Chemistry

ISOMERISM
OBJECTIVE QUESTIONS EXERCISE-I
1. CH 3 CHOHCH 2 CHO and CH 3 CH 2 CH 2 COOH 8. The simplest alkanol exhibiting optical activity is
constitute a pair of :- (1) n-butyl alcohol
(1) Position isomers (2) Isobutyl alcohol
(2) Metamers (3) s-butyl alcohol
(3) Optical isomers (4) t-butyl alcohol
(4) Functional isomers
H3C H
2. The minimum number of carbon atoms present in C C H
9. H3C C Exhibits :-
an organic compound to show chain isomerism is H3C COOH
(1) 2 (3) 3 (3) 5 (4) 4
3. The minimum number of carbon atoms present in (1) Tautomerism
an organic compound to be able to show position
isomerism is :- (2) Optical isomerism

(1) 3 (2) 4 (3) 2 (4) 5 (3) Geometrical isomerism

4. The simplest pair of compounds exhibiting (4) Geometrical and optical isomerism
functional group isomerism have a minimum of 10. The isomerism shown by Benzaldoxime
(1) Four carbons (2) Three carbons
(3) Five carbons (4) Two carbons CH=N–OH is :-
5. Which of the following compound is isomeric with
propanoic acid :- (1) Optical (2) Functional
(3) Metamerism (4) Geometrical
(1) CH3 C OC2H5
O
COOH
H OH
(2) CH2 CH2 C H 11. Meso–tartaric acid H OH is optically
OH O COOH
(3) CH 3 —CH(OH)—CH 3 inactive due to the presence of :-
(4) CH 3 O—CH 2 —CH 2OH (1) Molecular symmetry
6. The pair of structures represents :- (2) Molecular asymmetry
(3) External compensation
Me Me
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(4) Two asymmetric carbon atoms

H Me H H
12. Which is optically active molecule :-

(1) C6H5 C OH
H H H H O
Cl CH2Cl (2) CH3 CH C2H5

(1) Enantiomers (2) Position isomers OH

(3) Conformers (4) None (3) C6H5 CH OH

7. CH3—NH—C2H5 and (CH3)3N show which type of H
isomerism :-
(4) C6H5 CH CH3
(1) Position (2) Functional
CH3
(3) Chain (4) None
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13. Which conformation of butane will have the minimum 20. Identify R configuration :
energy :-
(1) Gauche (2) Anti staggered COOH CN
(3) Eclipsed (4) None
14. The number of stereo isomers of glucose (a six (1) H (2) CH3 C CH
carbon sugar) is :-
(1) 8 (2) 12 (3) 16 (4) 24 CH3 CH2OH
15. The pair of structures given below represents :
CH3 CH2 CH3 CHO COOH
H
C C (3) H OH (4) H2N H
HO
HO CH2 CH3 CH3
H CH2OH CH3
(1) Enantiomers (2) Diastereomers
(3) Homomers (4) Position isomers 21. Which of the following has Z-configuration :
16. Rotational angle require to get maximum stable
conformer from minimum stable conformer in
H3C CH
n-butane is : (1) C C 2 5
H H
(1) 360° (2) 180° (3) 120° (4) 240°
17. Number of chiral carbon atoms in the compound
X, Y and Z respectively would be : Br CH(CH3)2
(2) C C
HOCH2 CH2 CH3
Me
X
Me Cl H
Me (3) C C
Br D
Y
(4) All the above

Z 22. Which of the following pairs are diastereomers :

Me
(1) 2, 1, 1 (2) 1, 1, 1 H Br Br H
(3) 2, 0, 2 (4) 2, 0, 1 C C
(1) and
18. Which of the following compoundis optically inactive:
Me Me Me Me
COOH COOH
OH H CH3
(1) CH3 CH2 CH CH3 (2) C C
(2) Me and Me
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H3C CH3 H OH HO H
H CH3
(3) C C C (4) CHO CHO
H3C H
Cl H
19. CH3 CH CH2 C O and (3) H C NH2 and H C NH2
Cl H
CH3 CH2 CH2 C Cl CH2OH CH2 OH
O Me Me
are constitute a pair of :-
(4) HO H and HO H
(1) Position isomers
(2) Metamers H Cl Cl H
(3) Optical isomers
(4) Functional group isomers C6H5 C6H5

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23. Which of the following has E-configuration :- 28. If optical rotation produced by the compound [A] is
+65°, then produced by the compound [B] is–
(1)

[A] [B]
(2)
(1) +65° (2) –65°
(3) Zero (4) Unpredictable
29. Which of the following represents the structure
3 having cis arrangement around each double
(3)
bond :-

(1)
(4)
3
(2)
24. The minimum number of carbon atoms in ketone (3)
to show position isomerism :-
(1) 3 (2) 4
(3) 5 (4) 6
(4)
25. The correct statement about the compoud (A), (B)
and (C) is :–

30. Compound is :-
(A) , (B)

(1) D–form (2) L–form

(3) R–form (4) Z–form
31. Which is a pair of geometrical isomers :-

(C)
, ,

(1) (A) and (B) are identical

,
(2) (A) and (B) are diastereomers
(3) (A) and (C) are enantiomers (1) I and II (2) I and III
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(4) (A) and (B) are enantiomers (3) II and IV (4) III and IV
26. Which of the following does not contain any 32. Among the following structure I to III
asymmetric carbon but can show enantiomerism:-
(1) Lactic acid (2) 1,3-pentadiene , ,
(3) Tartaric acid (4) 2,3-pentadiene
27. The following two compounds are
It is true that :-

and (1) All three are chiral compounds

(2) Only I and II are chiral compounds

(1) Enantiomers (2) Diastereomers
(3) Only II is chiral compound
(3) Identical (4) Epimers
(4) Only I and III are chiral compounds
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33. The IUPAC name of compound 38. Which of the following show optical isomerism :-

(1) (2)
is :–

(1) (Z)–4,6,7–Trimethyl hept–4–en–3–one

(2) (E)–4,6–Dimethyl oct–4–en–3–one
(3) (Z)–4,6–Dimethyl oct–4–en–3–one
(3) (4) All the above
(4) (E)–4,6–Dimethyl oct–4–en–6–one
34. The absolute configuration of the compound is :
39. How many isomers of C 5H 11OH will be primary
CH3
Cl H alcohols :–
Cl H (1) 2 (2) 3
C2H5 (3) 4 (4) 6
(1) 2S, 3R (2) 2S, 3S 40. The minimum number of carbon atoms in ketone
(3) 2R, 3S (4) 2R, 3R to show metamerism :–
35. Which one of the following is a meso–compound. (1) 3 (2) 4
(3) 5 (4) 6
41. Which of the following is optically active :–
(1) (2)

(1) (2)

(3) (4)

36. The complete IUPAC name of the compound :- (3) (4)

42. BrCH 2 –CH 2–CH=O and CH 3 –CH 2– C=O are

Br
(1) (R)-1-Bromo-1-phenyl ethane (1) Functional isomers
(2) (S)-1-Bromo-1-phenyl ethane
(2) Position isomers
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(3) Chain isomers

(4) (Z)-1-Bromo-1-phenyl ethane
37. Which one of the following is the most stable (4) Metamers
conformation of 2, 3–butanediol :– 43. Which are metamers :-
CH3
(1) CH3–O–CH2CH 2CH3, CH3–O–CH
CH3
(1) (2)
(2) C2H5–O–C2H 5, CH3CH 2CH2CH2OH

(3) CH3–O–C2H5, CH3–CH2–O–CH3

(3) (4) (4) CH3– C –CH3, CH3–CH2– C –H

O O

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44. Which can show 'Geometrical isomerism' :-
CH2OH CH2OH
CH3 H H CH3 HO H
(1) C=C (2) C=C H HO
CH3 H H H (3) C=O (4) C=O
H OH H OH
CH3 CH3 CH3 CH3 CH2OH CH2OH
(3) C=C (4) C=C
H H H CH3
48. The total number of configurational isomers of the
45. Which is incorrect statement :-
given compound are :-
H H
Me Me H H CH 3–CH=CHCHOHCHOHCH 3
(1) and
(1) 2 (2) 4 (3) 6 (4) 8
H H H Me
H Me 49. Which is a pair of geometrical isomers :-
are conformations Cl Me Br H
(1) C=C and C=C
Cl Cl Br H Cl Me
(2) is a meso-compound Ph OH Me
H H (2) C=N and C=N
Me Ph OH
(3) and
Br
are Geometrical isomers (3) Br and Br Br

H H H H
(4) C=C=C and C=C=C Me Ph H Ph
Me Me Me Me (4) C=C and C=C
are Enantiomers H Me Me Me

46. The IUPAC name of the given compound is :-

COOH COOH
H OH H 3C OH
50. and are :-
HO CH3 HO H
Br Cl CHO COOH COOH

(1) (2R, 4Z)-4-bromo-2-chloro-2-methyl-4-hexenal (1) Enantiomers

(2) (2R, 4E)-4-bromo-2-chloro-2-methyl-4-hexenal (2) Position isomers
(3) Geometrical isomers
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(3) (2Z, 5R)-3-bromo-5-chloro-5-formyl-2-hexene

(4) (2S, 4E)-4-bromo-2-chloro-2-methyl-4-hexenal (4) Homomers

47. Amongst the following, which one could be the 51. The correct configuration assigned for compounds
structure of an optically inactive monosaccharide (I) and (II) respectively are :-
having the molecular weight 150 :- COOH HO CH3
C
H OH Br
CH2OH CH2CH3
CHO CH3
C=O (I) (II)
H OH
H OH
(1) HO H (2) (1) R, R (2) S, S
H OH
H OH
CH2OH (3) R, S (4) S, R
CH2OH
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52. Which similarity is necessary for isomerism– 55. Number of structural isomers of C 6H14 is -
(1) Molecular formula (2) Structure formula (1) 3 (2) 4
(3) Physical formula (4) Chemical formula (3) 5 (4) 6
53. Which compound is optical active –
56. Which of the following is not a metamer of C4H10O
H OH
(1) Diethyl ether
(1) CH3 C COOH (2) CH3 C COOH (2) Methyl n-propyl ether
H H
(3) 2–Methoxy propane
CH3 CH3
(4) Isobutyl alcohol
(3) CH3 C COOH (4) CH3 C COOH 57. The number of optically active isomers of Glucose
OH Cl is :-
54. Geometrical isomerism is shown by : (1) 4 (2) 8
H CH3 I CH3
(1) C C (2) C C (3) 12 (4) 16
H H CH3 H
CH3 CH3 Br CH3
(3) C C (4) C C
CH3 H Br H

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ANSWER KEY EXERCISE-I

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

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ISOMERISM
PREVIOUS EXAMS QUESTIONS EXERCISE-II
1. The configuration of the given compound is :- 6. Among the following L–serine is – [AIIMS-2006]
COOH
Br
(1) H2N CH2OH
C H
H3C Cl
H
CO2H
(1) E (2) R [AIP MT-2 00 5]

(3) S (4) Z (2) HOHC

2
H
2. Which one of the following pairs represents stereo
isomerism :- [AIP MT-2 00 5] NH2
NH2
(1) Linkage isomerism and Geometrical isomerism
(2) Chain isomerism and Rotational isomerism (3) H CO2H
(3) Optical isomerism and Geometrical isomerism
(4) Structural isomerism and Geometrical isomerism CH2OH
CH2OH
3. The geometrical isomerism is shown by :-
[ AI I MS -2 0 04 ] (4) H2N H

CH2
CH2 CO2H
(1) (2)
7. Which of the following is optically active :-
[ AI IM S- 2 0 1 0 ]
CHCl
CHCl
(3) (4) OH OH
CH3
(1) (2)
4. Correct configuration of the following is :-

CH3 OH
H OH

CH3 OH
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OH CH3
[ AI I MS -2 0 05 ]
H
(3) (4)

(1) 2S, 3S (2) 2S, 3R

(3) 2R, 3S (4) 2R, 3R 8. Which of the following is not chiral :-
5. Among the following which one can have a meso
(1) 2–Butanol [AIP MT-2 00 6]
form – [ AI IM S- 2 0 0 6 ]
(2) 2,3–Dibromo pentane
(1) CH3CH(OH)CH(Cl)C2H 5
(2) CH3CH(OH)CH(OH)CH3 (3) 3–Bromo pentane

(3) C2H5CH(OH)CH(OH)CH3 (4) 2–Hydroxy propanoic acid

(4) HOCH2CH(Cl)CH3

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9. CH3—CHCl—CH2—CH3 has a chiral centre which
one of the following represents its
CH3
R configuration :- [AI PM T- 20 07 ] CH3
H H CH3

C2H5 C2H5
(3) (4)
CH3 H H
(1) H (2) Cl CH3 H
H
CH3 H H
Cl H
13. Which of the following conformers for ethylene
CH3 C2H5 glycol is most stable :- [AI PM T- 20 10 ]

(3) H Cl (4) H3C Cl

OH OH
OH H
C2H5 H
(1) (2)
H HO
H H
10. How many stereoisomer does this molecule have H H H H

CH 3CH=CHCH 2CHBrCH 3 [AI PM T- 20 08 ]

(1) 8 (2) 2 OH OH
H OH H H
(3) 4 (4) 6
(3) H
(4) H
11. Which of the following compounds will exhibit H H
cis-trans (geometrical) isomerism ? [AIP MT-2009] H OH

(1) 1-Butanol (2) 2-Butene 14. Find the number of stereo isomers of
(3) 2-Butanol (4) 2-Butyne 1,2-dihydroxy cyclopentane: [AIIMS-201 1]
12. In the following the most stable conformation of (1) 1 (2) 2
n–butane is :- [ AI P M T- 2 0 1 0 ] (3) 3 (4) 4
CH3 15. Which of the following acids does not exhibit optical
CH3
H H CH3 isomerism? [A IP M T P r e .- 20 1 2 ]

(1) Lactic acid (2) Tartaric acid

(1) (2)
H H H (3) Maleic acid (4) -amino acids
H H3C H
H

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ANSWER KEY EXERCISE-II

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

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ISOMERISM
ASSERTION & REASON QUESTIONS EXERCISE-III
These questions consist of two stateme nts each, pr inted as Assertion and Reason.
W hile answering the se Que stions you are required to choose any one of the follow ing
fo ur r e s po ns e s .
A. If b ot h Asser t ion & Re as on a re Tr ue & the Re as on i s a c or re ct e x pl a na t i on of
the A ss er ti on .
B. If b ot h Asser t ion & Re as on a re Tr u e bu t Re as on i s n ot a c or re c t ex pl a n a t io n
of t he A s s er ti on .
C. If A ss er t i on is Tr u e bu t t he Rea s on i s Fal se .
D. If b ot h Asser t ion & Re as on are Fal se .

1. Assertion :– The melting point of fumaric acid 7. Assertion :– Benzaldehyde forms two oximes on
is higher than that of maleic acid. reacting with NH 2OH.
Reason :– The molecules of fumaric acid are more Re a son :– The two oximes arise due to
symmetric than those of maleic acid and hence it geometrical isomerism around C N bond.
gets closely arranged in the crystal lattice. 8. Assertion :– The boiling point of cis 1, 2–dichloro
2. Assertion :– trans–1–chloro propene has higher ethene is higher than that of corresponding trans–
dipole moment than cis–1–chloro propene. isomer.
Re a son :– The resultant of two vectors in Reason :– Cis– 1,2–dichloro ethene has higher
trans–1–chloropropene is more than in cis–1–chloro moment as compared to that of the trans–isomer.
propene. 9. As s e rti on : – Meso tartaric acid is optically
inactive.
3. Assertion :– Enol form of cyclohexane–1,3,5–
trione is more stable than its ketoform. Reason :– Its optically inactivity is due to external
compensation.
Reason :– It contains –hydrogen atoms.
10. Assertion :– 1,2–propadiene exhibits optical
4. As se r t ion :– Trihydroxyglutaric acid (HCOO— isomerism.
CHOH—CHOH—CHOH—COOH) exists in four
Re a son :– Its mirror image is non–
stereoisomeric forms, two of which are optically
superimposable.
active while the other two are meso–forms.
11. As s erti on : - Ethanol cannot show position
Reason :– It contains two asymmetric and one isomerism.
pseudo–asymmetric carbon atom.
Reason :- Ethanol cannot show isomerism.
5. Assertion :– Organic compounds which do not 12. Assertion :- CH 3–CH 2–Br and CH 3–CH 2–I are
contain chiral carbon atoms can not be optically Functional isomers.
active.
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Reason :- They have different Functional groups.

Reason :– An organic compound is optically active 13. Ass ertion :- Neopentyl bromide and
only when its mirror image is non–superimposable 2–bromo–2–methyl butane are enantiomers.
irrespective of the fact whether it contains a chiral
Reason :- Both have same molecular formula as
carbon atom or not.
well as structure formula.
6. As s e rti on : – All the hydrogen atoms in 14. Assertion :- Organic compounds which do not
CH2 C CH 2 lie in one plane. contain chiral carbon atoms can be optically active.
Re a son :– All the carbon atoms in it are Reason :- Presence of chiral carbon is an essential
sp 2–hybridized. condition for optical activity.

ANSWER KEY EXERCISE- III

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ans. A A B A D D A A C D C D D C

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