reaction is known as 13. n-propyl alcohol and isopropyl alcohol can
( i) CHCl3 /NaOH be chemically distinguished by which
Phenol  → Salicylaldehyde
( ii) H+
reagent:
(1) Gattermann aldehyde synthesis
(1) PCl5
(2) Sandmeyer’s reaction
(3) Kolbe’s reaction (2) Reduction
(4) Reimer-Tiemann reaction (3) Oxidation with Pot. Dichromate, followed
by I2/NaOH
9. HBr reacts fastest with
(4) Ozonolysis
(1) 2-methyl propan-2-ol
(2) Propan-2-ol
(3) Propan-1-ol 14. An alcohol gives turbidity with HCl/ZnCl2
(4) 2-methylpropan-1-ol (Lucas reagent) only on heating. The
alcohol is:
10. Arrange the following alkanols a, b and c in OH
order of their reactivity towards acid
catalysed dehydration: OH
(a) CH3–CH–CH2–CH2 (1) (2)
CH3 OH OH
OH
(3) OH (4)
(b) CH3–C–CH2–CH3
CH3
CH3 OH 15. The major product obtained in the following
(c) CH3–CH–CH–CH3 reaction is
(1) a > b > c (2) b > a > c OH
(3) b > c > a (4) c > b > a (i) NaOH
+ CHCl3 (ii) Slight H+
11. Conc H2SO4
CH3–CH–CH2–CH3 Heat
OH
OH
CHO
Product (I) + Product (II)
Which is not true regarding the products: HO
(1)
(1) Product (I) and (II) are position isomers
OH
(2) Product (I) and (II) contains the same
number of sp3 and sp2 carbon atoms COOH
(3) The yield of the product I and II is same OH
(4) Reaction obeys Saytzeff rule (2)
HO
12. Among the following compounds which can
OH
be dehydrated very easily ?
(1) CH3CH2CH2CH2CH2OH Cl
OH (3)
(2) CH3CH2CH2CHCH3 OH
CH3 OH
CHO
(3) CH3CH2CCH2CH3
(4)
OH Cl
(4) CH3CH2CHCH2CH2OH OH
CH3

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Cl 20. Identify the major product A and B

O
NaOH
⊕
H CH3–C–Cl
respectively in the following reaction of
16. X y Z; is Z
623K phenol
300atm
OH
O O
Conc. Conc.
CH3 (A) (B)
(1) (2) Cl H2SO4 HNO3

OH OH
O CH3 O Cl NO2
(3) (4) (1) and
O O
SO3H SO3H
17. Least soluble alcohol in water is: OH OH
O2N NO2
OH (2) and
(1) (2)
OH
SO3H SO3H
OH
(3) (4) OH
OH OH
SO3H O2N NO2
(3) and
18. Which of the following statement is
SO3H NO2
incorrect.
OH OH
(1) Ethanol and phenol can be
(4) and
differentiated by their reaction with NO2
bromine water. SO3H SO3H
(2) O-Nitrophenol and p-nitrophenol can
be separated by steam distillation 21. Which of the following options is not
method correctly showing the trend of the
(3) Vapours of t-butyl alcohol when passed properties mentioned?
over copper at 573K forms isobutylene (1) CH3–CH2–OH > CH3–CH2–CH2–OH >
CH3–CH2–CH2–CH2–OH (Solubility)
by dehydrogenation
(2) CH3–CH2–OH < CH3–CH2–CH2–OH <
(4) Primary alcohols do not produce
CH3–CH2–CH2–CH2–OH (Boiling point)
turbidity with Lucas reagent at room
(3) CH3–CH2–CH2–CH2–OH >
temperature
CH3
CH3–CH– CH2–OH > CH3–C–OH (Boiling
CH3
⊕ CH3 CH3
O2 H2O/H
19. H3C–CH 
Air
→ (X)  → (Y) + point)

(4) CH3–CH2–CH2–CH2–OH > CH3–CH– CH2–OH

CH3
carbolic acid; y is CH3
(1) Phenol (2) Benzyl alcohol > CH3–C–OH ( Solubility)
(3) Propanone (4) Acetophenone CH3

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22. Which of the following reaction will not 24. Among the following compounds which can
yield phenol? be dehydrated most difficulty.
Cl (1) CH3–CH2–CH–CH2–OH

(1) (i) Fusion with NaOH at 300 atm CH3

(ii) H2O/H+
(2) CH3–CH2–CH2–CH–CH3
NH2
OH
(2) (i) NaNO2/HCl
OH
(ii) H2O(warming)

(i) Oleum (3)

(3) (ii) NaOH,(heating)
(iii) H+ (4) CH3–CH2–CH = CH–CH2–OH
Cl
25. In the given reaction :
(4) Conc. H2SO4, ∆
OH

(i) NaOH/CO2/∆
[X]
23. In the following reactions, product ‘P’ is (ii) H⊕

OH Major product [X]is :

(i) NaOH ⊕
(ii) CO2 (CH3CO)2O, H (1) Salicylic acid
A (P)
(iii) H+
(Major) (Major) (2) m-hydroxybenzoic acid
(1) Salicylaldehyde (2) Benzoquinone (3) Mixture of 1 and 2
(3) Salicylic acid (4) Aspirin (4) Salicylaldehyde

Ether:
Classification: Ether are classified as simple or symmetrical, if the alkyl or aryl group attached to the
oxygen atom are the same, and mixed or unsymmetrical, if the two groups are different.

Symmetrical ether: C2H5OC2H5, O

Unsymmetrical ether: CH3–O–C2H5. O–C2H5

Naming of Ether: Common names of ethers are derived from the names of alkyl/aryl groups written as
separate words in alphabetical order and adding the word ether at the end.
Ex. CH3–O–C2H5 ethyl methyl ether
C2H5–O–C2H5 diethyl ether

•• ••
Structure of Functional Groups: O
CH3 CH3
111.7°
• In ethers, the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in
a tetrahedral arrangement. The bond angle is slightly greater than the tetrahedral angle due to the
repulsive interaction between the two bulky (–R) groups.
• The (C–O) bond length is almost the same as in alcohols.

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Preparation of ethers
(1) By intermolecular dehydration of alcohol:
Alcohols undergo dehydration in the presence of protic acids (conc. H2SO4, or cone. H3PO4). The
formation the product, alkene or ether depends on the reaction condition.
H2SO4
Ex. CH2=CH2
CH3–CH2–OH 443 K
H2SO4
CH3–CH2–O–CH2–CH3
413 K
The formation of ether is a nucleophilic bimolecular reaction, as indicated below
H
•• ⊕
Step I: CH3CH2–O–H+H
••
+
CH3CH2–O–H
••

•• ⊕ H ⊕
Step II: CH3CH2–O••+ CH3CH2 —O CH3CH2–O–CH2CH3 + H2O
H
H H
⊕
Step III: CH3CH2–O – CH2CH3 CH3CH2–O–CH2CH3 + H+
H
This method is suitable for the preparation of ethers having 1° alkyl groups only and the alkyl group
should be unhindered and temp. be kept low other wise reaction favour the formation of alkene. The
dehydration of 2° and 3° alcohol give alkenes as major product.

(2) Williamson's synthesis:

It is an important method to prepare all type of ether. Alkyl halide is allowed to react with sodium
alkoxide.
RO—Na+ + X —R' R—O—R'

This is a nucleophilic substitution reaction which undergoes via SN2

RO—Na+ Θ
CH2–X RO----CH2-----Xδ- ROCH2R
R R
Note. Better results are obtained if the alkyl halide is 1°. In case of 2° alkyl halide, elimination dominate over
substitution. If 3º alkyl halide is used, an alkene is the only product and no ether is formed.
Θ⊕
C2H5ONa
Ex. (i) CH3–CH2–CH2–X SN2
CH3–CH2–CH2–O–CH2–CH3(Major)
(1° Alkyl halide)
Θ⊕
C2H5ONa
(ii) CH3–CH2–X
E2
CH3–CH=CH2 (Major)
CH3
(2° Alkyl halide)
CH3
C2H5ONa
(iii) CH3–C–X E2
CH3–C=CH2 (Only product)
CH3 CH3
(3°Alkyl halide)

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• Phenols are also converted to ethers by this method
OH ONa O–R

R—X
+NaOH

Physical properties:
Due to large size of alkyl groups (than H) R–O–R bond angle in ethers is large (> 110°) therefore net
dipole-moment becomes less and therefore ethers show:
(i) Low boiling points
(i) Low solubility in water
Solubility in water further decreases with increasing molar mass of ether as oxygen of ether become
more hindered and become less available for inter-molecular H-bonding with water.

Chemical reactions of ether

(1) Reaction with conc. HX:
• Ether due to low polarity are less reactive functional group
• With cold & Conc. HX, they form alkyl halide and alcohols by cleavage of C–O bond
• reactivity of hydrogen halides is → HI > HBr > HCl

CH3 CH3
Conc.
Ex. CH3–C–O–CH2–CH3 CH3–C–I+CH3–CH2–OH
HI
CH3 CH3

Mechanism:
CH3 CH3
•• HI ⊕
CH3–C–O–CH
•• 2–CH3 CH3–C — O–CH2–CH3
CH3 CH3 H
(Oxonium ion)
Slow step

CH3 CH3
IΘ
CH3–C–I+CH3–CH2–OH Fast step CH3–C⊕ + CH3–CH2–OH
CH3 CH3
• Reaction with cold & conc. HI is known as Zeisel method.
⊕
 
Note. (i) If oxonium ion gives resonance stabilised carbocation  Ph − CH2⊕ ,CH2 =
CH − CH2  or 3° carbocation
 
then reaction follow SN1 mechanism.
(ii) If oxonium ion does not give resonance stabilised or 3° carbocation then reaction follow SN2
mechanism and IΘ attack at less hindered carbon

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Alcohols, Phenols & Ethers CHEMISTRY
Conc.
Ex. CH3 − O − CH2 − CH3 
HI
→ CH3 − I + CH3 − CH2 − OH

Θ
⊕
mechanism: CH3–O–CH2–CH3+H–I CH3–O–CH2–CH3 Ι

H
Less hindered CH3–I+CH3–CH2–OH

If conc. HI is taken in excess then alcohol is also converted to Alkyl Iodide

•• ⊕ Θ
CH3–CH2–O–H
••
+ H– I CH3–CH2–O–H + I → CH3–CH2–I + H2O
H
Note. (iii) Alkyl aryl ethers are cleaved at the alkyl oxygen bond due to the more stable aryl-oxygen bond.
O–CH3 Conc. H–I OH
Ex. + CH3–I

(2) Electrophilic substitution :

Aryl ethers show the reactions like phenols as –OR is also an activating group.
The electrophilic substitution shown by aryl ethers are

(i) Halogenation
OCH3 OCH3 OCH3
O
Br
Br2/CH3–C–OH
+

Br

(ii) Friedel–Crafts Reaction

OCH3 OCH3 OCH3
CH3
CH3CI/AICl3(anhyd)
+

CH3
This is known as Friedel-Craft alkylation
OCH3 O OCH3 OCH3
COCH3
CH3C–CI/AICl3 (anhydrous)
+

COCH3
This is known as Friedel-Craft acylation

(iii) Nitration
OCH3 OCH3 OCH3
NO2
(conc. H2SO4 + conc. HNO3)
+

NO2

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CHEMISTRY Alcohols, Phenols & Ethers
Exercise 1.3
1. Which of the following is not expected to 6. In the reaction
give ether on reaction with sodium heat
(CH3)3 C — O — CH2CH3 + HI →
(1mole)
methoxide:
the product(s) formed is (are):
(1) CH3CH2CH2Cl
CH3
(2) CH2=CHCH2Cl
(1) CH3–C–OH and CH3CH2I
(3) PhCH2Cl
CH3
(4) CH2=CHCl
CH3
2. The given reaction is called as: (2) CH3–C–I and CH3CH2OH
C2H5ONa + BrC2H5 → C2H5–O–C2H5 + CH3
NaBr CH3
(1) Frankland reaction (3) CH3–C–I and CH3CH2I
(2) Wurtz reaction CH3
(3) Williamson's Ether synthesis CH3
⊕
(4) Cannizzaro reaction (4) H3C–C–O–CH2CH3 ΙΘ
H3C H
3. Among the following sets of reactants
which one produces anisole? 7. What will happen when diethyl ether
(1) CH3CHO; RMgX treated with hot and conc. HI:
(2) C6H5OH; NaOH; CH3I (1) Ethyl alcohol (2) Ethyl iodide
(3) C6H5OH; netural FeCl3 (3) Methyl iodide (4) Methanol
(4) C6H5–CH3; CH3COCl; AlCl3
8. The major organic product in the given
4. The order of reactivity of halogen acids with reaction is.
ether is: CH3 – O – CH(CH3)2 + HΙ → Product
(1) HCl > HBr > HI (1) CH3OH + (CH3)2CHΙ
(2) HI > HBr > HCl (2) ΙCH2OCH(CH3)2
(3) HCl > HI > HBr (3) CH3OC(CH3)2
(4) HI > HCl > HBr Ι
(4) CH3Ι + (CH3)2CHOH
5. Consider the following transformation:
conc. HI ∆/373 K
CH3CH=CH–O–CH2CH3 → 9. + HI P
O (excess)
The major product(s) formed is (are):
Product P of the reaction
(1) CH3CH=CHI and CH3CH2I
(2) CH3CH=CHI and CH3CH2OH (1) (2)
HO I I I
(3) CH3CH2CHO and CH3CH2I
(4) CH3CH2CH–O–CH2CH3 (3) (4) No reaction
Ι OH OH

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Alcohols, Phenols & Ethers CHEMISTRY
18
10. HO O–CH3
HBr
The major ONa
Heat
Conc. H2SO4
product obtained in the following reaction is: 13. + H3C –I → (X) 
Conc. HNO3
→(Y)
18
(1) HO Br , CH3–OH
major product Y is:
OCH3 OCH3
(2) HO OH , CH3–Br NO2 SO3H
(1) (2)
18
(3) HO OH , CH3–Br

OCH3 OCH3
18
(4) Br OH , CH3–OH
(3) (4)

SO3H NO2
11. Mark the correct statement:
(1) Ethers behave as Lewis base
14. In the given reaction
(2) Ethers form coordinated complexes
Excess HΙ / ∆
C6H5 − O − CH2 − CH3  →[X] + [Y]
with Lewis acids
[X] and [Y] will be :
(3) With cold HI, diethyl ether gives ethyl
(1) C6H5I and CH3CH2I
alcohol & ethyl iodide (2) C6H5OH and CH3 − CH2 − I
(4) All are correct (3) C6H5I and CH3CH2OH
(4) C6H5OH and CH2 = CH2
12. In the reaction
15. Which of the following reaction follows SN2
Cold & conc. HΙ
O—CH2CH3 path.
Conc. Hl
(1) CH2–O–CH3
the product(s) formed is (are):
OH + CH3CH2—Ι Conc. Hl
(1) (2) O

(2) Ι + CH3CH2—OH (3) O Conc. Hl

(3) Ι + CH3CH2—Ι CH3

Conc. Hl
(4) CH3–CH–CH2–C–O-CH2–CH3
⊕ Θ
(4) O—CH2CH3 Ι CH3 CH3
H

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CHEMISTRY Alcohols, Phenols & Ethers
Exercise 2
1. In a set of the given reactions, acetic acid 5. Mark the correct decreasing order of
yielded a product C. reactivity of the following compounds with
PCl5 C6H6 Lucas reagent
CH3COOH  → A  →B
AICI 3
CH2OH CH2OH CH2OH
(i) C2H5MgBr/ether
 →C
⊕
(ii) H3O (a) (b) (c)
product C would be:
OH Br CH3
(1) CH3CH(OH)C2H5 (2) CH3COC6H5
C2H5 (1) c > b > a (2) c > a > b
(3) a > c > b (4) a > b > c
(3) CH3CH(OH)C6H5 (4) CH3–C(OH)C6H5
6. Identify the products (A) and (B) respectively
in the following reaction.
2. Which one among the following reactions is
an example of Williamson synthesis - (i)Cl2 /AlCl3
 → (A) →
2 2 7 Na Cr O
(B)
(ii) NaOH, 625K , 300atm H SO 2 4
Zn−Hg
(1) CH3COCH3 
Conc.HCl,heat
→ CH3CH2CH3 (iii) H+

H2SO4 (Conc.)
OH O–Na+
(2) C2H5OH 
170°C
→ CH2 = CH2
heat (1) [A] , [B]
(3) C2H5I + C2H5ONa → C2H5OC2H5
heat
(4) HCHO+NaOH → HCOONa+CH3OH OH

3. Mark the correct increasing order of (2) [A] , [B]

reactivity of the following compounds with O
HBr:
CH2OH CH2OH CH2OH (3) [A] , [B]

(a) (b) (c) OH O

NO2 Cl (4) [A] , [B]

(1) a < b < c (2) b < a < c
O
(3) b < c < a (4) c < b < a
SO3H

CH3 7. (i) NaOH(molten)

 → (X)
+
(ii) H
Cu
4. CH3–C–OH  → A + HBr →
300°C (i) CHCl3 + Aq. NaOH
 → (Y);
CH3 +
(ii) H
(iii) Zinc dust
Mg/dry ether H2O
B  → C → D Major Products (Y) is
What is D ? COOH CHO
CH2 CH3 OH
(1) CH3–C (2) CH3–C–Br (1) (2)

CH3 CH3 COOH CHO

CH3 OH
CH3
(3) CH3–C–MgBr (4) CH3–CH–CH3 (3) (4)

CH3
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Alcohols, Phenols & Ethers CHEMISTRY
8. Grignard’s reagent can be used to prepare 12. Arrange the following compounds in
tertiary alcohol by reaction with increasing order of boiling point.
(1) Acetone (2) Acetaldehyde Propan-1-ol, butan-1-ol, butan-2-ol, pentan-
(3) Benzaldehyde (4) Formaldehyde 1-ol
(1) Propan-1-ol, butan-2-ol, butan-1-ol,

9. Phenol can be prepared by pentan-1-ol

(2) Propan-1-ol, butan-1-ol, butan-2-ol,
Cl
pentan-1-ol
i) O2 NaOH
(1) ii) H3O⊕
(2) (3) Pentan-1-ol, butan-2-ol, butan-1-ol,
∆ & Pr
propan-1-ol
SO3H COOH
(4) Pentan-1-ol, butan-1-ol, butan-2-ol,
NaOH Zn
(3) (4) propan-1-ol
∆ & Pr ∆

OH 13. An organic compound (A) with molecular

formula C6H12O2 is hydrolysed with dil.
10. Which of the following conversion is H2SO4 to give a carboxylic acid (B) and an
incorrect? alcohol (C). ‘C’ gives white turbidity
NaI → CH − CH −CH − I
(1) CH3− CH2− CH2 − Cl  immediately when treated with anhydrous
Acetone 3 2 2

ZnCl2 and conc. HCl. The organic compound

Cl
(2)
H2O (A) is
25°c OH
O O
O OH (1) (2)
OCH3 OCH3 O O
NaBH4
(3) O O O
O
(3) (4) O
CH3 CH3 O
(4) CH3–C–Br + NaOCH3 Methanol CH3–C–OCH3
CH3 CH3 14. The final product formed in the following
multistep reaction is :
(I)B H
11. HI
→X+Y CH2 →
CH3−CH= 2 6
(II)H O ,NaOH
O–CH2 2 2 (aq.)

Cu/573K (I)CH3MgBr
Mg (A)  →(B) → ⊕ (C)
X 
dryether
→ Z (organometallic compound) (II)H2O/H

Y
(1) CH3–CH2–CH2– CH2–OH
Z  →P ; P is
(2) CH3–CH2–CH–CH3
CH3 OH
OH
(1) (2)
(3) CH3–C–CH3
O–Ph CH2–OH CH3
(4) CH3–CH–CH3
(3) (4)
CH2–OH

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CHEMISTRY Alcohols, Phenols & Ethers
15. Which of the following compounds give instant 18. Which of the following statement is correct?
turbidity with anhyd. ZnCl2/Conc.HCl(Lucas (1) The reaction of methyl magnesium
reagent) iodide with acetone followed by
CH2–CH3 hydrolysis gives secondary butanol.
CH3 OH
(a) CH3–C–OH (b) (2) Primary alcohols are dehydrated easily
H
than secondary and tertiary alcohols.
CH3 CH3–CH2
(3) Tertiary alcohol is more acidic than
OH
primary alcohol.
(c) OH (d) (4) Tertiary butyl alcohol gives turbidity
fastest with Lucas reagent.
(e) OH
19. Choose the correct X and Y in the given
(1) Only a and c (2) Only a, b and c reactions.
(3) Only a, c and d (4) Only e O
CH2–C–OCH3 NaBH4 (X)
(i) O
CH3
conc.H PO
16. CH3—C — CH—CH3 
∆

3 4
→A NaBH4
(ii) CH3CH2CH–CHO (Y)
CH3 OH
CH3
(i) O3

(ii) Zn/H2O
→ B ; Final product ‘B’ is O
CH2–CH2–OH ;
O (1) (X)
(1) 2CH3—C — CH3
O (Y) CH3CH2CH–CH2OH
(2) HCHO+ CH3—C—CH— CH3 CH3
CH3 OH
H3C CH2–CH–OCH3 ;
(3) CH3—CHO + C=O (2) (X)
H3C OH
CH3
(Y) CH3CH2–CH–CH2OH
(4) CH3—C — CHO + CHO
CH3
CH3
OH
CH2–C–OCH3 ;
17. A primary alcohol, C3H8O(A) on heating with (3) (X) O
sulphuric acid undergo dehydration to give
an alkene (B). ‘B’ when reacts with HCl gives (Y) CH3CH2–CH–CH2OH
C, which on treatment with aqueous KOH
CH3
gives compound D(C3H8O).
O
A and D are CH2–C–OCH2 OH ;
(1) Functional group isomers (4) (X)
O
(2) Position isomers
(3) Chain isomers (Y) CH3CH2CH2CH2CH2OH
(4) stereoisomers.
86 Sarvam Career Institute
Alcohols, Phenols & Ethers CHEMISTRY
20. Which reducing agent is used for the 23. Which of following is incorrect for reactivity
following conversion ? of 1°, 2° and 3° alcohols with given reagents.
RCOOH→RCH2OH (1) Esterification 3°>2°>1°
(1) LiAlH4 (2) NaBH4 (2) Lucas reagent 3°>2°>1°
(3) K2Cr2O7 (4) KMnO4 (3) With Na 1°>2°>3°
(4) All of these
21. In the following reactions,
CH3 24. Benzene is converted into para-nitrophenol
H+/heat by using reagents:
(i) CH3–CH–CH–CH3 A + B
Major Minor
(I) dil. NaOH (II) HNO3/H2SO4
OH product product (III) Br2/Fe
The most suitable sequence of these
HBr, dark reagents are respectively
(ii) A C + D
In absence of peroxide
Major Minor (1) I, II, III (2) I, III, II
product product (3) II, III, I (4) III, II, I
The major product A and C are respectively
CH3 CH3 25. In following reaction
Η⊕ O2 H⊕/H2O
(1) CH2=C–CH2–CH3 and CH2–CH–CH2–CH3 +CH3–CH=CH2 A B ∆
Br
OH
CH3 CH3 O
(2) CH3–C=CH–CH3 and CH3–C–CH2–CH3 CH3–C–CH3+
Br
The structure of ‘B’ is :-
CH3 CH3 CH3 CH3
(3) CH3–C=CH–CH3 and CH3–CH–CH–CH3 (1) CH–CH3 (2) CH3–C–O–O–H
Br OH
CH3 CH3
(4) CH2=C–CH2–CH3 and CH3–C–CH2–CH3 CH3 CH3
Br (3) CH3–C–O–H (4) CH3–C–OH
OH
OH

Br2/CS2
22. In the given reactions, 273 K X
26. In the following sequence of reaction
(i) CHCl3, aq.NaOH
Red P+Ι2 Mg HCHO
(ii) dil. HCl CH3–CH2OH A dry ether B C
Y H2O
X and Y are respectively D
(1) bromobenzene and acetophenone the compound D is-
(2) o-and p-bromophenol and salicylaldehyde (1) Butanal (2) n-butyl alcohol
(3) p-bromophenol and salicylic acid (3) n-propyl alcohol (4) propanal
(4) o-bromophenol and benzoic acid.

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CHEMISTRY Alcohols, Phenols & Ethers
27. An ether is more volatile then alcohol having 32. The reactant A in the following reactions is:
the same moleculer formula this is due to :- CH3
(1) Inter moleculer H -bonding in ether K Cr O (i) CH MgΙ
A dil.2 H22SO7 4 B (ii) H 3O/H⊕ CH3–C–CH3
2
(2) Inter moleculer H -bonding in alcohols
OH
(3) Dipolar character of ether
(1) CH3CHOHCH3 (2) CH3CH2CHO
(4) Alcohol has resonating structure
(3) C2H5OH (4) CH3COOH
28. n-propyl bromide reacts with aq. KOH to
33. In the given reaction
form:-
PCC
(1) Propane (2) Propene CH3 − CH
= CH − CH2 − OH → 'P'
(3) Propyne (4) 1-propanol Product 'P' is :
(1) CH3–CH2–CH2–CH2–OH
29. When phenol is treated with bromine/H2O, it (2) CH3–CH2–CH2–CHO
gives:- (3) CH3–CH=CH–CHO
(1) o-and p-dibromophenol (4) CH3–CH=CH–COOH
(2) 2,3,4-tribromophenol
(3) 2,4,6-tribromophenol 34. In the given reaction
ZnO + Cr2O3
(4) None CO + H2 →
573−673 K
Product
(1) CH3OH (2) CH2–OH
30. Among the following the one that gives
CH2–OH
positive iodoform test upon reaction with I2
(3) CHO (4) COOH
and NaOH is:-
(1) CH3–CH2–CH–CH2–CH3 CHO COOH

OH
35. The following reaction :
(2) C6H5–CH2–CH2–OH
(3) CH3–CH–CH2OH OH
conc. H2SO4
CH3 CH3–CH–CH2–CH3 ∆ CH3–CH=CH–CH3
(4) PhCHOHCH3 is known as :
(1) Friedel-Craft reaction
31. In the given reaction : (2) Acidic dehydration of alcohol
(i) CH3MgBr (3) Kolbe reaction
CH3–CH–CH2 [X]
(ii) HOH/H+ (4) Gatterman-Koch reaction
O
[X] will be :
36. Ethylene glycol on oxidation with periodic
(1) CH3–CH–CH2–OH acid gives :
CH3 (1) Oxalic acid (2) Glyoxal
(2) CH3–CH–CH2–CH3 (3) Formaldehyde (4) Glycolic acid

OH 37. Amongst the following alcohols which would

OH react fastest with conc. HCl and ZnCl2 ?
(3) CH3–C–CH3 (1) 2-Methylbutan-1-ol
(2) 2-Pentanol
CH3
(3) 1-Pentanol
(4) CH2=CH–CH3 (4) 2-Methyl-2-butanol

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Alcohols, Phenols & Ethers CHEMISTRY
38. Complete the missing links. 42. Which compound on reaction with Cu at
CH3CHBrCH3 300°C may not give aldehyde or ketone
alc. KOH
 HBr
→ X  CH3ONa
→ Y  →Z OH
Peroxide

X, Y, Z are respectively ? (1) (2)

OH
(1) CH3CH = CH2
CH3CH(Br)CH2Br, CH3CH(OH)CH3 (3) (2) All of these
OH
(2) CH3CH = CH2
CH3CH2CH2Br, CH3CH2OCH2CH3 43. R − OH Reagent
→R − Cl
(3) CH3CH = CH2 Which of the following reagents can be used
CH3CH(Br)CH3 , CH3CH2OCH2CH3 for this reaction.
(4) CH3CH = CH2 (a) HCl/ZnCl2 (b) NaCl
(c) PCl3 (d) PCl5
CH3CH2CH2Br, CH3CH2CH2OCH3
(e) SOCl2
(1) a, c, d, e (2) a, b, c, d
39. Butan-2-one is obtained, if following
(3) a, b, d, e (4) All of the above
compound is reacted with methyl
magnesium bromide followed by hydrolysis?
44. Which of the following alcohols is most
(1) CH3–CHO (2) CH3–C–Cl
reactive towards substitution with HBr ?
O CH2–OH
(3) CH3–C≡N (4) CH3–CH2–C≡N (1)

CH2–OH
CH3
(2)
40. CH–CH3 O 2N
1. O2
CH2–OH
A+B
2. H3O+ (3)
CH3O
A and B can't be distinguish by :-
CH2–OH
(1) Neutral FeCl3 test
(4)
(2) 2, 4-DNP test Br
(3) Bromine water test
(4) Tollen's test OH
Conc.
45. H2SO4
+
OH K CH3 −CH2 −Cl ∆
41. →( A)  →(B), B is :
A B
OH O-CH2–CH3 consider the above reaction, and choose the
correct statement :
(1) (2)
(1) The reaction is not possible in acidic
CH2–CH3
medium
OH (2) Both compounds A and B are formed
CH2–CH2–OH
equally
(3) (4) (3) Compound A will be the major product
CH2–CH3 (4) Compound B will be the major product
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CHEMISTRY Alcohols, Phenols & Ethers
46. Which compound give alcohol on reaction (i) Nal, H3PO4
with one mole of phenyl magnesium bromide 49. H3C–CH2–CH–CH3 (ii) Mg, Dry ether
(iii) D2O
[x]
(PhMgBr) followed by acidic hydrolysis ? Product
(1) CH3–C≡N (2) Ph–C≡N OH
Product [X] formed in the above reaction is :
O
H
(3) CH3–C–O–Ph (4) Ph–C–CH3
(1) H3C–CH2–CH–CH3 (2) H3C–CH2–C–CH3
O
D OH
(3) H3C–CH2–CH=CH2 (4) H3C–CH=CH–CH3
47. Consider the following reaction sequence.
OH
OH
NaOH (i) CO2 (CH3CO)2O
A (ii) H3O+ B C H3PO4 (BH3)2
(Major) H⊕ (Major)
50. 120°C A H2O2/OH, H2O P
Major Product Major Product
The product C is
Consider the above reaction and identify the
COOH COOH Product P:
OCOCH3 OH CH3 OH
(1) (2)

O CH3 (1) (2)

COOCH3 COOCH3
OH CH3
OH OCOCH3
(3) (4) OH OH
(3) (4)

48. 'A' and 'B' formed in the following set of

reactions are: CH2–OH
CrO3
X;X
51. anhydrous
OH OH
HBr HBr
∆ A ∆ B
(1) CH=O (2) COOH
CH2OH OCH3
OH OH (3) CH3 (4) O
(1) A= ,B=
Cl
CH2Br Br
OH Br (i) NaOH, ∆& Pressure (i) CO2, NaOH, ∆
52. X Y; Y
(ii) H⊕ (ii) H⊕
(2) A= ,B= Major
is :
CH2Br OCH3 OH OH
OH OH COOH
(3) A= ,B= (1) (2)

CH2Br OH COOH
OH Br COOH OH
(4) A= ,B=
(3) (4)
CH2Br Br

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Alcohols, Phenols & Ethers CHEMISTRY

53. conc. H2SO4

CH3 − CH2 − OH  → 58. In the given pair of alcohols, in which pair
second alcohol is more reactive than first
CH3 − CH2 − O − CH2 − CH3
towards hydrogen bromide?
Suitable temperature in the above reaction OH
to get ether with good yield is :
(1) and CH–CH3
(1) 443 K (2) 413 K
(3) 100 K (4) 470 K OH
CH3 OH
H OH
Na CH3 −CH2 −Cl
(2)
54. CH3 − CH2 − OH  → X  → Y;Y is : and
(1) CH3 − CH2 − O − CH2 − CH3
and OH
(2) CH3 − CH2 − CH2 − CH2 − Cl (3)
OH
(3) CH3 − CH2 − CH2 − CH2 − OH
(4) CH3 − CH2 − CH2 − CH2 − ONa (4) and
OH OH
HΙ / ∆ 59. Which of the following is correct
55. CH3 − O − CH2CH3 →
Excess
(A) + (B) HBr
(A) CH3 − CH2 − CH2 − O − CH3  →
Product A and B are
CH3Br + CH3CH2CH2OH
(1) CH3OH + CH3CH2I
OCH2CH3 Br
(2) CH3I + CH3CH2OH
(B)
(3) CH3I + CH3CH2I HBr
+CH3CH2OH
(4) CH3OH + CH3CH2OH
CH3
56. Ethyl chloride is converted into diethyl ether (C) CH3–C–O–CH2CH3 HI

by
CH3
(1) Hell-Volhard Zelinsky reaction
CH3
(2) Kucherov's reaction
(3) Wurtz reaction CH3–C–OH+CH3CH2I
(4) Williamson's synthesis CH3
OCH3 OCH3 OCH3
57. R–1
NO2
CH3 CH3 (D) HNO3
Θ H2SO4 +
⊕
CH3–C–ONa+CH3–Br CH3–O–C–CH3
CH3 CH3 NO2
HI
R–2 (E) CH2–O
CH3 CH3
⊕ Θ
CH3–C–Br+NaO–CH3 CH3–O–C–CH3 CH2–I+ OH

CH3 CH3 (1) Only A & B correct

Which reaction is appropriate for the (2) Only B & C incorrect
preparation of an ether (3) Only A & D incorrect
(1) Both R1 & R2 (4) Only D & E incorrect
(2) Only R1 60. Which of the following reagent can't be used
(3) Only R2 to distinguish between phenol and ethyl
(4) Both R1 & R2 wrong alcohol ?
(1) Na (2) Neutral FeCl3
(3) Br2/H2O (4) NaOH
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CHEMISTRY Alcohols, Phenols & Ethers
Exercise 3
1. Assertion: Phenol is stronger acid than 4. Assertion: Phenols give picric acid on
alcohols. nitration with conc. HNO3.
Reason: Conjugate base of phenol is stabilized by Reason: –OH group in phenol shows –M
resonance whereas alcohol are not. effect.
(1) If both Assertion & Reason are True & the (1) If both Assertion & Reason are True & the
Reason is a correct explanation of the Reason is a correct explanation of the
Assertion. Assertion.
(2) If both Assertion & Reason are True but (2) If both Assertion & Reason are True but
Reason is not a correct explanation of Reason is not a correct explanation of
the Assertion. the Assertion.
(3) If Assertion is True but the Reason is (3) If Assertion is True but the Reason is
False. False.
(4) If Assertion & Reason are false. (4) If Assertion & Reason are false.

2. Assertion: Anisole on reaction with HI gives 5. Assertion: o-nitrophenol is more volatile

phenol and CH3I. than p-nitrophenol
Reason: Phenol-oxygen bond is stronger Reason: Intramolecular hydrogen bonding is
present in o-nitrophenol while
than methyl-oxygen bond in anisole and
intermolecular H-bonding is present in p-
hence it is not cleaved by HI.
nitrophenol.
(1) If both Assertion & Reason are True & the
(1) If both Assertion & Reason are True & the
Reason is a correct explanation of the Reason is a correct explanation of the
Assertion. Assertion.
(2) If both Assertion & Reason are True but (2) If both Assertion & Reason are True but
Reason is not a correct explanation of Reason is not a correct explanation of
the Assertion. the Assertion.
(3) If Assertion is True but the Reason is (3) If Assertion is True but the Reason is
False. False.
(4) If Assertion & Reason are false. (4) If Assertion & Reason are false.

6. Assertion: CH3OCH3 and C2H5OH has

3. Assertion: In Lucas test, 3° alcohols react
comparable molecular weight but boiling point
immediately. of C2H5OH is more than dimethyl ether.
Reason: A mixture of anhydrous ZnCl2 and Reason: C2H5OH forms intermolecular H-
conc. HCl is called Lucas reagent. bonding while CH3OCH3 forms
(1) If both Assertion & Reason are True & the intramolecular H-bonding.
Reason is a correct explanation of the (1) If both Assertion & Reason are True & the
Assertion. Reason is a correct explanation of the
(2) If both Assertion & Reason are True but Assertion.
Reason is not a correct explanation of (2) If both Assertion & Reason are True but
Reason is not a correct explanation of
the Assertion.
the Assertion.
(3) If Assertion is True but the Reason is
(3) If Assertion is True but the Reason is
False. False.
(4) If Assertion & Reason are false. (4) If Assertion & Reason are false.
92 Sarvam Career Institute
Alcohols, Phenols & Ethers CHEMISTRY
7. Assertion: Ethers behave as bases in the 10. Assertion: Depending on condition alcohol
presence of mineral acids can react both as nucleophile and
Reason: It is due to the presence of lone pair electrophile
Reason: Alcohol react with active metals
of electrons on the oxygen.
such as sodium, potassium to yield
(1) If both Assertion & Reason are True & the
corresponding alkoxide and liberate
Reason is a correct explanation of the hydrogen
Assertion. (1) If both Assertion and Reason are correct
(2) If both Assertion & Reason are True but and the Reason is a correct explanation
Reason is not a correct explanation of of the Assertion.
the Assertion. (2) If both Assertion and Reason are correct
but Reason is not a correct explanation
(3) If Assertion is True but the Reason is
of the Assertion.
False.
(3) If the Assertion is correct but Reason is
(4) If Assertion & Reason are false. incorrect.
(4) If the Assertion is incorrect and Reason
8. Assertion: The major products formed by is correct.
heating C6H5CH2OCH3 with HI are C6H5CH2I
and CH3OH 11. Assertion: Treatment of bromine water with
propene yields 1–bromo Propan–2–ol.
Reason: Benzyl cation is more stable than
Reason: Attack of water on bromonium ion
methyl cation.
follows Markovnikov rule and results in 1–
(1) If both Assertion & Reason are True & the bromo Propan–2–ol.
Reason is a correct explanation of the (1) If both Assertion and Reason are correct
Assertion. and the Reason is a correct explanation
(2) If both Assertion & Reason are True but of the Assertion.
Reason is not a correct explanation of (2) If both Assertion and Reason are correct
but Reason is not a correct explanation
the Assertion.
of the Assertion.
(3) If Assertion is True but the Reason is
(3) If the Assertion is correct but Reason is
False. incorrect.
(4) If Assertion & Reason are false. (4) If the Assertion is incorrect and Reason
is correct.
9. Assertion: 2-Butanol on heating with H2SO4
gives 2-butene as major product. 12. Assertion: synthesis of ethyl phenyl ether
may be achieved by Williamson synthesis.
Reason: Dehydration of 2-butanol follows
Reason: Reaction of bromobenzene with
Saytzeff's rule.
sodium ethoxide yields ethyl phenyl ether.
(1) If both Assertion & Reason are True & the (1) If both Assertion and Reason are correct
Reason is a correct explanation of the and the Reason is a correct explanation
Assertion. of the Assertion.
(2) If both Assertion & Reason are True but (2) If both Assertion and Reason are correct
Reason is not a correct explanation of but Reason is not a correct explanation
the Assertion. of the Assertion.
(3) If the Assertion is correct but Reason is
(3) If Assertion is True but the Reason is
incorrect.
False. (4) If the Assertion is incorrect and Reason
(4) If Assertion & Reason are false. is correct.
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CHEMISTRY Alcohols, Phenols & Ethers
13. Assertion: The correct order of acidic 16. Assertion: Phenol forms 2, 4, 6-
strength for the alcohols is
CH3 tribromophenol on treatment with Br2–
CH3–CH2–OH > CH3–CH–OH > CH3–C–OH water at 273 K.
CH3 CH3
Reason: Phenol is o, p-directing group
Reason : Electron releasing group (–CH3)
increases electron density on oxygen (1) If both Assertion and Reason are correct
tending to decrease the polarity of O–H
bond. and the Reason is a correct explanation
(1) If both Assertion and Reason are correct
and the Reason is a correct explanation of the Assertion.
of the Assertion. (2) If both Assertion and Reason are correct
(2) If both Assertion and Reason are correct
but Reason is not a correct explanation but Reason is not a correct explanation
of the Assertion.
(3) If the Assertion is correct but Reason is of the Assertion.
incorrect. (3) If the Assertion is correct but Reason is
(4) If the Assertion is incorrect and Reason
is correct. incorrect.
14. Assertion: Anisole undergoes electrophilic
substitution at ortho and para-positions. (4) If the Assertion is incorrect and Reason
Reason: Anisole is less reactive than phenol is correct.
towards electrophilic substitution reaction.
(1) If both Assertion and Reason are correct
and the Reason is a correct explanation
of the Assertion. 17. Given below are two statement.
(2) If both Assertion and Reason are correct Statement I:- Bromination of phenol in
but Reason is not a correct explanation
of the Assertion. solvent with low polarity such as CHCl3 or
(3) If the Assertion is correct but Reason is
incorrect. CS2 requires lewis acid catalyst
(4) If the Assertion is incorrect and Reason
Statement II:- The lewis acid catalyst
is correct.
15. Assertion : When alkyl aryl ethers react with polarizes the bromine to generate Br+
excess of hydrogen halides, phenol and alkyl
(1) Statement I is true but Statement II is
halide are produced.
Reason: Alkylarylethers are cleaved at the false.
alkyl-oxygen bond due to more stable aryl-
oxygen bond. (2) Statement I is false but Statement II is
(1) If both Assertion and Reason are correct
and the Reason is a correct explanation true
of the Assertion. (3) Both Statement I and Statement II are
(2) If both Assertion and Reason are correct
but Reason is not a correct explanation true
of the Assertion.
(3) If the Assertion is correct but Reason is (4) Both Statement I and Statement II are
incorrect.
false.
(4) If the Assertion is incorrect and Reason
is correct
94 Sarvam Career Institute
Alcohols, Phenols & Ethers CHEMISTRY
18. Match the starting materials given in 20. Match the column I with column II and mark
Column-I with the products formed by these the appropriate choice.
(Column II) in the reaction with HI. Column-I
Column I Column II OH OH OH
OH CH3
a CH3–O–CH3 i (A) + CH3Cl
Anhyd. AlCl3
CS2 +

+ CH3Ι
CH3
b CH3 ii CH3 OH OH
CH–O–CH3 CHO
CH3 CH3–C–Ι + CH3OH (i) CHCl3/aq. NaOH
(B) (ii) H⊕
CH3
OH
c CH3 iii Ι OH
COOH
H3C–C–O–CH3 (i) NaOH
+ CH3OH (C) (ii) CO2
CH3 (iii) H⊕

d OCH3 iv CH3–OH + CH3–Ι COOH

NaOH/ CaO
(D) ∆

Column-II
v CH3
CH–OH+CH3Ι (i) Decarboxylation
CH3 (ii) Friedel-Crafts reaction
vi CH3 (iii) Reimer-Tiemann reaction
CH–Ι+CH3OH (iv) Kolbe’s reaction
CH3 (1) (A)→(i), (B)→(ii), (C)→(iii), (D)→(iv)
vii CH3
(2) (A)→(ii), (B)→(iii), (C)→(iv), (D)→(i)
CH3–C–OH + CH3Ι (3) (A)→(iii), (B)→(iv), (C)→(i), (D)→(ii)
CH3 (4) (A)→(iv), (B)→(iii), (C)→( ii), (D)→(i)
(1) a → v; b → iii; c → vi; d → i 21. Match the column I with column II and mark
(2) a → vii; b → vi; c → v; d → ii the appropriate choice.
(3) a → iv; b → v; c → ii; d → i Column I Column II
(4) a → iii; b → i; c → iv; d → ii (A) Catechol (i) OH

19. Match list I with list II:

List I List II OH
(Compound) (pKa value) (B) Resorcinol (ii) OH
A. Ethanol I. 10.0 OH

B. Phenol II. 15.9

C. m-Nitrophenol III. 7.1 (C) p-Cresol (iii) OH

D. p-Nitrophenol IV. 8.3

Choose the correct answer from the options OH
gives below: (D) Quinol (iv) OH

(1) A-I, B-II, C-III, D-IV

(2) A-II, B-I, C-IV, D-III
(3) A-III, B-IV, C-I, D-II CH3
(1) (A)→(ii), (B)→(iii), (C)→(iv), (D)→(i)
(4) A-IV, B-I, C-II, D-III
(2) (A)→(i), (B)→(ii), (C)→(iii), (D)→(iv)
(3) (A)→(iv), (B)→(iii), (C)→(ii), (D)→(i)
(4) (A)→(ii), (B)→(iv), (C)→(i), (D)→(iii)
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CHEMISTRY Alcohols, Phenols & Ethers
22. Match the column 24. Match the column
Column-I Column-II Column-I Column-II
(Reactions) (Product) (Reactions) (Product)

(a) OH (P) OH (a) OH (q) Aspirin

CHO Conc
Br2/H2O
HNO3

(b) OH (p) Picric acid

(b) OH (Q) OH
NaOH
(i) CHCl3+Aq.NaOH CO2/H⊕
(ii) H⊕

(c) OH (r) Salicylic acid

Br
(c) OH (R) OH Na2Cr2O7
H2SO4
Br Br
(i) CO2+NaOH
(ii) H+ (d) Salicylic acid (s) Benzoquinone
3 ( CH CO ) O
Br +
2
H

(d) OH (S) OH (1) a-p, b-r, c-s, d-q

COOH (2) a-q, b-s, c-r, d-p
Br2/CS2
(3) a-s, b-r, c-p, d-q
(4) a-r, b-s, c-q, d-p
(1) a-R, b-P, c-S, d-Q
(2) a-P, b-S, c-R, d-Q
25.
(3) a-P, b-Q, c-R, d-S
Column-I Column-II
(4) a-S, b-Q, c-P, d-R (Reactions) (Product)
(a) (q) C2H5–O–C2H5
CO+2H2
23. Match the following :
ZnO–Cr2O3
(a) Phenol + (q) No reaction 573-673K

Neutral FeCl3 (b) C6H12O6

Zymase (p) CH3–OH
(b) Phenol+Br2(aq.) (p) Violet
(c) (r) C2H5OH+CO2
colour CH3CH2–OH
Conc.H2SO4
(c) Phenol+NaHCO3 (r) White ppt. 443K

(d) Picric (s) CO2 gas is (d) (s) CH2=CH2

CH3CH2–OH
acid+NaHCO3 evolved Conc.H2SO4
413K
(1) (a) -p, (b) -r, (c) -q, (d) -s
(1) a-q, b-r, c-p, d-s
(2) (a) -r, (b) -p, (c) -s, (d) -q
(2) a-p, b-r, c-s, d-q
(3) (a) -p, (b) -s, (c) -q, (d) -r
(3) a-r, b-s, c-p, d-q
(4) (a) -s, (b) -q, (c) -r, (d) -p
(4) a-r, b-p, c-q, d-s

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Alcohol, Phenol & Ether CHEMISTRY

Exercise 4 (Previous Year's Questions)

1. Among the following ethers, which one will 5. The compound A on treatment with Na gives
B and with PCl5 gives C, B and C react
produce methyl alcohol on treatment with
together to give diethyl ether. A, B and C are
hot concentrated HI? [NEET-UG – 2013] in the order [NEET-2018]
(1) CH3–CH–CH2–O–CH3 (1) C2H5OH, C2H6, C2H5Cl
(2) C2H5OH, C2H5Cl, C2H5ONa
CH3 (3) C2H5Cl, C2H6, C2H5OH
(2) CH3–CH2CH2–CH2–O–CH3 (4) C2H5OH, C2H5ONa, C2H5Cl
(3) CH3–CH2–CH–O–CH3 6. Identify the major products P, Q and R in the
following sequence of reaction:
CH3
[NEET- 2018]
CH3 Anhydrous
AlCl3 (i) O2
(4) CH3–C–O–CH3 + CH3CH2CH2Cl P Q+R
(ii) H3O+/∆
CH3 CH2CH2CH3 CHO

(1) , , CH3CH2–OH
2. Identify Z in the sequence of reactions: P Q R
[AIPMT – 2014] CH2CH2CH3 CHO COOH

HBr/H2O2 C2H5ONa (2) , ,

CH3CH2CH = CH2  → Y  →Z
(1) CH3 – (CH2)3 – O – CH2CH3 P Q R
CH(CH3)2 OH
(2) (CH3)2 CH – O – CH2CH3 (3) , , CH3CH(OH)CH3
(3) CH3(CH2)4 – O – CH3 P Q R
(4) CH3CH2 – CH(CH3) – O – CH2CH3 OH
CH(CH3)2
(4) , , CH3–CO–CH3
3. Which of the following is not the product of
P Q R

dehydration of OH ? 7. The structure of intermediate A in the

following reaction is : [NEET - 2019]
CH3
[Re-AIPMT – 2015] CH OH
CH3 O
O2 H+
A +H3C CH3
(1) (2) H2 O

CH3
CH CH3
(3) (4) O CH3
(1) (2) H3C–C–O–O–H

4. The heating of methyl phenyl ether with HI

CH3
produces: [NEET – 2017] CH2–O–O–H
O–O–CH HC
(1) ethyl chlorides (2) Iodobenzene (3) (4) CH3
CH3
(3) phenol (4) benzene

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CHEMISTRY Alcohols, Phenols & Ethers
8. When vapours of a secondary alcohol is 11. Given below are two statements:
passed over heated copper at 573 K. the [NEET-2022]

product formed is: [NEET - 2019(ODISHA)] Statement I: The acidic strength of

monosubstituted nitrophenol is higher than
(1) a carboxylic acid (2) an aldehyde
phenol because of electron withdrawing
(3) a ketone (4) an alkene
nitro group.
Statement II: o-nitrophenol, m-nitrophenol
9. What is the IUPAC name of the organic and p-nitrophenol will have same acidic
compound formed in the following chemical strength as they have one nitro group
reaction? [NEET-2021] attached to the phenolic ring.
In the light of the above statements, choose
() 2 5
i C H MgBr,dry Ether
Acetone → Product
⊕
( ii) H2O,H the most appropriate answer from the
(1) 2-methyl propan-2-ol options given below:
(1) Both Statement I and Statement II are
(2) pentan-2-ol
correct.
(3) pentan-3-ol
(2) Both Statement I and Statement II are
(4) 2-methyl butan-2-ol incorrect.
(3) Statement I is correct but Statement II is
10. Given below are two statements:
incorrect.
[NEET-2022]
(4) Statement I is incorrect but Statement II is
Statement I: In Lucas test, primary, correct.
secondary and tertiary alcohols are
12. Match the reagent (List-I) with the product
distinguished on the basis of their reactivity
(List – II) obtained from phenol.
with conc. HCl + ZnCl2, known as Lucas
[Re-NEET-2022]
Reagent.
List-I List-II
Statement II: Primary alcohols are most a (i) NaOH I Benzoquinone
reactive and immediately produce turbidity (ii) CO2
at room temperature on reaction with Lucas (iii) H+
Reagent. b (i) Aqueous II Benzene
NaOH +CHCl3
(1) Both Statement I and Statement II are
(ii) H+
correct.
c Zn dust, ∆ III Salicylaldehyde
(2) Both Statement I and Statement II are
d Na2Cr2O7, H2SO4 IV Salicylic acid
incorrect. Choose the correct answer from the options
(3) Statement I is correct but Statement II is given below :
incorrect. (1) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(4) Statement I is incorrect but Statement II (2) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
is correct.
(4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

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Alcohols, Phenols & Ethers CHEMISTRY
13. Which one of the following reaction 16. Consider the following reaction:
sequence is incorrect method to prepare CH2–O
HI
A+B
phenol ? [Re-NEET-2022] ∆
+
(1) Cumene, O2, H3O Identify products A and B [NEET-2023]
Cl
(2) , NaOH, STP condition (1) A = CH3 and B = OH

(2) A = CH2OH and B = I

(3) , oleum, NaOH, H3O+
(3) A = CH2I and B= OH
(4) Aniline, NaNO2+HCl, heating
(4) A = CH3 and B= I
14. Consider the following reaction and identify
the product (P). [NEET-2023]
HBr
CH3–CH–CH–CH3 → Product (P) 17. Which one of the following alcohols reacts
instantaneously with Lucas reagent?
CH3 OH
[NEET-2024]
3-Methylbutan-2-ol (1) CH3 − CH − CH2OH
Br |
CH3
(1) CH3–C–CH2–CH3
CH3
CH3 |
(2) CH3 − C− OH
(2) CH3 CH=CH–CH3 |
CH3
(3) CH3–CH–CH–CH3
(3) CH3 − CH2 − CH2 − CH2OH
CH3 Br
(4) CH3 − CH2 − C H − OH
CH3 |
CH 3
(4) CH3–C–CH2 Br
CH3 18. Major products A and B formed in the following
reaction sequence, are : [NEET-2024]
15. Which amongst the following will be most
OH
readily dehydrated under acidic conditions ?
H3C
[NEET-2023] PBr3
A alc. KOH B
NO2 OH (major) ∆ (major)

OH OH
(1) Br
CH3 H3C H3C
H (1) A =
; B=
OH OH
OH O
(2) H3C H3C Br H3C
H (2) A =
NO2 ; B=
H
Br
(3) OH H3C H3C
OH (3) A =
; B=
NO2
Br
H3C H3C
(4) (4) A =
OH ; B=

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 CHEMISTRY Alcohols, Phenols & Ethers

19. Identify D in the following sequence of 20. given below are two statements:
[RE-NEET-2024]
reactions : [RE-NEET-2024] Statement-I: propene on treatment with
CH3CH2OH
P+ Ι2
A dryMg HCHO diborane gives an addition product with the
ether B C
H2O
formula (CH3)2–CH –B
3

D Statement II: oxidation of (CH3)2–CH –B

3
(major)
with hydrogen peroxide in presence of NaOH
(1) n-propyl alcohol
gives propan-2-ol.
(2) isopropyl alcohol in the light of the above statements, choose
(3) propanal the most appropriate answer from the
options given below :
(4) propionic acid
(1) Statement I is correct but Statement II is
incorrect
(2) Statement I is incorrect but Statement II
is correct
(3) Both Statement I and Statement II are
correct
(4) Both Statement I and Statement II are
incorrect

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Alcohols, Phenols & Ethers CHEMISTRY

ANSWER KEYS

Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 4 2 4 3 1 2 4 1 4 2 1 1 2 3 1

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 4 3 4 2 3 3 4 1 3 3 3 3 3 1 3 4 3 3 3
Que . 21 22 23 24 25
Ans. 4 4 4 3 1

Exercise 1.3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 4 3 2 2 3 2 2 4 2 3 4 2 4 2 2

Exercise 2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 3 3 4 3 4 4 1 1 4 2 1 2 2 3 1 2 4 3 1
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 2 2 1 4 2 3 2 4 3 4 2 1 3 1 2 3 4 4 4 4
Que . 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 1 1 3 3 4 2 3 1 4 1 1 2 1 3 4 2 4 2 1

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 1 2 3 1 3 1 1 1 2 1 3 1 2 1 2 2 3 2 2
Que . 21 22 23 24 25
Ans. 1 1 1 1 2

Exercise 4 (Previous Year's Questions)

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 1 4 3 4 4 2 3 4 3 3 2 2 1 2 3 2 3 1 2

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 Notes

Sarvam Career Institute

Aldehydes, ketones, and carboxylic acids CHEMISTRY

Chapter
Chapter
ALDEHYDES,
4
4 KETONES, AND
CARBOXYLIC ACIDS

Chapter Summary
Introduction
• Introduction This chapter includes knowledge of carbonyl compounds i.e.
• Preparation of Aldehydes The compound containing C = O as functional group. In this we
and ketones
• Physical properties of study aldehydes, ketones, carboxylic acids and their derivatives.
aldehyde and ketones In aldehydes carbonyl group is attached to either two H–atoms
• Chemical reaction of or one R–or Ar–group and one H-atom.
aldehydes and ketones In ketones carbonyl group is attached to two alkyl or aryl
• Uses of Aldehyde and groups. If both the groups are similar then this is a simple
ketones
ketone but if both the groups are different it is a mixed ketone.
• Methods of Preparation of
O O O
carboxylic acids
C C C
• Physical properties of R H R R R R'
Aldehyde Simple ketone Mixed ketone
carboxylic acids (R-may be H-or (Both R-groups are alkyl (Both R-groups are
• Chemical reactions of alky or aryl group) or both aryl groups) different)

carboxylic acids
The carbonyl compound in which carbonyl group is attached to
an –OH group, are called carboxylic acids
O
C
i.e., R OH where R– can be H, alkyl or aryl group.

The compounds in which–OH group of carboxylic acid is

replaced by other groups such as –X,–OR,–OCOR,–NH2 etc. are
called acid derivatives. All these derivatives form carboxylic
acid again on hydrolysis.

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
O
–OH Acid halide
+x C or Acyl halide
R X
O O
–OH
O +(–O–C-R') C C Acid anhydride
O
R O R'
C
R OH
O
–OH
+OR' C Ester (R & R'-may or may not be same)
R OR'
O
–OH Acid amide
C
+(–NH2)
R NH2
Some important common name
Structure Common name Structure Common name
HCHO Formaldehyde O Acetone
CH3–C–CH3
CH3CHO Acetaldehyde O Acetophenone

CH3

CH3–CH2–CHO Propionaldehyde O Propiophenone

CH3–CH3

CH3–CH–CHO Isobutyraldehyde O Benzo phenone

CH3

CH2=CH–CHO Acrolein
CHO Benzaldehyde

CHO Phthaldehyde
CHO

CH=CH–CHO Cinnamaldehyde

CHO Salicyldehyde
OH

Structure of the carbonyl group:

Carbon atom of carbonyl group is sp2 hybridised having triangular planar geomery. Carbon atom forms
three sigma bonds and one π(pi) bond, out of which 3σ bonds are located in the same plane whereas
fourth π bond. Which is formed by lateral or sideways overlapping, is situated above and below the
plane. Bond angles are approximately 120°
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120°
σ bond ••
sp 2
C O 120° C O
••

120°

Carbonyl group is polarized due to difference in electronegativity between carbon and oxygen. Oxygen
being more electronegative pulls the shared pair of electrons more towards itself making oxygen as a
nucleophilic centre and carbon as an electrophilic centre. Hence carbonyl compound have substantial
dipole moment and their polarity can be expressed on the basis of resonance.
O OΘ
C C⊕

Preparation of Aldehydes and ketones

(1) By oxidation of alcohols
PCC
R–CH2–OH or R–CHO
PDC
or
CrO3
(anhydrous)
OH O
PCC
R–CH–R or R–C–R
PDC
or
CrO3
(anhydrous)

(2) By catalytic dehydrogenation: This method is suitable for volatile alcohols and is of industrial
application. In this method alcohol vapours are passed over heavy metal catalyst such as Ag or copper.
Cu,573K
CH3CH2CH2OH CH3CH2CHO+H2
OH O
Cu,573K
CH3–CH–C2H5 CH3–C–C2H5+H2

(3) From Hydrocarbons

(i) By ozonolysis of alkenes:
ozonolysis of alkenes followed by reaction with zinc dust and water gives aldehyde, ketones or a
mixture of both depending on the substitution pattern of the alkenes
O Zn/H2O
O3 C C C=O + O=C + ZnO
C=C
O O

(ozonide)
(i) O3
Ex. CH3–CH=CH–CH3 2CH3–CHO
(ii) Zn/H2O

CH3 O
(i) O3
Ex. CH3–CH= C–CH3 CH3–CHO+ CH3–C–CH3
(ii) Zn/H2O

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
(ii) By hydration of Alkynes
Addition of water to ethyne in the presence of dil. H2SO4 and HgSO4 gives acetaldehyde. all other
alkynes give ketones in this reaction
dil. H2SO4 Tauto.
Ex. CH ≡ CH HgSO CH2=CH–OH CH3–CHO (mechanism: Hydrocarbon chapter)
4

OH O
dil. H2SO4 Tauto.
CH3–C ≡ CH HgSO4
CH3–C=CH2 CH3–C–CH3

OH O
dil. H2SO4 Tauto.
CH3–C ≡ C–CH3 HgSO4
CH3–C=CH–CH3 CH3–C–CH2–CH3

Specific methods of preparation of aldehyde

(1) From acyl chloride (acid chloride)
Acid chloride (Acyl chloride) is hydrogenated over catalyst, Palladium on barium sulphate. This
reaction is called Rosenmund reduction.
O
CHO
H2
Ex. Cl Pd-BaSO
4

O O
Ex. H2
Cl Pd-BaSO4
H

(2) From nitriles

Nitriles are reduced to corresponding imine with stannous chloride in the presence of HCl. Which on
hydrolysis give corresponding aldehyde. This reaction is called Stephen reaction.
SnCl2+HCl H2O/H⊕
Ex. R–CN R–CH=NH R–CHO

SnCl2+HCl H2O/H⊕
Ex. CH3–CH2–CH–C≡N CH3–CH2–CH–CH=NH CH3–CH2–CH–CHO
CH3 CH3 CH3

Note. Nitriles are selectively reduced by diisobutyl aluminium hydride (DIBAL-H) to imines followed by
hydrolysis to aldehydes
(i)DIBAL −H
Ex. R − CN →
(ii) H O
R − CHO
2

(i)DIBAL −H
CH3 − CH = CH − CH2 − CN →
(ii) H O
CH3 − CH = CH − CH2 − CHO
2

(3) From esters

O O
(i) DIBAL-H
Ex. CH3–CH2–CH–C–OC2H5 (ii) H O CH3–CH2–CH–C–H
2

CH3 CH3
O O
(i) DIBAl-H H
Ex. OCH3 (ii) H2O

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(4) From hydrocarbons

(i) By oxidation of methyl benzene
(a) By chromyl chloride (CrO2Cl2):
Chromyl chloride oxidizes methyl group to a chromium complex which on hydrolysis gives
corresponding benzaldehyde. This reaction is called Etard reaction.
CH3 CH(OCrOHCl2)2 CHO
CrO2Cl2 ⊕ H3 O
in CS2/CCl4

(Toluene) (Chromium complex) (Benzaldehyde)

(b) By chromic oxide (CrO3):

CHO
CH3 CH(OCOCH3)2 H3O⊕
273-283K
+CrO3 + (CH3CO)2O ∆

(Toluene) (Benzaldehyde)

(ii) By side chain chlorination followed by hydrolysis

CH3 CHCl2 H2O CHO

Cl2/hν
373K

(Toluene) (Benzal chloride) (Benzaldehyde)

(iii) By Gattermann – Koch reaction

CO, HCl
CHO
Anhyd. AlCl3/CuCl

Mechanism:
⊕ δ⊕ δΘ ⊕ ⊕ Θ
Θ AlCl3
O ≡ C + H – Cl [O ≡ C–H O = C–H]AlCl4
O
⊕ H
⊕ –H⊕ H
+ H–C ≡ O C–H
O
(iv) By Gattermann Aldehyde synthesis
O
(i) Anhyd. AlCl3 H
+HCN + HCl (ii) H2O

Mechanism:
AlCl3 ⊕ ⊕ Θ
H – C ≡ N + H – Cl [H – C ≡ N – H HC = N – H]AlCl4
O
⊕ H CH=NH H O/H⊕ C
⊕ ⊕ H
+ H–C ≡ NH CH=NH –H 2

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
Specific method of preparation of ketones
(1) From acyl chlorides
2R–Mg–X + CdCl2 R2Cd + 2Mg(X)Cl

2R'–C–Cl + R2Cd 2R'–C–R + CdCl2

O O

(2) From nitriles

NMgBr O
R'Mgx H3O⊕
R–C ≡ N in ether R–C–R' R–C–R'
NMgBr O
C6H5MgBr H3O⊕
CH3–CH2–C≡N in ether CH3–CH2–C–C6H5 CH3–CH2–C–C6H5

(3) From Friedel–craft acylation reaction

O
O
Anhyd. Ar/R
+Ar/R–C–Cl AlCl
3
O
O
Anhyd. CH2–CH3
+CH3–CH2–C–Cl AlCl
3

Exercise 1.1
1. In the following reaction product P is 4. Reduction of benzoyl chloride with Pd and
O BaSO4 produces–
H2
R–C–Cl Pd-BaSO4 P (1) Benzoic acid
(1) RCH2OH (2) RCOOH
(2) Benzaldehyde
(3) RCHO (4) RCH3
(3) Benzoyl cyanide
2. The reagent most suitable for converting a (4) Benzyl alcohol
primary alcohol into an aldehyde is:
(1) K2Cr2O7 and H2SO4 KCN (i) SnCl2 +HCl
(2) KMnO4 and NaOH 5. CH3–CH2–Br → X (ii) H O
→ Y; Y is
2

⊕ (1) CH3–CH2–CONH2
(3) NH , ClCrO3–
(2) CH3–CH2–CH2–NH2
(4) HNO3
(3) CH3–CH2–CHO
3. Stephen reaction is the reaction involving: (4) CH3–CH2–COOH
(1) Reduction of alkanoyl chloride with
Pd/BaSO4.
(2) Reduction of alkyl isocyanide with 6. Ozonolysis of Isobutylene gives-
sodium and alcohol. (1) Acetone
(3) Reduction of alkyl cyanide with SnCl2 (2) Formaldehyde
and HCl and hydrolysing the
(3) Acetaldehyde
intermediate aldimine.
(4) Reduction of carbonyl compound with (4) 1 and 2 both
zinc amalgam and HCl
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CHO x
11. CH2 CHO
‘A’+(CH3CO)2O H O+
CH3 [Intermediate] 3
273-283K ∆
X is :
7. CHO
(1) (a) BH3, THF (b) H2O2/OH–
‘B’+CS2 H3O⊕ (2) H3O+
[Intermediate]

In the above given reaction, identify the (3) (a) O3 (b) Zn/H2O
reagent ‘A’ and reagent ‘B’ respectively (4) (a) (i) BH3/THF (ii) H2O2/OH–
(1) CrO3 and CrO3 (b) Anhydrous CrO3
(2) CrO2Cl2 and CrO3
(3) CrO3 and CrO2Cl2 [0]
12. RCH2OH  →RCHO
(4) CrO2Cl2 and CrO2Cl2
Which of the following oxidising agent is
KmnO4 most suitable for above reaction?
8. A
Conc.H2SO4
B
(i) O3
C+D
∆ (ii) Zn/H2O (1) PCC (2) K2Cr2O7
(major)
In the above given sequence reaction (3) KMnO4 (4) All
product C & D are
O O 13. In which reaction ethyl phenyl ketone is
(1) H3C C CH3 and H3C C
H obtained
O () i EtMgBr
O (1) C6H5 − C ≡ N 
( ii) H O/H⊕
→
2
C
(2) H3C C OH and CH3 H (2) CH3–CH2–C–Cl + Ph2Cd→
O O O
(3) H3C C CH3 and H C H Anhyd.
(3) + CH3–CH2–C–Cl
O O AlCl3
O
(4) CH3–CH2–C–OH and H C H
(4) All of them
() i DIBAL −H
9. R − CN →
( ii)H O
R−Y
Baeyer s ′
HIO4
14. CH3 —=
CH CH2  → X  →Product
2

Reagent
Consider the above reaction and identify “Y”
(1) –CONH2 (2) –CHO Product is :
(3) –COOH (4) –CH2NH2 (1) CH3 − CH =
O
(2) H2C = O
10. The major product obtained in the following
reaction is : (3) CH3 − CO − CH3
O (4) (1) and (2) both
O

DIBAL–H
–78°C 15. Which of the following on heating with aq.
COOH KOH, produce acetaldehyde
OH
(1) CH2ClCH2Cl
CHO CHO (2) CH3CHCl2
(1) (2)
CHO COOH (3) CH3–C–Cl
OH
O
CHO CHO (4) CH3–CH2–Cl
(3) (4)
CHO COOH
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Physical properties of aldehyde and ketones
1. Physical state: Methanal is a pungent smelling gas, ethanal is volatile liquid with b.p.=294 K. Rest of
the aldehydes and ketones containing upto eleven carbon atoms are colouless liquids while higher
members are solids.
2. Solubility: Aldehydes and ketones upto four carbon atoms are miscible with water due to presence of
H-bonding between polar carbonyl group and water molecules.
δ–
δ– δ+ O δ+ δ–
C=O H H O=C
Solubility of these compounds decreases rapidly on increasing length of alkyl chain and higher
members are practically insoluble in water. All these compounds are quite soluble in organic solvents
such as CH3OH, ether, CHCl3, benzene etc. Ketones, themselves are good solvents.

3. Boiling points: Boiling points of aldehydes and ketones are higher than those of non-polar compounds
i.e., hydrocarbons and weakly polar compounds, such as ether, due to dipole-dipole interactions but
lower than that of alcohols of comparable molar masses, due to absence of intermolecular H-bonding.
Compound C4H10 (n-Butane) CH3OCH2CH3 CH3CH2CH2OH CH3CH2CHO CH3COCH3
Molar mass 60 60 60 58 58
B.P. (K) 273 281 370 322 329
4. Smell: Lower aldehydes have exceptionally unpleasant odour whereas other members generally have
pleasant odour. As the size of molecule increases smell becomes less pungent and more fragrant. Many
naturally occurring aldehydes and ketones are used for blending of many flavoring agents and
perfumes.

Chemical reaction of aldehydes and ketones

(1) Nucleophilic addition reactions:
Due to polar nature of carbonyl group, nucleophilic addition reaction are quite common to these
compounds in which nucleophile attacks first followed by attack of electrophile
Nu Nu
Θ
δ⊕ δΘ Nu Θ E⊕
C O Slow step C–O Fast
C–OE
(R.D.S.)
Tetrahedral
Trigonal planer Intermediate
• Stereochemistry of the reaction:
If chiral carbon is generated on carbonyl carbon during nucleophilic attack then equimolar mixture of
the compound having 'R' and 'S' configuration w.r.t. that chiral carbon is formed as product
• Reactivity:
There are two important factors which help in comparing relative reactivity of aldehydes and ketones
in nucleophilic addition reactions.
(i) Steric factor: Attack of nucleophile changes the trigonal planar carbonyl group to tetrahedral state.
Increase in size of alkyl or aryl groups around carbonyl 'C' atom produce steric hindrance to the
attacking nucleophile.
It is due to the crowding of space around carbonyl carbon atom. Hence greater the number of alky/aryl
groups or bigger the size of groups, more will be the steric hindrance and lesser will be the tendency
of nucleophile to attack the carbonyl carbon.

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O
(ii) Electronic factor: Alkyl groups present on —C— group show +l effect and tend to increase e–density on
it, due to which attacking tendency of nucleophile decreases. It means, more the +l effect lesser will
be reactivity of carbonyl compounds.
In aromatic aldehydes, due to –M effect of carbonyl group their reactivity towards nucleophilic addition
decreases.
O O– O– – O
H H H H O H
C C C C C
+ +

+
Electron density on the carbon atom of carbonyl group increases, hence the attacking tendency of
nucleophile decrease.
By considering both factors the comparative reactivity of different carbonyl compound is
H CH3 C6H5 CH3 C2H5 C6H5
C=O > C=O > C=O > C=O > C=O > C=O
H H H CH3 C2H5 C6H5

(I) Addition of HCN (hydrogen cyanide):

• Aldehydes & ketones react with HCN to yield cyanohydrin
• reaction occurs very slowly with pure HCN (weak acid) therefore reaction is carried out in basic medium
so that it can generate CNΘ ion for nucleophilic attack.
Θ Θ
HCN + OH :CN + H2O
Θ
δΘ O OH
δ⊕ Θ H⊕ C
C = O + CN C
CN CN
Tetrahedral (Cyanohydrin)
intermediate
Θ
Ex. C6H5 Θ
OH C6H5 * OH (racemic mix)
C = O+HCN C
CH3 CH3 CN

Ex. CH3 Θ
CH3 OH (racemic mix)
C = O+HCN OH *
C
H H CN

(II) Addition of NaHSO3 (Sodium hydrogen sulphite)

SO3H Proton transfer SO3H
C = O+NaHSO3 C C
ONa OH
Bisulphite addition Compound
(Crystalline)
• In most aldehydes equilibrium lies to the right hand side whereas for the most ketones, it lies to the
left hand side
• Hydrogen sulphite addition compound is water soluble and can be converted back to the original
carbonyl compound by treating it with dilute mineral acid or alkali. Therefore these are useful for
separation and purification of aldehydes & ketones.

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
• Limitation of the reaction:
(i) All aldehyde (Aliphatic & aromatic) show this reaction
(ii) Methyl ketones [CH3–C–R] show this reaction
O
Where R group should not be in resonance with –C– (i.e. R ≠ Ar)
O
(III) Addition of Grignard reagent: Aldehydes and ketones give addition products with Grignard reagent
which produce alcohol on being hydrolysed.
Θ ⊕ OMgX H2O/H+ OH
C=O + R MgX C C
R R
(a) HCHO gives 1° or primary alcohol
H H OMgI H2O/H+ H OH OH
C=O + CH3MgI C C +Mg
H H CH3 H CH3 I
(b) Higher aldehydes form 2° or secondary alcohol
R R OMgX R OH X
C=O + R'MgX H2O/H+
C C +Mg
H H R' H R' OH
(c) Ketones form 3° or tertiary alcohols
R R OMgX H O/H+ R OH
C=O + R'MgX C 2
C
R R R' R R'

(IV) Addition of alcohols:

••
••
••
R'–OH OR' R'–OH OR'
C = •O• C C
dry HCl (H⊕) dry HCl (H⊕) OR'
OH
If Aldehyde (Hemiacetal) (Acetal)
(ketal)
If ketone (Hemiketal)
Mechanism:
⊕ R'
••
•• H⊕ ⊕ R'–OH O ⊕ OR'
C=O
•• C O–H
••
C •• H –H C ••
OH
••
OH
••

H⊕

⊕ R'
O ••
OR' –H⊕ R'–OH
•• OR'
C C H C ⊕
OR' OR' OH2

R HO–CH2 R O CH2
dil. HCl
Ex. C O+ C + H2O
HCl gas
R HO–CH2 R O CH2
(ethylene glycol) ethylene glycol ketal

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• Dry HCl Protonate the oxygen of carbonyl compound and increases electrophilicity of carbonyl carbon
towards the attack of nucleophile
• Acetals and ketals are hydrolysed with aqueous mineral acids to yield corresponding aldehydes and
ketones respectively

(V) Addition of NH3 and its derivatives:

Ammonia and its derivatives such as H2N–Z add to the carbonyl group of aldehydes and ketones.
reaction is reversible and catalysed by acid. The equilibrium favours the product formation due to rapid
dehydration of intermediate to form C = N – Z .
Here Z can be alkyl, aryl, –OH, – NH2, –NHC6H5, –NHCONH2 etc.

H+ OH
C = O + H2N–Z C C = N–Z+H2O
NHZ
mechanism:
⊕ ••
⊕
••
H2N–Z NH2–Z -H⊕ NH–Z H⊕ NH–Z
•• H⊕ C C
C=O C = O–H C •• ⊕
OH OH OH2

-H⊕ ⊕ –H2O
C = N– Z C=NH–Z

• Reaction follows nucleophilic addition elimination mechanism

• Reaction is carried out in acidic medium (PH value = 4–6)
Group Z Reagent Carbonyl derivative Product Name
1. —H NH3 C = NH Imine
Ammonia
2. —R H2NR C = NR Substituted imine
Amine or Schiff's base
3. —OH H2NOH C = NOH Oxime
(Hydroxyl amine)
4. —NH2 H2N—NH2 C = N–NH2 Hydrazone
(Hydrazine)
5. —NH H2N—NH C = N—NH Phenylhydrazone

Phenyl hydrazine
6. —NHCONH2 H2N—NHCONH2 C = N–NHCONH2 Semi carbazone
Semi carbazide
7. O 2N O 2N O 2N 2,4–Dinitrophenyl
hydrazone
—NH NO2 H2N—NH NO2 C = N–NH NO2

2,4-Dinitrophenyl hydrazine

Note. 2,4–DNP derivatives (2,4–Dinitrophenyl derivatives) are yellow, orange or red crystalline compounds
and it can be used for the identification of aldehyde and ketones.
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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
–H2O
C = O + H2N–NH NO2 C=N–NH NO2

NO2 NO2
(Yellow or orange solid)
• Carbonyl compound react with primary amine to give Schiff's base (substituted imine)
• except formaldehyde, all other aldehydes react with ammonia to give imine as the final product but
formaldehyde react with ammonia to give urotropine [(CH2)6N4] as the final product.
N
H H ⊕
C = O + NH3 N
H 6 : 4 N N
urotropine
(Hexamethylenetetramine)

(2) Reduction Reactions:

(I) Reduction to alcohols: By catalytic hydrogenation or LiAlH4 or NaBH4, aldehydes and ketones are
reduced to 1° and 2° alcohols respectively
H2/Ni or Pt or Pd
C=O CHOH
LiAlH4 or NaBH4
H2/Ni or Pt or Pd
Ex. CH3CHO LiAlH or NaBH CH3CH2OH
4 4

OH
LiAlH4
Ex. CH3COCH3 CH3—CH—CH3
CHO CH2OH
Ex. LiAlH4

(II) Reduction to hydrocarbons:

C O group of aldehyde and ketones can be reduced to –CH2 group by use of following reagent
(i) Clemmensen reduction: reduction with Zn-Hg (amalgum) and concentrated HCl
Zn–Hg
C=O HCl
CH2 +H2O

O
Zn-Hg
Ex. +H2O
HCl

O O
CH3
Ex. H Zn-Hg
HCl

(ii) Wolf-kishner reduction: reduction with hydrazine (NH2–NH2) followed by heating with base (KOH or
NaOH) in high boiling solvent such as ethylene glycol.
OH OH
NH2–NH2
O –H2O
C = N–NH2 KOH/ CH2–CH2 CH2 + N2
∆

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O
Ex. (i) N2H4
Θ
(ii) OH/∆
O
O
Ex. (i) N2H4
O Θ
(ii) OH/∆

(iii) Reduction with Red P and HI

Red P/HI
Ex. CH3 − CHO ∆
→ CH3 − CH3
O
Red P/HI
Ex.
∆
O

3. Oxidation reaction:
• Aldehyde can easily be oxidised to carboxylic acids on treatment with common oxidising agent like
HNO3, KMNO4, K2Cr2O7
[ ] O
R − CHO  →R − COOH
• Ketones are oxidised only under vigorous condition, i.e. by strong oxidising agents and at elevated
temperature.
Popoff's rule: during the oxidation of unsymmetrical ketones, the carbon-carbon bond cleavage will
always occur such that keto group (C = O) remains attached with smaller alkyl groups.

O O O
Ex. KMnO4/H⊕
CH3—CH2—C—CH3 CH3—C—OH + HO–C–CH3
High Temp.
smaller
groups

O O
Ex. KMnO4/H⊕ OH +
High Temp. HO
O
(a) oxidation by Tollen's reagent → Tollen's Test
On warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollen's reagent), a
bright sliver mirror is produced due to formation of silver metal
⊕ Θ
R − CHO + 2  Ag (NH3 )2  + 3OH 
→R − COOΘ + 2Ag ↓ +2H2O + 4NH3
O O
Ex. (i) [Ag(NH3)2]OH
H (ii) H⊕ OH + Ag↓

O O
Ex. H (i) [Ag(NH3)2]OH OH + Ag↓
(ii) H⊕

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
(b) Oxidation by Fehling reagent → Fehling's Test
Fehling reagent is stored in two different bottles as two different solution
Fehling solution (A) : Aqueous CuSO4
Fehling reagent
Fehling solution (B): NaOH + Sodium Potassium tartarate
(Alkaline) (Rochelle salt)
• When Fehling solution (A) and Fehling solution (B) are mixed, dark blue color cu-complex obtained.
This cu-Complex (Cu2+) oxidises aldehyde into acid & gives reddish brown ppt of Cu2O
• Aromatic aldehydes do not respond to this test
Θ
R − CHO + 2Cu2+ + 5OH 
→R − COOΘ + Cu2O ↓ + 3 H2O
(Red −brownPPt )

(C) Oxidation of methyl ketones by Haloform reaction:

• Aldehyde and ketones having at least one methyl group linked to carbonyl carbon atom (i.e having
O
–C–CH3 group) are oxidised by sodium hypohalite solution (x2 in presence of NaOH) to give sodium
salts of carboxylic acid having one carbon atom less than that of carbonyl compound.
• If reagent is I2+NaOH or NaOI, then yellow precipitate of CHI3 is obtained. It is used for detection of
O OH
CH3–C– group or CH3–CH– group in compound

O O
NaOH+X2
R–C–CH3 or R–C–ONa+CHX3(x=Cl,Br,I)
NaOX

Mechanism:
O O O
Θ
NaOH Θ X—X
R–C–CH2 R–C–CH2 R–C–CH2–X
(rds.) (–Nax)
H (–H2O)
(–H2O) NaOH

O O O
x Θ NaOH x X—X Θ
R–C–C R–C–CH R–C–CH–X
x (–H2O) x (–Nax)
(–Nax) x–x

O O O
Θ
x NaOH Θ Θ⊕
R–C–C x R–C–O–H+CX3 R–C–ONa+CHX3
x

H CH3 H CH3
NaOH/I2 C=C
Ex. (i)
H3C ONa + CHI3
H3C CH3 O
O (yellow PPt)

O O
NaOH/I2
(ii) CH3 H CH3–C–ONa + CHI3
(yellow PPt)

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O
CH3–CH2–OH NaOH/I2
(iii) CH3–C–ONa + CHI3
(yellow PPt)

OH O
(iv) CH3–CH–CH–CH3 NaOH/I2
CH3–CH–C–ONa + CHI3
CH3 (yellow PPt)
CH3

4. Reactions due to α–hydrogen :

Acidity of α–hydrogen of aldehydes & ketones :
The acidity of α–H atoms of carbonyl compounds is due to strong electron withdrawing effect of the
carbonyl group and resonance stabilisation of the conjugate base
Θ
O O O
—C—C— —C—C— —C=C—
Θ
H •B
•

(a) Aldol condensation :

Aldehydes & Ketones having at least one α–hydrogen undergo a reaction in the presence of dilute alkali
(eg NaOH/KOH/Na2CO3/Ba(OH)2) as catalyst to form β–hydroxy aldehydes (aldol) or β–hydroxy
Ketones (Ketol), respectively. This is known as Aldol reaction
Ex. (i) CH3–CHO+CH3–CHO dil.NaOH ∆
CH3–CH–CH2–CHO –H O CH3–CH=CH–CHO
2

OH α,β–unsaturated Aldehyde
β–hydroxy aldehyde (Aldol condensation product)
(Aldol)
Mechanism :

O Θ
O Θ O O O
α OH Θ CH3–C–H
CH2–C–H CH2–C–H CH3–CH–CH2–C–H
(–H2O) r.d.s.
Θ
H Θ
O (-OH) H–OH
CH2=C–H
enolate ion

O OH O OH O
Θ
–OH Θ ∆
CH3–CH=CH–C–H CH3–CH–CH–C–H CH3–CH–CH–C–H
(E1CB)
Θ
OH H
Similarly
O O OH O O
Ba(OH)2 ∆
(ii) CH3–C–CH3+CH3–C–CH3 CH3–C–CH2–C–CH3 CH3–C=CH–C–CH3
–H2O
CH3 CH3
β–hydroxy ketone α,β–unsaturated ketone
(Ketol) (Aldol condensation product)

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• Carbanion (enolate) is intermediate
• Attack of carbanion is r.d.s.
• The Aldol and ketol readily lose water in the presence of heat to give α,β-unsaturated carbonyl
compounds, hence this reaction is called Aldol condensation
• Dehydration takes place by E1–CB mechanism
(b) Cross aldol condensation: When Aldol condensation is carried out between two different aldehydes or
ketones, it is called cross Aldol condensation. If both the molecules have α-H atom, then they can give
a mixture of four-condensation products.
Ex. CH3CHO + CH3CH2CHO  (i)NaOH
(ii) ∆
→
Four possible products are formed as following :
(i) NaOH
2CH3CHO CH3CH=CH—CHO (Self aldol product)
(ii) ∆
But-2-enal
(i) NaOH
2CH3CH2CHO CH3CH2CH=C—CHO (Self aldol product)
(ii) ∆
CH3
2-Methylpent-2-enal
(i) NaOH
CH3CHO+H2C–CHO CH3CH=C—CHO (Cross aldol product)
(ii) ∆
CH3 CH3
2-Methylbut-2-enal
(i) NaOH
CH3–CH2CHO+CH3CHO CH3CH2CH=CH—CHO (Cross aldol product)
(ii) ∆
Pent-2-enal
Hence cross aldol condensation becomes useful when one of the carbonyl compound does not contain
α-H atom.
OH
∆
Ex. HCHO+CH3CHO dil. NaOH CH2—CH2CHO –H2O
CH2=CH–CHO
OH
CHO CH–CH2COCH3 ∆ CH=CHCOCH3
+CH3COCH3 –H2O

5. Cannizaro reaction: Aldehyde which do not have an a-hydrogen atom, undergo self oxidation and
reduction (disproportionation) reaction on heating with concentrated alkali.
In this reaction , one molecule of the aldehyde is reduced to alcohol while other is oxidised to carboxylic
acid salt.
O O O
50%KOH Θ⊕
Ex. H–C–H+H–C–H ∆ CH3–OH+H–C–OK
reduction
oxidation
Mechanism:
Θ
O Θ
O O O
–OH slow
H–C–H H–C–H + H–C–H (r.d.s.) H–C–O–H
+
OH(Migration of Hydride) Θ
CH3–O

O
Θ
H–C–O + CH3–OH
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Note. Migration of hydride ion is r.d.s. step

∆
Ex. 2 CHO+Conc.NaOH CH2OH + COONa

Cross Cannizaro reaction :

• This reaction takes place between two different aldehydes (both not having any α-H atom) in which
aldehyde having more electrophilic carbonyl carbon get oxidised and other get reduced
O O O
Ex. C–H + H–C–H
Conc.KOH
CH2–OH + H–C–OK
∆

Mechanism:
Θ
O O O O
Θ
–OH slow Θ
H–C–H H–C–H + H–C H–C–O–H + O–CH2
(r.d.s.)
OH
O
H–C–OK + HO–CH2

Θ
Ex. CH3 CHO + O2N CHO OH
∆ CH3–O CH2–OH

+
Θ
O 2N COO

6. Electrophilic substitution reaction: Aromatic aldehydes and ketones undergo electrophilic

substitution reaction in the ring in which carbonyl groups act as a deactivating and meta directing
group.
CHO CHO

(i) Halogenation: + Cl2 FeCl3 + HCl

Cl
COCH3 COCH3

+ Cl2 FeCl3
Cl
(ii) Nitration: By conc. HNO3 and conc. H2SO4
CHO CHO

conc.H2SO4
+ HNO3(conc.)
NO2

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Sulphonation: By H2SO4 (fuming) or oleum (H2S2O7)
CHO CHO
Fuming
+ H2SO4
SO3H
Uses of Aldehyde and ketones
Aldehydes and ketones are used in chemical industries as solvents, starting material and reagents for
synthesizing the other products, e.g.
(1) 40% aq. HCHO known as formalin and is used in preserving biological specimens.
(2) Formaldehyde is also used as disinfectant.
(3) HCHO is used in manufacturing synthetic polymers like Bakelite.
(4) Acetaldehyde is used for silvering of mirrors.
(5) Benzaldehyde as a flavoring agent in perfume industry.
(6) Acetone is used as a solvent in industries and laboratories.
(7) Acetone is a common constituent of nail polishes. Many aldehydes and ketones such as butyraldehyde,
vanillin, acetophenone, camphor etc. are known for their odours and flavours.
Exercise 1.2
1. Which one of the following on treatment with 4. Which of the following is an example of
50% aq. NaOH yields the corresponding aldol reaction ?
alcohol and acid NaOH
(1) C6H5CHO + HCHO  →
∆
(1) C6H5CHO (2) CH3CH2CH2CHO
C6H5–CH2–OH
(3) CH3COCH3 (4) CH3CHO dil.NaOH
(2) CH3–CH=O  →
2. Which factors will increase the reactivity of OH
C = O group? CH3–CH–CH2–CH=O
O O
(i) Presence of a group with positive (3) H–C–H NaOH H–C–OΘNa⊕+CH3OH
inductive effect ∆

(ii) Presence of a group with (–ve) (4) All of these

inductive effect 5. Cannizzaro reaction does not take place
(iii) Acidic medium with:
(1) Only (i) (2) Only (ii) (1) (CH3)3CCHO (2) CHO
(3) Both (i) & (iii) (4) Both (ii) & (iii)
(3) CHO (4) CH3CHO
3. Which of the following compound has the
largest equilibrium constant for the
O
addition of water ? OHΘ
O 6. CHO + ‘X’ 293K CH=CH–C

(1) CH3O C Identify ‘X’:

O (1) CH2–CHO
(2) O2N C NO2
(2) CH2–CH2–OH
O
O
(3) C–Ph
(3) C–CH3
O
(4) CH2O C NO2
(4) CH=CH3

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7. Acetaldehyde reacts with semi carbazide, 13. Predict the product when given compound
product will be: reacts with NaBH4.
(1) CH3CH = NNH – CO – NH2 O O
(2) CH3CH = NCONHNH2
C C
(3) CH3CH = NHNH2 H OCH3
O (A)
(4) CH3–C–NH–CONH2 O

OH O
8. Acetone gives test with:
C–O–CH3
(1) 2,4 Dinitro phenyl hydrazine (1)
(2) Fehling solution
(3) Schiff’s reagent
OH
(4) All
OH O
9. Which is most difficult to oxidise: C–O–CH3
(1) HCHO (2) CH3CHO (2)
(3) CH3COCH3 (4) CH3CH2CHO
O
10. Which of the following can be used to
OH
differentiate between ethanal and
OH
propanal- (3)
(1) Ammoniacal AgNO3
(2) NaHSO3 OH
(3) I2 in presence of base
O
(4) 2,4-DNP
C OH
(4) H
11. The reaction
CHO
OH
heat
CHO + conc. NaOH →
produces 14. Which of the following pairs of carbonyl
CH(OH)2 compounds and Grignard reagent should be
(1) (2) O chosen as reactants to prepare 2, 4-
CH(OH)2 dimethyl-3-pentanol ?
CH2OH COO– (1) CH3CH2CH2CHO and CH3CH2CH2MgBr
(3) (4) (2) (CH3)2CHCHO and (CH3)2 CHMgBr
COO– COO–
(3) CH3CH2COCH3 and (CH3)2 CHMgBr
(4) (CH3)2CHCHO and CH3CH2CH2MgBr
12. Consider the following sequence of
reactions. 15. Which of the following forces is correctly
NaOH LiAlH4
CH3COCH3 
heat
→ A 
→B described about boiling point of Aldehydes
& ketones:
The final product (B) is :
(1) (CH3)2C(OH)CH2COCH3 (1) Hydrogen bond
(2) (CH3)2C=CHCOCH3 (2) Vander wall force
(3) (CH3)2CHCH2CHOHCH3 (3) Dipole-dipole attraction
(4) (CH3)2C=CHCHOHCH3 (4) None of these

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16. Consider the following sequence of O
Θ
OH →
reactions 21. → 3
(CH ) S
A 
∆
B; ‘B’ is –
3 2

+
H3O 1. CH MgΙ
CH3C≡CH 
Hg2+
→ A 
2. H O
→B
3
CHO OH
2
(1) (CH2)4 (2)
The final product (B) is : CHO
OH
(1) CH3C≡CCH3 OH
(2) CH3COCH2CH3 (3) CHO (4) CHO
(3) CH3CH2CHOHCH3
(4) (CH3)3 C – OH
22. The suitable reagent to convert
CH3CH=CH–CHO into CH3–CH2CH2–CH2–OH is :
17. Carbonyl compounds react with which of
(1) NaBH4 (2) LiAlH4
the following reagent to form a colorless
(3) Zn – Hg/HCl (4) H2/Pd
crystalline solid–
(1) PCl5 (2) HCN
23. Predict the products in the given reaction.
(3) NH2OH (4) NaHSO3
CHO
50% KOH
18. Which of the following statement is true for
Cannizzaro reaction– Cl
(1) The aldehyde is oxidised as well as CH2OH COO¯
(1) +
reduced
Cl Cl
(2) The aldehydes do not containing α-
CH2OH COO¯
Hydrogen atoms give the reaction (2) +
(3) The reaction is not given by aldehydes
OH OH
containing α-Hydrogen atoms
CH2OH CH2COO¯
(4) All of these (3) +

Cl Cl
19. Acetaldehyde and acetone can be
CH2OH OH
distinguished by all the following except- (4) +
(1) Iodine + alkali OH OH
(2) Tollen’s reagent
(3) Fehling solution 24. CH3CHO and C6H5CH2CHO can be
(4) Schiff reagent distinguished chemically by :
(1) Tollen’s reagent test
20. Which of the following statements is wrong
(2) Fehling solution test
(1) All methyl ketones give a positive
(3) Benedict test
iodoform test.
(4) Iodoform test
(2) Acetaldehyde is the only aldehyde that
gives iodoform test. 25. Which of the following is useful for
(3) All secondary alcohols give positive characterisation of aldehyde and ketones.
iodoform test. (1) Derivatives of Semi carbazide
(4) Any alcohol that can be oxidised to an (2) Derivatives of Phenylhydrazine
acetyl group gives a positive iodoform (3) 2, 4–DNP Derivatives
test. (4) Derivatives of Hydrazine
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26. Which of the following arrangements with 29. Consider the following reaction sequence:
respect to their reactivity in nucleophilic CN
addition reaction is correct? (I) AlH(i–Bu)2 CH3CHO
(II) H2O
A B
(1) benzaldehyde < acetophenone dil NaOH,∆

< p-nitrobezaldehyde < p-tolualdehyde The Major Product ‘B’ is

(2) acetophenone < benzaldehyde (1) N=CH–CH3
< p-tolualdehyde < p-nitrobezaldehyde
(3) acetophenone < p-tolualdehyde (2) CH=CH–CHO
< benzaldehyde < p-nitrobezaldehyde
(3) CH2–N=CH–CH3
(4) p-nitrobezaldehyde < benzaldehyde
< p-tolualdehyde < acetophenone O
(4) C–N=CH–CHO
(i) NaBH4
27. CH3–CH2– CH– CH=O P;
(ii) H2O P is
CH3 O

(1) (2) 30. Br2/AlBr3 ‘A’ (Major Product)

(3) (4) What is the product ‘A’?

OH HO O O

(1) (2)
28. Which cannot show simple aldol Br
condensation Br
(1) CH3 – CH = 0 (2) CH3–C–CH3 O O
O
(3) C6H5 – CH = 0 (4) Ph–C–CH3 (3) (4)
Br Br
O

Carboxylic acids:- Introduction

Nomenclature: Common names of carboxylic acids are based on their source of origin e.g., Formic acid
(HCOOH) was first obtained from red ants (Latin Formica means ants) similarly acetic acid (CH3COOH)
is so named because it was obtained from vinegar (latin acetum means vinegar), butyric acid
(CH3CH2CH2COOH) from rancid butter (Latin Butyrum means butter) etc.
In common system position of substituents is indicated by Greek letters α, β, γ, δ etc.
α β α
CH3–CH–COOH CH3CH–CH2–COOH
CH3 CH3
α–methyl propionic acid β–methylbutyric acid
(Isbutyric acid)
Common Name of some important carboxylic acids
Structure Common name
HCOOH Formic acid
CH3–COOH Acetic acid
CH3-CH2–COOH Propionic acid

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CH3–CH2–CH2–COOH Butyric acid
(CH3)2CHCOOH Isobutyric acid
HOOC–COOH Oxalic acid
HOOC–CH2–COOH Malonic acid
HOOC–(CH2)–COOH Succinic acid
HOOC–(CH2)3–COOH Glutaric acid
HOOC–(CH2)4–COOH Adipic acid
HOOC–(CH2)5–COOH Pimelic acid
Benzoic acid
COOH

Cinnamic acid
CH=CH–COOH

COOH Phthalic acid

COOH

Methods of Preparation of carboxylic acids

(1) From primary alcohol: Primary alcohols are easily oxidised to carboxylic acids with oxidants such as
KMnO4, in neutral, acidic or alkaline media or by CrO3 in acidic media or K2Cr2O7 in acidic media.
Ex. (i) CH3–CH2–OH (i) alkaline KMnO4
CH3COOH
(ii) H3O+

CH2OH COOH
(ii) (i) alk. KMnO4, ∆
(ii) H3O+

CrO3/H2SO4
(iii) CH3(CH2)8CH2–OH CH3(CH2)8COOH
Jones reagent
decanol decanoic acid

(2) From oxidation of aldehydes and ketones: Aldehydes can be oxidised to carboxylic acids even with mild
oxidising agent while ketones are oxidised under vigorous conditions i.e. strong oxidising agents & at
elevated temperatures and mixture of acids having lesser number of carbon atoms than parent ketone
is obtained.
KMnO4/H⊕
Ex. (i) CH3–CHO CH3COOH
KMnO4 /H⊕
(ii) CH3–COCH3 High temp CH3COOH+CO2+H2O
KMnO4/H⊕
(iii) CH3–COCH2 CH2CH3 CH3COOH+HOOC–CH2–CH3
High temp

(3) By hydrolysis of nitriles (cyanides) and amides : Nitriles are hydrolysed to amides and then to
carboxylic acids in the presence of H+ or OH– as catalyst. If conditions used are mild, then reaction can
be stopped even at amide stage
–
Ex. (i) R–CN H+ or OH
RCONH2 H+ or OH–
RCOOH + NH3
H2O ∆
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–
(ii) CH3C≡N H+ or OH
CH3CONH2 H+ or OH–
CH3COOH + NH3
H2O ∆
CN CONH2 COOH
⊕
(iii) H/H2O H+/H2O + NH3
∆

(iv) CH3CONH2 H3O

CH3COOH + NH3
∆

(4) From Grignard reagent: Reaction of Grignard reagent with dry ice(i.e. solid CO2) in dry ether followed
by acidic hydrolysis produces carboxylic acid containing one more carbon atom than Grignard reagent.
O
– + Br
Ex. (i) O=C=O+R– MgBr Dry ether +
R–C–OMgBr H2O/H R–COOH+Mg
OH
(ii) C2H5MgBr+CO2 Dry ether H2O/H+ OH
C2H5COOMgBr C2H5COOH+Mg
Br
MgΙ COOMgΙ COOH
(iii)
+ CO2
H2O/H+
+ Mg OH
Ι

(5) By hydrolysis of acid halides and anhydrides: Acidic hydrolysis of acid halide and anhydride directly
gives carboxylic acids while their basic hydrolysis first forms carboxylates, which on acidification form
corresponding carboxylic acids.

⊕
H2O/H
Ex. (i) RCOCl R–COOH+HCl
Θ Θ Θ
OH/H2O H+
R–COO+Cl R–COOH
COCl COOH
H2O
(ii) H+
+ HCl

H2O
(iii) (CH3CO)2O H+
2CH3COOH

CO COOH
(iv) H2O
O H+

2
COOH
H2O
(v) C6H5COOCOCH3 H+ +CH3COOH

(6) By hydrolysis of esters: Acidic hydrolysis of esters directly gives carboxylic acids while their basic
hydrolysis first forms carboxylates, which on acidification form corresponding carboxylic acids.

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Ex. (i) COOC2H5 COOH

H+/H2O
+ C2H5OH

NaOH
(ii) CH3CH2CH2COOC2H5 CH3CH2CH2COONa+C2H5OH
H3O+

CH3CH2CH2COOH
+
H2O/H
(iii) CH3COOCH3 CH3COOH+CH3OH

(7) From alkyl benzenes: Alkyl benzene containing benzylic H-atom on vigorous oxidation by H2CrO4 or
⊕ Θ
H/KMnO4 or OH/KMnO4 gives Aromatic carboxylic acid.
CH3 COOK COOH
KMnO4,OH– H3O+
∆

Toluene
COOH
CH2CH2CH3 COOK
KMnO4,OH– H3O+
∆

Note. Entire side chain containing primary and secondary alky group of benzene ring is oxidised to–COOH
group, irrespective to the length of side chain while tertiary alkyl group is not affected.
CH3 COOH

(i) KMnO4/OH–
(ii) dil. H2SO4

CH3 COOH
CH3
CH–CH3 (i) KMnO4/OH–
COOH
(ii) dil. H2SO4

(8) From oxidation of alkenes: Alkenes undergo oxidative cleavage by hot acidic KMnO4 or K2Cr2O7. to give
carboxylic acid
KMnO4/OH–,∆ H+
RCH=CHR 2RCOOK 2RCOOH

KMnO4/OH–,∆
CH3CH2CH=CH–CH3 CH3CH2COOH+CH3COOH

KMnO4/OH–,∆ COOH
COOH

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Exercise 1.3

1. NaCN
(CH3 )2 CO  H3O
→ A  →B
+
5. The final product S in the following reaction
(HCN)
sequence is
In the above sequence of reactions A and B
+
CH3Cl NBS KCN H3O
are:  →P → Q → R  →S
Anhy.AlCl 3

(1) (CH3)2C(OH)CN, (CH3)2C(OH)COOH

CH3
(2) (CH3)2C(OH)CNH2, (CH3)2C(OH)2 (1)
(3) (CH3)2C(OH)CN, (CH3)2CHCOOH COOH

(4) (CH3)2C(OH)CNH2, (CH3)2C = O CH3

(2)
COOH
2. By which of the following sequence of steps
CH2–COOH
the alcohol RCH2CH2OH(X) can be (3)
converted into RCH2CH2COOH(Y)?
PBr3 KCN H3O + CH3
(1) X → →  →Y
PBr3 KCN H2 /Pt
(2) X → →  → Y (4)
+
COOH
KCN H3O
(3) X →  →Y
+
HCN PBr3 H3O
(4) X → →  →Y 6. Consider the following reaction.
CH2OH PBr3 KCN H3O⊕ O
3. (A) (B) (C) 18 16
H +

CH3–C–O–C2H5 +H2O  →
CH2OH
The products formed in the reaction are
In the above sequence of reactions the end
O
product (C) is : 16 18
(1) CH3–C–OH and C2H5 OH
CH2CH2CN CH2CN O
(1) (2) 16 16
CH2CH2CN CH2CN (2) CH3–C–OH and C2H5 OH
O
CH2COOH CH2CH2–NH2 18 16
(3) (4) (3) CH3–C–OH and C2H5 OH
CH2COOH CH2CH2–NH2
O
18 18
(4) CH3–C–OH and C2H5 OH
4. Hg2+,H2SO4
A
HCN
B
H3O⊕
7. Consider the following sequence of
The end product 'B' will be:
reaction:
OH O KCN H3O +

(1) (2) HO C2H5Cl →

DMSO
X Δ
→Y
OH
CN The products (X) and (Y) are, respectively:
O (1) C2H5CN and C2H5CH2NH2
CN (2) C2H5CN and C2H5CONH2
(3) OH (4)
OH OH (3) C2H5NC and C2H5NHCH3
(4) C2H5CN and C2H5COOH

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8. The Final Product ‘B’ in the following 11. The end product (Z) in the given sequence
reaction sequence of reactions is
CH3—CH2—C—CH3 HCN Conc.H2SO4 (i) KCN
A ∆ B CH3CH=CHCHO NaBH4 X HCl Y (ii) H+/H O Z
ZnCl2 2
O
(1) CH3CH=CHCH2COOH
CH3
(2) CH3CH2CH2COOH
(1) CH3—CH=C—COOH (3) CH3CH=CHCOOH
(2) CH3—CH=C—CN (4) CH3CH(Cl)CH2COOH
CH3 C
(3) CH3—CH=C—CONH2
12. (i) Mg(1 Mole)/THF Product (major)
(ii) CO2
CH3
(iii) H⊕
OH
B
(4) CH3—CH2–C—COOH Product is:-
COOH Cl
CH3
9. Study the following sequence of reactions (1) (2)
and identify the product (Y).
dil NaOH HCN COOH COOH
CH3CHO + HCHO  Heat
→ x 
H O⊕
→y
3 COOH
CN
Cl
(1) CH2=CH—CH—COOH (2) CH3—C—COOH (3) (2)
OH OH
Br
(3) CH3CH2—CH—COOH (4) CH2=CH—CH—COOH

OH CN 13. Consider the following reaction,

Zn dust CH Cl alkaline KMnO4
10. Identify (X) and (Y) in the given reaction Phenol X Anhyd.3 AlCl3 Y Z
sequence. The product Z is
COCH3 (1) benzaldehyde (2) benzoic acid
(3) benzene (4) toluene.
KMnO4 H3 O +
KOH
X Y
14. → X ( gas )
BaCO3 + H2SO4 
COOH CHO (i) Mg, THF
Br (ii) X
y; y will be
(1) X= , Y= (iii) H3O⊕
O
CH2OH
OH
CH=CH2 (1) (2) O H
O
(2) X= , Y= Br H
(3) (4)
O O
COOK COOH
Θ →(A)
( i)BH3 /THF
15. CH3–CH2–C≡CH 
( ii)H2O2 /OH
(3) X= , Y=
→ (B ) ; final product 'B' is-
CrO3 /H2 SO4

CH2OH CHO O
(1) CH3CH2C–CH3 (2) CH3CH2CH2–COOH
(4) X= , Y= OH
(3) CH3CH2CH–CH3 (4) CH3CH2CH2–CHO

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Physical properties of carboxylic acids

1. Physical state: First 3 members are colorless liquids with pungent smell, next 6 members are only
liquids with unpleasant odour. Higher acids are colorless waxy solids.
Benzoic acid and its homologues are colorless solids.

2. Solubility: Among aliphatic acids first four members are very soluble in water due to the formation of
H-bond with water. The solubility decreases with increasing number of carbon atoms. Higher carboxylic
acids are practically insoluble in water due to the increased hydrophobic interaction of hydrocarbon
part. Benzoic acid, the simplest aromatic carboxylic acid is nearly insoluble in cold water.
O
C
R O—H O–H
O O=C
H–O H H R
C=O
R

3. Boiling Points: boiling points of carboxylic acids are higher than corresponding aldehydes, ketones,
ethers or even alcohols of comparable molecular masses.
Compound CH3COOH CH3CH2CH2OH CH3CH2CHO CH3COCH3 CH3CH2CH2CH3
Boiling Point (K) 390 370 322 329 309

This is due to more extensive association of carboxylic acid molecules through intermolecular H–
bonding. The H-bonds are not broken completely even in the vapour phase. In fact, most carboxylic
acids exist as dimer in the vapour phase or in the aprotic solvent
O H–O
R–C C–R
O–H O
Dimer In vapour phase or in aprotic solvent
Chemical reactions of carboxylic acids
Chemical properties of carboxylic acids can be discussed under the following heads:
1. Reactions involving cleavage of O–H bond i.e. acidic nature
Carboxylic acids dissociate in water to give resonance stabilized carboxylate ion and hydronium ion.
Θ
O O O
+
R–COOH+H2O H3O + R–C R–C ≡ R–C –
O O O
Θ
For the above reaction
[H3 O+ ][RCOO− ]
K eq =
[H2 O][RCOOH]
[RCOO− ][H3O+ ]
=K a K=
eq [H2 O]
[RCOOH]
Where Ka = acid dissociation constant and
Keq = Equilibrium constant and
pKa = –log Ka
Generally comparative acidic strength is studied by comparing their pKa values, higher the pKa, weaker
is the carboxylic acid and vice versa.

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Strong acids have pKa < 1
Moderately strong acids have pKa values between 1 to 5.
Weak acids have pKa value between 5 to 15.
Extremely weak acids have pKa > 15
Carboxylic acid as like alcohols react with electropositive metals, evolve hydrogen gas and form salts
with alkalies as like phenols. Carboxylic acids can also react with weaker bases such as carbonates and
bicarbonates to evolve CO2 which phenols cannot. This reaction is used to detect presence of carboxyl
group in an organic compound.
2RCOOH + 2Na → 2RCOO–Na+ + H2
Sodium carboxylate
RCOOH + NaOH → RCOONa + H2O
RCOOH + NaHCO3 → RCOONa + H2O + CO2
Compound HCl CF3COOH C6H5COOH CH3COOH C6H5OH C2H5OH
pKa values -7 0.23 4.19 4.76 10 16

Carboxylic acids are more acidic than phenols as carboxylate ion is stabilized by two equivalent
resonating structures in which -ve charge is at the more electronegative oxygen atom. In comparison
although phenoxide ion is also resonance stabilised but resonating structures are non equivalent in
which -ve charge is at less electronegative carbon atom, hence resonating structures of phenoxide are
less important than that of carboxylate ions.
Effect of substituents on acidity of carboxylic acid:
Any substituent which can increase stability of conjugate base of carboxylic acid i.e., carboxylate ion
will increase its acidity and a substituent decreasing the stability of carboxylate ion will decrease the
acidity of carboxylic acid.it means electron withdrawing group i.e., EWG will increase and electron
donating group i.e., EDG will decrease the acidity of carboxylic acid.
Therefore the effect of the various substituents in increasing acidity order must be
–CF3 > –NO2 > –CN > –F > –Cl > –Br > –I > –Ph
Similarly effect of various groups in decreasing acidity order is
CH3 –> C2H5– > (CH3)2CH– > (CH3)3C–
Based on pKa values the order of decreasing acidity of carboxylic acid are as follows
CF3COOH > CCl3COOH > HCCl2COOH > NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH >
BrCH2COOH > HCOOH > ClCH2CH2COOH > C6H5COOH > C6H5CH2COOH > CH3COOH > CH3CH2COOH.

2. Reactions involving cleavage of C–OH group

O
–OH group of carboxylic acids can be replaced by many group such as–Cl, –NH2; –OR and —O–C—R' to
form acid derivatives i.e., acid halides, acid amides, ester and acid anhydrides.
(a) Reaction with PCl3, PCl5 or SOCl2 forming acid chlorides:
3RCOOH + PCl3 → 3RCOCl + H3PO3
RCOOH + PCl5 → RCOCl + PCl3 + HCl
RCOOH + SOCl2 → RCOCl + SO2 + HCl
Uses of SOCl2 is preferrable over the use of PCl3 and PCl5, because in this reaction the other two
products obtained are gaseous i.e., SO2 and HCl, which escape itself from the container, leaving behind
the acid halide.

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(b) Reaction with NH3: Reaction of carboxylic acid with NH3 forms ammonium salts first which form amide
on heating.
∆
RCOOH + NH3  RCOO−NH4+ →RCONH2 + H2O
Θ ⊕
COOH COONH4 CONH2
∆
+ NH3 –H2O

Benzamide

Θ ⊕ O
COOH COONH4 CONH2
∆ C
+ NH3 Strong heating
Θ ⊕ –2H2O –NH3 NH
COOH COONH4 CONH2 C
Phthalamide
O
Phthalimide

(c) Formation of anhydride: Carboxylic acid on heating in presence of strong dehydrating agent i.e., P2O5
or conc. H2SO4 give corresponding acid anhydride

O
RCOO H R–C
∆
+ P2O5
O + H2O
RCO OH R–C
O
O O
∆
2CH3COOH P2O5 CH3–C–O–C–CH3+H2O

O O
COOH C C
∆ O
2 P2 O5

(d) Esterification: Reaction of carboxylic acid with alcohols or phenols in the presence of mineral acid i.e.,
conc. H2SO4 or HCl forms esters.
H+
RCOOH+R'OH RCOOR'+H2O
H2SO4
CH3COOH+HOC2H5 CH3COOC2H5+H2O

COOH H2SO4
COOC2H5
+ C2H5OH + H2O

Reactivity of alcohols toward esterification is

3° alcohol < 2° alcohol < 1° alcohol < CH3OH
and reactivity of carboxylic acids towards esterification is
R3C–COOH < R2CH–COOH < RCH2COOH < CH3COOH < HCOOH

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
Mechanism of esterification: It is a nucleophilic acyl substitution which is acid catalysed reaction,
which can be studied in following steps.
⊕
OH OH
O R'–OH ⊕
R–C + H+ R–C–OH R–C—O–R
OH
(carboxylic acid) OH H
(Tetra hedral intermediate)

Proton transfer
⊕
O OH OH
–H⊕ –H2O
R–C OR' R–C–OR' R–C—O–R'
(Ester) ⊕OH
2

3. Reactions involving –COOH group

(a) Sodalime decarboxylation: Carboxylic acids lose CO2 to form hydrocarbon on heating sodium or
potassium salts of carboxylic acid with sodalime (NaOH + CaO).
CaO, ∆
R COONa + NaO H  →R − H + Na2CO3
CaO, ∆
CH3 COONa + NaO H  → CH4 + Na2CO3

(b) Kolbe's electrolysis: Electrolysis of aq. Solution of sodium or potassium salts of carboxylic acids leads
to decarboxylation to give alkanes.
electrolysis
Ex. 2CH3 − COONa( aq ) → CH3 − CH3 + 2CO2 + 2NaOH + H2
• •
At anode: 2CH3COOΘ 
−2eΘ
→ 2CH3COO 
−2CO
→ 2CH3 
→ CH3 − CH3
2 Alkane

At cathode: 2H2O + 2e  2OH +H2↑

– Θ

2Na⊕ + 2OHΘ  2NaOH

(c) distillation: Heating of calcium salts of carboxylic acids forms aldehydes and ketones by distillation.
(i) Distillation of calcium formate gives formaldehyde
HCO O
distillation
Ca HCHO + CaCO3
∆
H COO
(ii) Distillation of calcium formate with calcium acetate gives acetaldehyde.
HCO O OOC CH3
distillation
Ca + Ca ∆ 2CaCO3 + 2CH3CHO
HCO O OOC CH3

(iii) Distillation of calcium acetate forms acetone.

CH3CO O
distillation
Ca CH3COCH3 + CaCO3
∆
CH3 COO

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(d) Reduction:
(i) Carboxylic acids are reduced to primary alcohols by LiAlH4 or B2H6 in T.H.F.
LiAlH /ether
RCOOH 4
orB H /THF
→RCH2 OH
2 6

• Diborane does not reduce functional groups such as ester, nitro, halo etc.
• NaBH4 does not reduce carboxyl group.

(ii) Reduction of carboxylic acids with Red P and HΙ gives alkane.

Red P
RCOOH + 6HΙ 
473K
→R − CH3 + 2H2O + 3Ι2

4. Substitution reaction in the hydrocarbon part:

(a) Halogenation: Carboxylic acids having atleast one or more α-H atoms react with Cl2 or Br2 in the
presence of a small amount of Red P to give α-halocarboxylic acids. The reaction is known as Hell-
Volhard-Zelinsky reaction
(i) Red P/X2
R–CH2COOH R–CH–COOH(X=Cl,Br)
(ii)H2O
X
α-halocarboxylic acid

(i) Br2/Red P
Ex. CH3CH2COOH (ii) H2O
CH3–CH–COOH
Br

(b) Ring substitution in aromatic acids: –COOH group in aromatic carboxylic acids is a ring deactivating
& meta directing group. They do not undergo Friedel–Craft reaction, because carboxyl group is
deactivating and catalyst AlCl3 (Lewis acid) gets bonded to carboxyl group.
Some common electrophilic substitution reactions of benzoic acid are as following:
COOH COOH

Conc.(HNO3+H2SO4)
(i) Nitration:
NO2

COOH COOH

FeBr3
(ii) Halogenation: + Br2
Br

COOH COOH

SO3
(iii) Sulphonation: + H2SO4
SO3H

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids

Exercise 1.4
O 4. Let us consider an esterification of
NaBH4 LAH
1. (B) COOH (A) isotopically labelled carboxylic acid :
Me
O
18
The products (A) and (B) are : CH3–C–OH + CH3CH2 OH  H ⊕
→ (X) and (Y);
OH
(X) and (Y) respectively are :
COOH & Me COOH
(1) Me O
18
(1) CH3–C–OC2H5 ;H2O
OH
COOH O
COOH & Me 18
(2) Me (2) CH3–C–OC2H5 ;H2O
OH OH O
COOH CH –C–OC2H5 ;H2O
(3) Me OH & Me (3) 3
OH OH (4) both (1) and (2)
COOH & Me OH
(4) Me 5. The rate of esterification of HCOOH (I),
CH3COOH (II), (CH3)2CHCOOH (III) and
(CH3)3CCOOH (IV) with ethanol follows in
Cl2 Alc.KOH
2. CH3CH2COOH 
red P
→ (A)  → (B) the order :
The compound (B) is : (1) IV > III > II > I (2) I > II > III > IV
(1) CH3CH2OH (3) II > I > IV > III (4) III > IV > I > II
(2) CH3CH2COCl
(3) CH2=CHCOOH 6. Ethanol on heating with acetic acid in the
(4) CH3–CHCl–COOH presence of a few drops of sulphuric acid
gives:
3. In a set of reactions acetic acid yielded a (1) Oil of wintergreen
product D. (2) Oil of mustard
⊕
SOCl2 benzene HCN H2O/H
CH3COOH  → A  →B → C → D (3) Fruity smell
Anhy.AlCl 3
(4) Oil of bitter almonds
The structure of D would be
COOH 7. redP
X ← LiAlH4
 CH3COOH  → Y. What does
HI

(1) C–CH3 not true for X and Y –

OH (1) X is hydrocarbon of general formula
COOH CnH2n+2 while Y belong to alkanol
(2) X can be obtained by reducing
(2) C–CH3
CH3CH2Cl while Y by its hydrolysis
OH (3) X gives positive litmus test but Y does
OH not
(4) X and Y both belong to different
CH2–C–CH3
(3) homologeous series
COOH
OH 8. Acetic acid exists as dimer in C6H6 due to:
(CH2)2–C–CH3 (1) Condensation reaction
(4) (2) Hydrogen bonding
COOH (3) Presence of a carboxyl group
(4) Presence of hydrogen at α-carbon
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9. In the following sequence of reactions 13. Hell Volhard Zelinsky reaction involves
KCN H3O+
LiAlH4 (1) Halogenation of carbonyl compound at
CH3 − Br → A 
→B 
ether
→ C,
α–C
the end product (C) is :
(2) Halogenation of carboxylic ester at α–C
(1) Acetaldehyde (2) Ethyl alcohol
(3) Halogenation of alcohol at α–C
(3) Acetone (4) Methane
(4) Halogenation of carboxylic acid at α–C
+
CuCN H2O | H NH3
10. C6H5N2Cl  → A → B →
∆
C; 14. Identify the compounds (X), (Y) and (Z) in
final product is the following reaction:
(1) C6H5–NH2 (2) C6H5CH2NH2 Mg/ether (i) CO2 CH3OH,H+
CH3Br X Y ∆ Z
(ii) water
(3) C6H5–CONH2 (4) C6H5 –COOH
(1) (X)=CH3MgBr, (Y)=CH3COOH,
(Z)=CH3COOCH3
11. Which of the following acid has Lowest Pka
(2) (X)=CH3CH2Br, (Y)=CH3CH2OH,
value?
(Z)=CH3CH2CH2CH3
O O
(3) (X)=CH3CH2MgBr, (Y)=CH3CH2COOH,
(1) CH2–C–OH (2) CH2–CH2–C—OH
(Z)=CH3CH2COCH3
NO2 NO2
(4) (X)=CH3MgBr, (Y)=CH3CH2COOH
O Cl O (Z)= CH3CH2COOCH3
(3) CH2 –C–OH (4) H—C —C—OH
Cl Cl 15. Reagent
CH3 − COOH  → CH3 − CH2 − OH
∆

The reagent of the given reaction will be :-

CrO3 −H2 SO4 (i)B2H6
12. C6H5CH2–OH 
→ (A) 
(ii)H3O⊕
→ (B) (1) LiAlH4 /H3O+
In the above given sequence reaction, final (2) B2H6 /H3O+
product ‘B’ is
(3) Both (1) and (2)
(1) C6H5–CHO (2) C6H5–COOH
(4) NaBH4 /H3O+
(3) C6H5–CH3 (4) C6H5–CH2–OH

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids

Exercise 2
1. HO–CH2–CH2–OH on heating with periodic 4. In a set of reactions, ethyl benzene yielded
acid gives :-
a product D
H
(1) 2 C=O (2) 2CO2 CH2CH3
H KMnO4 Br2 C2H5OH

H⊕
→B 
FeCl
→ C H+
→D
3

CHO
(3) 2HCOOH (4) ‘D’ would be:
CHO
COOC2H5
2. Which of the following compounds would
undergo the Cannizzaro’s reaction– (1)
(1) Acetaldehyde Br
(2) Benzaldehyde
(2) CH2–CH–COOC2H5
(3) Propionaldehyde
Br
(4) Anisole
COOC2H5
3. Predict the product when given compound
reacts with LiAlH4 : (3)
O O
Br
C C
H OCH3
CH2–COOC2H5
(A)
O (4)
O Br
OH
C–O–CH3
NH3 Heat
5. In a reaction, (A) → (B)  → (C)
(1)
P2O5
OH → C2H5CN; (A), (B) and (C) are :-
O (1) CH3COOH, CH3COONH4 and CH3CONH2
OH
C–O–CH3
(2) CH3COCl, CH3CONH2 and CH3COONH4
(2)
(3) C2H5COOH, C2H5COONH4 and C2H5CONH2
O
(4) C2H5CONH4, C2H5CONH2 and C2H5COOH
OH
OH 6. Formic acid and formaldehyde can be
(3)
OH distinguished by treating with –
O (1) Benedict’s solution
C (2) Tollen’s reagent
H OH
(4)
(3) Fehling’s solution
OH
(4) NaHCO3

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7. Consider the following compounds : +

H3O HBr
10. CH3MgBr +  → (A)  → (B)
(i) COCl
O
(ii)O2N COCl Mg/ether
 HCHO
→ (C)  → (D)
+
H3O

(iii) H3C COCl The end product (D) is :

(iv) OHC COCl

(1) (2) CH2OH
The correct order of reactivity towards
CH3 CH2OH CH3
hydrolysis is:
(1) i > ii > iii > iv OH
(2) iv > ii > i > iii
(3) CH3 (4)
(3) ii > iv > i > iii
(4) ii > iv > iii > i CH3 CH2OH

⊕
11. Methyl acetate on treating with excess of
.
HBr KCN H3 O
8. CH2OH  → A → B →
∆
C;
C2H5MgBr produces :
C is–
CH3 C2H5
(1) Phenyl ethanoic acid
(1) CH3–C–OH (2) CH3–C–OH
(2) Benzoic acid
CH3 C2H5
(3) Benzyl amine
(4) Aniline C2H5 CH3
(3) C2H5–C–OH (4) CH3–C–OH
Br
⊕
C2H5 C2H5
Mg CO2 H3O
9. 
dryether
→P 
(dry ice)
→ Q  →R ;
NO2 12. Which gives a ketone on treating with a
R is– Grignard’s reagent–

COOH
(1) Formaldehyde
(1) (2) Ethyl alcohol
NO2 (3) Methyl cyanide

HOOC (4) Methyl iodide

(2)
NO2
13. Benzaldehyde and formaldehyde differ
CH2COOH from acetaldehyde in their reaction with–
(3)
(1) NaOH
NO2
(2) HCN
HOOCH2C (3) 2,4-Dinitrophenyl hydrazine
(4)
NO2 (4) Semi carbazide

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
[O] 18. Consider the following sequence of
14. C3H8O → C3H6O
K Cr O /H SO
2 2 7 2 4 reactions:
I2 +NaOH(aq.)
 → CHI3, Ketone ( A ) 
1. C2H5MgBr
2. H O
H2 SO4 , heat
→B →−H O
C 1. O3
2. Zn, H O
→
2 2 2

In this reaction the first compound is – H

(1) CH3CH2CH2OH (2) CH3–CH–CH3 +
O O
OH The ketone (A) is :
(3) CH3OCH2CH3 (4) CH3CH2CHO
(1) (2)
Dil.
O O
15. CH3–CH2–CHO  → Product.
alkali (3) (4)
The product in the above reaction is– O O
(1) CH3–CH2COOH 19. Which of the following reactions will not
(2) CH3–CH2–CH2OH result in the formation of carbon-carbon
(3) CH3–CH2–CH–CH2–CHO bonds?
(1) Friedel-Crafts acylation
OH (2) Reimer-Tieman reaction
CH3 (3) Cannizzaro reaction
(4) Wurtz reaction
(4) CH3–CH2–CH–CH–CHO
OH 20. In a set of the given reactions, acetic acid
yielded a product C.
PCl5 C H
16. An organic compound C3H6O does not give a CH3COOH  → A →
6 6
AlCl
B
3

precipitate with 2,4-dinitrophenyl hydrazine (i)C2H5MgBr/Ether

(ii) H O+
→C
reagent also does not react with metallic 3

product C would be:

sodium. It could be: (1) CH3CH(OH)C2H5
(1) CH3–CH2–CHO (2) CH2=CH–OCH3 (2) CH3COC6H5
(3) CH3–COCH3 (4) CH2=CH–CH2–OH (3) CH3CH(OH)C6H5
C2H5
(4) CH3–C(OH)C6H5
Br
CH3MgBr
17.  → (A) O
1 (eq.)
CH=O
H2CrO4
CH3
(1) 21. ∆ Product is:
O

(2) O
O COOH C
H
CH3 (1) (2)
O
(3) CH3
OH
(3) CH3 (4)
(4)
O CH3

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22. An organic compound contains 69.77% 25. From the given below, number of
carbon, 11.63% hydrogen and rest oxygen.
compounds which give positive Fehling’s
The molecular mass of the compound is 86.
It does not reduce Tollen’s reagent but test are.
forms an addition compound with sodium Benzaldehyde, Acetaldehyde, Acetone,
hydrogensulphite and give positive
iodoform test. On vigorous oxidation it acetophenone, methanal.
gives ethanoic and propanoic acid. Possible (1) 2 (2) 3
structure of the compound is.
(1) CH3 – CH2 –C–CH2 – CH3 (3) 4 (4) 5

O
(2) CH3 – C – CH2 – CH2 –CH3 26. 3-chloro benzaldehyde on treatment with
O
50% KOH solutions gives product
CH3
(3) CH3 – C – CH – CH3 OH O
CH—C
CH3 (1)
(4) –CH – CH3
OH Cl Cl
23. The final product A formed in the following
COOΘ CH2OH
multistep reaction sequence is: (2)
(i)H2O,H⊕ +
(ii)CrO3

(iii)H2N–NH2 ,KOH/ ∆
→ A
Cl OH
OH O
O
(1)
O C–H C–OH
(3) +
N–NH2

(2) OH OH

COOΘ CH2OH
N (4)
(3) NH2 +

Cl Cl
(4)

27. The product A and B obtained in the

24. (i)CrO2Cl2 ,CS2

 HCl(dil.)
→ (A)  →(B) following reaction respectively are :
(ii)H O⊕
3
(iii)NaHSO3 (Crystal)
O
Toluene
(Excess) R–COOH + PCl5 → R–C–Cl + POCl3 + A
A and B formed in the above sequence O
reaction respectively are : R–COOH + SOCl2 → R–C–Cl + HCl + B
(1) Ph-CHO and Ph-COOH
(2) Ph–CH(OCrOHCl2) and Ph-CHO (1) H3PO3 and SO2
(3) Ph-CHO and Ph-COONa (2) HCl and SO3
OH
(4) Ph–CH and Ph-CHO (3) H3PO4 and SO2

SO3Na (4) HCl and SO2

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
28. Positive Tollen’s test is observed in which of 31. Which of the following compounds give
the following compound yellow precipitate with I2/NaOH?
CH2OH O OH
C=O CHO (a) (b)
(i) H OH (ii) H
OH H (c) (d)
CH2OH O O

O (e) CH3–CHO
(iii) CH3–CHO (iv) CH3–C–CH3 (1) (a) and (e) only
CH2OH (2) (a) and (b) only
O
HH OH (3) (a), (b) and (e) only
OH H (4) (a) only
(v) HO H
H OH
(1) Only (ii) and (iii) 32. Which of the following reaction is correct?
(2) Only (i), (ii), (iii) and (v) Br
∆
(3) Only (ii), (iii) and (v) + Br2
(1) sunlight
(4) Only (ii), (iii) and (iv) Br
CH3 CHO
29. An organic compound (X) with molecular
formula C8H8O give positive 2, 4-DNP test. CO, HCl
(2) anhyd.AlCl3/CuCl
It give a yellow precipitate on reaction with
iodine and sodium hydroxide solution. (X) NO2 NO2
does not reduce Fehling’s reagent but on NaOH & CaO
(3) R–COONa  → R–CH3 + Na2CO3
drastic oxidation with chromic acid it gives ∆

a carboxylic acid (Y) with molecular O O

formula C7H6O2. (X) and (Y) respectively are CH3 (i) DIBAL-H CH3
(1) C6H5COCH3, C6H5COOH (4) (ii) H2O
(2) C6H5CH2CHO, C6H5CH2COOH
O OC2H5 O H
(3) C6H5COCH3, CH3COOH
(4) C6H5CH2CHO, C6H5COOH
33. In the given reaction, final product is
30. The structure of product ‘C’ formed by the
Ph–CH–CH3 anhyd. A
Mg,ether
B
(i) HCHO
following sequence of reactions is: C
ZnCl2/HCl (ii) H3O⊕
Benzene HCN OH
CH3COOH + SOCl2 → A 
AlCl
→B 
Θ →C 3
OH (1) Ph–CH2–CH2–CH2–OH
O (2) Ph–CH–CH3
NC OH
CH3 CHO
(1) CH3 (2)
CN (3) Ph–CH–CH3

H CH3
COOH
C CH2–CH2CN (4) Ph–CH–CH3
(3) CH3 (4)
CH2OH

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34. Which of the following reaction represent 37. A compound X (C8H10O) upon treatment with
incorrect product? alkaline solution of iodine gives a yellow
O precipitate. The filtrate on acidification gives a
CH2 white solid Y(C7H6O2). Write the structures of X
+ Cl (i) Anhyd. AlCl3
(1) (ii) N2H4 and Y.
Θ
(iii) OH/∆ X Y
CH3 CHO
(1) CH2–CH2OH CH2COOH
(2) (i) Cl2/hν
(ii) H3O⊕
(2) CH–CH3 COOH
CH2-OH CHO
OH
(3) CrO3-H2SO4

(3) CH2CH3 COOH

O

COOH C–OH O OH
OH O–C-CH3
(4) (CH3CO)2, H⊕ (4) CH–CH3 CHO
OH

35. The major product of the following 38. C6H5CHO HCHO CH3COCH3
chemical reaction is: I II III
(1)H3O+ ,∆ In context with above compounds, consider
(2)SOCl2
CH3CH2CN 
( 3)Pd/BaSO4 ,H2
→ the following statements.
(1) CH3CH2CH3 a. Compound I reduces Tollens' reagent
and Fehling reagent both.
(2) CH3CH2CH2OH
b. Compound II reduces Tollens' reagent
(3) (CH3CH2CO)2O
but not Fehling reagent.
(4) CH3CH2CHO c. Compound III reduces neither Tollens'
reagent nor Fehling reagent.
36. Consider the following reaction sequence. Select the correct statement(s).
Conc. KOH
(1) a and b only (2) a and c only
CO,HCl
A B+C (3) band c only (4) c only
Anhyd. AlCl3 (Major) ∆
Major products B and C are 39. Which of the following product is formed
CH3 CHO when cyclopentanone undergoes aldol
condensation reaction on heating with
(1) dilute alkali?
and
O
CH2OH COOK (1)
(2) O O
and
(2)
CH2OH OK
O
(3)
and
(3)
CH2OH CHO

(4) (4)
and
OH OH

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
40. Alkene (X) (C5H10) on ozonolysis gives a 43. A compound 'A' having the molecular
mixture of two compounds (Y) and (Z). formula C5H12O, on oxidation gives a
Compound (Y) gives positive Fehling's test compound 'B' with molecular formula
and iodoform test. Compound (Z) does not C5H10O. Compound 'B' gave a 2,4-
give Fehling's test but give iodoform test. dinitrophenylhydrazine derivative but did
Compounds (X),(Y) and (Z) are not give haloform test or silver mirror test.
X Y Z The structure of compound 'A' is
(1) C6H5COCH3 CH3CHO CH3COCH3 (1) CH3–CH2–CH2–CH2–CH2–OH
(2) CH3—CH=C—CH3 CH3CHO CH3COCH3 (2) CH3—CH2—CH2—CH—CH3

CH3 OH
(3) CH3CH2CH= CH2 CH3CH2CHO HCHO (3) CH3—CH2—CH—CH2—CH3
(4) CH3–CH=CH–CH3 CH3CHO CH3CHO OH
(4) CH3—CH2—CH—CH2—OH
41. Find the product of the given reaction:
CH3
O
C H+ 44. An organic compound of molecular formula
CH3+CH3CH2NH2
C3H6O did not give a silver mirror with
CH2–CH2CH3 Tollens' reagent but give an oxime with
hydroxylamine. It may be
(1) (1) CH2=CH—CH2—OH (2) CH3COCH3
(3) CH3CH2CHO (4) CH2=CH—OCH3
CH3
C=NCH2CH3 45. Identify (X), (Y) and (Z) in the given
(2) reaction.
OH
Z
X+Y CH3—CH—CH2—CHO
CH2CH2CH2CH3 3-Hydroxybutanal
(3)
X Y Z
H3C (1) HCHO CH3CHO KOH
CH–NH–CH2CH3 (2) CH3CHO CH3CHO NaOH
(4) (3) CH3CH2OH HCHO H2SO4
(4) CH3CH2CHO HCHO Dry ether

42. In the following sequence of reaction, the 46. CHO NaOH

final product (Z) is X
CHO
Hg2+ CH3MgX [O]
CH≡CH H2SO4 X H2 O Y Z The product (X) will be
(1) CH2ONa (2) COOH
(1) ethanal
COONa COOH
(2) propan-2-ol
(3) CH2OH (4) CH2OH
(3) propanone
COONa CH2OH
(4) propan-1-ol

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47. An organic compound (X) with molecular 52. Which of the following reactions will yield
formula C9H10O gives positive 2,4-DNP and benzaldehyde as a product?
Tollens' tests. It undergoes Cannizzaro COOH (i) SOCl , quinoline
reaction and on vigorous oxidation it gives (A)
2

(ii) H2/Pd/BaSO4
1,4 -benzenedicarboxylic acid. Compound
(X) is CH2OH
(1) benzaldehyde
CrO3/H2SO4
(2) o-methylbenzaldehyde (B)
(3) p-ethylbenzaldehyde
(4) 2, 2-dimethylhexanal. O
C —OCH3
48. The final product (C) obtained in the (C) (i) NaBH4
reaction sequence is (ii) PCC
C6H6 NH2NH2 CH3
CH3CH2COOH PCl3 (A) AlCl3 (B) OH–, heat (C)
O (i) CrO3, (CH3CO)2O
(D) (ii) H3O+,∆
(1) C–OCH2CH3
(1) (B) and (C) (2) (C) and (D)
O
(3) (A) and (D) (4) (A) and (C)
(2) C–CH2CH3
53. The major product of the following reaction
(3) CH2CH2CH3
is
OH (1) AlH(i-Bu)2
R–C≡N (2) H2O
(4) CH–CH2CH3
(1) RCOOH (2) RCHO
(3) RCONH2 (4) RCH2NH2
49. The reagent which does not react with both
acetone and benzaldehyde is______. CHO (i) 50% NaOH
(1) sodium hydrogensulphite +HCHO (ii) H3O+
(2) phenyl hydrazine 54.
(3) Fehling's solution COOH
(1) CH3OH and
(4) Grignard reagent
50. Cannizzaro's reaction is not given by_____. CH2OH
CHO (2) HCOOH and

(1) (2) CHO CH2OH COOH

CH3 (3) and
(3) HCHO (4) CH3CHO
(4) CH3OH and HCO2H
51. Compounds (A) and (C) in the following
KMnO4 PCl5 C
reactions are 55. Cumene 
Boil
→ A → B →
(i) CH3MgBr
CH3CHO (ii) H2O
(A) H2SO4,∆ (B) Benzaldehyde
Find “C” :
Hydroboration oxidation
(C) H /Pd − BaSO4
(1) identical (1) 2
Quinoline
(2) positional isomers (2) CO /HCl / AlCl3
(3) functional isomers (3) CH3NH2 /HCl / AlCl3
(4) optical isomers.
(4) CrO3 / Ac2O
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56. Which of the following is correct order: 58. Which of the compound have most acidic αH:
O
O O O
(1) (2)
(p) CH3–CHO>Ph–CHO>CH3–C–CH3>Ph–C–Ph
O
Reactivity in nucleophilic addition reaction O O O
(NAR)
(3) (4)
O
O O O O O
O
(q) Ph–C–Cl>Ph–C–O–C–CH3>Ph–C–OR>Ph–C–NH2
59. How many compounds gives Cannizzaro
Reactivity in hydrolysis reaction and aldol condensation respectively:
CHO CHO CHO CHO O O
H–C–H ; CH3–C–H ;
(r) > > >
O O
Cl OCH3 CH3 NO2 C–H C–CH3
; ;
Reactivity in nucleophilic addition reaction
(NAR) (1) 2,2 (2) 1,3 (3) 3,1 (4) 2,3
COCl COCl COCl COCl 60. Which reaction shows decrease in number
of carbon atoms in product :
(s) > > > (1) Decarboxylation reaction
(2) HVZ reaction
CH3 Cl CN NO2 (3) Hoffmann's hypobromamide reaction
Reactivity in hydrolysis (4) 1 & 3 both
(t) HCHO > CH3CHO > ( CH3 )2 CO > Ph − CHO 61. H3O
A 
+
→B + EtOH ; B ∆
→ Acetone + CO2
Reactivity towards Grignard reagent A is :
COOH COOH COOH O O O
(1) (2)
(u) > > O O
O O O
O O O
(3) (4) O
Rate of decarboxylation O O
(1) p, q (2) r, s, t
62. Correct statement for nucleophilic addition of
(3) p, q, r, s (4) p, q, u sodium bisulphite on carbonyl compounds is:
(1) This addition is highly sensitive to
57. Which of the following statement is not crowding, steric crowding increases, then
true about carbonyl group : addition decreases, so aldehydes are more
reactive than ketones in this reaction
(1) Carbon-oxygen bond is polarised due to
(2) This reaction used for separation of
higher electronegativity of oxygen aldehyde and ketone from a mixture
(2) Carbonyl carbon is electrophilic while containing some other compounds
oxygen is nucleophilic center because addition product is a
crystalline salt
(3) Carbonyl group have more dipole
(3) This reaction can be reversed in
moment than ether presence of acid or base, so bisulphite
(4) Polarity of carbonyl group as given addition product can be converted into
O O⊕ corresponding carbonyl compound.
C CΘ (4) All of these
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63. Which of the following compound does not 68. dil. H2 SO4
CH3 − C ≡ CH  Ba(OH)2
→ X  → Y ; Y is
Hg2+ ∆
form idoform with I2 + NaOH or NaOI:
(1) CH3–CH2–CH=CH–C–CH3
(1) CH3–CHO (2) CH3–CH2–CHO
O
(3) CH3–CH2–OH (4) CH3–C–CH3 CH3
O (2) CH3–C=CH–C–CH3
O
64. Which of the following compound can form CH3
hydrazone and will give iodoform test but
(3) CH3–C=CH–CH2–CH=O
will not give Tollens' test?
(4) CH3–CH2–CH=CH–CH2–CH=O
O
C 69. Which of the following is correct –
(1) CH2–CH3 (1) CH3–COCl>CH3–COOCO–CH3>
CH3–COOC2H5>CH3–CONH2
(reactivity towards hydrolysis)
(2) H3C–C–CH2–CH3
(2) C6H5–CH=O>CH3–CH=O>CH3–CO–CH3
O (reactivity towards cyanohydrin formation)
(3) H3C–C–CH2–CHO OMe OH Cl NO2
(3) > > >
O
(4) CH3–CH2–OH (reactivity towards electrophilic substitution
reaction)
65. What will the product, when salicylic acid
(4) CH3–CH2–CH2–CH2–OH>
obtained by kolbe's rection is oxidized with
OH
Zn powder?
OH H3C–CH2–CH–CH3>(CH3)3C–OH
CHO CHO (reactivity towards Lucas reagent)
(1) (2) 70. Which of the following reaction is not
COOH OH correct according to major product
CHO CH2–OH
(1) conc. NaOH
(3) (4) CHO COONa
Alc. KOH
(2) H3C–CH2–CH–CH3
∆
66. Which product of given reaction gives F
instant turbidity with Lucas reagent :- CH3–CH=CH-CH3
1. CH3MgBr O
(1) HCHO →
2.H O⊕
Product (3) CH3
OH
3
HI + CH3–I
1.CH3Mgr
(2) CH3 − CHO 
2.H3O+
→Product NaOH
(4) CCl3–C–H CHCl3+HCOONa
O
O
1. CH3MgBr
(3) CH3–C–CH3 → Product
2. H O+ 3 71. In which reaction racemic mixture is
(4) 1. (CH3 )2 CH-Br
CH3 − CHO → Product obtained as a product?
2.H3O⊕
(1) Ph–CHO+HCN→
pyridine
67. Acetaldehyde and benzaldehyde can not be (2) H3C–CH–CH2–OH+SOCl2
distinguished by - C2H5
(1) Tollens' reagent Θ⊕ ROH
(3) H3C–CH–CH2–CH3+C3H7ONa
(2) Fehling solution ∆

(3) Benedict solution Cl

(4) All of these
(4) Iodoform test
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72. Which of the following do not give aldol 77. Alkene (X) (C5H10) on ozonolysis gives a
reaction with dil. NaOH ? mixture of two compounds (Y) and (Z).
(1) CH3CHO (2) CD3CHO Compound (Y) and (Z) gives positive
(3) CH3COCH3 (4) PhCHO Fehling's test, (Z) does not give idoform
test. Compounds (X), (Y) and (Z) are:-
X Y Z
73. How many aldol product will be obtained in
(1) CH2=C–CH2CH3 HCHO CH3–C–CH2CH3
the following reaction (Structural only)
CH3 O
NaOH
CH3CHO + PhCHO  →
(2) CH3–CH=C–CH3 CH3CHO CH3COCH3
(1) 2 (2) 3
CH3
(3) 4 (4) 1 (3) CH3CH2CH=CH2 CH3CH2CHO HCHO
(4) CH3CH= CHCH2CH3 CH3CHO CH3CH2CHO
1.DIBAL −H
74. CH3 − CH =CH − CH2 − CH2 − CN 
→? 2.H2O

78. Buta-1,3-diene was subjected to ozonolysis

(1) CH3 ( CH2 )4 CN
to prepare aldehydes. Which of the
(2) CH3 ( CH2 )5 NH2 following aldehydes will be obtained during
(3) CH3 − CH =CH − CH2 − CH2 − CHO the reaction?
CHO
(4) CH3 ( CH2 )3 − CH2 − CHO (1) +2HCHO
CHO
(2) CH3CHO + 2HCHO
75. Which of the following compound when
(3) CH3CH2CHO + CH3CHO
reacts with CH3MgBr form 3° alcohol ?
(2) 2CH3CH2CHO
O O
(1) H–C–H (2) CH3–C–H
79. How many Aldol condensation products
O O would you expect for the given reaction ?
(3) CH3–C–CH3 (4) C–H (Exclude stereo isomers)
(i) NaOH
CH3CHO + CH3CH2CHO (ii) ∆
→
OH (1) 4 (2) 3
PCC (i)HCHO (3) 1 (4) 2
76. → ( A) 
(ii) dil. KOH/ ∆
→(B)

CH2–Br alc. KOH NaNH2

P Q
(B) is :
80. CH2–Br
O O OH CH3Cl Red hot iron
S Anhydrous AlCl3 R
(1) OH (2) CH3
tube (873K)

S is :-
O (1) Chlorobenzene
O
(2) p-chloro toluene
(3) (4)
(3) Toluene
(4) Benzochloride
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O O OH O
81. CH3–C–CH2–CH2–CH2–CH2–CH2–C–H 85. CHO OH
(i) dil. NaOH
OH
 → (A), product A is – (I) (II) (III) (IV)
(ii) ∆

O Which among the above compound/s

does/do not form Silver mirror when
C–CH3 H3C–C=O treated with Tollen's reagent
(1) (2) (1) (II), (III) and (IV) only
(2) Only (IV)
H3C–C=O O (3) Only (II)
OH OH
(3) (4) (4) (II) and (III) only

86. An aromatic compound 'X' with molecular

OH O OH
formula C9H10O gives the following
82. CH3–CH–CH2–CH2–C–CH3→CH3–CH–(CH2)3–CH3 chemical tests
Above conversion can be achieved by :- (i) forms 2,4-DNP derivative
(1) Wolff-Kishner reduction
(ii) Reduce Tollens reagent
(2) Clemmensen reduction
(iii) Undergoes Cannizzaro reaction and
(3) LiAlH4
(4) NaBH4 (iv) On vigorous oxidation -1,2-
benzenedicarboxylic acid is obtained
83. Compound (X) C4H8O, which reacts with 2, CH2–CHO CHO
4-DNP but gives negative haloform test is :- (1) CH3
(2)
O
C2H5
(1) CH3–C–CH2–CH3 O
(2) CH3–CH–CHO CHO
C C2H5
CH3 CH3
(3) (4)
OH CH3
(3)
87. A carbonyl compound 'A' reacts with
OH hydrogen cyanide to form a cyanohydrin 'B'
(4) CH3–CH2–CH–CH3 which on hydrolysis gives an optically
active alpha hydroxy acid 'C', 'A' gives a
84. Compound A (molecular formula C3H8O) is positive iodoform test, identify A, B, C
treated with anhydrous CrO3 to from a product.
product B (molecular formula C3H6O). B H OH H OH
form a shining silver mirror on warming (1) HCHO; C ; C
with ammoniacal silver nitrate. B when H CN H COOH
treated with an aqueous solution of CH3 OH CH3 OH
H2NCONHNH2, gives a product C. Identify (2) CH3CHO; C ; C
the structure of C :- H CN H COOH
(1) CH3CH2CH=NNHCONH2 C2H5 OH C2H5 OH
(2) CH3–C=NNHCONH2 (3) CH3CH2CHO; C ; C
H CN H COOH
CH3
CH3 CH3 OH CH3 OH
(3) CH3–C=NCONHNH2 (4) C=O ; C ; C
CH3 CH3 CH3 CN CH3 COOH
(4) CH3CH2CH=NCONHNH2
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 CHEMISTRY Aldehydes, ketones, and carboxylic acids


88. CH3 − CH2 − CHO  OH

→ A  ∆
→B 90. Which one of the following reactions does
Product 'A' & 'B' will be respectively :- not represent correct combination of
(1) CH3–CH2–CH–CH2–CHO substrate and product under the given
conditions ?
OH O O
and CH3–CH2–CH=CH–CH2–CH2–CHO (1) Cl H2 H
(2) CH3–CH–CH2–CH2–CH2–CHO Pd. BaSO4

OH O
CN
and CH3–CH=CH–CH2–CH2–CHO (2) (i) DIBAL-H H
(ii) H2O
OH O
CO2C2H5
(3) CH3–C–CH2–CH2–CHO (3) (i) AlH(iso BU)2 H
(ii) H2O
CH3
O
and CH3–CH–CH=CH–CHO
(4) OH (i) Na2Cr2O7 H
CH3 (i) H2SO4,H2O
(4) CH3–CH2–CH–CH–CHO
OH CH3
and CH3–CH2–CH=C–CHO
CH3
Θ
OH
89. CHO+H–CHO ∆
Product (s)
(Major) will be.
CH2OH Θ
(A) (B) COO

(C) HCOOΘ (B) CH3–OH

(1) A, B and C are correct
(2) A & B are correct
(3) C & D are correct
(4) A & C are correct

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Aldehydes, ketones, and carboxylic acids CHEMISTRY

Exercise 3
1. Assertion: Acetaldehyde on treatment with 4. Assertion: α-Hydrogen atoms in aldehydes
alkaline gives aldol. and ketones are acidic.
Reason: Acetaldehyde molecules contains α Reason: The anion left after the removal of
hydrogen atom. α-hydrogen is stabilized by resonance.
(1) If both assertion and reason are true and (1) If both assertion and reason are true and
the reason is the correct explanation of the reason is the correct explanation of
the assertion. the assertion.
(2) If both assertion and reason are true but (2) If both assertion and reason are true but
reason is not the correct explanation of reason is not the correct explanation of
the assertion. the assertion.
(3) If assertion is true but reason is false. (3) If assertion is true but reason is false.
(4) If the assertion and reason both are (4) If the assertion and reason both are
false. false.

2. Assertion: Acetophenone and benzophenone 5. Assertion: 2, 2–Dimethyl propanal

can be distinguished by iodoform test. undergoes Cannizzaro reaction with
Reason: Acetophenone and benzophenone concentrated NaOH.
both are carbonyl compounds. Reason: Cannizzaro is a disproportionation
(1) If both assertion and reason are true and reaction.
the reason is the correct explanation of (1) If both assertion and reason are true and
the assertion. the reason is the correct explanation of
(2) If both assertion and reason are true but the assertion.
reason is not the correct explanation of (2) If both assertion and reason are true but
the assertion. reason is not the correct explanation of
(3) If assertion is true but reason is false.
the assertion.
(4) If the assertion and reason both are
false. (3) If assertion is true but reason is false.
(4) If the assertion and reason both are
3. Assertion: Benzaldehyde is more reactive false.
than ethanal towards nucleophilic attack.
Reason: The overall effect of –I and +R effect 6. Assertion: Carboxylic acid exist as dimer.
of phenyl group decreases the electron Reason: Carboxylic acid shows hydrogen
density on the carbon atom of >C=O group in bonding.
benzaldehyde. (1) If both assertion and reason are true and
(1) If both assertion and reason are true and the reason is the correct explanation of
the reason is the correct explanation of the assertion.
the assertion. (2) If both assertion and reason are true but
(2) If both assertion and reason are true but reason is not the correct explanation of
reason is not the correct explanation of the assertion.
the assertion. (3) If assertion is true but reason is false.
(3) If assertion is true but reason is false. (4) If the assertion and reason both are
(4) If the assertion and reason both are false.
false.

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
7. Assertion: Carboxylic acids have higher 10. Assertion: Aromatic aldehydes and formaldehyde
boiling points than alkanes. undergo Cannizzaro reaction.
Reason: Carboxylic acids have hydrogen Reason: Aromatic aldehydes and
bonding. formaldehyde have zero α-H.
(1) If both assertion and reason are true and (1) If both Assertion and Reason are correct
the reason is the correct explanation of and the Reason is a correct explanation
the assertion. of the Assertion.
(2) If both assertion and reason are true but (2) If both Assertion and Reason are correct
reason is not the correct explanation of but Reason is not a correct explanation
the assertion. of the Assertion.
(3) If assertion is true but reason is false. (3) If the Assertion is correct but Reason is
(4) If the assertion and reason both are incorrect.
false. (4) If the Assertion is incorrect and Reason
is correct.
8. Assertion: Acetals and ketals are unstable in
aqueous acidic medium. 11. Assertion: Acetaldehyde is more reactive
Reason: Acetals and ketals are hydrolyzed in than acetone in nucleophilic addition
acidic medium to yield corresponding reactions.
aldehydes and ketones. Reason: Two alkyl groups in acetone reduce
(1) Both Assertion and Reason are correct the electrophilicity of the carbonyl carbon.
and Reason is the correct explanation for (1) If both Assertion and Reason are correct
the Assertion and the Reason is a correct explanation
(2) Both Assertion and Reason are correct of the Assertion.
but Reason is not correct explanation for (2) If both Assertion and Reason are correct
Assertion but Reason is not a correct explanation
(3) Assertion is correct but Reason is of the Assertion.
incorrect (3) If the Assertion is correct but Reason is
(4) Assertion and Reason both are incorrect. incorrect.
(4) If the Assertion is incorrect and Reason
9. Assertion: Isopropyl alcohol shows iodoform is correct.
test
12. Assertion: Benzaldehyde and methanal
Reason: Sodium hypoiodite oxidises
undergo Cannizzaro reaction.
isopropyl alcohol into acetone
Reason: Those aldehydes which have α-H
(1) If both Assertion and Reason are correct
atom undergo Cannizzaro reaction.
and the Reason is a correct explanation (1) If both Assertion and Reason are correct
of the Assertion. and the Reason is a correct explanation
(2) If both Assertion and Reason are correct of the Assertion.
but Reason is not a correct explanation (2) If both Assertion and Reason are correct
of the Assertion. but Reason is not a correct explanation
of the Assertion.
(3) If the Assertion is correct but Reason is
(3) If the Assertion is correct but Reason is
incorrect.
incorrect.
(4) If the Assertion is incorrect and Reason
(4) If the Assertion is incorrect and Reason
is correct.
is correct.
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13. Given below are two statements 16. Given below are two statements:
Statement-I: Acidity of α-Hydrogen of Statement-I: In Aldol reaction, name aldol is
aldehydes and ketones is responsible for
Aldol reaction derived from the names of the two functional
Statement-II: Reaction between groups, aldehyde & alcohol, present in the
benzaldehyde and ethanal will not give
products.
cross-Aldol product
Choose the most appropriate answer from Statement-II: The aldol & ketol do not
the options given below: readily lose water to give aldol condensation
(1) Both statement I and statement II are
true product.
(2) Statement I is false but Statement II true In the light of the above statements, choose
(3) Statement I is true but Statement II is the most appropriate answer from the
false
(4) Both Statement I and Statement II are options given below:
false (1) Both statement I and II are correct.
(2) Both statement I and II are incorrect.
14. Given below are two statements:
Statement I: The nucleophilic addition of (3) Statement I is correct but statement II is
sodium hydrogen sulphite to an aldehyde or incorrect
a ketone involves proton transfer to form a
stable ion. (4) Statement I is incorrect but statement II
Statement II: The nucleophilic addition of is correct.
hydrogen cyanide to an aldehyde or a ketone
yields cyanohydrin as final product.
(1) Both statement I and statement II are 17. Given below are two statements:
true. Statement-I: The lower members of
(2) Statement I is true but statement II is
aldehydes and ketones such as HCHO,
false.
(3) Statement I is false but statement II is CH3CHO and CH3COCH3 are miscible with
true. water in all proportions because they form
(4) Both statement I and statement II are
false. H-bond with water.
Statement-II : All aldehydes and ketones are
15. Statement I: Addition of ammonia and its
fairly soluble in organic solvents like
derivatives to carbonyl group of aldehyde is a
nucleophilic addition-elimination reaction. benzene, ether, methanol, chloroform etc.
Statement II: Addition of hydrazine to In the light of the above statements, choose
acetaldehyde is catalysed by base.
the most appropriate answer from the
In the light of the above statements, choose
the correct answer from the options given options given below:
below. (1) Both statement I and II are incorrect.
(1) Statement I is incorrect but statement II
is correct. (2) Statement I is correct but statement II is
(2) Both statement I and statement II are incorrect
correct.
(3) Both statement I and II are correct.
(3) Both statement I and statement II are
incorrect. (4) Statement I is incorrect but statement II
(4) Statement I is correct but statement II is is correct.
incorrect.
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18. Given below are two statements: 21. Match list-I with list-II
Statement-I: Nitriles are reduced to
List-I List-II
corresponding imine with stanous chloride
in the presence of HCl , which on hydrolysis (A) OH (i) NaOH (i) Etard
give corresponding aldehyde. (ii) CO2
(iii) H2O/H+ reaction
Statement-II: Nitriles are selectively
reduced by DIBAL-H to imines followed by (B) O (ii) Gatterman-
hydrolysis to aldehydes.
Cl H2 Koch
In the light of the above statements, choose Pd-BaSO4
the most appropriate answer from the reaction
options given below:
(C) CH3 (iii) Rosenmund
(1) Statement I is correct but statement II is (i) CrO2Cl2,CS2
incorrect (ii) H3O⊕ reduction
(2) Both statement I and II are incorrect.
(3) Both statement I and II are correct. (D) CO,HCl (iv) Kolbe’s
anhyd. AlCl3/CuCl
(4) Statement I is incorrect but statement II reaction
is correct.
Choose the correct answer from the options
19. Given below are two statements:
Statement-I: Electronically, Aldehydes are give below:
less reactive than ketones towards (1) (A-iv), (B-iii), (C-i), (D-ii)
Nucleophilic Addition Reaction.
Statement-II: Two Alkyl groups increase the (2) (A-iii), (B-iv), (C-ii), (D-i)
electrophilicity of the carbonyl more (3) (A-i), (B-iii), (C-ii), (D-iv)
effectively.
In the light of the above statements, choose (4) (A-iii), (B-iv), (C-i), (D-ii)
the most appropriate answer from the
options given below: 22. Match the following
(1) Statement I is incorrect but statement II
is correct. Column-I Column-II
(2) Both statement I and II are incorrect. (p) CH3 − CH NaOH
= O  → (a) Cannizzaro’s
aq.
(3) Statement I is correct but statement II is
reaction
incorrect.
(4) Both statement I and II are correct. (q) Conc.H2 SO4
CH3 − CH2 − OH → (b) Aldol
413K
20. Given below are two statements condensation
Statement-I: Ketones react with ethylene
glycol to form cyclic products known as (r) C6H5 − CH 50%
= O  → (c) Fittig
KOH,∆
ethylene glycol ketals in presence of dry HCl . reaction
Statement-II: Dry hydrogen chloride
protonates the oxygen of the carbonyl (s) Na,∆ (d) Willianson’s
Cl 
dryether
→
compounds and therefore, increases the Continuous
electrophilicity of the carbonyl carbon
ether
facilitating the nucleophilic attack of glycol.
In the light of the above statements, choose the Synthesis
most appropriate answer from the options given
(1) (p–b), (q–d), (r–a), (s–c)
below:
(1) Both statement I and II are correct. (2) (p–d), (q–c), (r–b), (s–a)
(2) Both statement I and II are incorrect. (3) (p–a), (q–d), (r–b), (s–c)
(3) Statement I is correct but statement II is
incorrect (4) (p–c), (q–d), (r–b), (s–a)
(4) Statement I is incorrect but statement II is

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23. Match the reactions given in list-l with their 25. Match the column I with column II and mark
characteristic reactions/reagents given in the appropriate choice.
list-II and select the correct option. Column I Column II
List-I List- II (A) Clemmensen (i) Conc. KOH

a. Zn–Hg (i) Reduction reduction

C=O HCl CH2
(B) Rosenmund (ii) Zn/Hg+conc. HCl
by DIBAl-H
reduction
b. C=O
(i) NH2NH2
CH2
(ii) Reduction
(ii) KOH, ∆ (C) Iodoform (iii) H2/Pd-BaSO4
by LiAIH4
reaction
c. –CN→–CHO (iii) Clemmensen
(D) Cannizzaro (iv) NaOH+I2
reduction reaction
d. –COOH→–CH2OH (iv) Wolff- (1) (A)→ (i), (B) → (iii), (C) → (ii), (D) → (iv)
kishner (2) (A) → (iii), (B) → (iv), (C) → (i), (D) → (ii)
reduction (3) (A) → (ii), (B) → (iii), (C) → (iv), (D) → (i)
(1) a(iv), b(iii), c(ii), d(i) (4) (A) → (iv), (B) → (i), (C) → (ii), (D) → (iii)
(2) a(ii), b(i), c(iv), d(iii)
(3) a(iii), b(iv), c(ii), d(i) 26. Match List-I with List-II.
(4) a(iii), b(iv), c(i), d(ii) List-I
(A) COCl CHO

24. Match the reagents in column I with H2

Pd-BaSO4

products formed by reactions with acetone SnCl2/HCl

(B) CH3–CN H3O+
CH3–CHO
in column II and mark the appropriate
(C) CH3 CHO
choice.
CrO2Cl2
Column I H3O+

(A) Hydrazine (D) CHO

(B) Semicarbazide CO,HCl

AlCl3(anhyd)

(C) Phenylhydrazine
List-II
(D) Hydroxylamine
(I) Gatterman-Koch reaction
Column II
(II) Etard reaction
(i) (CH3)2C=NNHCONH2
(III) Stephen reaction
(ii) (CH3)2C=NOH
(IV) Rosenmund reaction
(iii) (CH3)2C=NNH2
Choose the correct answer from the options
(iv) (CH3)2C=NNHC6H5
given below:
(1) (A) (i), (B) (ii), (C) (iii), (D) (iv) (1) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(2) (A) (iv), (B) (iii), (C) (ii), (D) (i) (2) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(3) (A) (iii), (B) (i), (C) (iv), (D) (ii) (3) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(4) (A) (ii), (B) (iv), (C) (i), (D) (iii) (4) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
27. Match the following column. 29. Match the column
Column-I Column-II Column-I Column-II
(Pair of (Differentiate (Compound) (Boiling point)
compound) method) (a) Propanal (p) 370
(A) Acetone and (P) Br2+H2O (b) Acetone (q) 322
acetaldehyde (Bromine water) (c) 1-Propanol (r) 273
(B) Ethanol and (Q) Dil. alkaline KMnO4 (d) n-Butane (s) 329
methanol (Bayer’s reagent) (1) a–q, b–s, c–p, d–r
(C) Phenol and (R) NH4Cl+AgNO3 (2) a–q, b–p, c–s, d–r
Cyclohexanol (Tollen’s reagent) (3) a–p, b–q, c–r, d–s
(D) n-Pentane (S) NaOI (4) a–p, b–q, c–s, d–r
and 2- (Iodoform test)
pentene 30. Match the column:
(1) A→S, B→R, C→P, D→Q Column I Column I
(2) A→R, B→S, C→P, D→Q H⊕ 1º A O
PhMgBr+(P)
(3) A→R, B→S, C→Q, D→P alcohol H–C–H
(4) A→R, B→P, C→S, D→Q ⊕
PhMgBr+(Q) H 2º B O
alcohol CH3–C–H
28. Match the column
⊕
Column-I Column-II PhMgBr+(R) H 3º C O
(Transformations) (Reagents) alcohol CH3–C–CH3
(a) R–COOH→R–CH2OH (p) NH3/∆ D EtOH
H⊕
(b) O (q) Cl2/Red P, H2O PhMgBr+(S)
Ph–COOH → Ph–C–NH2
P Q R S
(c) R–CH2COOH → (r) P2O5, ∆ (1) D A C B
R–CH–COOH (2) B D A C
Cl (3) B C A D
(d) CH3COOH→(CH3CO)2O (s) B2H6/H3O⊕ (4) A B C D
(1) a–p, b–s, c–r, d–q
(2) a–r, b–p, c–s, d–q
(3) a–s, b–q, c–r, d–p
(4) a–s, b–p, c–q, d–r

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Exercise 4 (Previous Year's Questions)

1. Reaction by which Benzaldehyde cannot be 5. The correct statement regarding a carbonyl
prepared :- [NEET UG-2013] compound with a hydrogen atom on its alpha
COOH
(1) + Zn/Hg and conc. HCl carbon, is : [NEET-I- 2016]
(1) a carbonyl compound with a hydrogen
CH3
(2) +CrO2Cl2 in CS2 followed by H3O⊕ atom on its alpha-carbon never
equilibrates with its corresponding enol.
COCl (2) a carbonyl compound with a hydrogen
(3) + H2 in presence of Pd + BaSO4
atom on its alpha-carbon rapidly
(4) +CO+HCl in presence of anhydrous AlCl3 equilibrates with its corresponding enol
and this process is known as aldehyde-
ketone equilibration.
2. An organic compound ‘X’ having molecular
formula C5H10O yields phenyl hydrazone and (3) a carbonyl compound with a hydrogen
gives negative response to the iodoform test atom on its alpha-carbon rapidly
and Tollen’s test. It produces n-pentane on equilibrates with its corresponding enol
reduction. ‘X’ could be :- [AIPMT-2015]
(1) 2-pentanone (2) 3-pentanone and this process is known as
(3) n-amyl alcohol (4) pentanal carbonylation.
(4) a carbonyl compound with a hydrogen
CHO
COOH atom on its alpha-carbon rapidly
(i)[Ag(NH3 )2 ]⊕ OHΘ
3.  (ii)H⊕ / ∆
→ product equilibrates with its corresponding enol
and this process is known as keto-enol
Product of above reaction is [AIIMS-2015]
O tautomerism.
C O
C O O
CH3 CH=O
(1) (2)
O Zn–Hg/HCl
6. P
Q
COOH COOH
COOH P & Q respectively are : [AIIMS-2016]
(3) (4)
CH2–OH

4. The correct structure of the product A (1) CrO3/H3O⊕,

formed in the reaction
O CH3
H2 (gas, 1 atmosphere) (2) CrO2Cl2/H3O⊕,

Pd/carbon, ethanol
→ A is:
[NEET-II- 2016] CH3
OH OH
⊕ (3) NaOCl/∆,
(1) (2)
OH CH2–OH
O
(4) KMnO4/H⊕,
(3) (4)

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
C–Cl CHO 11. Compound A, C8H10O, is found to react with NaOI
7. P
O (Produced by reacting Y with NaOH) and yields a
P will be :- [NEET-2017] yellow precipitate with characteristic smell. A
(1) Pd/BaSO4 + H2 (2) LiAlH4 and Y are respectively: [NEET(UG) 2018]
(3) NaBH4 (4) Pd + H2 (1) H3C CH3–OH and I2

8. Which of the following product is formed (2) CH2–CH2–OH and I2

when cyclohexanone undergoes aldol
(3) CH3–CH3 and I2
condensation followed by heating? :
[NEET-2017] OH
CH3
O
(4) CH3 OH and I2
(1) (2)

OH O 12. The reaction that does not give benzoic acid

(3) (4) as the major product is:
[NEET(UG) (Odisha) 2019]
OH O O CH2OH
K2Cr2O7/H⊕
(1)
9. Consider the reactions : [NEET-2017] COCH3
(i) NaOCl
(2) (ii) H3O+
Cu/573K [Ag(NH3)2]+
X A – silver mirror observed CH2OH
(C2H6O) OH,∆
PCC
–
OH,∆
(3)
Y
O CH2OH
KMnO4/H+
NH2–NH–C–NH2 (4)
Z
Identify A, X, Y and Z
13. Reaction between benzaldehyde and
(1) A-methoxymethane, X-Ethanoic acid,
acetophenone in presence of dilute NaOH is
Y-Acetate ion, Z-hydrazine known as: [NEET(UG) 2020]
(2) A-methoxymethane, X-Ethanol, (1) Cross Aldol condensation
Y-Ethanoic acid, Z-Semicarbzide. (2) Aldol condensation
(3) A-Ethanal, X-Ethanol, Y-But-2-enal, (3) Cannizzaro’s reaction
Z-Semicarbazone. (4) Cross Cannizzaro’s reaction
(4) A–Ethanol, X–Acetaldehyde, 14. Which of the following acid will form an (a)
Y–Butanone, Z–Hydrazone. Anhydride on heating and (b) Acid imide on
strong heating with ammonia ?
10. Carboxylic acid have higher boiling points
[NEET(UG) (Covid-19) 2020]
than aldehydes, ketones and even alcohols COOH
of comparable molecular mass. It is due to COOH
their [NEET(UG) 2018] (1) (2)
COOH COOH
(1) formation of intramolecular H-bonding COOH OH
(2) formation of carboxylate ion
(3) more extensive association of carboxylic (3) (4)
COOH
acid via van der Waals force of attraction COOH
(4) formation of intermolecular H-bonding
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Aldehydes, ketones, and carboxylic acids CHEMISTRY

15. Identify compound (A) in the following 18. The intermediate compound ‘X’ in the
reaction: [NEET(UG) (Covid-19) 2020] following chemical reaction is:
CHO
[NEET(UG) 2021]
H2/Pd/BaSO4
A
O
(1) Benzoyl chloride (2) Toluene CH3 C
(3) Acetophenone (4) Benzoic acid CS2 H3O+ H
+CrO2Cl2 X
16. The product formed in the following
chemical reaction is : [NEET(UG) 2021] CH(OCrOHCl2)2
O (1)
O
CH2–C–OCH3
CH(OCOCH3)2
NaBH4
CH3 C2H5OH ? (2)
OH H
Cl
CH2–C–OCH3 CH
(1) OH (3) Cl
CH3
O
Cl
CH2–CH2–OH
CH
(2) (4) H
CH3
OH H
CH2–C–CH3
(3) 19. Match List-I with List-II.
OH
CH3 [NEET(UG) 2022]
OH O
List-I List - II
CH2–C–OCH3
A Cyanohydrin i NH2OH
(4)
CH3 B Acetal ii RNH2
17. Match List - I with List - II. [NEET(UG) 2021] C Schiff’s base iii alcohol
List-I List - II d Oxime iv HCN
a CO, HCl i Hell-Volhard
Anhyd, Choose the correct answer from the options
Zelinsky
AlCl3/CuCl
reaction given below:
b O ii Gattermann- (1) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
R–C–CH3 + NaOX → Koch
(2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
reaction
c R–CH2–OH + RCOOH iii Haloform (3) (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
Conc.H2 SO4
→ reaction (4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
d R–CH2–COOH iv Esterification
(i)X 2 /Red P
→ dry H3O+
(ii) H O
2
20. RMgX + CO2 Y  → RCOOH
Choose the correct answer from the options ether
given below. What is Y in the above reaction?
(1) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
[NEET(UG) 2022]
(2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
– +
(3) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) (1) RCOO Mg X (2) R3CO–Mg+X
(4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (3) RCOO–X+ (4) (RCOO)2Mg
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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
21. Which one of the following is not formed OH
when acetone reacts with 2-pentanone in (3) A= ;
the presence of dilute NaOH followed by
heating? [NEET(UG) 2022] H
O CH3 N
B= CH3
(1) H3C CH3

OH
CH3 (4) A= ;

(2) CH3 CH3

O CH3
N
CH3 B= H
CH3 O
(3) C H3 CH3 24. Complete the following reaction :

HCN OH
O
CH3 CN
[A] [B]
(4) CH3 CH3
conc. H2 SO4
→ [C]
Δ
O
CH3 [C] is. [NEET (UG) 2023]
22. Compound X on reaction with O3 followed by
Zn/H2O gives formaldehyde and 2-methyl (1) OH (2) COOH
propanal as products. The compound X is:
[NEET(UG) 2022]
(3) CHO (4) COOH
(1) 3-Methylbut-1-ene
(2) 2-Methylbut-1-ene
(3) 2-Methylbut-2-ene 25. Identify product (A) in the following
(4) Pent-2-ene reaction. [NEET (UG) 2023]
23. The products A and B in the following reaction O
Zn-Hg
sequence are: [Re-NEET (UG) 2022] (A) +2H2O
conc.HCl

(i) HBr (i) SOCl

O
Ph (ii) Mg, dry ether
A (ii) CH3–NH
2

2
B
(iii) CO2, H3O+
O
(1)
(1) A= OH ;
OH
O (2)
CH3 OH
B= N
H OH
OH (3) CH2
CH2OH
(2) A= ;
O
H (4) CH3
N
B= CH3 CH3
O

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26. Identify the final product [D] obtained in the 30. Identify the product in the following reaction
following sequence of reactions. Cl (i) KCN
i) LiAlH4 H2 SO4 HBr (ii) H2O/HCl,∆
CH3CHO 
ii) H3O+
→ [A]  ∆
→ [B] →[C] Product
(iii) Br2/red phosphorus
(iv) H2O
Br
[NEET (UG)-Manipur-2023]
[D] COBr COOH
Na/dry ether
[NEET (UG) 2023] (1) (2) Br

(1) (2) Cl Br
(3) C4H10 (4) HC ≡ C Na
 +
(3) Br (4)

27. Identify the major product obtained in the

following reaction: 31. The major product formed in the following
O
conversion is_________.
+2[Ag(NH3)2] + 3OH¯+ ∆
[NEET (UG)-Manipur-2023]
H
O O O
(i)NaBH4
Major product [NEET (UG) 2023] CH2–C–CH3 
(ii)H2 SO4 ,∆
→ Major
product
OH O

(1) (2)
OH OH (1)
O OH
(2)
(3) (4)
COO¯ COO¯
O
(i)"X"
28. R–COOH  →R–CH 2OH
(ii)H O/HCl
2 (3)
(i)"X"
R–CH=CH2 →
(ii)H2O,NaOH, H2O2
R–CH2 –CH2 –OH
Identify ‘X’ in above reactions (4)
[NEET (UG)-Manipur-2023]
(1) B2H6 (2) LiAIH4
(3) NaBH4 (4) H2/Pd
32. Select the incorrect reaction among the
29. The following conversion is known as: following : [RE-NEET-2024]
O H2O
(1) CH3COCl ∆
→ CH3COOH
C H2 CHO
Cl CONH2
Pd–BaSO4 COOH
(2) (i) LiAIH4
[NEET (UG)-Manipur-2023] (ii) H2O
(1) Stephen reaction
–
(i) KMnO4 /OH
(2) Gattermann-Koch reaction (3) CH3CH2OH →
(ii) H O⊕
CH3COOH
3

(3) Etard reaction CrO −H SO

(4) CH3CH2CH2OH  → CH3CH2COOH
3 2 4

(4) Rosenmund reaction

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 CHEMISTRY Aldehydes, ketones, and carboxylic acids
33. Fehling's solution 'A' is [NEET-2024] 35. The major product of the following reaction
(1) alkaline solution of sodium potassium is: [NEET - 2025]
tartrate (Rochelle's salt) O (i)CH3MgBr
(2) aqueous sodium's citrate (excess)
CN
(ii) H3O+
(3) aqueous copper sulphate
(4) alkaline copper sulphate H3C OH
OH
CH3
(1)
34. Identify the suitable reagent for the CH3
following conversion. [NEET - 2025] O
O CH3
(2)
CHO O
OCH3
H3C OH
+
(1) (i) NaBH4 (ii) H /H2O (3) CN
(2) H2/Pd-BaSO4
(3) (i) LiAlH4 (ii) H+/H2O H3C OH
CH3
(4) (i) AlH(iBu)2 (ii) H2O (4)
O

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ANSWER KEYS
Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 3 3 2 3 4 3 3 2 4 4 1 4 4 2

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 4 2 2 4 3 1 1 3 3 3 4 1 2 3 4 4 4 1 3
Que . 21 22 23 24 25 26 27 28 29 30
Ans. 4 4 1 4 3 3 4 3 2 1

Exercise 1.3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 1 1 3 2 3 1 4 1 1 3 1 2 2 1 2

Exercise 1.4
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 3 1 1 2 3 3 2 2 3 4 4 4 1 3

Exercise 2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 2 3 1 3 4 3 1 1 1 2 3 1 2 4 2 2 2 3 4
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 2 4 4 1 4 4 2 1 1 3 4 4 3 4 2 2 4 3 2
Que . 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 3 3 2 2 3 3 3 3 4 2 3 2 2 1 4 4 3 1 4
Que . 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. 2 4 2 2 3 3 1 2 1 2 1 4 1 3 3 4 4 1 1 3
Que . 81 82 83 84 85 86 87 88 89 90
Ans. 2 1 2 1 3 4 2 4 4 4

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 2 4 1 2 1 1 1 1 1 1 3 3 1 4 3 3 3 2 1
Que . 21 22 23 24 25 26 27 28 29 30
Ans. 1 1 4 3 3 1 2 4 1 4

Exercise 4 (Previous Year's Questions)

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 2 1 4 4 2 1 2 3 4 3 3 1 1 1 4 4 1 4 1
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Ans. 2 1 2 4 1 1 3 1 4 2 1 2 3 4 4

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 Notes

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Amines CHEMISTRY

Chapter
AMINES
5
Chapter Summary
Introduction
Definition:
• Introduction Amines can be considered as derivatives of ammonia, obtained
by replacement of one, two or all the three hydrogen atoms by
• Preparation of amines alkyl and/or aryl groups.
• Physical properties For example:
CH3
• Chemical Reactions CH3 –NH2, C6H5–NH2, CH3–NH–CH3, CH3–N
CH3
• Diazonium Salts Structures of Amines-
Like ammonia, nitrogen atom of amines is trivalent and carries
an unshared pair of electrons. Nitrogen orbitals in amines are
therefore, sp3 hybridised and the geometry of amines is
pyramidal. Each of the three sp3 hybridised orbitals of nitrogen
overlap with orbitals of hydrogen or carbon depending upon the
composition of the amines. The fourth orbital of nitrogen in all
amines contains an unshared pair of electrons. Due to the
presence of unshared pair of electrons, the angle C-N-E, (where
E is C or H) is less than 109.5°; for instance, it is 108° in case of
trimethylamine as shown in Fig.
unshared
electron
pair

CH3
CH3

108°
CH3

Fig. Pyramidal shape of trimethylamine

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Classification
Amines are classified as primary (1°), secondary (2°) and tertiary (3°) depending upon the number of 'H'
atoms replaced by alkyl or aryl groups in ammonia (NH3) molecule.
R R
Ex. NH3 R–NH2 N–H N–R''
R' R'
Primary (1°) Secondary (2°) Tertiary (3°)

Preparation of amines
Amines are prepared by following methods
(1) Reduction of nitro compounds:
Nitro compounds are reduced to amines by passing H2 gas in the presence of finely divided Ni, Pd or Pt
and also by reduction with metal in acidic medium
Ex. NO2 NH2
H2/Pd
ethanol
or or
R–NO2 R–NH2
NO2 NH2
Ex. Sn+HCl
or Fe+HCl
or or
R–NO2 R–NH2

Note. Reduction with Fe scrap and HCl is Preferred because FeCl2 formed gets hydrolysed to release
hydrochloric acid during the reaction thus only a small amount of HCl is required to initiate the reaction
(2) Ammonolysis of alkyl halides:
The process of cleavage of the C–X bond by NH3 molecule is known as ammonolysis. The reaction is
carried out in a sealed tube at 373 K. The 1° amine thus obtained behaves as a nucleophile and can
further react with alkyl halide to form 2° and 3° amines and finally 4° ammonium salt
•• SN2 ⊕ Θ
NH3 + R–X R–NH3X
(Nucleophile) (Substituted
ammonium salt)
The nucleophilic R–NH2 reacts with R–X to give 2°, 3° amines and some 4° salt resulting in a complex
mixture.
R–NH2 R–X R2NH R–X R–X ⊕ Θ
R3N R4NX
(1°) (2°) (3°) (4° ammonium salt)

Note. (i) Ammonolysis has the disadvantage of yielding a mixture of primary, secondary, tertiary amines and
also a quaternary ammonium salt
(ii) Primary amine is obtained as a major product by taking large excess of ammonia

3. Reduction of nitriles:
Nitriles on reduction with liAlH4 or catalytic hydrogenation produce 1° amines. This reaction is used for
ascent of amine series
H2/Ni
R–C≡N R–CH2–NH2
or
Na(Hg)/C2H5OH
or
(i) LiAlH4 (ii) H2O

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4. Reduction of amides:
The amides on reduction with LiAlH4 yield amine
O
(i) LiAlH4
R–C–NH2 (ii) H O R–CH2–NH2
2

5. Gabriel phthalimide synthesis:

Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating
with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amines
O O O
Θ⊕ R–X
N–H KOH NK N–R
(SN2)
O O O
Phthalimide N–alkyl phthalimide
O O
Θ⊕

N–R
NaOH(Aq.) ONa
Θ⊕ + R–NH2
ONa
O O (1° amine)
Note. (i) Gabriel synthesis is used for the preparation of primary amines
(ii) Aromatic primary amines can not be prepared by this method because aryl halide do not undergo
nucleophilic substitution with the anion formed by phthalimide

6. Hoffmann bromamide degradation reaction:

Amides on reaction with Bromine in an aqueous or ethanolic solution of sodium hydroxide give primary
amine with one 'C' atom less than that present in the amide
O
R–C–NH2 + Br2 + 4NaOH → R–NH2 + Na2CO3 + 2NaBr + 2H2O

Mechanism:
O O O
⊕ Θ
NaOH Θ ⊕
R–C–NH2 R–C–NHNa Br–Br C Br
(–H2O) (–NaBr)
R NH

(–H2O) NaOH

O
Θ
R–NH2 2 NaoH(Aq.)
R–N=C=0 (–NaBr) R–C–N–Br
••
+
Na2CO3 (Alkyl
isocyanate
Intermediate
Note. In this reaction migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the
nitrogen atom.

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Exercise 1.1
1. C2H5NH2 cannot be prepared by the 8. The best reagent for converting,
reduction of: 2-phenylpropanamide into
(1) C2H5NO2 (2) CH3CH = NOH 1- phenylethanamine is ____.
(3) C2H5NC (4) CH3CN
(1) excess H2/Pt
(2) NaOH/Br2
CO
NH →
KOH C2H5Br H2O (3) NaBH4/methanol
2. A  →B  →
CO (4) LiAlH4/ether
C + D, C and D in the sequence are:
(1) Benzoic acid + aniline 9. Which of the following will not give a
(2) phthalic acid + ethylamine
primary amine?
(3) Phthalic acid + aniline
Br2 ,KOH
(4) Benzoic acid + ethylamine (1) CH3CONH2 →
LiAlH4
(2) CH3CN  →
3. Which of the following is obtained by LiAlH4
(3) CH3NC  →
reducing methyl cyanide with Na + C2H5OH:
LiAlH4
(1) Methyl alcohol (2) Acetic acid (4) CH3CONH2  →
(3) Ethyl amine (4) Methane

10. Amine produced by Hoffmann degradation

4. Which of the following would undergo
of benzamide is
Hoffmann bromide reaction to form primary
amine: (1) Benzyl amine
(1) RCONHCH3 (2) RCOONH4 (2) Aniline
(3) RCONH2 (4) RCONHOH (3) Aromatic primary amine
(4) 2 & 3 Both
5. Gabriel reaction for the synthesis of
amines, involves the use of
11. The product (A) formed in the following
(1) 1° amide (2) 2° amide
(3) Imides (4) Aliphatic amide reaction sequence is
2+
(i)Hg ,H2 SO4
CH3 − C ≡ CH 
(ii)HCN
→ ( A )Pr oduct
6. In the Hoffmann bromamide degradation (iii)H2 /Ni

reaction, the number of moles of NaOH and NH2

Br2 used per mole of amine produced are -
(1) CH3–C–CH2–OH
(1) Four moles of NaOH and two moles of Br2
CH3
(2) Two moles of NaOH and two moles of Br2
(3) Four moles of NaOH and one mole of Br2 OH
(4) One mole of NaOH and one mole of Br2 (2) CH3–C–CH2–NH2
CH3
7. Which of the following is a 3° amine?
(3) CH3–CH2–CH– CH2–OH
(1) 1-methylcyclohexanamine
NH2
(2) Triethylamine
OH
(3) Tert-butylamine
(4) CH3–CH2–CH– CH2–NH2
(4) N-methylaniline
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Amines CHEMISTRY
12. Which of following amines cannot 14. The major products A and B from the
synthesized by Gabriel Phthalimide following reactions are:
synthesis method H
CH3 N Br2/AcOH
(A) CH–CH2–NH2 B LiAIH4 A
CH3
O
(B) NH2 H H
Br
N N
(C) CH2 –NH2 (1) A= B=
O OH
Br ,
(D)CH3–CH2–NH–CH3 Br
H H
H3C
N N
(E) H3C C–NH2
(2) A= B=
H3C O OH
,
Br
(1) only B (2) Only B and D H
Br
(3) Only A and C (4) Only B, D and E N NH2
(3) A= B=
O
,
13. Which of the following reaction is an Br Br
H H
example of ammonolysis?
N N
(1) C6H5COCl + C6H5NH2 → C6H5CONHC6H5 B=
(4) A= O
(I) LiAlH
(2) C6H5CH2CN  → C6H5CH2 − CH2NH2
 4 ,
(II)H O 2
Br

HCl
(3) C6H5NH2  → C6H5 NH3 Cl
⊕ Θ 15. Reduction of CH3CH2NC with hydrogen in
presence of Ni or Pt as catalyst gives
(4) C6H5CH2Cl NH
→3
C6H5CH2 − NH2 (1) CH3CH2NH2 (2) CH3CH2NHCH3
(3) CH3CH2NHCH2CH3 (4) (CH3)3N

Physical properties
(a) Physical state: The lower aliphatic amines are gasses with fishy odour. Primary amines with 3 or more
'c' atoms are liquid and higher ones are solid. Aniline and other arylamines are usually colourless but
get colored on storage due to atmospheric oxidation.

(b) Solubility: Lower aliphatic amines are soluble in H2O because they form H-bonding with H2O. But
solubility decreases with increase in molecular mass due to increase in size of the hydrophobic alkyl
part. Higher amines are insoluble in H2O.
Amines are soluble in organic solvents such as alcohol, benzene, and ether.
Butan-1-ol is more soluble in H2O than butan-1-amine due to greater EN of O atom than that of N atom
(EN of O atom in alcohol is 3.5, whereas EN of N atom in amine is 3.0). therefore alcohols are more
polar than amines and form stronger intermolecular H-bonds than amines.
1
Solubility ∝
Molecular weight (Hydrophobic part )
Solubility ∝ Branching (When MW & types of amines is same)

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 CHEMISTRY Amines
Ex. order of solubility:-
(i) CH3–CH2– CH2–NH2>CH3–CH2–NH–CH3 > CH3–N–CH3
(1° amine) (2° amine) CH3 (3° amine)
CH3
(ii) CH3–CH2–CH2–CH2–NH2<CH3–CH–CH2–NH2< CH3–C–NH2
CH3 CH3
(1° amine) (1° amine) (1° amine)
(c) Boiling Point: 1° amines form stronger intermolecular H-bonding between N atom of one and H atom
of another molecule, than 2° amines, since 1° amines have two H atoms for H-bonding; that is why
boiling points of 1° amines are greater than those of 2° amines of the same molar masses. Similarly, 2°
amines form stronger intermolecular H-bonding than 3° amines, since 2° amines have one H atom for
H-bonding, whereas 3° amines do not form H-bonding due to the absence of H atom available for H-
bond. Thus, the boiling points of 2° amines are higher than those of 3° amines of the same molar masses.
The order of boiling points of isomeric amines is: 1° > 2° > 3° amines.
The intermolecular H-bonding in 1° and 2° amines is shown in Fig.
R H R R R
••
R–N–H :N–H :N–R R–N–H
••
N–H N–H
H H H R R
••
H–N–R Intermolecular H-bonding in 2° amines
H
Intermolecular H-bonding in1° amines
Table: Comparison of boiling points of amines, alcohols, and alkanes of comparable molar masses.
S.NO. Compound Molar mass Boiling point (K)
1. n–C4H9NH2 (1º amine) 73 350.8
2. Et2NH (2º amine) 73 329.3
3. EtNMe2 (3º amine) 73 310.5
4. Me 72 300.8
Me (Alkane)
Me
5. n-C4H9OH (Alcohol) 74 390.3
(d) Basic Nature of Amines: Due to basic nature, they react with acid to form salts.
•• ⊕ Θ
R–NH2 + H – X R–NH3 X (Salt)
•• ⊕ Θ
Ph–NH2 + HCl Ph NH3 Cl (Anilinium chloride)
Amine salts on treatment with bases such as NaOH regenerate the parent amine.
⊕ Θ Θ •• Θ
RNH3X + OH→RNH2 + H2O + X
Amines have an unshared pair of electrons on N atom due to which they behave as Lewis base. Basic
characters of amines are expressed in terms of their Kb and pKb values.
⊕ Θ

R − NH2 + H2O  RNH3 + OH
RNH
⊕
 OHΘ
 RNH
⊕
 OHΘ

       
3 3
= , = eq [H2 O]
K b K=
K eq pKb = – log Kb
[RNH2 ][H2O] [RNH2 ]
The larger the value of Kb or smaller the value of pKb, the stronger is the base.
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i. Aliphatic amines are stronger bases than NH3 (pKb = 4.75) due to +I effect of the alkyl groups leading
to high e– density on the N atom. Their pKb values lie between 3 and 4.22.
ii. Aromatic amines are weaker bases than NH3 due to e withdrawing nature of the aryl group.

iii. Besides the inductive effect, there are other effects such as solvation effect, steric hindrance, etc.,
which affect the basic strength of amines.
iv. Comparison of basic strength of amines.
(a) Decreasing order of basicity of amines in gaseous phase:
[3° amine (R3N) > 2° amine (R2NH) > 1° amine RNH2 > NH3]
This is due to +I effect of alkyl groups.
•• ⊕ ⊕
R NH2 + H RNH3
•• ⊕ ⊕
H3N + H NH4
The alkyl group (R) pushes e– density towards N atom and thus makes the LP more available for sharing
with the H+ of the acid. Moreover, RNH3+ is stabilised due to dispersal of the positive charge by the +I
effect of the alkyl group. Thus, the basic character of the aliphatic amines increases with the increase
in the number of alkyl groups.
R R
R N•• •• ••
R NH R NH2 NH3
••
→ Order of basic strength of amines in gaseous phase.
R
3° amine 2° amine 1° amine

(b) Basic character of aliphatic amine in polar solvent:

In the aqueous phase, the substituted ammonium cations get stabilised not only by +I effect of alkyl
group but also by solvation with H2O molecule.
The greater the size of the ion, the lesser will be the solvation and less stabilised is the ion.
OH2

H R
⊕
R ⊕ H OH2 ⊕
R–N – H OH2 N R–N–H OH2
R H OH2
H R

OH2 primary secondary

Tertiary
The decreasing order of extent of H-bonding in water and order of stability of ions by solvation

The greater the stability of substituted ammonium cation, the stronger is the base. So, the order of
basicity should be 1° > 2° > 3° amine but according to +I effect of R group, the basicity order of amine
is 3° > 2° > 1°.
Both the results do not match with the observed basicity of amine. So, some other factor, steric
hindrance, is also responsible in determining the basic character of amines.

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When the alkyl group is small [such as (–CH3) group], there is no steric hindrance to H bonding. If the
alkyl group is bigger than (–CH3) group, there would be steric hindrance to H bonding.
Hence, inductive effect, solvation effect, and steric hindrance of alkyl group decide the basic strength
of alkyl amines in the aqueous phase.
Result :
Decreasing order of basic character of alkyl substituted amines in aqueous solution:
(R)– Relative basic strength
CH3– 2° > 1° > 3° > NH3
C2H5– 2° > 3° > 1° > NH3

Chemical Reactions
(1) Alkylation: Reaction of an amine with alkyl halide, result in alkylation of the nitrogen. The reaction
takes place by SN2 mechanism with the amine acting as a nucleophile.
••
CH3–NH2
Ex. CH3–CH2–Cl CH3–CH2–NH–CH3
SN2
CH2–Br CH2–NH–CH2–CH3
CH3–CH2–NH2
SN2

(2) Acylation: Primary and secondary amines undergo nucleophilic acyl substitution reaction when
treated with acyl halides or anhydrides forming N-substituted amides
CH3 H CH3
..
C2H5–N–H + C–Cl Base C2H5–N–C–Cl
⊕
C2H5–N–C–CH3 + H–Cl
H O H O H O
Θ
Ethanamine N-Ethylethanamide

H
Base
C2H5–N: + CH3–C–Cl C2H5–N — C–CH3 + H–Cl
C2H5 O C2H5O
N-Ethylethanamine N,N-Diethylethanamide
.. ..
C6H5–NH + CH3–C–O–C–CH3 C6H5–N – C – CH3 + CH3COOH
H O O H O
Benzenamine Ethanoic anhydride N-Phenyl ethanamide
or Acetanilide

Points to ponder:
• Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters
by nucleophilic acyl substitution reaction. This reaction is known as acylation.
• You can consider this reaction as the replacement of hydrogen atom of –NH2 or >N–H group by the acyl
group.
• The products obtained by acylation reaction are known as amides.
• The reaction is carried out in the presence of a base stronger than the amide, like pyridine, which
removes HCl and shifts the equilibrium to the right hand side.
• Amines also react with benzoyl chloride (C6H5COCl). This reaction is known as benzoylation.
CH3NH2 + C6H5COCl → CH3NHCOC6H5 + HCl
Methanamine Benzoyl chloride N–Methylbenzamide
• Benzoylation is also known as Schotten-Baumann reaction.

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(3) Carbyl amine reaction or Iso cyanide Test :
• 1° Aliphatic or aromatic amines on heating with chloroform and alcoholic KOH form isocyanide or
carbyl amines which is foul smelling substance.
• 2° and 3° amines do not show this reaction therefore it is used in detection of 1°amines
∆ ⊕ Θ
Ex. i. R–NH2+CHCl3 + 3KOH(Alcoholic) 
→ R − N ≡ C + 3KCl + 3H3O
(Foul smelling substance)
∆ ⊕ Θ
ii. Ph–NH2+CHCl3+3KOH(Alcoholic) 
→ Ph– N ≡ C + 3KCl + 3H2O

4. Reaction with Nitrous Acid:

(Test to distinguish between 1°, 2° and 3° amines)
• Nitrous acid (HNO2) is prepared in situ by the reaction of sodium nitrite (NaNO2) with dil. HCl
NaNO2 + HCl → HNO2 + NaCl
(a) Primary aliphatic amines:
1° aliphatic amines react with HNO2 to form aliphatic diazonium salts, which, being unstable, liberate
N2 gas quantitatively and alcohols. Quantitative evolution of N2 is used in the estimation of amino acids
and proteins.
NaNO2 + HCl ⊕ Θ H2 O
R–NH2 + HNO2 [R–N≡N]Cl [R–OH+N2+HCl
273–278 K
(1° amine) (Alkyl diazonium salt)
Mechanism:
NaNO2 + HCl → HNO2 + NaCl
⊕ ⊕
••
O=N–OH + H–Cl Θ
O=N–OH2 N=0 + H2O
–Cl
(act as base) (act as acid) (nitrosonium ion)
H
•• ⊕ ⊕ –H⊕
R–NH2+ N=0 R–N–N = O R–NH–N = 0
(electrophile)
H tautomerism
⊕ •• ⊕ ⊕ H⊕ ••
R–N≡N R–N=N R–N=N–OH2 R–N=N–OH
••

⊕ H2O
–N2↑ R R–OH
(carbocation)
(b) Primary aromatic amines:
NaNO2 + HCl ⊕ Θ
C6H5–NH2 C6H5–N2Cl + NaCl + 2H2O
273–278 K
Aniline Benzene diazonium
chloride
The reaction for converting 1° aromatic amines to diazonium salts by treatment with a cold (273 – 278K)
solution of nitrous acid is called Diazotisation. If however, the temperature, rises above 278 K, the
diazonium salts decompose to form phenols.
⊕ Θ >278 K
C6H5–N2Cl + H2O C6H5–OH + HCl +N2

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(c) Secondary amines (Aromatic and Aliphatic):
(i) R2NH + HNO2 R2N–N=O + H2O
N-nitroso amine
(yellow oily liquid)
NH–R R–N–N=0
(ii)
+ HNO2 + H2O

(yellow oily liquid)

(d) Tertiary amines (Aromatic and Aliphatic):
⊕
(i) R3N + HNO2 R3N–N= O
(Salt)
R–N–R R– N– R

(ii) + HNO2

N=O
(5) Reaction with aryl sulphonyl chloride: Hinsberg' test
• This reaction is used to distinguish 1°, 2° and 3° amine
(a) Reaction with primary amines:

O O
S–Cl + H– N–C2H5 S–N–C2H5 + HCl
O H (1° amine) O H
Benzene sulphonyl N–ethyl benzene
chloride sulphonamide
(Hinsberg's reagent) NaOH(base)

O
Θ ⊕
S–N–C2H5 Na + H2O
O
(Soluble)
• The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong
electron withdrawing sulphonyl group. Hence it is soluble in alkali
(b) Reaction with secondary amines:

O O
S–Cl + H–N–C2H5 S–N–C2H5 + HCl
O C2H5 O C2H5
(2° amine) N,N–Diethyl benzene
sulphonamide
NaOH

No reaction

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• Since N,N-diethyl benzene sulphonamide does not contain any hydrogen atom attached to nitrogen
atom therefore it is not acidic and insoluble is alkali

(c) Reaction with Tertiary amines:

O
S–Cl + (C2H5)3N → No reaction
O (3° amine)
• Tertiary amines do not react with benzene sulphonyl chloride (Hinsberg reagent)

Note. However these days benzene sulphonyl chloride is replaced by P-toluene sulphonyl chloride

6. Electrophilic substitution reactions:

(i) Bromination: Aniline reacts with Br2 water at room temperature or with brominating reagent to give a
white precipitate of 2,4,6-tribromo aniline.
NH2
NH2 Br
Br
Br2 in CS2
0–5°C
(White PPt)

Br
NH2
Br
Br2/H2O Br
(White PPt)

Br
Since aromatic amines have high reactivity, so to prepare mono-substituted aniline derivative, (–NH2)
group is protected by acetylation with acetic anhydride (Ac2O), then the desired substitution is carried
out followed by hydrolysis of the substituted amide to obtain the substituted amine.
O O
NH2
NH2 NH–C–Me NH–C–Me
Br2/AcOH HOΘ
Ac2O
or H⊕
Pyridine

N-Phenyl Br
Br
ethanamide (Major) 4-Bromo aniline
(Acetanilide)

The lone Pair on N atom of acetanilide interact with O atom due to resonance (i.e., cross conjugation).
Hence, lone pair on N atom are less available for donation to benzene ring by resonance, thereby
decreasing the activating effect of (–NH2) group.
•• • •• Θ
•O O
• •
• •
⊕
N–C–Me
••
↔ N = C – Me

(ii) Nitration: The direct nitration of aniline gives tarry oxidation products along with nitro derivatives, In
the acidic medium, aniline is protonated to give anilinium ion which is meta directing that's why
m-derivative along with o- and p-products are obtained.
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 CHEMISTRY Amines

NH2 NH2 NH2 NH2

NO2
HNO3 +
+
+H2SO4
288 K NO2
NO2 (47%) (2%)
(51%)
If (–NH2) group is protected by acetylation, p-nitro derivative is obtained as the major product.
NH2 NHCOMe NHCOMe NH2
Θ
Ac2O HNO3 + H2SO4 OH
Pyridine 288 K or H⊕

Acetanilide NO2 NO2

p-Nitroacetanilide p-Nitroaniline

(iii) Sulphonation: Aniline on sulphonation with concentrated H2SO4 first gives anilinium hydrogen
sulphate, which on heating with H2SO4 at 453-473 K yields p-aminobenzene sulphonic acid (sulphanilic
acid) as the major product.
⊕
⊕
NH2 [NH3]HSO4
Θ
NH2 NH3

H2SO4 453- 473K

Anilinium SO3H SOΘ3

hydrogen Sulphanilic Zwitter ion
sulphate acid
(iv) Friedel-Crafts Reaction: Aniline does not undergo F.C. reaction (both alkylation and acetylation) due
to salt formation with AlCl3 (Lewis acid) which acts as a catalyst. Due to this, N atom of aniline acquires
positive charge and hence acts as a strong deactivating group for further reaction.

Exercise 1.2
1. Among the following, the strong base is – 3. The presence of primary amines can be
(1) C6H5NH2 confirmed by:
(2) p − NO2C6H4NH2 (1) Reaction with HNO2
(3) m − NO2 − C6H4NH2 (2) Reaction with CHCl3 and alc. KOH
(4) C6H5CH2NH2 (3) Reaction with Hinsberg reagent
(4) All
2. Which of the following should be most
volatile? 4. A compound (X) having the molecular
(I) CH3CH2CH2NH2 (II) (CH3)3N formula C3H9N reacts with benzenesulphonyl
CH3–CH2
chloride to form a solid that is insoluble in
(III) NH (IV) CH3CH2CH3
alkalies. The compound (X) is:
CH3
(1) II (2) IV (1) CH3CH2CH2NH2 (2) (CH3)2CHNH2
(3) I (4) III (3) CH3CH2NHCH3 (4) (CH3)3N

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5. Which one of the following will not react 10. In the reaction the structure of the product
with the Grignard reagent (C2H5MgBr): [X] is:
C2H5 O
(1) NaOH/Br2
N–H H3C C O
(1) C2H5 – NH2 (2) NH2 (2) C
C2H5 Cl
C2H5 O [X]
O O
(3) C2H5–N (4) CH3–C–NH2
C2H5 (1) H3C C—O—C
O
6. The end–product in the reaction sequence
(2) NH—C CH3
would be:
HNO2 PCl5 NH3 O
Ethyl amine  → A → B → C
(1) Ethyl cyanide (2) Ethyl amine (3) H3C NH—C
(3) Methyl amine (4) Acetamide O O

7. Which one of the following tests can be (4) H3C C—NH—C

used to identify primary amino group in a
given organic compound:- 11. Aniline on treatment with bromine water
yields white precipitate of:-
(1) Iodoform test
(1) o–Bromoaniline
(2) Victor Meyer’s test
(2) p–Bromoaniline
(3) Carbylamine reaction
(3) 2, 4, 6–Tribromoaniline
(4) Schiff’s Test (4) m–Bromoaniline

8. Amine whose product is insoluble in NaOH 12. Acetanilide when treated with bromine in
(in Hinsberg test)- acetic acid mainly gives:-
(1) CH3CH2NH2 (2) ( CH3 )2 CHNH2 (1) o–Bromoacetanilide
(2) N–Bromoacetanilide
(3) ( CH3 )2 NH (4) ( CH3 )3 N
(3) p–Bromoacetanilide
(4) m–Bromoacetanilide
NH–CO–CH3
HNO3 /H2 SO4 H2O/H⊕ 13. Reaction of nitrous acid with aliphatic
9.  → X → Z,
heat primary amine in the cold gives
(1) A diazonium salt (2) An alcohol
Z is: (3) A nitrite (4) A dye
NH2 NH2
14. The reagents needed to convert:
(1) (2) O
NO2 O NH2 HN CH3
NO2
is/are
NHCOCH3 NH2
O2 N NO2 (1) KOH + Br2 ;LiAlH4
(3) (4) (2) KOH + Br2 ;CH3COCl /Py
NO2
(3) HONO,Cu2Cl2 , ( CH3CO )2 O
NO2
(4) KOH + CHCl3 ,CH3COCl /Py

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15. Gabriel phthalimide synthesis is used for 21. Which of the transformation given below
the preparation of would not result in aromatic amines?
(1) Primary aromatic amines O
(2) Primary aliphatic amines NH2
(3) Both (1) and (2) (1) + Br2 + NaOH →
(4) Secondary amines
O
Cl
CH3 Θ⊕ (i)
(2) NK
HNO2 (ii)NaOH(aq.)
16. CH3–C–CH2NH2  → (major product) is
CH3 O
CH3 CH3 NO2
H2 /Pd
(1) CH3–C–CH2OH (2)CH3–C–CH2CH3 (3) →
Ethanol

CH3 OH NHCOCH3
CH3 CH3 Θ
⊕ OH/H2O
(3) CH3–C–CH2N2 (4)CH3–C–CH=CH2 (4) →
CH3
22. Which among the following compounds
gives foul smelling products on heating
17. The compound 'A' when treated with HNO3
with Chloroform and ethanolic potassium
(in presence of H2SO4) gives compound 'B'
hydroxide?
which is then reduced with Sn and HCl to
NH2
aniline. The compound 'A' is (A) CH3–CH2–NH2 (B)
(1) Toluene (2) Benzene
(3) Ethane (4) Acetamide (C) CH3– CH2– NH–CH3
(D) CH3–CH2–N–CH3
18. Which of the following amine will not react CH3
with benzoyl chloride - (1) Only A (2) Only A & B
(1) isopropyl amine (3) Only A,C and D (4) Only A & C
(2) p-toludine
NH2
(3) N, N-Diethyl aniline
(4) N-methyl aniline HNO3 ,H2 SO4
23. 
288K
→
O NH2 NH2 NH2
LiAlH4
19. CH3–C–NH2 
→P NO2
NaNO2
 → Q + Gas; Q is + +
HCl (aq.)
(Major )
NO2
(1) CH3–CH2–Cl (2) CH3–CH2–OH NO2
(3) CH3–CH2–NO2 (4) CH3–CH2–NH2 (A) (B) (C)
Incorrect statement about the given
20. Which Statement is true chemical reaction is :
(1) Aniline undergoes Friedel craft (1) In strongly acidic medium, aniline is
alkylation at faster rate than benzene protonated to form the anilinium ion
(2) Aniline reacts with bromine water at which is meta directing.
room temperature to give 2,4,6 – (2) % yield of ‘B’ is greater than ‘C’
tribromo aniline with yellow ppt. (3) % yield of ‘B’ is greater than ‘A’
(3) Anilinium ion is meta directing (4) Reaction is possible and compound A
(4) All will be major product.
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24. Conder the following reaction sequence O
NO2 NH2
28.
H2/Pd (CH3CO2)O
A Pyridine
B
Ethanol HNO3, H2SO4 Br2 +KOH(aq) CHCl3 +KOH
 → X →∆
Y ; Y is
288 K
Θ CH2–CN CH2–NC
OH
D C (major) (1) (2)
∆
Final Product ‘D’ is- NC
NH2
NH2 NC
(3) (4)
(1) (2) Br

O CH3 NO2 NH2

NH2
NHCOCH3 29. Br2/CS2
A
(3) (4) OH
NO2
NO2
Br2/H2O
B
Heat
25. R-NH2+CHCl3+KOH  → (Y)+KCI+ H₂O A & B respectively are -
OH NH2
Product (Y) has foul smell. The reaction is
known as (1)
(1) Wurtz reaction
(2) Kolbe's reaction Br Br
(3) Hell-Volhard-Zelinsky reaction NH2 OH
Br Br Br Br
(4) Carbylamine reaction
(2)
26. The correct order of boiling points of the
Br Br
following isomeric amines is
NH2 OH
C4H9NH2, (C2H5)2NH, C2H5N(CH3)2 Br Br
(1) C2H5 N ( CH3 )2 > ( C2H5 )2 NH > C4H9NH2 (3)
(2) ( C2H5 )2 NH > C2H5 N ( CH3 )2 > C4H9NH2 Br
Br
(3) C4H9NH2 > ( C2H5 )2 NH > C2H5 N ( CH3 )2 NH2 OH
Br Br
(4) ( C2H5 )2 NH > C4H9NH2 > C2H5 N ( CH3 )2
(4)

27. Ethyl amine and aniline can be Br Br

distinguished by :- 30. Primary, secondary and tertiary amines can
(1) Isocyanide test be separated using :-
(2) Azo dye test (1) Para-Toluene sulphonyl chloride
(3) Hinsberg's reagent (2) Chloroform and KOH
(3) Benzene sulphonic acid
(4) Grignard reagent (4) Acetyl amide

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Diazonium Salts
Introduction: 1° aromatic amines form arene diazonium salts which are stable for a short time in
solution at low temperature (273-278 K) unlike 1° aliphatic amines which form highly unstable alkyl
diazonium salts. The stability of arenediazonium ion is due to resonance stabilisation.
⊕ •• ⊕ Θ ⊕ Θ ⊕ Θ ⊕ ••
N≡N N = N•• N = N•• N = N•• N≡N
⊕ ⊕

Diazotisation: The conversion of 1° aromatic amine into diazonium salt with HNO2 (nitrous acid) is
known as diazotisation. HNO2 is produced in the reaction mixture by the reaction of NaNO2 and HCl.
Due to instability, diazonium salts are used immediately after their preparation.
273–278 K ⊕ Θ
PhNH2 + NaNO2 + HCl [Ph–N ≡ N]Cl + NaCl + 2H2O
Benzene diazonium
chloride (I)
Physical properties of benzene diazonium salt :
Benzenediazonium chloride is a colourless crystalline solid. It is readily soluble in water and is stable
in cold but reacts with water when warmed. It decomposes easily in the dry state. Benzenediazonium
fluoroborate is water insoluble and stable at room temperature.
Chemical properties of benzene diazonium salt :

Cu/HCl
C6H5Cl+N2
Cu/HBr
(Gattermann reaction)
C6H5Br + N2
Cu2Cl2/HCl
C6H5Cl + N2
Cu2Br2/HBr
C6H5Br + N2 Sandmeyer
CuCN/KCN reaction
C6H5CN + N2
KI
∆ C6H5I + N2 + KCl
⊕ Θ
⊕ Θ HBF4
C H – N BF ∆ C6H5–F + BF3 + N2
C6H5N2Cl 6 5 2 4

NaNO2
H2O Cu, ∆ C6H5–NO2 + NaBF4 + N2
∆ C6H5OH
C2H5OH
C6H6 + CH3CHO + N2 + HCl
H3PO2+H2O
C6H6 + H3PO3 + N2 + HCl
N=N–C6H5
OH
β-Naphthol
10% NaOH (Red orange dye)
Ph–OH
PH 8 to 9
C6H5—N=N—C6H4OH
p-Hydroxy azobenzene (Orange dye) Coupling
reaction
C6H5NH2 C6H5N=N—C6H4NH2
PH 5 to 6
p-Amino azobenzene or (yellow dye)

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Exercise 1.3
1. Which of the following amines give N- 5. Compound A has a molecular formula
nitroso derivative with NaNO2 and HCl:- C7H7NO. On treatment with Br2 and KOH, A
gives an amine B which gives carbylamine
(1) C2H5NH2 (2) NH2 test. B upon diazotisation and coupling
with phenol gives an azo dye. A can be -
R (1) C6H5CONHCOCH3
(3) N–H (4) NH2 (2) C6H5CONH2
(3) C6H5NO2
(4) o, m or p–C6H4 (NH2)CHO
2. Which statement is incorrect -
(1) –NH2 group is present in both ethylamine 6. Identify S in the following sequence of
and aniline reactions
CuCN LiAlH4 HNO2
(2) Both ethylamine and aniline give C6H5 N2Cl 
KCN
→  → Y  → S.
(1) Benzoic acid (2) Phenyl acetic acid
isocyanide test
(3) Benzyl alcohol (4) Benzamide
(3) Ethylamine and aniline give unpleasant
smell with CHCl3 and KOH CH3
(i) Br2,Fe
(4) Both ethylamine and aniline give 7. (ii) Sn,HCl (A)
hydroxy compound with HNO2 at 0°C (iii) NaNO2, HCl, 273K
(iv) H3PO2, H2O
NO2
Final product (A) in the above sequence
3. Identify ‘Z’ in the reaction given below
reaction is
NH2 CH3
1. HNO2 (275K ) NaOH CH3 I Br
 →X 
→ Y 
→Z
1. H2O;Boil (1) (2)
Br CH3
NH—CH3 N2Cl CH3 CH3
H3 C CH3 H3 C CH3 Br
(1) (2) (3) (4)

CH3 CH3 OH Br

OCH3 OCH3 8. Consider the following reaction sequence

OH OH NH2
(3) (4) NaNO2
A HBF4
B NaNO2
C(Major)
HCl Cu/∆
(273K–278K)
OH
Product ‘C’ is –
+
NO2 F
Cu/HCl
4. ArN2 Cl −  → ArCl+N2+ CuCl ;
(1) (2)
the reaction is named as _________. F
(1) Sandmeyer reaction F
(2) Gattermann reaction
(3) (4)
(3) Balz Schiemann reaction
(4) Carbylamine reaction NO2 NO2

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9. The Major Product ‘X’ formed in the 12. 'Z' in the following sequence of reactions is
HNO3 /H2 SO4 Sn/HCl NaNO2 /HCl
following reaction sequence is C6H6  ∆
→ W  → X →
300K
Y
Zn/ ∆
CH3  →Z
COOH
(i) Cl2/Fe
‘X’
(ii) Sn/HCl
(iii) NaNO2/HCl ; O–5°C (1) (2)
(iv) KI
NO2
OH Cl
CH3 CH3
Cl I (3) (4)
(1) (2)
13. Which is Sandmeyer reaction among the
I Cl following :
HBF4
CH3 CH3 (1) C6H5 N2Cl 
∆
→ C6H5 F
Cu/HCl
(2) C6H5 N2Cl  → C6H5Cl
(3) (4) Cu2Cl2 /HCl
(3) C6H5 N2Cl → C6H5Cl
Cl I
I Cl (4) All of these
OH
NH2
10. Identify the compound Y in the following
HNO
reaction. 14.  2
273 −278 K
→( A) 
pH− 8 to 9
→(B)
⊕ Θ
NH2 N2Cl Final product ?
NaNO2+HCl Cu2Cl2
Y + N2
273-278 K HCl (1) Ph–N=N NH2

(2) Ph–N=N–Ph
Cl
(1) (2) (3) Ph–N=N OH
OH
Cl O2N NO2
Cl
(4)
(3) (4) NO2
Cl Cl ⊕ Θ
NH2 N2Cl
[Y]
[X]

11. The product 'D' in the following sequence of 15. CH3 CH3
reactions is ⊕
NO2 N2BF4Θ
Br2 NaNO2 HBF4 Heat
C6H5NH2 →
(aq)
A 
HCl
→B  → C  →D [Z]

CH3 CH3
(1) 2, 4, 6-tribromofluorobenzene
X, Y, Z are-
(2) fluorobenzene (1) NaNO2/Cu, ∆, HBF4, NaNO2+HCl/0–5°C
(3) p-bromofluorobenzene (2)NaNO2+HCl/0–5°C,B2H6/THF,NaNO2/Cu, ∆
(3) NaNO2+HCl/0–5°C, AgF, NaNO2/Cu, ∆
(4) tribromobenzene. (4) NaNO2+HCl/0–5°C, HBF4, NaNO2/Cu, ∆

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Exercise 2
1. Which one of the following compound gives 6. (i)CO2
PhMg Br  → ( A ) 
Br2 ,NaOH
→ (B )
(ii)H O⊕
a secondary amine on reduction 3
(iii)NH3 ,∆
(1) Nitromethane What would be the final product of above
(2) Nitrobenzene
sequence reaction?
(3) Methyl isocyanide
O
(4) Methyl cyanide
CH2–NH2 C–NH2
NH3 P2O5 Ni
2. C6H5COCl →× → Y 
H
→ Z, Z is (1) (2)
2

The end product in the above sequence of

reaction is - NH2 C≡N
(1) Benzoic acid (2) Aniline
(3) (4)
(3) Benzyl amine (4) Benzonitrile

3. Amongst the given set of reactants, the 7. Hoffmann bromamide degradation of

most appropriate for preparing 2° amine is benzamide gives product ‘A’ which upon
_____. heating with CHCl3 and NaOH gives product
(1) 2° R—Br + NH3 ‘B’. The structure of ‘A’ and ‘B’ respectively
(2) 2° R—Br + NaCN followed by H2/Pt are :
(3) 1° R—NH2 + RCHO followed by H2/Pt NH2 OH
(4) 1° R—Br (2 mol) + potassium CHO
(1) and
phthalimide followed by H3O+/heat.
NH2 NC
4. Which among the following compounds
(2) and
react with Hinsberg’s reagent?
NH2 N(CH3)2
CH2NH2 CH2NC
(A) (B)
(3) and
(C) CH3–NH2 (D) (CH3)3 N
H CH2NH2 NC
N (4) and
(E)

Choose the correct answer from the options O O

(I) Zn-Hg/HCl
given below: 8. NH2 (II) LiAlH4
Product
(1) B and D only (2) C and D only (III) H3O⊕
(3) A, B and E only (4) A, C and E only major product is–
5. Compound A from the following reaction OH
sequence is: (1) OH
(CH3CO)2 O Br2 (i)H2 O/OH–
(A) 
Pyridene
→ (B) 
CH3COOH
→(C) 
(ii)NaNO2 /HCl;300K
→ OH

Br (2) NH2

(3) NH2

OH OH OH
(1) Benzoic acid (2) Aniline (4) NH2
(3) Salicylic acid (4) Phenol

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9. Which of the following is least basic 12. Identify the product A and D formed in the
NH2 following reaction
(1) NH2 (2)
CH3

O Br2 (i)NaNO2/HCl
(A) Sn/HCl (B) (C)
FeBr3 0–5°C
NH (ii) H3PO2/H2O
(3) (4)
N NO2
O Θ
(i) KMnO4/OH
(C) (D)
(ii)H+
⊕ Θ
Me COOH
Sn+HCl C6H5 N2 Cl
10. C6H5NO2 → ‘A’ H⊕
→ 'P' Br Br
( Yellow coloured )
(1) A= , D=
Consider the above reaction, the product ‘P’ Br Br
is : NO2

N=N Me COOH
(1) Br Br
NH2 (2) A= , D=

N=N NH2
(2) NO2
Me Me
NH Br Br
(3) (3) A= , D=

NO2
N=N–NH
(4) Me COOH
Br Br
(4) A= , D=
11. A compound ‘X’ having molecular formula
C5H13N on reaction with Hinsberg’s reagent NO2 OH

gives compound ‘Y’ which does not dissolve 13. Identify the Major Product ‘C’ formed in the
following reaction sequence :
in aqueous alkali. The structure of the Ι KCN Θ
OH/H2O NaOH
compound ‘X’ could be. A B Br2 C
Partial
Hydrolysis

N Br
(1) (1)

NH NH2
(2) (2)

NH2
(3) (3)
NH2

(4) All of these Br

(4)
O

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14. Select the incorrect product among the 19. Which of the following amines does not
following reactions: react with Hinsberg's reagent?
O (1) CH3CH2 − NH2 (2) CH3 − NH − CH3
(1) H2 O
CH3—C—Cl ∆
CH3COOH (3) ( CH3CH2 )3 N (4) All of these
OH O
CrO3–H2SO4
(2) 20. Which of the following compounds cannot
be identified by carbylamine test?
CH2—CH2OH CH2—COOH
(1) CH3CH2NH2

(3)
(i) KMnO4/KOH (2) ( CH3 )2 CHNH2
(ii) H⊕
(3) C6H5NH2
CONH2 CH2NH2
(4) C6H5NHC6H5
(i) LiAlH4
(4) (ii) H2O
21. Which of the following methods can be used
to carry out the following conversions?
15. Hoffmann bromamide degradation reaction
O
is shown by which of the following?
NHCH3
(1) ArNH2 (2) ArCONH2 NH2
(3) ArNO2 (4) ArCH2NH2
(i) (ii) (iii)
16. Amine that cannot be prepared by Gabriel
phthalimide synthesis is (1) Br2/KOH CHCl3/KOH H2/Pd
(1) aniline (2) benzyl amine (2) KCN H2/Pd Sn/HCl
(3) methyl amine (4) iso-butylamine. (3) CuCN H2O/H+ H2/Pd
(4) HNO3/H2SO4 (CH3CO)2O Fe/HCl
17. The decreasing order of boiling points of
isomeric amines is p-amine >s-amine >t- 22. The end product Z of the reaction :
amine. This trend of boiling point can be Ethyl amine HNO2 PCl5
→ X → KCN
Y → Z is
explained as (1) propanenitrile (2) triethylamine
(1) boiling point increases with increase in (3) diethylamine (4) propylamine.
molecular mass
(2) tertiary amines have highest boiling
23. Which of the following tests is suitable to
point due to highest basicity
differentiate between aniline and
(3) intermolecular hydrogen bonding is
maximum in primary amines and absent benzylamine?
in tertiary amines (1) Aniline gives dye test on diazotisation
(4) intramolecular hydrogen bonding is and reaction with β-naphthol while
present in tertiary amines. benzylamine gives alcohol.
(2) Benzylamine gives green dye with β-
NaCN Ni/H2 Acetic
18. CH3CH2Cl  → X  → Y 
anhydride
→Z naphthol after diazotisation while
Z in the above reaction is aniline gives orange dye.
(1) CH3CH2CH2NHCOCH3 (3) Aniline gives carbylamine reaction
(2) CH3CH2CH2NH2 while benzylamine does not.
(4) Benzylamine gives carbylamine
(3) CH3CH2CH2CONHCH3
reaction while aniline does not.
(4) CH3CH2CH2CONHCOCH3

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24. Identify X and Y in the reaction. 27. Which of the following reactions is not
correctly matched?
CH3–CH–CONH2 Br2/NaOH X HNO2 Y (1) Reaction used to Hoffmann
CH3 convert amide into Bromamide
primary amine reaction
(1) X=CH3–CH–CH2NH2;Y=CH3–CH–CH2–OH with one carbon
CH3
atom less
CH3
(2) Reaction used to Carbylamine
(2) X=CH3CH=CHNH2; Y=CH3CH=CHOH convert primary reaction
amines into
(3) X=CH3–CH–NH2;Y=CH3–CH–OH isocyanides
CH3 (3) Reaction used to Hinsberg’s
CH3
distinguish reaction
OH primary,
secondary and
(4) X=CH3–CH–CH2–NH2;Y=CH3–C–CH3
tertiary amines
CH3 CH3 (4) Preparation of Victor
primary amines Meyer’s
25. The reagent required to convert using phthalimide synthesis
O
O NH2 28. Mark the correct route of the conversion of
NH CH3
p-chloroaniline to p-chlorobenzylamine.
NH2 CH2NH2
is
(i) (ii) (iii)
A B
(1) KOH /Br2 ,AC2O /Py
Cl Cl
(2) NaOBr,CH3COCl /Py
(i) (ii) (iii)
(3) HNO2 , ( CH3CO )2 O (1) Alkylation KCN H2/PT
(2) Diazotisation CuCN H2/PT
(4) 1 and 2 both (3) Oxidation H2/Pt Hydrolysis
(4) Diazotisation H2O/H+ Sn/HCl
26. An organic compound (X) was treated with 29. The major organic product formed from the
following reaction
sodium nitrite and HCl in ice cold
O →( i)CH3NH2
product
( ii)LiAIH ( iii)H O
4 2
conditions. Bubbles of nitrogen gas were
(1) CH3
seen coming out. The compound (X) may be
OH
(1) a secondary aliphatic amine (2)
NHCH3
(2) a primary aromatic amine O–NHCH3
(3)
(3) a primary aliphatic amine OH
(4) NHCH3
(4) a tertiary amine. OH

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30. The best reagent for converting 34. The correct stability order of the following
2-phenylpropanamide into diazonium salt is
2-phenylpropanamine is _________.
+ + + +
(1) excess H2 N2Cl– N2Cl– N2Cl– N2Cl–
(2) Br2 in aqueous NaOH
(3) iodine in the presence of red phosphorus
(4) LiAlH4 in ether.
OCH3 NO2 CN
31. What is the product obtained in the (A) (B) (C) (D)
following reaction: (1) (A) > (B) > (C) > (D)
NO2 (2) (A) > (C) > (D) > (B)
Zn
NH4Cl Product?
(3) (C) > (A) > (D) > (B)
NH2 (4) (C) > (D) > (B) > (A)
(1)

NHOH
35. The correct sequential addition of reagents
(2) in the preparation of 3 -nitrobenzoic acid
from benzene is
N
(3) N (1) Br2 / AlBr3 ,HNO3 /H2SO4 ,Mg / ether,

O– CO2 ,H3O+
N=N
(4) + (2) Br2 / AlBr3 ,NaCN,H3O+ ,HNO3 /H2SO4
(3) Br2 / AlBr3 ,HNO3 /H2SO4 ,NaCN,H3O+
32. Best method for preparing primary amines (4) HNO3 /H2SO4 ,Br2 / AlBr3 ,Mg / ether,
from alkyl halides without changing the
CO2 ,H3O⊕
number of carbon atoms in the chain is
(1) Hoffmann bromamide reaction
(2) Gabriel phthalimide synthesis 36. CH2–NH2 →
NaNO2 +HCl

(3) Carbyl amine reaction

(4) Ammonolysis of alkyl halide Find out major product :
OH
33. Which of the following methods of CH2OH
preparation of amines will not give same (1) (2)
number of carbon atoms in the chain of
OH CH3
amines as in the reactant? (3) CH3 (4)
(1) Reaction of nitrite with LiAlH4.
(2) Reaction of amide with LiAlH4 followed SOCl2 NH3 Br2
37. CH3CH2COOH  → A → B  →C
by treatment with water. ∆ KOH

(3) Heating alkyl halide with potassium Structure of compound C is –

salt of phthalimide followed by (1) CH3CH2CH2NH2
hydrolysis. (2) CH3CH2NH2
(4) Treatment of amide with bromine in
(3) CH3CH2NHCH3
aqueous solution of sodium hydroxide.
(4) CH3CH2CONH2

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38. Arrange the following in decreasing order 43. Which of the following compound will give
of boiling point: Hoffmann bromamide reaction?
(I) CH3CH2CH2NH2 (II) CH3–N–CH3 O

CH3 (1) Ph–C–NH–CH3

(III) CH3CH2 − NH − CH3 CH3–N–CH3
(2)
(IV) CH3CH2CH2CH3 CH3
(1) I > III > II > IV (2) II > III > I > IV (3) Ph–NH2
(3) IV > I > II > III (4) I > II > III > IV
O
39. A compound reacts with Hinsberg's reagent
(4) CH3–CH2–C–NH2
and gives a product which is soluble in
alkali solution (NaOH solution). Compound
NaBH4 H /Pd
is – 44. B←  CH=CH–CHO 
2
→A
(1) C2H5NH2 (2) C2H5NHC2H5
A and B are respectively:-
(3) C6H5NHC2H5 (4) C2H5–N–C2H5
(1) CH2CH2CHO,
C2H5
40. Which of the following compound gives
CH=CH–CH2OH
yellow oily liquid nitrosoamine when
reacted with nitrous acid :
CH2–NH2 NH–CH3 (2) CH2CH2CH2OH,
(1) (2)
CH2–NH2 CH=CH–CH2OH
N–C6H5
(3) (4)
NH2 (3) CH=CH–CHO in both case
( CH3CO )2 O
(i) Nitrating mixture
41. (A) 
py
→(B)  (ii) H3O+
→
(4) CH2CH2CH2OH in both case
NO2
Find out (A) :
N2Cl NH2
45. Which of the following are incorrect
(1) (2) statement?
NHCOCH3 (A) Aniline does not give nitration reaction
NH2
(ESR).
(3) (4) (B) Aniline give Friedel-Craft reaction.
(C) Diazonium ion of aromatic amine are
COCH3
more stable than aliphatic amine.
42. Identify 'Z' in the sequence : (D) pKb of Aniline is more than
⊕
NaNO2 +HCl CuCN H/H2O
C6H5NH2 →
273 K
X 
→Y 
Boil
→Z Methylamine.
(1) C6H5CN (2) C6H5CONH2 (1) A, B, C (2) A, B, D
(3) C6H5COOH (4) C6H5CH2NH2
(3) A, C (4) A, B

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46. Incorrect reaction is :- 49. What is A in the following reaction
O
O
Br2+KOH CH2Br (i)
NΘK⊕
CH3–CH2–C–CH3
(1) O
A
CHBr3+CH3CH2COOK (ii) ΘOH/H2O (Major Product)

O
OH OH
(1) NH–CH2
CHO
(2) CHCl3+KOH

CH2OH
(2)
O
Br2+KOH O
CH3–CH2–C–NH2
(3)
CH3–CH2–NH2 (3) NH

O
NH2 NH2
CHO CH2NH2
(4) CHCl3+KOH (4)

NH2
47. Amine is obtained as major product in
NaNO2/ HCl HBF4 NaNO2 Sn+HCl
O A B C D
50. Cu, ∆
NaBH4
(1) CH3–C–NH2
Which statement is correct
(2) CH3–CH2–CH–CH3 NaNH2
NH3 (1) A gives orange color with phenol
Cl (2) B is fluorobenzene
O (3) D is benzylamine
LiAlH4
(3) CH3–C–NH–CH3 (4) D is more basic than pyridine
(4) All of these
51. 1° Amine is not formed by which reaction.
O
NH2 NH2 NH2 NH2 (i) KOH
NO2 (1) NH (ii) C6H5–Cl
48. HNO3,H2SO4
+ (iii) NaOH
288 K +
NO2 O
NO2 Na(Hg)
(2) CH3–C≡N
(A) (B) (C) C2H5OH

Consider the given reaction, O

(i) LiAlH4
percentage yield of products: (3) CH3–C–NH2
(ii) H2O
(1) C > A > B (2) B > C > A
CH3 O
(3) A > C > B (4) C > B > A (4) CH3–C — C—NH2 KOH+Br2

CH3

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52. Which of the following amine among C3H9N 55. In a set of reactions m-bromobenzoic acid
gave a product 'C'. Identify the product 'C'
will not react with hinsberg reagent
COOH
(1) CH3–CH2–CH2–NH2 SOCl2 NH3 NaOH
 → A → B 
Br
→C
2
NH2
Br
(2) CH3–CH–CH3
SO2NH2 COOH
(3) CH2–CH2–NH–CH3
(1) (2)
(4) CH3–N–CH3
Br NH2
CH3

NH2 CONH2
53. Some statements are given about following
(3) (4)
order of basic strength in aqueous medium. Br Br
(CH3)2NH>CH3–NH2>(CH3)3N>NH3
(a) More substituted ammonium ion gets 56. An organic compound ‘A’ on treatment with
NH3 gives ‘B’ which on heating gives ‘C’. ‘C’
stabilized due to +I effect of alkyl group. when treated with Br2 in the presence of KOH
(b) The greater the size of ammonium ion, produces ethylamine. Compound ‘A’ is:
(1) CH3CH2COOH
more will be solvation, so more stable (2) CH3COOH
ion so more basicity. (3) CH3CH2CH2COOH
(4) CH3—CHCOOH
(c) More the alkyl groups, more steric
CH3
hindrance to H -bonding in ammonium
ion. O O
C
(d) Basic strength in aqueous medium is (i) KOH OH +Primary
57. NH
(ii) Z OH amine
explained by more than one type of C
O (iii) H2O
effects. Correct statements is/are O
Z may not be :
(1) a, b, c, d H
(2) a, b, d only
(1) CH3–Cl (2) CH3–C–Cl
(3) a, c, d only
CH3
(4) a and d only CH2–Cl Cl
(3) (4)
54. In the given reaction sequence:
Θ
(i)KMnO4 /OH/ ∆ (i)SOCl2 58. The end product (B) formed in the reaction
CH3–CH2–OH  → (A)  → 1. Br2 /CCl 4 1. LiAlH4
(ii)H⊕ (ii)NH3 /∆ CH2 = CH2 
2. KCN (excess)
→ A 
2. H O+
→B
3

Br2 /KOH
(B)  → (C) will be - (1) CH2 = CHCH2NH2
(2) H2N(CH2)4NH2
(1) Methylamine (2) Ethylamine (3) CH3NH(CH2)2NHCH3
(3) Propylamine (4) Acetamide (4) NC(CH2)2CN

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59. Aniline in a set of the following reactions 60. Among the following compounds which one
yielded a coloured product 'Y'
will produce a Schiff base on reaction with
NH2
cyclohexanone:
NaNO2/HCl N,N-dimethylaniline
X Y
(273-278K)
(1) N
The structure of 'Y' would be
H
(1) CH3
N=N N
CH3 NHCH3
CH3 CH3 (2)
(2) HN NH NH
NH2
(3)
(3) H C N=N NH2
3
N(CH3)2
CH3 CH3 (4)
(4) HN N=N NH

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Exercise 3
1. Assertion: Hoffmann's bromamide reaction 4. Given below are two statements; one is
is given by primary amines. labelled as Assertion (A) and the other is
Reason: Primary amines are more basic
labelled as Reason(R).
than secondary amines.
(1) Both Assertion and Reason are correct Assertion: The correct boiling point order
and Reason is the correct explanation of isomeric amines is -
of Assertion Primary > Secondary > Tertiary
(2) Both Assertion and Reason are correct
Reason: The intermolecular H -bonding in
but Reason is not the correct
explanation of Assertion primary amine is maximum due to two
(3) Both Assertion and Reason are hydrogen available for more association
incorrect. but in tertiary amine there is no hydrogen
(4) Assertion is not correct but Reason is
available for H -bond.
correct.
In the light of the above statements, choose
2. Assertion: N-Ethylbenzene sulphonamide the most appropriate answer from the
is soluble in alkali. options given below :
Reason: Hydrogen attached to nitrogen in
(1) Both (A) and (R) are correct and (R) is
sulphonamide is strongly acidic.
(1) Both Assertion and Reason are correct the correct explanation of (A).
and Reason is the correct explanation (2) Both (A) and (R) are correct but (R) is
of Assertion not the correct explanation of (A).
(2) Both Assertion and Reason are correct
but Reason is not the correct (3) (A) is correct but (R) is not correct.
explanation of Assertion (4) (A) is not correct but (R) is correct.
(3) Both Assertion and Reason are
5. Given below are two statements; one is
incorrect.
labelled as Assertion (A) and the other is
(4) Assertion is not correct but Reason is
labelled as Reason(R).
correct.
Assertion: The alkyldiazonium salt are
unstable but the arenediazonium salt are
3. Assertion: Only a small amount of HCl is
stable for a short time at low temperature
required in the reduction of nitro
(0°–5° C)
compounds with iron scrap and HCl in the
Reason: Due to resonance in
presence of steam.
arenediazonium salt carbon-nitrogen bond
Reason: FeCl2 formed gets hydrolysed to
becomes stronger than alkyldiazonium
release HCl during the reaction.
salt.
(1) Both Assertion and Reason are correct
In the light of the above statements, choose
and Reason is the correct explanation
the most appropriate answer from the
of Assertion
options given below :
(2) Both Assertion and Reason are correct
(1) Both (A) and (R) are correct and (R) is
but Reason is not the correct
the correct explanation of (A).
explanation of Assertion
(2) Both (A) and (R) are correct but (R) is
(3) Both Assertion and Reason are
not the correct explanation of (A).
incorrect.
(3) (A) is correct but (R) is not correct.
(4) Assertion is not correct but Reason is
(4) (A) is not correct but (R) is correct.
correct.

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6. Assertion : tert-Butyl amine can be formed 9. Assertion : Carbylamine reaction is used in
by Gabriel phthalimide synthesis. identification of primary amines but not
Reason: It follows SN1 mechanism. secondary and tertiary amines.
(1) If both assertion and reason are true Reason: Primary amines have boiling points
higher than isomeric secondary and tertiary
and reason is the correct explanation of
amines.
assertion.
(1) If both Assertion and Reason are
(2) If both assertion and reason are true but correct and the Reason is a correct
reason is not the correct explanation of explanation of the Assertion.
assertion. (2) If both Assertion and Reason are
(3) If assertion is true but reason is false. correct but Reason is not a correct
(4) If both assertion and reason are false. explanation of the Assertion.
(3) If the Assertion is correct but Reason is
7. Assertion: Gabriel phthalimide synthesis incorrect.
cannot be used to prepare aromatic primary (4) If the Assertion is incorrect and Reason
is correct.
amines.
Reason: Aryl halides do not undergo 10. Assertion: N,N–di ethyl benzene sulphonamide
nucleophilic substitution reaction with the is insoluble in alkali.
anion formed by phthalimide. Reason: sulphonyl group attached to
(1) If both Assertion and Reason are nitrogen atom is strong electron
correct and the Reason is a correct withdrawing group.
explanation of the Assertion. (1) If both Assertion and Reason are
(2) If both Assertion and Reason are correct and the Reason is a correct
correct but Reason is not a correct explanation of the Assertion.
explanation of the Assertion. (2) If both Assertion and Reason are
(3) If the Assertion is correct but Reason is correct but Reason is not a correct
explanation of the Assertion.
incorrect.
(3) If the Assertion is correct but Reason is
(4) If the Assertion is incorrect and Reason
incorrect.
is correct. (4) If the Assertion is incorrect and Reason
8. Assertion: Experimental reaction of CH3Cl is correct.
with aniline and anhydrous AlCl3 does not
11. Given below are two statements:
give o and p-methylaniline.
Statement-I: Aryl fluoride should not be
Reason: The-NH2 group of aniline becomes prepared by direct halogenation of
deactivating because of salt formation with benzene.
anhydrous AlCl3 and hence yields m-methyl Statement-II: This is easily obtained by
aniline as the product. reaction of Benzene diazonium salt with
(1) If both Assertion and Reason are fluoroboric acid/∆.
correct and the Reason is a correct In the light of the above statements, choose
explanation of the Assertion. the most appropriate answer from the
(2) If both Assertion and Reason are options given below:
(1) Statement-I and Statement-II are
correct but Reason is not a correct
correct.
explanation of the Assertion.
(2) Both Statement-I and II are incorrect.
(3) If the Assertion is correct but Reason is (3) Statement-I is correct but Statement-II
incorrect. is incorrect.
(4) If the Assertion is incorrect and Reason (4) Statement-I is incorrect but Statement-
is correct. II is correct.
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12. Statement-I: Butan–1–ol has boiling point 15. Given below two statement:
higher than Butan–1–amine Statement-I: Aniline react with conc. H2SO4
Statement-II: Solubility of ethanol is followed by heating at 453-473 K gives p-
greater than ethanal in polar solvent. amino benzene sulphonic acid which gives
(1) Both statement I and statement II are true blood red color in the “Lassaigne’s test”.
(2) Both statement I and statement II are false Statement-II : In Friedel-craft alkylation
(3) Statement I is true, but statement II is and acylation reactions, aniline forms salt
false with AlCl3 catalyst due to this nitrogen of
(4) Statement I is false but statement II is aniline acquires a positive charge and act
true as deactivating group.
Choose the correct answer from the option
13. Given below are two statements : given below:
Statement I : –NHCOCH3 group is also o/p- (1) Statement I is true but Statement II is
directing like –NH2 group but it is less false.
activating than –NH2 group. (2) Statement I is false but Statement II is
Statement II: Lone pair of nitrogen in – true
NHCOCH3 group is in resonance with (3) Both Statement I and Statement II are
carbonyl group. true
In the light of the above statements, choose (4) Both Statement I and Statement II are
the most appropriate answer from the false.
options given below:
(1) Statement I is correct but statement II 16. Match the following
is incorrect. Column-I Column-II
(2) Both statement I and statement II are Cannizzaro
a) R–NH2 + CHCl3 + KOH→ p)
correct. reaction
(3) Statement I is incorrect and statement Stephen’s
b) ArN2X + Cu2Cl2 /HCl → q)
II is correct. reaction
(4) Both statement I and statement II are ∆ Carbylamine
c) 2Ar–CHO+Conc.NaOH → r)
incorrect. reaction
(i)SnCl2 +HCl Sandmeyer’s
d) R–CN 
(ii)H3O+
→ s)
14. Given below are two statements: reaction
Statement-I: Mixing the solution of freshly (1) (a-r), (b-s), (c-p), (d-q)
prepared diazonium salt with cuprous (2) (a-s), (b-r), (c-q), (d-p)
chloride or cuprous bromide results in the (3) (a-q), (b-s), (c-p), (d-r)
replacement of the diazonium group by -Cl (4) (a-p), (b-q), (c-r), (d-s)
or -Br .
Statement-II: Replacement of the 17. Match the column
diazonium group by iodine does not require Column-I Column-II
the presence of cuprous halide and is done (Compound) (Pkb value)
simply shaking the diazonium salt with
(i) Methan amine (p) 4.75
potassium iodide.
N-Methyl
In the light of the above statements, choose (ii) (q) 3.38
methanamine
the most appropriate answer from the
N, N-Dimethyl
options given below: (iii) (r) 3.27
methanamine
(1) Statement-I is correct but Statement-II
(iv) Ammonia (s) 4.22
is incorrect.
(2) Both Statement-I and II are incorrect (1) (i)–(p), (ii)–(q), (iii)–(r), (iv)–(s)
(3) Statement-I is incorrect but Statement- (2) (i)–(p), (ii)–(r), (iii)–(s), (iv)–(q)
II is correct. (3) (i)–(s), (ii)–(r), (iii)–(q), (iv)–(p)
(4) Both statements-I and II are correct. (4) (i)–(q), (ii)–(r), (iii)–(s), (iv)–(p)

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18. Match the list-I with list-II 20. Watch the compounds given in column I
list-I with column II and mark the appropriate
+ – Cl choice.
N2Cl
(a) Cu2Cl2
+N2 Column I Column II
HCl
(A) Benzenesulphonyl (i) Zwitter ion
+ – Cl
N2Cl chloride
(b) Cu/HCl
+N2 (B) Sulphanilic acid (ii) Hinsberg's
CHO reagent

(c) CO, HCl (C) Alkyldiazonium (iii) Dyes

Anhyd. AlCl3/CuCl
salts
(i) Br2/Red P
(d) R–CH2–COOH R–CH–COOH (D) Aryldiazonium (iv) Conversion to
Br salts alcohols
list-II (1) (A) → (iv), (B) → (iii), (C) → (i), (D) → (ii)
(p) Gatterman-koch reaction
(2) (A) → (ii), (B) → (iv), (C) → (iii), (D) → (i)
(q) Sandmeyer reaction
(r) Hell-volhard-zelinsky reaction (3) (A) → (ii), (B) → (i), (C) → (iv), (D) → (iii)
(s) Gatterman reaction (4) (A) → (ii), (B) → (iii), (C) → (iv), (D) → (i)
(1) (a-p), (b-q), (c-s), (d-r)
(2) (a-q), (b-p), (c-s), (d-r) 21. Match the column I with column II and mark
(3) (a-s), (b-q), (c-p), (d-r) the appropriate choice.
(4) (a-q), (b-s), (c-p), (d-r)
Column I Column II
19. Match the list-I with list-II (A) + (i) Coupling
N2Cl–
List-I List-II reaction
CuCl
Name of reaction Reagent used

A. Decarboxylation (i) CHCl3/KOH (B) + (ii) Balz-

N2Cl–
Schiemann
B. Finkelstein (ii) AgF Cu
HCl
reaction
reaction
(C) + (iii) Gatterman
C. Swarts reaction (iii) NaI/dry N2Cl–
reaction
acetone HBF4
∆
D. Carbyl amine (iv) NaOH &
(D) + (iv) Sandmeyer
N2Cl–
reaction CaO,∆ reaction
C6H5OH
(1) (A-iv), (B-ii), (C-iii), (D-i) Alkaline medium

(2) (A-iv), (B-iii), (C-ii), (D-i) (1) (A)→ (iv), (B) → (iii), (C) → (ii), (D) → (i)
(3) (A-i), (B-iii), (C-ii), (D-iv) (2) (A)→ (iii), (B) → (iv), (C) → (i), (D) → (ii)
(4) (A-iii), (B-ii), (C-i), (D-iv) (3) (A)→ (ii), (B) → (iii), (C) → (iv), (D) → (i)
(4) (A)→ (i), (B) → (ii), (C) → (iii), (D) → (iv)

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22. Match the column :- 24. Match the column-I with Column-II
Column-I Column-II
Column-I Column-
(Reaction) (Process)
Compound II pKb (a) NH2 (p) Acetylation

value +CH3–CH2–Cl
(A) Benzenamine (p) 9.38
(b) NH2 (q) Alkylation
(B) Phenyl methanamine (q) 4.70
+CH3–C–Cl
(C) Ethanamine (r) 3.29
O
(D) N, N-dimethyl aniline (s) 8.92 (c) NH2 (r) Diazotization

(1) A-p, B-q, C-r, D-s +CHCl3+Alc.KOH

(2) A-q, B-p, C-r, D-s
(d) NH2 (s) Carbylamine
(3) A-p, B-q, C-s, D-r reaction
0°–5°C
+HNO2
(4) A-r, B-s, C-q, D-p
(t) Halogenation
23. Match List -I and List-II Correct match is
(1) (a) → (q), (b) → (p), (c) → (s), (d) → (r)
List-I List-II
(2) (a) → (t), (b) → (p), (c) → (s), (d) → (r)
(a) O (i) Br2/NaOH (3) (a) → (q), (b) → (p), (c) → (t), (d) → (s)
R–C–Cl→R–CHO (4) (a) → (t), (b) → (q), (c) → (s), (d) → (r)

(b) R–CH2–COOH→ (ii) H2/Pd– 25. Match the column

R–CH–COOH Column-I Column-II
BaSO4
Reaction Name/Process
Cl
(a) + – Cu Cl
(P) Gattermann
(c) O (iii) Zn(Hg)/Conc. ArN2X HCl2 2 reaction
R–C–NH2→R–NH2 HCl (b) + – Cu/HCl
(Q) Coupling
ArN2X reaction
(d) O (iv) Cl2/Red P, H2O (c) (R) Sandmeyer
+ – H PO
R–C–CH3→R–CH2–CH3 ArN2X H32O 2 reaction
Choose the correct answer from the options (d) + – Ph–OH
(S) Reduction
ArN2X 0°–5°C
given below : Correct match is –
(1) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) (1) (a) → (R), (b) → (P), (c) → (S), (d) → (Q)
(2) (a) → (R), (b) → (P), (c) → (Q), (d) → (S)
(2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(3) (a) → (P), (b) → (R), (c) → (S), (d) → (Q)
(3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) (4) (a) → (Q), (b) → (R), (c) → (P), (d) → (S)

(4) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

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Exercise 4 (Previous Year's Questions)

1. In the reaction 5. The following reaction
NO2 NO2 H
NH2 N
A + NaOH
A is Cl
Br Br O
O
N2Cl
⊕ Θ
is known by the name: [RE-AIPMT - 2015]
[NEET UT - 2013]
(1) Acetylation reaction
(1) H+/H2O (2) HgSO4/H2SO4
(3) Cu2Cl2 (4) H3PO2 and H2O (2) Schotten-Baumann reaction
(3) Friedel-Craft’s reaction
2. Nitrobenzene on reaction with conc.
(4) Perkin’s reaction
HNO3/H2SO4 at 80–100°C forms which one of
the following products? [NEET UT - 2013]
(1) 1, 2, 4-Trinitrobenzene 6. Method by which Aniline cannot be prepared
(2) 1,2-Dinitrobenzene is: [RE-AIPMT - 2015]
(3) 1,3-Dinitrobenzene
(1) reduction of nitrobenzene with H2/Pd in
(4) 1,4-dinitrobenzene
ethanol

3. In the following reaction, the product (A) (2) potassium salt of phthalimide treated
[AIPMT - 2014] with chlorobenzene followed by
+
N≡NCl– NH2 hydrolysis with aqueous NaOH solution

is:- (3) hydrolysis of phenylisocyanide with

+ H+ (A)
(Yellow dye) acidic solution

(1) N=N—NH (4) degradation of benzamide with bromine

NH2
in alkaline solution

(2) N=N ⊕ Θ
N2BF4
NH2
NaNO2
(3) N=N 7. Cu/∆
product,

(4) N=N NH2 product of given reaction is [AIIMS - 2015]

F NO2
4. Which of the following will be most stable
(1) (2)
diazonium salt? [AIPMT - 2014]
+ –
(1) CH3N2 X
F F
(2) C6H5N2+X–
(3) CH3CH2N2+X–
(3) (4)
(4) C6H5CH2N2+X– NO2
NO2

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8. Which of the following reactions is 11. Which of the following amine will give the
carbylamines test? [NEET(UG)-2020]
appropriate for converting acetamide to
NHC2H5 NH2
methanamine? [NEET - 2017]
(1) (2)
(1) Carbylamine reaction
NHCH3 N(CH3)2
(2) Hoffmann hypobromite reaction
(3) (4)
(3) Stephens reaction

(4) Gabriel phthalimide synthesis

12. Reaction of propanamide with sodium
hydroxide and bromine will give
[NEET(UG)-2020(Covid-2019)]
9. The major product of the following reaction (1) Ethylamine (2) Methylamine
is: [NEET(UG)-2019] (3) Propylamine (4) Aniline

COOH Strong heating 13. Identify the compound that will react with
+NH3 Hinsberg’s reagent to give a solid which
COOH
dissolves in alkali. [NEET(UG)-2021]
O CH2 • •
(1) CH3 NO2
COOH NH
(1) (2) CH2 •• CH3
(2) CH3 NH
CONH2 O
CH2 ••
COOH NH2 (3) CH3 NH2
(3) (4) CH2 •• CH2
NH2 NH2 (4) CH3 N CH3
CH3

10. The amine that reacts with Hinsberg’s 14. Given below are two statements :
reagent to give an alkali insoluble product is: [NEET-2022]
Statements I : Primary aliphatic amines react
[NEET(UG)-(Odisha)2019] with HNO2 to give unstable diazonium salts.
Statements II: Primary aromatic amines
(1) CH3–CH–NH–CH–CH3
react with HNO2 to form diazonium salts
CH3 CH3 which are stable even above 300 K.
In the light of the above statements, choose
CH2CH3
the most appropriate answer from the
(2) CH3–CH2–N–CH2CH3 options given below –
NH2 (1) Statement I is incorrect but Statement II
is correct
(3) CH3–C–CH2CH2CH3
(2) Both Statement I and Statement II are
CH3 correct
(3) Both Statement I and Statement II are
CH3
incorrect
(4) CH3–CCH–NH2
(4) Statement I is correct but Statement II is
CH3 CH3 incorrect

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15. The product formed from the following 18. The major product X formed in the following
reaction sequence is [NEET(UG)-2022] reaction sequence is :
CN (i) LiAIH4, H2O [RE-NEET-2024]
(ii) NaNO2, +HCl C2H5
(iii) H2O (i) Cl2/FeCl3
(ii) Sn/HCl
X
O (iii) NaNO2/HCl 273-278K
(iv) KI
⊕ Θ
N2Cl NO2
NH2
(1) (2)
C2H5 C2H5
Ι
(1) (2)
Cl OH Ι
(3) (4)
Cl Cl
16. Which of the following reaction will not give C2H5 C2H5
primary amine as the product – Cl
[NEET(UG)-2023] (3) (4)
Br2 /KOH
Cl
(1) CH3CONH2  → Product Ι Ι
(i)LIAlH4
(2) CH3CN 
(ii)H O⊕
→ Product
19. The compound that does not undergo
3

(i)LIAlH4
(3) CH3NC 
(ii)H3O⊕
→ Product Friedel-Crafts alkylation reaction but gives a
(i)LIAlH4
(4) CH3CONH2  → Product positive carbylamines test is :
(ii)H O⊕ 3
[RE-NEET-2024]
(1) aniline
17. Given below are two statements:
(2) pyridine
Statement I: Aniline does not undergo
(3) N-methylaniline
Friedel-Crafts alkylation reaction.
Statement II: Aniline cannot be prepared (4) triethylamine
through Gabriel synthesis. In the light of the
20. Which one of the following compounds does
above statements, choose the correct answer
from the options given below: not decolourize bromine water?

[NEET-2024] [NEET-2025]
(1) Statement I is correct, but Statement II is
(1) CH=CH2
false.
(2) Statement I is incorrect, but Statement II NH2
(2)
is true.
(3) Both Statement I and Statement II are
(3)
true.
(4) Both Statement I and Statement II are
(4) OH
false.

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21. Predict the major product 'P' in the following 22. Given below are two statements:
sequence of reactions – Statement I: Benzenediazonium salt is
prepared by the reaction of aniline with
[NEET-2025] nitrous acid at 273-278 K. It decomposes
(i) HBr, benzoyl peroxide easily in the dry state.
CH3 (ii) KCN
P Statement II: Insertion of iodine into the
(iii) Na (Hg)/C2H2OH (Major) benzene ring is difficult and hence
iodobenzene is prepared through the
CH3 reaction of benzenediazonium salt with KI.
(1) In the light of the above statements, choose
NC the most appropriate answer from the
NC options given below : [NEET-2025]
(2) CH3 (1) Statement I is correct but Statement II is
incorrect
(2) Statement I is incorrect but Statement II
CH3
is correct
(3) (3) Both Statement I and Statement II are
CH2NH2
correct
CH3 (4) Both Statement I and Statement II 4 are
(4)
CH2NH2 incorrect

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ANSWER KEYS

Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 2 3 3 3 3 2 2 3 4 2 4 4 4 2

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 2 4 3 3 2 3 3 2 3 3 3 2 2 2 2 2 3 2 3
Que . 21 22 23 24 25 26 27 28 29 30
Ans. 2 2 3 2 4 3 2 3 2 1

Exercise 1.3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 4 3 2 2 3 2 3 1 1 1 2 3 3 4

Exercise 2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 3 3 4 2 3 2 3 3 2 2 2 3 3 2 1 3 1 3 4
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 1 1 3 4 3 4 2 2 4 2 2 4 2 4 2 2 1 1 2
Que . 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 3 4 2 4 4 3 4 4 1 1 4 3 1 3 1 4 2 1 3

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 1 1 1 1 4 1 3 2 2 1 1 2 4 3 1 4 4 2 3
Que . 21 22 23 24 25
Ans. 1 1 3 1 1

Exercise 4 (Previous Year's Questions)

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 3 4 2 2 2 2 2 2 1 2 1 3 4 4 3 3 3 1 3
Que . 21 22
Ans. 3 3

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 Notes

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Chapter
Biomolecules CHEMISTRY

6 BIOMOLECULES

Chapter Summary Introduction

Complex organic compound which governs the common
• Introduction
activities of the living organism are called biomolecules. Living
• Classification of
systems are made up of various complex biomolecules like
Carbohydrates
carbohydrates, proteins, nucleic acids, lipids, etc. In addition,
• Monosaccharides
some simple molecules like vitamins and mineral salts also play
• Chemical reactions of
an important role in the functions of organisms.
Glucose
• Cyclic structure of Glucose Carbohydrates:
• Fructose (Ketohexose) Carbohydrates are primarily produced by plants and form a very
• Disaccharides large group of naturally occurring organic compounds.
• Polysaccharides Most of the carbohydrates have general formula Cx(H2O)y and
• Proteins considered as hydrates of carbon.
• Structure of Proteins Ex. C6(H2O)6 (Glucose, Fructose)
• Enzymes C12(H2O)11 (Maltose, Lactose, Sucrose)
• Vitamins But some compounds which have formula according to Cx(H2O)y
• Nucleic Acids are not known as carbohydrate

• Hormones Ex. CH2O (Formaldehyde)

C2(H2O)2 (Acetic acid)
C3(H2O)3 (Lactic acid)
Some carbohydrates are also known which do not have formula
according to Cx(H2O)y
Ex. C6H12O5 (Rhamnose)

Modern definition of carbohydrates/saccharides: poly hydroxy

aldehyde/ketones are known as carbohydrate

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Classification of Carbohydrates

Carbohydrate

Sugar Non sugar

• Sweet in taste • Taste less
• Crystalline • Amorphous
• Soluble in H2O • Insoluble in H2O

Polysaccharide
Mono Saccharide Oligo Saccharide
On hydrolysis it gives
• Simple sugar
large no. of mono
• Can't be hydrolysed Di Tri------Deca saccharide (> 10) Units
• Smallest Unit of sugar
On hydrolysis it gives Ex. (i) Starch
Ex. (i) Glucose '2' to '10' unit of mono
(ii) Fructose (ii) Cellulose
Saccharide
(iii) Galactose 2 ≤ Mono ≤ 10 (iii) Glycogen
(iv) Mannose Saccharide (iv) Gums
Ex. (i)Sucrose
etc
etc (ii) Maltose
(iii) Lactose

etc

Monosaccharides
A carbohydrate that cannot be hydrolysed further to give simplest unit of polyhydroxy aldehyde or
ketone is called monosaccharide.

Ex. Glucose, fructose, ribose, etc.

Monosaccharides are further classified on the basis of number of carbon atoms and the functional
group present in them.
If Aldehyde group is present in monosaccharide, then it is known as aldose.
If keto group is present in monosaccharide, then it is known as ketose.
Different types of Monosaccharides
Carbon Atom General term Aldehyde Ketone
3 Triose Aldotriose Ketotriose
4 Tetrose Aldotetrose Ketotetrose
5 Pentose Aldopentose Ketopentose
6 Hexose Aldohexose Ketohexose
7 Heptose Aldoheptose Ketoheptose

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Glucose (Aldohexose): Preparation of Glucose:
1. From sucrose (Cane sugar): If sucrose is boiled with dilute HCl or H2 SO4 in alcoholic solution, glucose
and fructose are obtained in equal amounts.
+
H
C12H22O11 + H2O  → C6H12O6 + C6H12O6
Sucrose Glucose Fructose

2. From starch: Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute H2 SO4
at 393 K under pressure.
H+
( C6H10O5 )n + nH2O nC6H12O6
Starch or cellulose 393K , 2 − 3 atm Glucose

Chemical reactions of Glucose

1. On prolonged heating with HI or Red/HI, it form n-hexane which confirm linear carbon chain in glucose
CHO
HΙ
(CH–OH)4 CH3 –CH2– CH2– CH2–CH2–CH3
∆
CH2OH n–hexane
2. Glucose react with NH2–OH (Hydroxylamine) to form an oxime
CHO CH=N–OH
(CHOH)4 H2N–OH (CHOH)4
CH2OH CH2OH

3. It react with HCN (Hydrogen cyanide) to give cyanohydrin

CN
CHO CH
OH
(CHOH)4 HCN (CHOH)4
CH2OH CH2OH
These reactions confirm the presence of a carbonyl group ( O ) in glucose

4. On reaction with Br2/H2O, gluconic acid is obtained since Br2/H2O is mild oxidizing agent can oxidize
only Aldehyde (Aldose)
CHO COOH
(CHOH)4 Br2/H2O (CHOH)4
CH2OH CH2OH
(Gluconic acid)
It means fructose doesn't react with Br2/H2O therefore glucose & fructose can’t be differentiated by
bromine water

5. On Acylation of glucose, glucose Penta acetate is obtained which confirm the presence of Five–OH
groups
CHO O CHO O
CH3–C–Cl/Py
(CHOH)4 (CH–O–C–CH3)4
or
CH2OH CH3–C–O–C– CH3 CH2–O–C–CH3
O O
O
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 CHEMISTRY Biomolecules
6. On reaction of glucose and gluconic acid with Nitric acid (HNO3), Both gives saccharic acid.
CHO COOH COOH
HNO3
(CHOH)4 HNO3 (CHOH)4 (CHOH)4
CH2OH 1°OH COOH CH2OH 1°OH
(D-Glucose) (Saccharic acid) (D-Gluconic acid)
It confirm the presence of 1°–OH
7. Reaction with phenyl hydrazine : (formation of osazone):
CHO C=N–NH–Ph
3 Ph-NH–NH2
CH–OH C=N–NH–Ph
(CH–OH)3 (CH–OH)3
CH2OH CH2OH
Glucose Glucosazone
Mechanism:
CH=O CH=N–NH–Ph CH–NH–NH–Ph
••
CH=NH
H2N-NH–Ph Tautomerism
CH–OH (–H2O)
CH–OH C–O–H C=O
•• –Ph–NH2
(CH–OH)3 (CH–OH)3 (CH–OH)3 (CH–OH)3
CH2OH CH2OH CH2OH CH2OH

CH=N–NH–Ph
2H2N–NH–Ph
C=N–NH–Ph –NH3
(CH–OH)3 –H2O

CH2OH
(Osazone)
Configuration in monosaccharides:
For assigning the configuration of monosaccharides, it is the last asymmetric carbon atom (as shown
below) which is compared. As in (+) glucose, -OH on the last asymmetric carbon is on the right hand
side which is comparable to (+) glyceraldehyde, so it is assigned D-configuration.
CHO
H–C–OH
HO–C–H
CHO H–C–OH
H OH *
H–C–OH
CH2OH CH2OH
D-(+)-Glyceraldehye D-(+)-Glucose
• 'D' and 'L' are relative configuration hence they don't have relation with the optical activity of the
compound.
• 'D' and 'L' are also not related with 'd' (+) and 'l'(-)
d→ dextrorotatory /(+) Direction of optical
l→ Laevorotatory /(–) rotation of the compound

• Glucose is a dextrorotatory
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Cyclic structure of Glucose
The following reactions and facts can’t be explained by open chain structure it can only explained by
cyclic structure of glucose
(i) Glucose doesn't give schiff's test
(ii) It doesn't react with NH3 (ammonia) and NaHSO3(sodium hydrogen sulphite)

6
CH2OH
5 schiff's
O No reaction
H reagent
H
NaHSO3 it means free –CHO
4 1
No reaction
OH H group is absent
HO NH3
3 2 OH No reaction
H OH
(iii) Glucose Penta acetate doesn't react with H2N–OH (hydroxyl amine) which indicate the absence of free
–CHO group
6 6
CH2OH CH2–OAC
5
O O 5
H O
H H
CH3–C–Cl H H2N–OH
4 1
(excess) 4 1 No reaction
OH H (excess)
HO OAC H
3 2 OH ACO 3 2 OAC
H OH H OAC
Note: It was found that glucose forms a six-membered ring in which -OH at C-5 is involved in ring formation.
This explains the absence of -CHO group and also existence of glucose in two forms as shown below.
These two cyclic forms exist in equilibrium with open chain structure.

O
1
1 1
H–C HO–C–H
H–C–OH 2
2
H
2
OH H OH
H OH O 3 O
HO 3 H HO H
HO 3 H 4
H 4 OH H 4 OH H OH
5
H 5 H 5 OH H
6
6
6
CH2OH CH2OH CH2OH
α-D–(+)–Glucose D–Glucose β-D–(+)–Glucose

36% 0.2% 63.8%

specific rotation +112° +52.7° +19°
(eq. mixture)
Anomers: The two sugars that differs in configuration only on the carbon that was the carbonyl carbon
in the open chain form is called as anomers and that carbon is called anomeric carbon.
α-glucose and β-glucose are anomers of glucose and their equilibrium mixture contains 36% α–D–
glucose, 63.8% β–D–glucose and 0.2% open chain form.
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Epimers: A pair of diastereomers that differ only in the configuration about of a single carbon atom are
said to be epimers. D(+)-glucose is epimeric with D(+) - Mannose and D(+)-galactose.
Epimers Epimers
O H O H O H
C C C
H–C–OH H–C–OH HO–C–H
HO–C–H HO–C–H HO–C–H
HO–C–H H–C–OH H–C–OH
H–C–OH H–C–OH H–C–OH
CH2OH CH2OH CH2OH
D - (+) - Galactose D - (+) - Glucose D - (+) - Mannose
Haworth Projection: The six membered cyclic structure of glucose is called pyranose structure (α- or β-), in
analogy with pyran. The cyclic structure of glucose is more correctly represented by Haworth structure
as given below.
6 6
CH2OH CH2OH
5
5
O O
O H H H OH
H H
4 1
4 1

OH H OH HO OH H H
pyran HO 3 3 2
2

H OH H OH
α-D–(+)–Glucopyranose β-D–(+)–Glucopyranose

Note: (i) Tollen's reagent and Fehling solution is used in Basic condition in which α & β– Form (Hemi acetal)
get decomposed and increase open chain aldehyde and oxidise this form into gluconic acid or we can
say that sugar reduce Tollen's and Fehling solution, hence it is known as reducing sugar
(ii) all monosaccharides are reducing sugar

Overview of reaction of glucose:

COOH
(CHOH)4 Br2/H2O Red P/HI
n-Hexane
CH2OH
CH3COCl
(Gluconic acid) Pentaacetate
NH2–OH
COONa Oxime
Tollen's
(CHOH)4 Reagent C6H12O6 HCN
Glucose cyanohydrin
Salts of CH2OH Glucose
gluconic 5HIO4
COONa HCHO + 5HCOOH
acids Fehling's
(CHOH)4 Schiff's reagent
Solution No reaction
CH2OH
NH3
COOH No reaction
HNO3 NaHSO3
(CHOH)4 No reaction
COOH
(Saccharic acid)

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Fructose (Ketohexose)
Structure of Fructose:
Fructose also has the molecular formula C6H12O6 and on the basis of its reactions it was found to
contain a ketonic functional group at carbon number 2 and six carbons in straight chain as in the case
of glucose. It belongs to D-series and is a laevorotatory compound. It is appropriately written as D-(-)-
fructose. Its open chain structure is as shown below.
CH2OH
C=O
HO–C–H
H–C–OH
H–C–OH
CH2OH
D-(–)-Fructose
It also exists in two cyclic forms which are obtained by the addition of -OH at C–5 to the C = O group.
The ring, thus formed is a five membered ring and is named as furanose with analogy to the compound
furan. Furan is a five membered cyclic compound with one oxygen and four carbon atoms.
1
CH2OH
2 2 1
1 2 C=O HO–C–CH2OH
O HOH2C–C–OH 3
HO
3
H O HO 3
H HO H O
H 4 OH H
4 H 4 OH
OH
Furan H 5 5
H 5
6 H OH 6

CH2OH 6 CH2OH
CH2OH
••
D-(–)-Fructose β-D–(–)–Fructofuranose
α-D–(–)–Fructofuranose

Haworth projections: The cyclic structures of two anomers of fructose are represented by Haworth
structures as given.
6 O 1 6 O
HOH2C CH2OH HOH2C OH
5
2 5
H HO H HO
2
H OH H CH2OH
4 3 1
4 3

OH H OH H
α-D–(–)–Fructofuranose β-D–(–)–Fructofuranose
(α–Furanose ring) (β–Furanose ring)

Disaccharides
The two monosaccharides are joined together by an oxide linkage formed by the loss of a water
molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic
linkage.
(i) Sucrose: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of
D- (+)-glucose and D-(-) fructose.
C12H22O11 + H2O 
→ C6H12O6 + C6H12O6
Surose D(+)-Glucose D-(-) Fructose

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These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose and C2
of β-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond
formation hence, sucrose is a non reducing sugar.

6
CH2OH
5
O 1

H H H HOH2C O H
4 1 2 5

O H HO CH2OH
HO OH
3
H
2 3 4
6
Glycosidic
H OH linkage
OH H
α-D–Glucose β-D–Fructose
Sucrose

Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose.
Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture
is laevorotatory. Thus, Hydrolysis of sucrose brings about a change in the sign of rotation, from dextro
(+) to laevo (-) and the product is named as invert sugar.
(ii) Maltose: Another disaccharide, maltose is composed of two α-D-glucose units in which C1 of one
glucose (I) is linked to C4 of another glucose unit (II). The free aldehyde group can be produced at C1 of
second glucose in solution and it shows reducing properties so it is a reducing sugar.

6 6
CH2OH CH2OH
5 O 5 O
H H H H H H
4 1 4 1

HO OH H O OH H OH
3 2 3 2

H OH H OH
(I) (II)
α-D–Glucose α-D–Glucose
Maltose
(iii) Lactose: It is more commonly known as milk sugar since this disaccharide is found in milk. It is
composed of β-D-galactose and β-D-glucose. The linkage is between C1 of galactose and C4 of glucose.
Hence it is also a reducing sugar.

6 6
CH2OH CH2OH
O 5 O
HO H
5
H H OH
4 1 O 4 1

H OH H H OH H H
3 2 3 2

H OH H OH
(I) (II)
β-D–Galactose β-D–Glucose
Lactose

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Polysaccharides
Polysaccharides contain a large number of monosaccharide units joined together by glycosidic
linkages. They mainly act as the food storage or structural materials.

(i) Starch: Starch is the main storage polysaccharide of plants. It is the most important dietary source for
human beings. High content of starch is found in cereals, roots, tubers and some vegetables. It is a
polymer of α-glucose and consists of two components-amylose and amylopectin
Amylose: Amylose is hot water soluble component which constitutes about 15-20 % of starch.
Chemically amylose is a long unbranched chain with 200-1000 α-D-(+)-glucose units held by C1 − C4
glycosidic linkage.
6
CH2OH CH2OH CH2OH
O 5 O O
H H H H H H H H H
4 1
4 1 4 1
O OH H O OH H O OH H O
3 2
H OH H OH H OH
α-Link α-Link
Amylose
Amylopectin: Amylopectin is insoluble in hot water and constitutes about 80–85% of starch. It is a
branched chain polymer of α-D-glucose units in which chain is formed by C1 − C4 glycosidic linkage
whereas branching occurs by C1 − C6 glycosidic linkage.
CH2OH CH2OH
O O
H H H H H H
4 1 4 1
O OH H O OH H

H OH H OH α-Link
O
Branch at C6

CH2OH 6
CH2 CH2OH
O O O
H H H H H H H H H
4 1 4 1 4 1
O OH H O OH H O OH H O

H OH H OH H OH
α-Link α-Link
Amylopectin
(ii) Cellulose: Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant
kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain
polysaccharide composed only of β-D-glucose units which are joined by glycosidic linkage between C1
of one glucose unit and C4 of the next glucose unit.

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HOH2C
H O
O
H
OH H
HOH2C H
O O H OH
H
H
OH H
HOH2C H
O OH
H O H
H β-Link
OH H
O
H
H OH Cellulose
(iii) Glycogen: The carbohydrates are stored in animal body as glycogen. It is also known as animal starch
because its structure is similar to amylopectin and is rather more highly branched. It is present in liver,
muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose.
Glycogen is also found in yeast and fungi.

Exercise 1.1
1. Carbohydrates may be: 6. A disaccharide which give only glucose on
(1) Sugars hydrolysis:
(2) Starch (1) Lactose (2) Fructose
(3) Polyhydroxy aldehyde/ ketones (3) Sucrose (4) Maltose
(4) All
2. Sucrose on hydrolysis gives: 7. Which of the following is not a reducing
(1) Glucose and galactose sugar-
(2) Maltose and galactose (1) Sucrose (2) Galactose
(3) Glucose and fructose (3) Glucose (4) Lactose
(4) None of these
8. All the following are composed exclusively
3. When glucose is heated with nitric acid, the of glucose except:
product is : (1) Lactose (2) Amylose
(1) Gluconic acid (2) Saccharic acid
(3) Cellulose (4) Maltose
(3) Glycolic acid (4) Oxalic acid
9. Glycogen is a branched chain polymer of α-
4. Ring structure of glucose is due to
D-glucose units in which chain is formed by
formation of hemiacetal and ring formation
C1—C4 glycosidic linkage whereas
between:
branching occurs by the formation of C1-C6
(1) C1 and C5 (2) C1 and C4
(3) C1 and C3 (4) C2 and C4 glycosidic linkage. Structure of glycogen is
similar to.
5. Which of the following is a ketohexose :
(1) Amylose (2) Amylopectin
(1) Glucose (2) Fructose
(3) Cellulose (4) Glucose
(3) Sucrose (4) Starch

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10. Which of the following monosaccharides 14. Which of the following statement is not
are present as five membered cyclic true for glucose?
structure (furanose structure)? (1) Glucose exists in two crystalline forms
(i) Ribose (ii) Glucose α-D-glucose and β-D-glucose
(iii) Fructose (iv) Galactose (2) Glucose gives Schiff’s test for aldehyde
(1) iii, iv only (2) i, ii, iii only (3) Glucose reacts with hydroxylamine to
(3) i, iii only (4) i, iv only form oxime
(4) The pentaacetate of glucose does not
11. Benedict’s solution is not reduced by :
react with hydroxylamine to give oxime
(1) Glucose (2) Fructose
(3) Maltose (4) Sucrose
15. Which of the following statements is not
12. Which of the following polymer is stored in true about glucose?
the liver of animals? (1) It is an aldohexose.
(1) Amylose (2) Cellulose (2) On heating with HI it forms n-hexane.
(3) Amylopectin (4) Glycogen (3) It is present in furanose form.
(4) It does not give 2,4-DNP test.
13. Select the incorrect statement.
(1) Starch is a homopolymer of glucose
forming an -glycosidic chain.
(2) Maltose and Lactose are
disaccharidase.
(3) Cellulose has only D-glucose unit
which are joined by glycosidic linkage
between C1 of one glucose unit and C4
of another glucose unit.
(4) The structure of glycogen is similar to
amylose.

Proteins
• The word protein is derived from Greek word, "proteios" which means primary or of prime importance.
• Proteins are the most abundant biomolecules of the living system. Chief sources of proteins are milk,
cheese, pulses, peanuts, fish, meat etc.
• All proteins are polymers of α-amino acids.
Amino acids : Amino acids contain amino ( −NH2 ) and carboxyl (-COOH ) functional groups. Depending
upon the relative position of amino group with respect to carboxyl group, the amino acids can be
classified as α, β, γ, δ and so on. Only α-amino acids are obtained on hydrolysis of proteins. They may
contain other functional groups also.

side chain R—CH—COOH Carboxylic group

NH2 amino group
α-amino acid

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COOH
Table : Natural Amino Acids H2N H
R
Name of the Characteristic Three letter One letter code
amino acids feature of side Symbol
chain, R
1. Glycine H Gly G
2. Alanine –CH3 Ala A
3. Valine* (H3C)2 CH– Val V
4. Leucine* (H3C)2 CH–CH2– Leu L
5. Isoleucine* H3C–CH2–CH– Ιle I
CH3
6. Arginine* HN=C–NH–(CH2)3– Arg R
NH2
7. Lysine* H2N–(CH2)4– Lys K
8. Glutamic acid HOOC–CH2–CH2– Glu E
9. Aspartic acid HOOC–CH2– Asp D
10. Glutamine O Gln Q
H2N–C–CH2–CH2–
11. Asparagine O Asn N
H2N–C–CH2–
12. Threonine* H3C–CHOH– Thr T
13. Serine HO–CH2– Ser S
14. Cysteine HS–CH2– Cys C
15. Methionine* H3C–S–CH2–CH2– Met M
16. Phenylalanine* C6H5–CH2– Phe F
17. Tyrosine (p) HO–C6H4–CH2– Tyr Y
18. Tryptophan* –CH2 Trp W

N
H
19. Histidine* –CH2 His H
NH

N
(a)
20. Proline a Pro P
COOH
HN H
CH2

* essential amino acid, a= entire structure

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Classification of Amino Acids: Amino acids are classified as acidic, basic or neutral depending upon
the relative number of amino and carboxyl groups in their molecule.
• The amino acids, which can be synthesized in the body, are known as non-essential amino acids.

Ex. (1) Glycine (2) Alanine (3) Tyrosine (4) Serine

(5) Asparagine (6) Proline (7) Glutamine (8) Cysteine
(9) Aspartic acid (10) Glutamic acid

• Those amino acids which cannot be synthesized in the body and must be obtained through diet, are
known as essential amino acids.

Ex. (1) Valine (2) Leucine (3) Isoleucine (4) Phenylalanine (5) Tryptophan
(6) Threonine (7) Methionine (8) Lysine (9) Arginine (10) Histidine

• Amino acids are usually colourless, crystalline solids. These are water-soluble, high melting solids and
behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence
of both acidic (carboxyl group) and basic (amino group) groups in the same molecule.

O O
R–CH–C–O–H R–CH–C–O–
+
• NH3
•NH2
(Zwitter ion)
• In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving
rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative
charges.

• In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids and bases.
• Except glycine, all other naturally occurring α-amino acids are optically active, since the a carbon atom
is asymmetric. These exist both in 'D' and 'L' forms. Most naturally occurring amino acids have
L-configuration. L-Aminoacids are represented by writing the −NH2 group on left hand side.

Structure of Proteins
Proteins are the polymers of α-amino acids and they are connected to each other by peptide bond or
peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and −NH2
group.
H2N–CH2–COOH + H2N–CH–COOH
–H2O CH 3

H2N–CH2–CO–NH –CH–COOH
Peptide linkage CH3
Glycylalanine [Gly-Ala]

The reaction between two molecules of similar or different amino acids, proceeds through the
combination of the amino group of one molecule with the carboxyl group of the other. This results in
the elimination of a water molecule and formation of a peptide bond −CO − NH − .

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Proteins can be classified into two types on the basis of their molecular shape.
(a) Fibrous proteins:
When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds,
then fibre- like structure is formed. Such proteins are generally insoluble in water. Some common
examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc.

(b) Globular proteins:

• This structure results when the chains of polypeptides coil around to give a spherical shape. These are
usually soluble in water. Insulin and albumins are the common examples of globular proteins.
• Structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary
and quaternary, each level being more complex than the previous one.
(i) Primary structure of proteins:
Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked
with each other in a specific sequence and this sequence of amino acids that is said to be the primary
structure of that protein. Any change in this primary structure (i.e., the sequence of amino acids)
creates a different protein.
(ii) Secondary structure of proteins:
• The secondary structure of protein refers to the shape in which a long polypeptide chain can exist.
• They are found to exist in two different types of structures viz. α-helix and β-pleated sheet structure.
• These structures arise due to the regular folding of the backbone of the polypeptide chain due to
O
hydrogen bonding between –C– and –NH- groups of the peptide bond.
(A) α-Helix: α-Helix is one of the most common ways in which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed screw (helix) with the - NH group of each amino acid
residue hydrogen bonded to the C = O of an adjacent turn of the helix as shown in figure.

C
O •
•
••
C
C •
•
O
••
•
H O
••
•
C H
O • N
••
• •• CN
H O•• ••
•• C
H C N •• •
• H O••
•

N O ••
•
N H
••

H N
••
N
••

Fig. -Helix structure of proteins

(B) β-pleated sheet: In β-pleated sheet structure, all peptide chains are stretched out to nearly maximum
extension and then laid side by side which are held together by intermolecular hydrogen bonds. The
structure resembles the pleated folds of drapery and therefore is known as β-pleated sheet.

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N N
RCH RCH N
H C O C
O
RCH
N
HCR
H N H N C O
HCR HCR
C C
O O
N H O C
RCH N H N H
RCH
C O C RCH
H N O
H N H N C O
HCR HCR
C HCR
C C

(β–Pleated sheet structure of proteins)

(iii) Tertiary structure of proteins: The tertiary structure of proteins represents overall folding of the
polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular
shapes viz. fibrous and globular. The main forces which stabilise the 2° and 3° structures of proteins
are hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction.

(iv) Quaternary structure of proteins: Some of the proteins are composed of two or more polypeptide
chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other
is known as quatenary structure.
A diagrammatic representation of all these four structures is given in figure.

Primary Secondary Tertiary Quaternary

structure structure structure structure

Denaturation of Proteins:
• Physical change like change in temperature or chemical change like change in pH, the hydrogen bonds
are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological
activity. This is called denaturation of protein.
• The coagulation of egg white on boiling is a common example of denaturation. Another example is
curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.
• During denaturation secondary and tertiary structures are destroyed but primary structure remains
unchanged/intact.

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Exercise 1.2
1. Peptide linkage is - 6. Each polypeptide in a protein has
O O aminoacids linked with each other in a
(1) –C–O– (2) –C–NH2 specific sequence. This sequence of amino
O O acids is said to be.
(3) –C–NH– (4) –C–NH–NH2 (1) Primary structure of proteins.
(2) Secondary structure of proteins.
2. Select the correct statement. (3) Tertiary structure of proteins.
(1) Proteins upon hydrolysis gives α-amino (4) Quaternary structure of proteins.
acid.
(2) Except glycine, all other naturally 7. Out of the following which type of interaction
occurring α-amino acids are optically is responsible for the stabilisation of α-helix
active. structure of proteins ?
(3) In fibrous proteins polypeptide chains (1) Ionic bonding
are held together by hydrogen and (2) Hydrogen bonding
disulphide bonds. (3) Covalent bonding
(4) All (4) van der Waals forces

3. Which of the following is an essential 8. Which of the following statement is/are

correct for denaturation of proteins ?
amino acid?
(1) Can be caused by change in
(1) Serine (2) Cysteine
temperature.
(3) Glycine (4) Phenylalanine
(2) Can be caused by change in pH.
4. α-Helical structure refers to the: (3) During denaturation secondary and
(1) Primary structure of protein tertiary structure are destroyed but
(2) Secondary structure of protein primary structure remains intact.
(3) Tertiary structure of protein (4) All of the above are correct.
(4) Quaternary structure of proteins 9. Proteins can be classified into two types on
5. Amino acids are classified as acidic, basic the basis of their molecular shape i.e.,
or neutral depending upon the relative fibrous proteins and globular proteins.
number of amino and carboxyl groups in Examples of globular proteins are :
their molecule. Which of the following are (i) Insulin (ii) Keratin
acidic? (iii) Albumin (iv) Myosin
(i) (CH3)2CH–CH2–COOH (1) i, iii only (2) i, ii, iii only
NH2 (3) iii, iv only (4) ii, iv only

10. Lysine, H2N—(CH2)4—CH—COOH

(ii) HOOC–CH2–CH2–CH–COOH
NH2
NH2
is.
(iii) H2N–CH2–CH2–CH2–COOH (i) α-Amino acid
(iv) HOOC–CH2–CH–COOH (ii) Basic amino acid
(iii) Amino acid synthesised in body
NH2
(iv) β-Amino acid
(1) i, ii, iii (2) ii, iii (1) i, ii only (2) i, ii, iii only
(3) ii, iv (4) ii only (3) ii, iv only (4) iii, iv only
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11. Which of the following is non-essential 13. Amino acids generally exist in the form of
amino acid? zwitter ions. This means they contain
(1) Glycine (2) Valine (1) basic —NH2 group and acidic -COOH
group
(3) Leucine (4) Threonine ⊕
(2) the basic − NH3 group and acidic—COOΘ
12. Identify the essential amino acids from group
below: (3) basic —NH2 and acidic —H⊕ group
(A) Valine (B) Proline ⊕
(4) basic —COOΘ group and acidic − NH3
(C) Lysine (D) Threonine
group.
(E) Tyrosine
Choose the correct answer from the options 14. Keratin, a structural protein is present in
given below: (1) hair (2) wool
(3) silk (4) all of these.
(1) (A), (C) and (D) only
(2) (A), (C) and (E) only 15. Most common types of secondary
structures of proteins are
(3) (B), (C) and (E) only
(1) α-helix and β-helix structures
(4) (C), (D) and (E) only
(2) α-helix and β-pleated sheet structures
(3) right and left hand twisted structures
(4) globular and fibrous structures.

Enzymes
Enzymes are biological catalysts. almost all enzymes are globular proteins. enzyme are very specific
for a particular reaction and for a particular substrate. Some important enzymes and their functions
are given below:
Enzyme Reaction catalysed
(i) Invertase or sucrase Sucrose 
→ Glucose + Fructose
(ii) Maltase Maltose 
→ Glucose + Glucose
(iii) Lactase Lactose 
→ Glucose + Galactose
(iv) Amylase Starch 
→ n  Glucose
(v) Pepsin Proteins 
→ Amino acids
(vi) Trypsin Proteins 
→ Amino acids
(vii) Nucleases DNA or RNA  → Nucleotides
(viii) DNA polymerase Deoxynucleotide triphosphates DNA
(ix) RNA polymerase Ribonucleotide triphosphates RNA

Vitamins
• Certain organic compounds are required in small amounts in our diet but their deficiency causes
specific diseases. These compounds are called vitamins.
• Most of the vitamins cannot be synthesised in our body but plants can synthesise almost all of them,
so they are considered as essential food factors. However, the bacteria of the gut can produce some of
the vitamins required by us.
• All the vitamins are generally available in our diet. Different vitamins belong to various chemical
classes and it is difficult to define them on the basis of structure. They are generally regarded as organic

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compounds required in the diet in small amounts to perform specific biological functions for normal
maintenance of optimum growth and health of the organism.
• Vitamins are designated by alphabets A, B, C, D, etc. Some of them are further named as sub-groups
e.g. B1, B2, B6, B12, etc. Excess of vitamins is also harmful and vitamin pills should not be taken without
the advice of doctor.
• The term “Vitamine” was coined from the word vital + amine since the earlier identified compounds
had amino groups. Later work showed that most of them did not contain amino groups, so the letter ‘e’
was dropped and the term vitamin is used these days.
• Vitamins are classified into two groups depending upon their solubility in water or fat.
(i) Fat soluble vitamins: Vitamins which are soluble in fat and oils but insoluble in water are kept in this
group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues.
(ii) Water soluble vitamins: B group vitamins and vitamin C are soluble in water so they are grouped
together. Water soluble vitamins must be supplied regularly in diet because they are readily excreted
in urine and cannot be stored (except vitamin B12) in our body.

Some important vitamins, their sources and their deficiency diseases

S.No. Vitamin Sources Deficiency diseases
1 Vitamin A Fish liver oil, carrots, Xerophthalmia (hardening of
butter and milk cornea of eye) Night blindness
2 Vitamin B1 Yeast, milk, green Beri-Beri (loss of appetite,
(Thiamine) vegetables and cereals. retarded growth
3 Vitamin B2 Milk, egg white, liver, Cheilosis (fissuring at corners
(Riboflavin) kidney. of mouth and lips), digestive
disorders and burning
sensation of the skin
4 Vitamin B6 Yeast, milk, egg yolk, Convulsions
(Pyridoxine) cereals and grams
5 Vitamin B12 Meat, fish, egg and curd Pernicious anaemia (RBC
deficient in haemoglobin)
6 Vitamin C Citrus fruits, amla and Scurvy (bleeding gums)
(Ascorbic acid) green leafy vegetables
7 Vitamin D Exposure to sunlight, fish Rickets (bone deformities in
and egg yolk children) and osteomalacia
(soft bones and joint pain in
adults)
8 Vitamin E Vegetable oils like wheat Increased fragility of RBCs
germ oil, sunflower oil, and muscular weakness
etc.
9 Vitamin K Green leafy vegetables Increased blood clotting time

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Nucleic Acids
Introduction:
It has been observed that nucleus of a living cell is responsible for the transmission of inherent
characters, also called heredity. The particles in nucleus of the cell, responsible for heredity, are called
chromosomes which are made up of proteins and another type of biomolecules called nucleic acids.
These are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Since
nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides.

Chemical composition of Nucleic Acids:

Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric acid and nitrogen containing
heterocyclic compounds (called bases). In DNA molecules, the sugar moiety is β-D-2-deoxyribose
whereas in RNA molecule, it is β-D-ribose.

5 O 5 O
HOH2C OH HOH2C OH
4 1 4 1
H H H H H H H H
3 2 3 2

OH OH OH H
β-D–ribose β-D–2-deoxyribose

DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also contains
four bases, the first three bases are same as in DNA but the fourth one is uracil (U).

NH2 O
N C N C
HC C N C NH
HC
N C CH
N C C–NH2
H N N
H
Adenine(A) Guanine(G)
NH2 O O
H3C
C C C
HC C NH HC NH
N
HC C HC C HC C
N O N O N O
H H H
Cytosine (C) Thymine (T) Uracil (U)

Structure of Nucleic Acids:

Nucleoside: A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside.
It contain only two components: a pentose sugar and a nitrogenous base.
Nucleotide: When nucleoside is linked to phosphoric acid at 5' -position of sugar moiety, we get a
nucleotide.
It contain all the three components: a pentose sugar, a nitrogenous base and a phosphoric acid group.

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O
5'
HOH2C O Base – 5'
O
O–P–O–H2C Base
4' 1' –O 4' 1'
H H H H H H H H
3' 2'
3' 2'

OH OH OH H
nucleoside nucleotide
Nucleotides are joined together by phosphodiester linkage between 5' and 3' carbon atoms of the
pentose sugar. The formation of a typical dinucleotide is shown in below Fig.

5' end of chain

O
5' BASE
O –O–P–O–CH2
5' BASE O
–O–P–O–CH2 O O– 1'
SUGAR
–
O SUGAR 1' 3'

3' O
OH Phosphodiester –O–P=O
+ linkage
O BASE
O
5' BASE 5' CH2 O
–O–P–O–CH2 O 1'
O– SUGAR
1'
SUGAR
3'
3'
OH
OH

A simplified version of nucleic acid chain is as shown below.

5' 3'
Base Base Base
Sugar Phosphate Sugar G≡C
Phosphate Sugar A=T
n T=A
A=T
Information regarding the sequence of nucleotides in the chain of a nucleic C≡G
acid is called its primary structure. Nucleic acids have a secondary T=A
G≡C
structure also. James Watson and Francis Crick gave a double strand helix C≡G
structure for DNA (Fig.). Two nucleic acid chains are wound about each
other and held together by hydrogen bonds between pairs of bases. The two T=A
G≡C
A=T
strands are complementary to each other because the hydrogen bonds are T=A
formed between specific pairs of bases. Adenine forms hydrogen bonds
with thymine whereas cytosine forms hydrogen bonds with guanine. A=T
T=A
In secondary structure of RNA, helics are present which are only single C≡G
G≡C
stranded. Sometimes they fold back on themselves to form a double helix
structure. RNA molecules are of three types and they perform different
functions. They are named as messenger RNA (m-RNA), ribosomal RNA 5' 3'
(r-RNA) and transfer RNA (t-RNA). Double strand helix
structure for DNA

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DNA Fingerprinting:
It is known that every individual has unique fingerprints. These occur at the tips of the fingers and have
been used for identification for a long time but these can be altered by surgery. A sequence of bases on
DNA is also unique for a person and information regarding this is called DNA fingerprinting. It is same for
every cell and cannot be altered by any known treatment. DNA fingerprinting is now used
(i) in forensic laboratories for identification of criminals.
(ii) to determine paternity of an individual.
(iii) to identify the dead bodies in any accident by comparing the DNA’s of parents or children.
(iv) to identify racial groups to rewrite biological evolution.

Biological Functions of Nucleic Acids:

DNA is the chemical basis of heredity and may be regarded as the reserve of genetic information. DNA is
exclusively responsible for maintaining the identity of different species of organisms over millions of
years. A DNA molecule is capable of self duplication during cell division and identical DNA strands are
transferred to daughter cells. Another important function of nucleic acids is the protein synthesis in the
cell. Actually, the proteins are synthesised by various RNA molecules in the cell but the message for the
synthesis of a particular protein is present in DNA.

Hormones
Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands
in the body and are poured directly in the blood stream which transports them to the site of action.
Chemical Nature of Hormones:
1. Steroids:- Estrogens, androgens
2. Polypeptides:- Insulin, endorphins
3. Amino acid derivatives:- Epinephrine, norepinephrine

• The role of insulin is keeping the blood glucose level within the narrow limit. Insulin is released in
response to the rapid rise in blood glucose level. Hormone glucagon tends to increase the glucose level
in the blood. The two hormones together regulate the glucose level in the blood.
• Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex
hormones play role in growth and development.
• Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine. Abnormally
low level of thyroxine leads to hypothyroidism which is characterised by lethargyness and obesity
Increased level of thyroxine causes hyperthyroidism. Low level of iodine in the diet may lead to
hypothyroidism and enlargement of the thyroid gland. This condition is largely being controlled by
adding sodium iodide to commercial table salt (“Iodised” salt).
• Steroid hormones are produced by adrenal cortex and gonads (testes in males and ovaries in females).
Glucocorticoids control the carbohydrate metabolism, modulate inflammatory reactions, and are
involved in reactions to stress. The mineralocorticoids control the level of excretion of water and salt
by the kidney. If adrenal cortex does not function properly then one of the results may be Addison’s
disease characterised by hypoglycemia, weakness and increased susceptibility.
• Hormones released by gonads are responsible for development of secondary sex characters.
Testosterone is the major sex hormone produced in males. It is responsible for development of
secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is
the main female sex hormone. It is responsible for development of secondary female characteristics
and participates in the control of menstrual cycle. Progesterone is responsible for preparing the uterus
for implantation of fertilised egg.

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Exercise 1.3
1. Which of the following bases is not present 9. Bases common to RNA and DNA are
in DNA ? (1) adenine, guanine, cytosine
(1) Cytosin (2) Gaunine (2) adenine, uracil, cytosine
(3) Uracil (4) Adenine (3) adenine, guanine, thymine
2. In DNA, the complimentary bases are : (4) guanine, uracil, thymine.
(1) Adenine and thymine; guanine and uracil 10. A unit in nucleic acid which contains 'base-
(2) Adenine and guanine; thymine and sugar-phosphate' unit is called
cytosine
(1) nucleotide (2) nucleoside
(3) Uracil and adenine; cytosine and
(3) phosphotide (4) polypeptide.
guanine
(4) Adenine and thymine; guanine and 11. Which of the following statements is not true
cytosine about RNA?
3. Nucleic acids are the polymers of. (1) It is present in the nucleus of the cell.
(1) Nucleosides (2) Nucleotides (2) It has always double stranded α-helix
(3) Bases (4) Sugars structure.
(3) It controls the synthesis of protein.
4. Which one of the following vitamins is water
(4) It usually does not replicate.
soluble :
(1) Vitamin A (2) Vitamin B 12. A metal present in vitamin B-12 is:
(3) Vitamin E (4) Vitamin K (1) Aluminium (2) Zinc
5. Deficiency of vitamin B1 causes the disease: (3) Iron (4) Cobalt
(1) Cheilosis (2) Sterility 13. Which of the following is/are incorrect?
(3) Convulsions (4) Beri-Beri
(1) Nucleoside is linked to phosphoric acid
6. Which one of the following is a peptide at 5 position of sugar moiety get
hormone: nucleotide.
(1) Glucagon (2) Testosterone (2) DNA contain four bases adenine
(3) Thyroxin (4) Adrenaline guanine, cytosine and thymine.
7. Which of the following vitamin is helpful in (3) RNA contain four bases adenine
delaying the blood clotting: guanine, cytosine and uracil.
(1) Vitamin C (2) Vitamin B (4) Glucose form the hydrogen sulphite
(3) Vitamin E (4) Vitamin K addition product with NaHSO3
8. Vitamin C must be supplied regularly in diet 14. The monomer of nucleic acids are held
because together by :
(1) it is water soluble hence excreted in (1) Phosphodiester linkage
urine and can't be stored in the body (2) Amide linkage
(2) it is fat soluble hence stored in the body
(3) Glycosidic linkage
and cannot be used on regular basis
(4) Peptide linkage
(3) it is required in a large amount by the
body hence supplied regularly 15. Insulin is secreted by:
(4) it is water soluble hence used by the (1) Pancreas (2) Stomach
body on daily basic and it to be supplied (3) Thyroid (4) Adrenal medulla
regularly.

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Exercise 2
1. Starch molecules are polymer with
H OH HO H
repeating glucose units. Select the correct H OH H OH
(3) HO H O HO H O
statement(s).
(1) Glucose units are joined through - H OH H OH
H H
glycosidic linkage.
CH2OH CH2OH
(2) The branches of amylopectin are linked
to the chain with -1,6’-glycosidic H OH HO H
H OH HO H
linkages. (4) HO H O H H O
(3) Amylose has an unbranched skeleton of H OH HO OH
glucose molecules with -1,4’ glycosidic H H
linkages. CH2OH CH2OH
(4) All of the above 4. In disaccharides, if the reducing groups of
monosaccharides i.e. aldehydic or ketonic
2. Which of the following reactions of glucose groups are bonded, these are non-reducing
can be explained only by its cyclic structure. sugars. Which of the following disaccharide
(1) Glucose forms penta acetate is a non-reducing sugar?
(2) Glucose reacts with hydroxylamine to CH2OH CH2OH
H O H H H
form an oxime
H H
(3) Penta acetate of glucose does not react (1) OH H OH H
HO O OH
with hydroxylamine
H OH H OH
(4) Glucose is oxidised by nitric acid to
CH2OH
gluconic acid H O HOH2C O H
H
H
3. Which of the following pairs represents (2) OH H OH H
HO O CH2OH
anomers?
H OH OH H
CHO CHO
CH2OH CH2OH
H OH HO H
HO H HO O O OH
HO H HO
H O H
(1) H OH H OH OH H OH H
(3) H
H OH H OH H H
CH2OH CH2OH H OH H OH

CHO CHO CH2OH CH2OH

H OH HO H H O O OH
H
HO H H OH H O H
(2) H OH HO H (4) OH H OH H
H H
HO
H OH HO H
H OH H OH
CH2OH CH2OH

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5. Three cyclic structures of monosaccharides 10. D(+) glucose react with hydroxyl amine and
are given below which of these are anomers. yield an oxime. The structure of the oxime
would be
H OH HO H HO H
H OH H OH HO H CH=N–OH CH=N–OH
HO H O HO H O HO H O H OH
H OH HO H
H OH H OH HO H H OH HO H
H H H (1) (2)
HO H H OH
CH2OH CH2OH CH2OH H OH H OH
(I) (II) (III) CH2OH
CH2OH
(1) I and II
CH=N–OH CH=N–OH
(2) II and III H OH HO H
(3) I and III HO H HO H
(4) III is anomer of I and II (3) (4)
HO H H OH
H OH H OH
6. Which of the following carbohydrates are
branched polymer of glucose? CH2OH CH2OH
(i) Amylose (ii) Amylopectin
11. Identify the incorrect statement from the
(iii) Cellulose (iv) Glycogen
following
(1) ii, iv only (2) i, ii, iv only
(1) Amylose is a branched chain Polymer of
(3) iii, iv only (4) ii, iii, iv only glucose
7. Amylopectin is composed of : (2) Starch is a Polymer of α-D glucose
(1) α-D–Glucose, C1–C4 and C1–C6 (3) β–Glycosidic linkage makes cellulose
glycosidic linkage polymer
(2) α-D–Glucose, C1–C4 glycosidic linkage (4) Glycogen is called as animal starch
(3) β-D–Glucose, C1–C4 and C1–C6 12. Which of the following is non reducing sugar
glycosidic linkage (1) Starch (2) Cellulose
(4) β-D–Glucose, C1–C4 glycosidic linkage (3) Sucrose (4) All

8. Identify the incorrect statement from the 13. Which of the following statement is not true
following: about glucose?
(1) Glycogen is called as animal starch (1) It is an aldohexose
(2) Glycosidic linkage makes cellulose (2) On heating with HI, it forms n-hexane
polymer (3) It does not give schiff’s test
(3) Amylose is a branched chain polymer of (4) It is present in furanose form
glucose 14. Which of the glycosidic linkage between
(4) Starch is a polymer of D-glucose galactose and glucose is present in lactose?
9. When glucose is heated with nitric acid, the (1) C–1 of galactose and C-4 of glucose
product is (2) C–1 of glucose and C-6 of galactose
(1) Gluconic acid (2) Saccharic acid (3) C–1 of glucose and C-4 of galactose
(3) n-hexane (4) oxalic acid (4) C–1 of galactose and C-6 of glucose

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15. On oxidation with a mild oxidising agent like 21. Starch is composed of two polysaccharides
Br2/H2O, the glucose is oxidised to which are
(1) saccharic acid (2) glucaric acid (1) amylopectin and glycogen
(3) gluconic acid (4) valeric acid. (2) amylose and glycogen
(3) amylose and amylopectin
16. During acetylation of glucose it needs x (4) cellulose and glycogen.
moles of acetic anhydride. The value of x
22. Carbohydrates are stored in human body as
would be
the polysaccharide
(1) 3 (2) 5 (3) 4 (4) 1 (1) starch (2) glycogen
17. Five-membered ring structures of fructose (3) cellulose (4) amylose.
are given below. Mark the incorrect 23. Amino acids are classified as acidic, basic or
statement. neutral depending upon the relative number
HOH2C O CH2OH HOH2C O OH
of amino and carboxyl groups in their
molecule. Which of the following are basic?
H HO H HO (i) (CH3)2CH—CH—COOH
H OH H CH2OH
OH H OH H NH2
(ii) NH2—CH2— CH2— CH—COOH
(1) The five-membered ring structures are
named as furanose structures. NH2
(2) The cyclic structures represent two (iii) H2N—CH2—CH2—CH2—COOH
(iv) HOOC—CH2— CH—COOH
anomers of fructose.
(3) Five-membered ring structures are NH2
(1) (ii) only (2) (iii) and (iv)
named as pyranose structures.
(3) (i) and (ii) (4) (ii) and (iii)
(4) These are also called Haworth
structures. 24. Which of the following statements is not
correct?
18. Which one of the following sets of (1) Only α-amino acids are obtained on
monosaccharides forms sucrose? hydrolysis of proteins.
(1) α–D-Galactopyranose and α-D-glucopyranose (2) The amino acids which are synthesised in
the body are known as non-essential
(2) α-D-Glucopyranose and β-D-fructofuranose
amino acids.
(3) β-D-Glucopyranose and α-D-fructofuranose (3) There are 20 essential amino acids.
(4) α-D-Glucopyranose and β-D-fructopyranose (4) L-amino acids are represented by writing
the —NH2 group on the left side.
19. Invert sugar is
(1) a type of cane sugar 25. Which of the following statements is not
(2) optically inactive form of sugar correct?
(1) Proteins are polyamides formed from
(3) mixture of glucose and galactose
amino acids.
(4) mixture of glucose and fructose in
(2) Except glycine, all other amino acids
equimolar quantities. show optical activity.
(3) Natural proteins are commonly made up
20. The glycosidic linkage involved in linking the
of L-isomer of amino acids.
glucose units in amylose part of starch is
(4) In α-amino acids, —NH2 and -COOH
(1) C1-C4 β-linkage (2) C1-C6 β-linkage groups are attached to different carbon
(3) C1-C6 α-linkage (4) C1-C4 α-linkage atoms.
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26. The peptide linkage formed between glycine 31. Which of the statements about
(NH2CH2COOH) & alanine (NH2CH—COOH) to “Denaturation” given below are correct ?
CH3 (a) Denaturation of proteins causes loss of
give glycylalanine can be shown as secondary and tertiary structures of the
(1) NH2—CH2—NH—CH—COOH) protein
CH3 (b) Denaturation leads to the conversion of
(2) NH2—CH2—CONH—CH—COOH) double strand of DNA into single strand.
CH3 (c) Denaturation affects primary structure
(3) H2NCOCH2—CH—CONH2 which gets distorted.
CH3 (1) (a), (b) and (c) (2) (b) and (c)
(4) HOOC—CH 2 —NH—NH—CH—COOH (3) (a) and (c) (4) (a) and (b)
CH3
32. Each polypeptide in a protein has amino
27. In fibrous proteins, polypeptide chains are acids linked with each other in a specific
held together by sequence. This sequence of amino acids is
(1) van der Waals' forces said to be _____.
(2) electrostatic forces of attraction
(1) primary structure of proteins
(3) hydrogen bonds
(2) secondary structure of proteins
(4) covalent bonds.
(3) tertiary structure of proteins
28. Globular proteins are present in (4) quaternary structure of proteins
(1) blood (2) eggs
(3) milk (4) all of these. 33. DNA and RNA contain four bases each.
Which of the following bases is not present
29. Denaturation of protein leads to loss of its
in RNA?
biological activity by
(1) formation of amino acids (1) Adenine (2) Uracil
(2) loss of primary structure (3) Thymine (4) Cytosine
(3) loss of both primary and secondary
34. Optical rotations of some compounds along
structure
with their structures are given below. Which
(4) loss of both secondary and tertiary
structures. of them have D-configuration?
CHO CH2OH
30. Mark the wrong statement about H OH C=O
denaturation of proteins. HO H HO H
CHO H OH H OH
(1) The primary structure of the protein does
not change. H OH H OH H OH
(2) Globular proteins are converted into CH2OH CH2OH CH2OH
fibrous proteins. (+) rotation (+) rotation (–) rotation
(3) Fibrous proteins are converted into (I) (II) (III)
globular proteins.
(1) I, II, III (2) II, III
(4) The biological activity of the protein is
(3) I, II (4) III
lost.

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35. Structure of a disaccharide formed by 39. Which of the following disaccharide is a
glucose and fructose is given below. Identify nonreducing sugar :
anomeric carbon atoms in monosaccharide CH2OH CH2OH
units. HO O O OH
H
6 (1) H O H
CH2OH OH H OH H
1 H H H
H 5 O HOH2C O H
H
4
H 1 2 5
H OH H OH
OH H OH HO
HO 3 2 O CH2OH CH2OH CH2OH
3 4 6
H OH OH H H O H H H
(2) H H
(1) '1' carbon of glucose and '1' carbon of OH H OH H
HO O OH
fructose.
H OH H OH
(2) '1' carbon of glucose and '5' carbon of
fructose. OH
CH2OH
(3) '1' carbon of glucose and '2' carbon of H O CH2 O H
H
fructose. (3) H
OH H H HO
(4) '6' carbon of glucose and '6' carbon of HO O CH2OH
fructose. H OH OH H
CH2OH CH2OH
36. The structure of protein that is unaffected by
HO O O OH
HO
heating is (4) H O H
OH H OH H
(1) secondary structure H H
H
(2) tertiary structure H OH H OH
(3) primary structure
40. Relation between (A) and (B) is :
(4) quaternary structure.
CH2OH CH2OH
37. The secondary structure of protein is H O H O OH
H
stabilised by H H
OH H OH OH H
(1) van der Waals' forces OH OH H
(2) peptide bond H OH H OH
(3) hydrogen bonding (1) Enantiomers
(4) glycosidic bond. (2) Anomers
(3) Positional isomers
38. Which of the given statements is incorrect
(4) Functional isomers
about glycogen?
CHO
(1) It is present in animal cells.
HIO4
41. (CH–OH)4 Product
(2) Only α-linkages are present in the
molecule. CH2–OH
(3) It is a straight chain polymer similar to (1) 5 HCHO+HCOOH
(2) 5HCOOH+HCHO
amylose.
(3) CH3COOH+5HCHO
(4) It is present in some yeast and fungi.
(4) 5CH3CHO+HCHO
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42. Which of the following is sulphur containing 44. Which of the following are non reducing in
amino acid : nature-
(A) Cellulose (B) Sucrose
(a) Methionine (C) Lactose (D) Starch
(b) Cysteine (1) A and B are correct
(c) Leucine (2) A, B and C are correct
(3) A, B and D are correct
(d) Glutamic acid (4) A, C and D are correct
(1) a, b, c (2) a, b
45. Which one of the following sets of
(3) b, c (4) c, d
monosaccharides forms sucrose ?
(1) D-Glucofuranose and -
43. Which of the following are essential
D-fructofuranose
aminoacids : (2) D-Glucopyranose and -
(1) Tryptophane, Lysine and Histidine D-fructopyranose
(3) D-Galactopyranose and -
(2) Serine, Lysine and Tryptophane
D-glucopyranose
(3) Serine, Histidine and Tryptophane (4) D-Glucopyranose and -
(4) Serine, Histidine and Lysine D-Fructofuranose

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Exercise 3
1. Assertion: -glycosidic linkage is present in 4. Assertion: Maltose and lactose are examples
maltose. of reducing sugars.
CH2OH CH2OH Reason: Maltose and lactose reduce
H O H H H Fehling's solution and Tollens' reagent.
H H (1) If both Assertion and Reason are correct
OH H OH H
HO O OH and the Reason is a correct explanation
H OH H OH of the Assertion.
Reason: Maltose is composed of two α-D- (2) If both Assertion and Reason are correct
glucose units in which C–1 of one glucose but Reason is not a correct explanation
unit is linked to C–4 of another glucose unit. of the Assertion.
(1) If both Assertion & Reason are True & the (3) If the Assertion is correct but Reason is
Reason is a correct explanation of the
incorrect.
Assertion.
(2) If both Assertion & Reason are True but (4) If the Assertion is incorrect and Reason
Reason is not a correct explanation of is correct.
the Assertion.
5. Assertion: Glucose is correctly named as D-
(3) If Assertion is True but the Reason is
False. (+)-glucose.
(4) If Assertion & Reason are false. Reason: 'D' before the name of glucose
represents its dextrorotatory nature.
2. Assertion: Oxidation of glucose by Br2– (1) If both Assertion and Reason are correct
water gluconic acid. and the Reason is a correct explanation
Reason: Br2–water oxidizes –CHO group of the Assertion.
into–COOH group. (2) If both Assertion and Reason are correct
(1) If both Assertion & Reason are True & the but Reason is not a correct explanation
Reason is a correct explanation of the of the Assertion.
Assertion. (3) If the Assertion is correct but Reason is
(2) If both Assertion & Reason are True but incorrect.
Reason is not a correct explanation of (4) If the Assertion is incorrect and Reason
the Assertion. is correct.
(3) If Assertion is True but the Reason is
6. Assertion: Hydrolysis of sucrose brings
False.
about a change in sign of rotation from
(4) If Assertion & Reason are false.
dextro to laevo.
Reason: Hydrolysis always changes the
3. Assertion: Glucose does not give 2,4 -DNP
optical rotation of a compound.
test.
(1) If both Assertion and Reason are correct
Reason: Glucose exists in cyclic hemiacetal
and the Reason is a correct explanation
form,
of the Assertion.
(1) If both assertion and reason are true and
(2) If both Assertion and Reason are correct
reason is the correct explanation of
but Reason is not a correct explanation
assertion,
of the Assertion.
(2) If both assertion and reason are true but
(3) If the Assertion is correct but Reason is
reason is not the correct explanation of
incorrect.
assertion.
(4) If the Assertion is incorrect and Reason
(3) If assertion is true but reason is false.
is correct.
(4) If both assertion and reason are false.

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7. Assertion: Sucrose is a disaccharide and a 10. Given below are two statements:
non-reducing sugar. Statement-I: Amylose is water soluble
Reason: Sucrose involves glycosidic linkage
component which constitutes about
between C1 of β-glucose and C2 of α-fructose.
Choose the most appropriate answer from 15- 20% of starch. Chemically amylose is a
the options given below: long unbranched chain with 200-1000
(1) If both Assertion and Reason are correct α–D–(+)–glucose units held together by
and the Reason is a correct explanation C1–C4 glycosidic linkage.
of the Assertion.
Statement-II: Amylopectin is insoluble in
(2) If both Assertion and Reason are correct
but Reason is not a correct explanation water and constitutes about 80–85% of
of the Assertion. starch. It is a branched chain polymer of α–
(3) If the Assertion is correct but Reason is D–glucose units in which chain is formed by
incorrect.
C1–C4 glycosidic linkage whereas branching
(4) If the Assertion is incorrect and Reason
is correct. occurs by C1–C6 glycosidic linkage.
In the light of the above statements, choose
8. Given below are two statements the most appropriate answer from the
Statement-I: D–(+)–Glucose and D–(–)– options given below:
Fructose are formed on hydrolysis of sucrose
Statement-II: Sucrose is an invert sugar (1) Statement-I and statement-II both are
Choose the most appropriate answer from correct.
the options given below: (2) Statement I is correct and statement II is
(1) Both statement I and statement II are incorrect
true
(3) Statement I is incorrect and statement II
(2) Statement I is false but Statement II true
(3) Statement I is true but Statement II is is correct.
false (4) Statement I and statement II both are
(4) Both Statement I and Statement II are incorrect
false
11. Given below are two statements:
9. Given below are two statements:
Statement-I: The pentaacetate of glucose Statement-I: Proteins are the polymer of α-
does not react with hydroxylamine amino acids.
indicating the absence of free –CHO group. Statement-II : The amino acids which can be
Statement-II: Despite having the aldehyde
synthesized in body are known as essential
group, glucose does not give schiff's test and
it does not form the hydrogen-sulphite amino acids.
addition product with NaHSO3. In the light of the above statements, choose
In the light of the above statements, choose the most appropriate answer from the
the most appropriate answer from the
options given below:
options given below:
(1) Statement I is correct and statement II is (1) Both statement I & II are correct.
incorrect (2) Both statement I & II are incorrect.
(2) Statement-I and statement-II both are (3) Statement I is incorrect but statement II
correct. is correct
(3) Statement I and statement II both are
(4) Statement I is correct but statement II is
incorrect
(4) Statement I is incorrect and statement II incorrect
is correct.
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12. Statement-I: Enzymes which catalyze the 15. Given below are two statements:
oxidation of one substrate with Statement-I: Only α-Amino acids are
simultaneous reduction of another substrate
obtained on hydrolysis of proteins.
are named as oxidoreductase enzymes.
Statement-II: Enzymes are needed only in Statement-II: The amino acids which can be
small quantity for the progress of a reaction. synthesized in body are known as essential
(1) Statement I is correct but statement II is amino acids.
incorrect In the light of the above statements, choose
(2) Statement I is incorrect but statement II
the most appropriate answer from the
is correct.
(3) Both statements I and statements II are options given below:
correct. (1) Both statement I & II are correct.
(4) Both statements I and statements II are (2) Both statement I & II are incorrect.
incorrect. (3) Statement I is incorrect but statement II
13. Given below are two statements: is correct
Statement-I: All carbohydrates which (4) Statement I is correct but statement II is
reduce Fehling's solution and Tollen's
reagent are referred to as reducing sugars. incorrect
Statement-II: All disaccharides whether
aldose or ketose are reducing sugars. 16. Match the column.
In the light of the above statements, choose Column-I Column-II
the most appropriate answer from the
(Reaction of Glucose) (Product)
options given below:
(1) Statement I is incorrect and statement II a Glucose 
→
HI, ∆ p Oxime
is correct.
b Glucose 
2
→
Br Water q n-Hexane
(2) Statement-I and statement-II both are
correct. c Glucose 
Nitric acid
→ r Gluconic acid
(3) Statement I is correct and statement II is
d NH OH
Glucose  →2 s Saccharic acid
incorrect H ⊕

(4) Statement I and statement II both are

(1) a – q, b – s, c – p, d – r
incorrect
(2) a – q, b – p, c – r, d – s
14. Given below are two statements:
(3) a – q, b – s, c – r, d – p
Statement-I: Sucrose is levorotatory but
after hydrolysis gives dextrorotatory glucose (4) a – q, b – r, c – s, d – p
and levorotatory fructose.
Statement-II: Hydrolysis of sucrose brings 17. Match List-I with List-II
about a change in the sign of rotation, from List-I List-II
dextro (+) to laevo (-) and the product is
named as invert sugar. (Saccharides) (Glycosidic-linkages)
In the light of the above statements, choose (A) Sucrose (I) α1 − 4
the most appropriate answer from the (B) Maltose (II) α1 − 4 and α1 − 6
options given below: (C) Lactose (III) α1 − β2
(1) Statement I is correct and statement II is
incorrect (D) Amylopectin (IV) β1 − 4
(2) Statement I is incorrect and statement II Choose the correct answer from the options
is correct. given below
(3) Statement-I and statement-II both are (1) (A-III), (B-I), (C-IV), (D-II)
correct.
(2) (A-IV), (B-II), (C-I), (D-III)
(4) Statement I and statement II both are
incorrect (3) (A-II), (B-IV), (C-III), (D-I)
(4) (A-I), (B-II), (C-III), (D-IV)
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18. Match the vitamins given in column I with 21. Match the vitamin in column I with their
the deficiency diseases caused by it given in scientific name in column II and choose the
column II and mark the appropriate choice. correct option-
Column I Column II Vitamin Chemical name
(A) Vitamin B1 (i) Convulsions (a) Vitamin B6 (i) Riboflavin
(B) Vitamin B2 (ii) Pernicious (b) Vitamin C (ii) Pyridoxine
anaemia (c) Vitamin B2 (iii) Ascorbic acid
(C) Vitamin B12 (iii) Beri-beri (d) Vitamin B1 (iv) Thiamine
(D) Vitamin B6 (iv) Cheilosis
(1) a(ii), b(iii), c(i), d(iv)
(1) (A) → (iv), (B) → (iii), (C) → (i), (D) → (ii)
(2) a(ii), b(i), c(iv), d(iii)
(2) (A) → (i), (B) → (iv), (C) → (iii), (D) → (ii) (3) a(iii), b(i), c(iv), d(ii)
(3) (A) → (ii), (B) → (i), (C) → (iv), (D) → (iii) (4) a(i), b(ii), c(iii), d(iv)
(4) (A) → (iii), (B) → (iv), (C) → (ii), (D) → (i)
22. Match the following column-I with product
19. Match the column I with column II and mark
given in the column-II and select the correct
the appropriate choice.
option from the code given below :-
Column I Column II
(A) Nucleoside (i) Sugar + base + Column- I Column- II
phosphoric acid (A) CHO (p) COOH
group HI ; ∆
(CHOH)4 (CHOH)4
(B) Nucleotide (ii) Cytosine + uracil Red P

(C) DNA (iii) Sugar + base CH2OH CH2OH

(D) RNA (iv) Cytosine +
(B) CHO (q) CN
thymine
NH2OH CH
(1) (A) → (iii), (B) → (i), (C) → (iv), (D)→ (ii) (CHOH)4 OH
(2) (A) → (i), (B) → (iv), (C) → (iii), (D)→ (ii) (CHOH)4
CH2OH
(3) (A) → (ii), (B) → (iii), (C) → (i), (D)→ (iv) CH2OH
(4) (A) → (iv), (B) → (ii), (C) → (i), (D)→ (iii)
(C) CHO (r) CH=N–OH
20. Match the column I with column II and mark HCN
the appropriate choice. (CHOH)4 Alkaline (CHOH)4
Column I Column II CH2OH CH2OH
(A) Peptide (i) Inversion
(D) CHO (s) CH3–(CH2)4–CH3
linkage
Br2 water
(B) Nucleic (ii) Polysaccharide (CHOH)4
acid
CH2OH
(C) Hydrolysis (iii) Proteins
of cane Codes :
sugar A B C D
(D) Starch (iv) Nucleotides (1) (s) (r) (q) (p)
(1) (A) → (ii), (B) → (i), (C) → (iii), (D) → (iv) (2) (s) (r) (p) (q)
(2) (A) → (iv), (B) → (i), (C) → (ii), (D) → (iii) (3) (r) (s) (q) (p)
(3) (A) → (iii), (B) → (iv), (C) → (i), (D) → (ii)
(4) (s) (q) (p) (r)
(4) (A) → (i), (B) → (iii), (C) → (iv), (D) → (ii)

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23. Match the column I with II and mark the 25. Match the following:
appropriate choice: Column-I Column-II
(i) Riboflavin (A) Beriberi
Column I column II (ii) Thiamine (B) Scurvy
(A) Nucleoside (i) Sugar + base + (iii) Pyridoxine (C) Cheilosis
phosphoric (iv) Asciorbic (D) Convulsions
Acid
acid group
(1) i  C; ii  A; iii  D; iv  B
(B) Nucleotide (ii) Cytosine +
(2) i  C; ii  D; iii  A; iv  B
Uracil (3) i  A; ii  D; iii  C; iv  B
(C) DNA (iii) Sugar + base (4) i  D; ii  B; iii  A; iv  C
(D) RNA (iv) Cytosine
+thymine 26. Match the column
Column-I Column-II
(Carbohydrates) (Source or name)
(1) A - iii; B - i; C - iv; D - ii a Sucrose p Animal starch
(2) A - i; B - iv; C - iii; D - ii b Lactose q Cereals, roots,
tubers
(3) A - ii; B - iii; C - i; D - iv
c Glycogen r Milk sugar
(4) A - iv; B - ii; C - i; D - iii d Starch s Cane sugar
(1) a – r, b – s, c – q, d – p
24. Match the column. (2) a – s, b – r, c – p, d – q
(3) a – s, b – p, c – r, d – q
Column-I Column-II (4) a – p, b – q, c – r, d – s

(Carbohydrates) (Source or name)

27. Match the column.
(a) Sucrose (p) Animal starch Column-I Column-II
(b) Lactose (q) Cereals, roots, (Carbohydrates) (Monosaccharide unit)
a Maltose p β-D-Glucose
tubers
b Lactose q Two α-D-Glucose
(c) Glycogen (r) Milk sugar c Cellulose r β-D-Galactose
(d) Starch (s) Cane sugar & β-D-Glucose
d Sucrose sα-D-Glucose &
β-D-Fructose
(1) a-r, b-s, c-q, d-p
(1) a – q, b – s, c – r, d – p
(2) a-s, b-r, c-p, d-q (2) a – q, b – r, c – p, d – s
(3) a-s, b-p, c-r, d-q (3) a – r, b – s, c – p, d – q
(4) a-p, b-q, c-r, d-s (4) a – r, b – p, c – q, d – s

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28. Match the column. 29. Match the column I & II:
Column-I Column-II Column-I Column-II
(Structure) (name) (Vitamins) (Sources)
a CH2OH p α-D-Mannose a Vitamin-A p Exposure to
H O H sunlight
H b Vitamin-B12 q Yeast, milk
OH H
OH OH c Vitamin-B6 r Meat, fish, curd
H OH d Vitamin-D s Fish liver oil
b q β-D-Fructose (1) a - s, b - r, c - q, d – p
CH2OH
(2) a - s, b - q, c - p, d – r
H O H
H (3) a - q, b - s, c - r, d – p
OH HO (4) a - r, b - p, c - q, d – s
OH OH
H H 30. Match the column:
c CH2OH OH r α-D-Glucose Column-I Column-II
O
a Fat soluble p Fingerprinting
H HO vitamin
H CH2OH b Water soluble q Vitamin A and
OH H vitamin vitamin D
d CH2OH CH2OH s α-D-Fructose c DNA r Uracil
O
d RNA s Vitamin B and
H HO vitamin C
H OH (1) a - s, b - s, c - r, d - p
OH H (2) a - q, b - s, c - p, d - r
(1) a – p, b – r, c – q, d - s (3) a - q, b - r, c - p, d - s
(2) a – r, b – p, c – q, d – s (4) a - q, b - s, c - r, d – p
(3) a – p, b – r, c – s, d – q
(4) a – r, b – p, c – s, d – q

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Exercise 4 (Previous Year's Questions)

1. D(+) glucose reacts with hydroxylamine 5. The central dogma of molecular genetics
and yields an oxime. The structure of the states that the genetic information flows
oxime would be: [AIPMT - 2014] from [NEET-II-2016]
CH=NOH CH=NOH (1) DNA  Carbohydrates  Proteins
H–C–OH HO–C–OH (2) DNA  RNA  Proteins
HO–C–H HO–C–H (3) DNA  RNA  Carbohydrates
(1) (2)
HO–C–H H–C–OH
(4) Amino acids  Proteins  DNA
H–C–OH H–C–OH
CH2OH CH2OH 6. Which of the following statements is not
CH=NOH CH=NOH correct? [NEET- 2017]
HO–C–H H–C–OH (1) Insulin maintains sugar level in the
H–C–OH HO–C–H blood of a human body
(3) (4) (2) Ovalbumin is a simple food reserve in
HO–C–H H–C–OH
H–C–OH H–C–OH egg white
CH2OH CH2OH (3) Blood proteins thrombin and fibrinogen
are involved in blood clotting
2. In a protein molecule various amino acids
(4) Denaturation makes the proteins more
are linked together by [NEET-I- 2016]
(1) Glycosidic bond active.
(2) Glycosidic bond 7. The difference between amylose and
(3) Peptide bond
amylopectin is [NEET(UG) 2018]
(4) Dative bond
(1) Amylopectin have 1,4 -linkage and
3. The correct statement regarding RNA and 1,6 -linkage
DNA, respectively is [NEET-I- 2016] (2) Amylose have 1,4 -linkage and
(1) The sugar component in RNA is 1,6 -linkage
arabinose and the sugar component in
(3) Amylopectin have 1,4 -linkage and
DNA is 2’-deoxyribose.
1,6 -linkage
(2) The sugar component in RNA is ribose
and the sugar component in DNA is (4) Amylose is made up of glucose and
2’-deoxyribose. galactose
(3) The sugar component in RNA is
arabinose and the sugar component in 8. Which of the following compound can form
DNA is ribose. a zwitterion? [NEET(UG) 2018]
(4) The sugar component in RNA is (1) Aniline (2) Acetanilide
2’-deoxyribose and the sugar (3) Benzoic acid (4) Glycine
component in DNA is arabinose.
9. The non-essential amino acid among the
4. Which one given below is a non-reducing
following is [NEET(UG) 2019]
sugar? [NEET-I- 2016]
(1) Maltose (2) Lactose (1) Valine (2) Leucine
(3) Glucose (4) Sucrose (3) Alanine (4) Lysine

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10. Which structure(s) of proteins remains(s) (1) Both Statement I and Statement II are
intact during denaturation process ? true.
[NEET(UG) 2019(Odisha)] (2) Both Statement I and Statement II are
(1) Both secondary and tertiary structures false.
(2) Primary structure only (3) Statement I is true but Statement II is
(3) Secondary structure only false.
(4) Tertiary structure only
(4) Statement I is false but Statement II is
11. Sucrose on hydrolysis gives true.
[NEET(UG) 2020]
16. Cheilosis occurs due to deficiency of
(1) D-Fructose + -D-Fructose
[NEET(UG)-Manipur 2023]
(2) D-Glucose + -D-Fructose
(1) Thiamine (2) Nicotinamide
(3) D-Glucose + -D-Glucose
(4) D-Glucose + -D-Fructose (2) Pyridoxamine (4) Riboflavin

12. Which of the following is a basic amino 17. The reagents with which glucose does not
acid: [NEET(UG) 2020] react to give the corresponding
(1) Lysine (2) Serine tests/products are [NEET-2024]
(3) Alanine (4) Tyrosine A. Tollen's reagent B. Schiff's reagent
C. HCN D. NH2OH
13. The RBC deficiency is deficiency disease of:
E. NaHSO3
[NEET(UG) 2021]
(1) Vitamin B12 (2) Vitamin B6 Choose the correct options from the given
(3) Vitamin B1 (4) Vitamin B2 below:
(1) B and E (2) E and D
14. The incorrect statement regarding
(3) B and C (4) A and D
enzymes is: [NEET(UG) 2022]
(1) Enzymes are biocatalysts. 18. Given below are two statements :
(2) Like chemical catalysts enzymes [RE-NEET-2024]
reduce the activation energy of bio Statements I: Glycogen is similar to
processes. amylose in its structure.
(3) Enzymes are polysaccharides. Statements II: Glycogen is found in yeast
(4) Enzymes are very specific for a and fungi also.
particular reaction and substrate. In the light of the above statements, choose
15. Given below are two statements: the correct answer from the options given
Statement I: a unit formed by the below :
attachment of a base of 1’ position of sugar (1) Statement I is true but Statement II is
is known as nucleoside. false.
Statement II: When nucleoside is linked to (2) Statement I is false but Statement II is
phosphorous acid at 5' -position of sugar true.
moiety, we get nucleotide. (3) Both Statement I and Statement II are
In the light of the above statements, choose
true.
the correct answer from the option given
(4) Both Statement I and Statement II are
below : [NEET(UG) 2023]
false.

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19. Sugar 'X' [NEET-2025]
A. is found in honey
B. is a keto sugar
C. exists in a and b anomeric forms
D. is laevorotatory
'X' is :
(1) Maltose (2) Sucrose
(3) D-Glucose (4) D-Fructose

20. Match List I with List II. [NEET-2025]

List I List II
(Name of (Deficiency
Vitamin) disease)
A. Vitamin B12 I. Cheilosis
B. Vitamin D II. Convulsions
C. Vitamin B2 III. Rickets
D. Vitamin B6 IV. Pernicious
anemia
Choose the correct answer from the options
given below:
(1) A-II, B-III, C-I, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-I, B-III, C-II, D-IV
(4) A-IV, B-III, C-I, D-II

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ANSWER KEYS
Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 4 3 2 1 2 4 1 1 2 3 4 4 4 2 3

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 4 4 2 3 1 2 4 1 1 1 1 4 4 2

Exercise 1.3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 4 2 2 4 1 4 1 1 1 2 4 4 1 1

Exercise 2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 3 3 2 1 1 1 3 2 2 1 3 4 1 3 2 3 2 4 4
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 3 2 1 3 4 2 3 4 4 3 4 1 3 1 3 3 3 3 3 2
Que . 41 42 43 44 45
Ans. 2 2 1 3 4

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 1 1 1 3 3 3 1 2 1 4 3 3 2 4 4 1 4 1 3
Que . 21 22 23 24 25 26 27 28 29 30
Ans. 1 1 1 2 1 2 2 2 1 2

Exercise 4 (Previous Year's Questions)

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 3 2 4 2 4 1 4 3 2 4 1 1 3 3 4 1 2 4 4

238 Sarvam Career Institute

 Notes

Sarvam Career Institute

 Notes

Sarvam Career Institute