GENERAL ORGANIC CHEMISTRY
7XWRULDO²10
Optical Isomerism : Chiral Compounds with Chiral Carbons, Chirality in Cyclic Compounds, Resolution,
Racemization, Number of Optical Isomers, Conformers in Cyclic Compounds
1.
Draw structures of all stereoisomers of the compound 4-chloro-2-pentene and select pairs of enantiomer and diastereomers.
2.
Draw all possible structures of chloro bromo cyclopropane and label them as optically active, if any.
3.
Between the two conformational isomers of cyclohexane, i.e., chair and boat forms, which one is more stable and why ?
4.
(2R, 3R)-2, 3-butanediol is
(a)
5.
(b)
(c)
(d)
(i) In each of the following molecules, indicate the presence of a centre of chirality with an asterisk (*)
(a)
(b)
(d)
(e)
(c)
(ii) Star (*) each asymmetric carbon atom in the following structure :
6.
Pure enantiomeric acid + optically active alcohol having chiral C atom
?
The product will be:
7.
(a) An optically active compound
(b) A meso compound
(c) A racemic mixture
(d) A pure enantiomer
For the resolution of a racemic mixture, convertion into a mixture of suitable diasteriomers helps because
(a) boiling points of diastereomers are always different
(b) boiling points of diastereomers are always same
(c) boiling points of diastereomers are same in some cases but different in other cases
(d) diastereomers have same reactivity with a chiral reagent
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8.
9.
10.
The specific rotation of (R)-(–)-2-bromooctane is –36°. What is the percentage composition of a mixture of enantiomers of 2bromooctane whose rotation is +18° ?
Which of the following compounds has asymmetric centre ?
(a)
(b)
(c)
(d)
Among chair, boat and twist boat conformers of cyclohexane, stability order is
(a) chair > boat > twist boat
(b) chair > twist boat > boat
(c) boat > chair > twist boat
(d) twist boat > boat > chair
11.
Isopentane, (CH3)2CH.CH2.CH3 can form four isomeric monochloro derivatives. How many of these are optically active ?
12.
Out of the following which are chiral
(a) 1
(a) I, II
13.
(d) 4
(c) I, III
(d) only III
(b) 2
(c) 3
Number of stereoisomeric forms of the compounds CH3–CH=CH–CHBr–CH3 is
(a) 3
15.
(b) II, III
(c) 3
How many optically active stereoisomers are possible for butane – 2, 3 – diol ?
(a) 1
14.
(b) 2
(b) 6
(d) 4
(c) 2
(d) 4
(c)
(d)
Which of the following is optically active ?
(a)
(b)
Answer Key
4. (a)
5. (ii) There are three asymmetric carbons, marked with asterisks.
1. The (CHOH) carbon of the side chain is asymmetric. Its four substituents are the ring, a hydrogen atom, a hydroxyl group,
and a methyl group.
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2. Carbon atom C1 of the ring is asymmetric. Its four substituents are the side chain, a hydrogen atom, the part of the ring
closer to the chlorine atom (–– CH2 –– CHCl ––), and the part of the ring farther from the chlorine atom (–– CH 2 –– CH2
–– CH2 –– CHCl ––).
3. The ring carbon bearing the chlorine atom is asymmetric. Its four substituents are the chlorine atom, a hydrogen atom, the
part of the ring closer to the side chain, and the part of the ring farther from the side chain.
Notice that different groups might be different in any manner. For example, the ring carbon bearing the chlorine atom is
asymmetric even though two of its ring substituents initially appear to be –– CH 2 –– groups. These two parts of the ring
are different because one is closer to the side chain and one is farther away. The entire structure of the group must be
considered.
6. (a)
7. (a)
8. Let x = mole fraction of R, 1 – x = mole fraction of S. x (–36°) + (1 – x) (36°) = 18° or x = ¼
The mixture has 25% R and 75% S; it is 50% racemic and 50% S.
9. (b)
10. (b)
11. (b)
12. (b)
14.
13. (b)
15. (b)
Dream on !!
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