D P P - Da i ly Practice P roblems

C h a pter-wise S h eets

Date : L..�-----l Start Time : End Time :

C H E M I STRY (Cc1 �
SYLLABUS : Organic chemistry - Some basic principles and Techniques

Max. Marks : 180 Marking Scheme : + 4 for correct & (-1 ) for incorrect Time : 60 min.

INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQ's. For each question only one option is correct.

Darken the correct drcle/ bubble in the Response Grid provided on each page.

1. The correct decreasing order of priority for the ftmctional (c) CH3-CH=CH-OH
groups of organic compmmds in the IUPAC system of (d) C�-C(CHpH)=C�
nomenclature is
(a) - COOH,- S03H,-CONI\,-CHO
(b) -SO3H._, -COOH- - , COl\Jl.l
,1.'11&,� ' -CHO
4. The compound 0 is

(c) -CHO, -COOH,-S03H,-CO� known by which of the following names ?

(d) - CON�.- CHO, - S03H, - COOH (a) Bicyclo [2.2.2] octane
2. The IUPAC name ofthe compound (b) Bicycl o [2.2.1 ] octane
(c) Bicycl o [ 1 .2.1 ] octane

(X)
(d) Bicyclo [ 1 . 1 . 1 ] octane
5. The stability of the compounds

6�00
(a) 1 , 2, 3 - trifonnyl propane
(b) Propane- !, 2, 3 - tricarbaldehyde
(c) 3 - formyl - 1 , 5 - pentanedial
(d) Propane- 1, 2, 3 - trial (i) (ii) (iii) (iv)
3. Vinylcarbinol is
(a) (iv) > (iii)> (i) > (ii) (b) (i)> (iii)> (ii)>(iv)
(a) HO-C�-CH=C�
(b) CH3C(OH)= C� (c) (ii)> (iii)>(i)> (iv) (d) (iv) > (i)>(iii)>(ii)

1. ®®@@ 2. ®®@@ 3. ®®@@ 4. ®®@@ 5. ®®@@

------ Spacefor Rough Work ------
lc46 �---- DPP/ C C 1 2
6. Fractional distillation is used when C diethyl ether & ID. Position isomers
(a) there is a large difference in the boiling point ofliquids methyl propyl ether
(b) there is a small difference in the boiling points of D. dimethyl ether and ethanol lV Chain isomers
liquids (a) A - I; B - III; C - II; D - IV
(c) boiling points ofliquids are same (b) A - ll; B - ill; C - IV; D - I
(d) liquids form a constant boiling mixture (c) A - II; B - TV; C - I; D - TTI
7. Rate of the reaction (d) A - ll; B - I; C - IV; D - lll

�
�0 <S> �0 <S> 16. The accepted IUPAC name ofthe camphor is
R - � + Nu ----7 R - C.. + Z
Z Nu 0
is fastest when Z is
(a) OC2lfs (b) � (c) CI- (d) ococ�
8. The order of activity of the various o- and p-director is
(a) -Q->-OH>-OCOC� >-COC� (a) 1 , 7, 7 -trimethyl bicyclo (2. 2. l ) heptan- 2 - one
(b) -OH>-Q- >-OCOC� >-COCH3 (b) 1 , 7, 7, -trimethyl bicyclo (2. 1 . 2] heptan - 2 - one
(c) -OH>-Q- >-COCH,> -OCOC� (c) 1 , 2, 2 -trimethyl bicyclo (2. 2. l ) heptan- 6- one
(d) -o- > -COCH, > -OCOCH, >-OR (d) None of these
9. 17. The pair of structures given below represent
(b) diols
The general formula C11�1102 could be for open chain
(a) carboxylic acids H3 CH3
(c) dialdehydes (d) diketones
H
I
C6H5CHO + HCN � C6H5 - C - CN
I
10.
H H H
The product would be OH
(a) a racemate
(b) optically active CH3
(c) a meso compound (a) enantiomers
(d) a mixture of diastereomers (b) diastereomers
11. The number of possible open chain (acyclic) isomeric (c) structural isomers
compounds for molecular formula C5H10 would be (d) two molecules ofthe same compound.
(a) 8 (b) 7 (c) 6 (d) 5 18. Which of the following compounds exhibits geometrical
12. Which one of the following is the stablest structure of isomerism?
cyclohexatriene?
(b) Boat form
(a) C2HsBr (b) (CHMCOOH)2
(a) Chair fonn (c) CH,CHO (d) (C�)z(COOH)2
(c) Halfchair form (d) Planar form 19. Chlorine in vinyl chloride is less reactive because
13. The compound formed in the positive test for nitrogen with (a) sp23 - hybridised carbon bas more acidic character than
the Lassaigne solution of an organic compound is sp - hybridised carbon
(a) Fe4[Fe(CN)J3 (b) Na3[Fe(CN)J
(b) C - Cl bond develops partial double bond character
(c) Fe(CN)3 (d) Na4[Fe(CN)5NOS]
(c) of resonance
14. 2. 79 g ofan organic compound when heated in Carius tube
(d) All are correct
with cone. HN03 and H3P0 4 formed converted into 20. The best method for the separation of naphthalene and
MgNH4 .P04 ppt. The ppt. on heating gave 1.332 g of benzoic acid from their mixture is:
(a) distillation (b) sublimation
Mg2P207 . The percentage ofP in the compound is (c) chromatography (d) crystallisation
(a) 23.33% (b) 13.33% (c) 33.33% (d) 26.66% 21. The percentage ofsulphur in an organic compound whose
15. Match the columns 0.32 g produces 0.233 g ofBaS04 [At. wt. Ba = 137, S = 32]
Column-1 Column-fJ is
A. �COOH & HCOOCH, I. Functional isomers (a) 1.0
B. !-butene & 2-butene U. Metamers (b) 10.0 (c) 23.5 (d) 32.1

6. ®®@@ 7 ®®@@ 8. ®®@@ 9. ®®@@ 10. ®®@@

11. @®@@ 12.@®@@ 13 . @®@@ 14. @®@@ 15. ®®@@
16.@®@@ 17.@®@@ 18.@®@@ 19. @®@@ 20. ®®@@
21.
Spacefor Rough Work -------
DPP/ C C 1 2 --------� c47l
22. Indicate the wrongly named compound 27. The IUPAC name of
CH 3 0
I I II
(a) CH 3-CH-CH 2 -CH 2 -CHO
CH3 CH 3- CH -C-CH 2 -CH 2 0H is
(a) 1 - Hydroxy -4- methyl -3- pentanone
(4-methyl -1- pentanal) (b) 2- Methyl -5- hydroxy -3- pentanone

I
(b) CH 3-CH-C = C- COOH (c) 4- Methyl -3- oxo -1- penanol
t
(d) Hexanol -1 - one -3
Cl-h 28. An important chemical methodto resolve a racemic mixture
makes use ofthe formation of
(4- methyl -2- pentyne -1- oic acid) (a) a meso compound (b) enantiomers

I
(c) CH3CH 2CH2-CH-COOH (c) diastereomers (d) racemate
29. The Lassaigne's extract is boiled with dil. HN03 before
CH 3 testing for halogens because
(a) Silver halides are soluble in HNO.'
(2- methyl -1- pentanoic acid) (b) Na2S and NaCN are decomposed by HNO.'

II (d) A.gCN is soluble is HN03

0 (c) Ag S is soluble in HN03

(d) CH3 -CH2 -CH= CH-C-CH3 30. What is the decreasing order of strength of the bases
OH - , NH2, HC = c - and CH3CH2 ?
(3-hexen -5- one)
23. The (R)- and (S)- enantiomers of an optically active (a) CH3CH2 > HC = C > NH2 > OH-
compound differ in (b) 1-IC = c- > CH3CH2 > NH2 > OH-

0
(a) their reactivity with achiral reagents (c) 01-C > NH2 > HC = c - > CH3CI-f2
(b) their optical rotation ofplane polarized light
(c) their melting points (d) NH2 > HC = c - > OH- > CH1Cl·I2
(d) None of these 31. Given
24. But-2-ene exhibits cis-trans-isomerism due to

=
CH3 CH3

(l
CH3

Q
(a) rotation armmd c3 - c4 sigma bond
(b) restTicted rotation around C C bond

�
CH3 / CH3
(c) rotation around c , - c2 bond
I
(d) rotation around C2 - C3 double bond CH3
25. Which ofthe following compounds undergoes nucleophilic

Q
0 0 0
substitution reaction most easily ? (I) (II) (III)
Wl1ich of the given compounds can exhibit tautomerism?
(a) I and ITT (b) nand Ill
(a) Y'n
� �)
(c) I, 11 and Ill (d) 1 and II
32. Cyclohexanol (1), acetic acid (ll), 2, 4, 6-tri.nitrophenol (ill)
N02 and phenol (lV) are given. In these the order of decreasing

Q
CH3 acidic character will be:
(a) ill> IT> IV> I (b) II> ill> I > IV

��
(c) U > ill>IV > I (d) ill>IV>ll>I
c1 33. Some meta-directing substituents in aromatic substitution
(c) (d) are given. Wbich one is most deactivating?
v (a) -S03H (b) -COOH (c) -N02 (d) -C=N
34. Consider thiol anion (RS- ) and alkoxy anion (RO- ). Which
OCH3 of the following statements is correct ?
26. An organic compound contains C = 40%, H = 13.33% and (a) Rs- is less basic but more nucleophilic than Ro­
N = 46.67%. Its empirical formula would be (b) RS- is more basic and more nucleophilic than RO­
(a) CHN (b) C2�N (c) CH4N (d) C3�N (c) RS- is more basic but less nucleophilic than Ro­
(d) RS- is less basic and less nucleophilic than Ro-

RESPO'iSE
22.@@ @@ 23.@@ @@ 24.@@@@ 25.@@@@ 26. ®®@@
GRID
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32.@@@@ 33.@@@@ 34.@@@@
Spacefor Rough Work -------
lc48 �---- DPP/ C C 1 2

-@-
35. Consider the following compounds. 39. (C�)4N+ is neither an electrophile, nor a nucleophile
because it
02N coci
(i) C6IfsCOCl (ii) (a) does not have electron pair for donation as well as can

-@- -@- coci

not attract electron pair
(iii) H3c coci (iv) OI·Ic (b) neither has electron pair available for donation nor can
accommodate electron since all shells ofN are fully
The correct decreasing order of their reactivity towards occupied
(b) (iv) > (ii) > (i) >(iii)
hydrolysis is (c) can act as Lewis acid and base
(a) (i) > (ii) >(iii)> (iv) (d) None of these
(c) (ii) > (iv)> (i) >(iii) (d) (ii) > (iv)> (iii) > (i)
36. The order ofstability ofthe following tautomeric compounds 40. In Kjeldahl's method for the estimation of N 2 , potassium
is: sulphate and copper sulphate are used. On the basis of
OH
0 0
- -- -
0
II II
their functions which of the following statement(s) is/are
I II
CH 2 = C CH2 C CH3 � CH3 C- CH2 - C- CH3
correct?
(i) Potassium sulphate raises the b.pt. and ensures
T U complete reaction.
II
0
(ii) Copper sulphate acts as catalyst.
I II
OH (iii) Potassium sulphate acts as catalyst and copper
CH3-C = CH-C-CH3 sulphate raises the b.pt.
Ill (a) Only (iii) is correct (b) (i) and (ii) are correct
(a) ill>II>I (b) II>I>ill (c) Only(ii)is correct (d) Noneis correct
(c) ll>ID>I (d) I > ll > ID 41. What is the relationship between open chain forms of
37. Match the columns D-glucose and D-altrose ?
Column - 1 Column - ll (a) enantiomers
80xm1 x lOO (b) constitutional isomers
A. Duma's method I.
(c) diastereomers
I88x m
3 1 x m1 x i O O % (d) different conformations of the same compound
B. Kjeldahl 's method n. 42. The production of an optically active compound from a
1 877 x m

( )
symmetric molecule without resolution is termed
(a) Walden inversion (b) Partial racemisadon
1 .4 x M x 2 v - �
c. Carius method m. 2 (c) Asymmetric synthesis (d) Partial resolution
% 43. The number of asymmetric C-atom created and number of
for bromine m possible stereoisomers when benzil (Ph CO CO Ph) is reduced
28x V x l OO with LiAIH4.
D. Percentage of IV %
22400x m (a) 2,3 (b) 2,2 (c) 2,4 (d) 3,2
phosphorus 44. Among the following, the true property about
(a) A - IV; B - III; C - I; D - II CH3 +
(b) A - ill; B - IV; C - II; D - I
(c) A - IV; B - I; C - IT; D - TIT CH3: C - CH3 is

(d) A - I; B - ill; C - II; D - IV (a) it is non-planar

2
38. Which of the following represents the correct order of (b) itsC+ issp -hybridized
stability of the given carbocations ? (c) an electrophile can attack on its c+
(d) it does not undergo hydrolysis
+ I+ + 45. The compound which contains all the four I0, 2°, 3° and 4°
(a) F3C > F3C -C > CH3 carbon atoms is
I (a) 2, 3-dimethylpentane
I+ + + I+ + + (b) 3-chloro-2, 3-dimethylpentane
(c) F3C - C > F3C > H3C (d) F3C - C > H3C > F3C (c) 2, 3, 4-trimethylpentane
I I (d) 3, 3- dimethylpentane

RI-.SPO:\Sl-. 35 @@@@ 36.@@@@ 37.@@@@ 38. @@@@ 39. ®®@@

GRID
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Spacefor Rough Work -------
 DAILY PRACTICE C H E M I STRY
PROBLEMS SOLUTIONS

1. (a) The correct order of priority for the given functional ..

group is 8. (a) :o: : QH

6 6
0 0
II II >
-COOH > -S03H > -C- NI-1 2 > -C- H
Most electron· Electron rel asing
e
2. (b) When - CHO is treated as side chain, its name is releasing due to due to -OH group
-ve charge on 0

6
carbaldehyde.
3 2 I
C H 2 - CH - C H 2 0
.. II
I I I -CH3
Hence CHO CHO CHO is propane-

1, 2, 3-tri carbaldehyde. > >

3. (a) Carbinol is met hyl alcohol (CH30H), hence
3 2 I Less electron releasing due to presence of Electron­
vinylcarbinol should be CH2 = CHCH20H -co grouping which shows resonance withdrawing group
with tbe electrons on 0
(prop-2-enol)
Cn l-f
(a) ·'2n02 is oeneral
9. (a) t:>
fonnula for carboxylic acid.
4.
Since during the reaction, a chiral carbon is created
(a) Cyclohexane (iv) is non-planar and has chair
10. (a)
5.
and tilrther since the CN- ion can attack the planar
conformation. In this conformation, the bond angle is
aldehyde group both from the top and the bottom face
the normal tetrahedral angle (1 09°, 28') and thus has
of the aldehyde group with equal ease, therefore, a
no angle strain and hence is most stable. The rest of 50 : 50 mixture of the two enantiomers, i.e. a racemic
the molecules are nearly planar and hence their stability
mixture is obtained.
depends upon the angle strain in accordance with
11. (c) C5H 10 has 1° degree ofunsaturation since the isomers
Baeyer �strain theory. Since cyclopropane has h igher ..
are acyclic, all of these are alkenes. For wntmg the
109°28'-60° isomers, first introduce the double bond at different
angle strain ( = 24°, 44 ') than
2 possible positions, and then consider the possibility

cyclopentane (
I 09° 2R '- I ORO ·)
= 00, 44 .
ofbranching in the alkyl group.
C�C�C�CH = C� C�C�CH = CHCif:;
2 l-pentene (i) 2- pentene, (cis,- trans) (ii), (iii)
Therefore cyclopentane (iii) is more stable than
CH3 CH3
cyclopropane (i). Further, because of the presence of I I
a double bond in a three membered ring, cyclopropene CH3-CH -CH = CH2 CH3CH2 C= CHz
(ii) is the least stable. Thus the order of stability is 3-methyl-l -butene, (iv) 2-methyl- 1 -butene, (v)
(iv) > (iii) > (i) > (ii).
6. (b) Ifthere is a small difference (1 0 or less) in the boiling CH3
I
points of liquids fractional distillation is used e.g. CH3 -C= CHCH3
acetone b.p. 333 K and methanol b.p. 338 K.
2-methyl-2-butene, (vi)
7. (c) CJ- is the best leaving group among the given option.
12. (d) Since all the six carbon atoms of cyclohexatriene
As it has high electronegativity thereby generating
(benzene) are sp2 hybridized, therefore its stablest
positive charge on ca rbonyl carbon atom.
structure is planar form.
ls-34 ...,._---- DPP/C C 1 2
13. (a) Prussian blue Fe4[Fe(CN)6 h is formed in Iassaigne 25. (a) ln SN Ar reactions, a carbanion is formed as an
intermediate, so any substituent that increases the
test for nitrogen.
stability of carbanion and hence the transition state
3
3Na4[Fe(CN)6] + 2Fe + � leading to its formation will enhance the SNAr reactions.
To compare the rates of substitution in chlorobenzene,
Fe4[Fe(CN)6h+ 12Na +
Prussian blue chlorobenzene having electron-withdrawing group,
and chlorobenzene having electron-releasing group,
62 wt.of Mg2P207
14. (b) PercentageofP = x I O0 we compare the structures carbanion I (from
222 x wt.of compound chlorobenzene), I I (from chlorobenzene containing
62 1.332 electron-withdrawing group) and I I I (from
X 100= 13.33%
= 222 X 2.79 chlorobenzene containing electron-releasing group).

X 0,
15. (a)
Z Cl Z Cl
16. (a) It is a bridge compound

:Ett 4
u � G
TT
1, 7, 7-trimethyl bicyclo [2. 2. 1] heptan-2-one G withdraws electrons, neutralises (disperses) -ve
17. (c) Convert these Newmann projections into open chain charge of the ring, stabilises carbanion, facilitates SN
structures. reaction (activation effect)

+ +
CH3 CH3 Z Cl

�H H� H
H Cl H H
H
CH3 CH2CI
Both structures have same molecular formula C4H9CI,
thus these are isomers. However, the two have different
III
groups, viz CH3 and CH2CI, so these are neither G releases electrons, intensifies -vc charge,
enantiomers nor diastereomers. Hence these are destabilizes carbanion, retards SN reaction
structural isomers.
(deactivation)
18. (b) (CHMCOOH)2 is HOOCCH = CHCOOH, hence here N02 is activating group and CH3 and OCH3 are
geometrical isomerism is possible.
deactivating group towards nucleophilic aromatic

(c) CH2 = CH- <;,�: B CH2 - CH = <;I : a double bond

.. + substitution reactions.
19.

¢ > ((1 >9 >

Hence, the correct order of nucleophilic substitution
is formed between C and Cl. Hence it is less reactive reactions
due to resonance Cl
20. (b) Among the given compounds naphthelene is volatile
but benzoic acid is non-volatile (it forms a dimer). So,
the best method for their separation is sublimation,
which is applicable to compounds which can be
converted directly into the vapour phase from its solid N02 CH3

(c) As in above question,

state on heating and back to the solid state on cooling.
26.
Hence it is the most appropriate method.
13.33 46.6
= 3.33'" H = = 13.33'" N =
40 7
0·233 C= = 3 .34
21. (b) % of S = � x x l00 = 1 0 % 12 I 14
233
.

0.32
Relative No. ofatoms,
3 2 11 I
0
6 5 4 3.33 1 3.33 . N 3.34
CH 3 CH2CH =CH-C-CH C= =4
22. (d)
3.33
= I'· H =
3.33 ' = 3.33 = l
(hex - 3- en - 2 -one)
23. :. Empirical formula= CH4N
(b) 24. (b)
DPP/CC1 2 -------t s-351

*
CH3 0
5 4 1 . 31 1 2 J OH
27. (a) CH3 - CH - C - CH2 - CH20H
32. (a) 02N NO, > CH,COOH >
1-Hydroxy-4-methyl-3-pentanone
28. (c) Diastereomers have different solubility, m.p. and b.p.,
hence they can be separated by fractional N02
crystallisation.
29. (b) Na2S and NaCN, formed during ntsion with metallic (Ill) (IT)
sodium, must be removed before adding AgN03,
otherwise black ppt. due to N3zS or white precipitate OH

6
due to AgCN will be formed and thus white precipitate
ofAgCl will not be identified easily.

Na2S + 2AgN03 � 2NaN03 + Ag2S .J,

Black (IV) (I )

NaCN + AgN03 � NaN03 + AgCN .J, Explanation : Presence of three - N02 groups in
o-, p-
White
positions to phenolic groups (in Ill) makes
NaCl + AgN03 � NaN03 + AgCI .J, phenol strongly acidic because its corresponding
conjugate base is highly stabilised due to resonance.
White
Na2S+ 2HN03
boil
2NaN03 + H2S t Conjugate base ofC�COOH, II (i.e. CH3Coo- ) is

NaCN +HN03
boil
NaN03 + HCN t resonance hybrid of two equivalent structures. The

(a) Stronger the acid, weaker the conjugate base. Since

conjugate base of phenol, IV is stabilized due to
30. resonance (note that here all resonating structures are
acidic character follows the order not equivalent). The conjugate base of cyclohexanol, 1
H20 > NH3 > HC = CH > CH3 -CH3 does not exhibit resonance, hence not formed.
the basic character oftheir conjugate bases decreases 33. (c) Decreasing order of deactivating effect of the given
in the reverse order, i.e., m-directing group is
- N02 > - CN > - S03H > - COOH
- N02 group is most deactivating group due to strong
31. (c) All of these compounds show tautomerism - E, - I and - M effects.
34. (a) On moving down a group, the basicity and
nucleophilicity are inversely related, i.e. nucleophilicity
increases while basicity decreases. i.e RS- is more
nucleophilic but less basic than Ro-. This opposite
behaviour is because of the fact that basicity and
nucleophilicity depends upon different factors.
Basicity is directly related to the strength of the
CH3

Q
H-element bond, while nucleophilicity is indirectly
CH3 �/==:;; related to the electronegativity of the atom to which
proton is attached.
(c) The degree ofhydrolysis increases as the magnitude
H
35.
0
of positive charge on carbonyl group increases.
Electron withdrawing group increases the positive

(1
charge and electron releasing group decreases the
CH3 =
/
positive charge. Among these N02 & CHO are electron

Y
·H3
=
:::::;; /C
withdrawing group from which N02 has more -1 effect
CH3 7 CH3
·· than -CHO. On the other band CH3 is a electron
OH releasing group therefore the order of reactivity
towards hydrolysis is
ls-36 ...,._---- DPP/C C 1 2
41. (b) Constitutional isomers

CHO
I
CHO H - C - OH
I I
Altrose is H O - C - H , Glucose is HO- C - H
I I
(a) Enolic form predominates in compounds containing (CHOH)J (CHOHh
I
36.
I
two carbonyl groups separated by a - C� group. This CH2 0H CH20H
is due to following two factors.
(I) Presence ofconjugation which increases stability. 42. (c) It is definition ofasymmetric synthesis.
(ii) Formation of intramolecular hydrogen bond OH OH
between enolic hydroxyl group and second I I.
43. PhCOCOPh Ph - C * - C - Ph
I
(a)
molecule. Hence the correct answer is lli > ll > 1.
carbonyl group which leads to stablisation ofthe
I
H H
37. (a) The product contains two similar asymmetric carbon
38. (b) - I group destablises carbocation and since inductive atoms and two optically active and one optically
effect decreases with increasing length ofcarbon chain. inactive meso form.
Therefore (b) is the correct option. 44. (b) In carbocations, carbon bearing positive charge is
2
39. (b) The octet around N is complete, hence it has no always sp -hybridised
electrophilic character. N has no unshared pair of ]0 10
4°1 ?:.
CH3 CH3
electrons to act as nucleophile.
I" I 1°

3•
CH3 - CH - C - cH - CH ·
I 2 3,
40. (b) K2S04 raises b.pt. and CuS04 acts as catalyst. 45. (b)
Cl
Thus all the four types ofcarbon atoms are present in
this compound.