JEE Advanced Chemistry: Stereochemistry questions with solutions

Q1. Two Fischer projections are drawn for a compound with adjacent stereocentres bearing CHO, OH, H and CH3 substituents. One projection is obtained from the other by rotating the entire formula 180 degrees in the plane of the paper. What is the stereochemical relationship between the two representations?

1. Identical compounds
2. Enantiomers
3. Diastereomers
4. Conformers

Answer: Identical compounds

In Fischer projection conventions, rotating the entire projection by 180 degrees in the plane of the paper does NOT change the configuration at any stereocentre. This is because each 90 deg rotation swaps horizontal and vertical substituents at every carbon (inverting configuration), and two such swaps restore the original. Therefore the two structures represent exactly the same compound with identical configurations at all chiral centres. They are identical, not enantiomers or diastereomers.

Q2. Consider the compound X: CH3-CH(OH)-CH=CH-CH(OH)-CH3. Which of the following statements about the stereoisomers of X are correct?

1. The total number of stereoisomers possible for X is 6
2. The total number of diastereomers possible for X is 3
3. If the double bond in X has trans (E) configuration, the number of enantiomers possible for X is 4
4. If the double bond in X has cis (Z) configuration, the number of enantiomers possible for X is 2

Answer: If the double bond in X has cis (Z) configuration, the number of enantiomers possible for X is 2

X has two chiral centres and a C=C double bond (E/Z). Total stereoisomers = 2 (E/Z) x 2² (chiral centres) = 8, but meso forms reduce this. The cis isomer has one meso form, giving fewer stereoisomers for that geometry.

Q3. Which of the following compounds shows geometrical isomerism?

1. Benzene
2. Decalin
3. O-Xylene
4. None of these

Answer: Decalin

Benzene and o-xylene are planar aromatic compounds with no restricted rotation giving rise to geometric isomers. Decalin has two fused six-membered rings sharing a C-C bond (the ring junction), and the two hydrogen atoms at the junction can be cis or trans, giving cis-decalin and trans-decalin — classic geometric (cis-trans) isomerism.

Q4. Which of the following compounds does NOT show geometrical isomerism?

1. ClHC=CClH (each double-bond carbon bears Cl and H)
2. ClHC=C(Cl)(CH3) (one carbon bears Cl and H; the other bears Cl and CH3)
3. A cyclopropane ring fused to a double bond such that one double-bond carbon belongs to the ring and bears two ring-carbons as substituents, while the other bears Cl and H
4. One double-bond carbon bears CH3 and cyclopropyl; the other bears CH3 and H

Answer: A cyclopropane ring fused to a double bond such that one double-bond carbon belongs to the ring and bears two ring-carbons as substituents, while the other bears Cl and H

For a compound to show geometrical isomerism, both doubly-bonded carbons must each carry two different substituents. In an endocyclic double bond within cyclopropane, one carbon of the double bond is attached to two equivalent ring carbons, making its two substituents identical and eliminating the possibility of geometrical isomerism.

Q5. Two acyclic molecules with molecular formula C8H8 are such that one shows enantiomerism and the other shows diastereoisomerism. The structural relationship between these two molecules is:

1. Chain Isomers
2. Positional Isomers
3. Geometrical Isomers
4. Conformation isomers

Answer: Positional Isomers

Both molecules are acyclic constitutional isomers of C8H8. They have the same type of functional groups but differ in the position of double bonds or substituents — making them positional isomers. This arrangement allows one to possess a chiral centre (enantiomerism) while the other has restricted rotation about a double bond (geometric/diastereoisomerism).

Q6. Which of the following is a meso compound? (A) HOOC-CH(Cl)-CH(Cl)-COOH, with the two Cl atoms on opposite sides (anti configuration in Fischer projection) (B) A methyl-substituted alkane (structure not specified here) (C) HOOC-CH(CH3)-CH(Cl)-COOH with CH3 and Cl as substituents (D) cis-1,3-dibromocyclobutane

1. HOOC-CH(Cl)-CH(Cl)-COOH, with the two Cl atoms on opposite sides as shown in option (A)
2. Me-substituted alkane shown in option (B)
3. HOOC-CH(CH3)-CH(Cl)-COOH, with CH3 and Cl as shown in option (C)
4. 1,3-dibromocyclobutane shown in option (D)

Answer: HOOC-CH(Cl)-CH(Cl)-COOH, with the two Cl atoms on opposite sides as shown in option (A)

Option A is meso-2,3-dichlorosuccinic acid: the molecule has two identical chiral centres in R,S (anti) configuration, giving an internal plane of symmetry and hence no net optical activity. Option C has two different substituents (CH3 and Cl) so no such internal symmetry exists.

Q7. Which of the following compounds can be resolved into enantiomers (i.e., is optically active and resolvable)? (A) 3-chlorocyclohexene (achiral due to ring-flip averaging) (B) N-methyl-N-ethylamine (lone pair on N inverts rapidly, so nitrogen is not a stable chiral center) (C) trans-1,2-dichlorocyclopropane (D) A symmetric dioxolane ring bearing two identical chloro substituents

1. 3-chlorocyclohexene
2. N-methyl-N-ethylamine (H, Me, Et on N)
3. trans-1,2-dichlorocyclopropane
4. Symmetric dichloro dioxolane

Answer: trans-1,2-dichlorocyclopropane

trans-1,2-dichlorocyclopropane has C1 (R) and C2 (S) — but because both substituents and the ring connections are different, there is no internal mirror plane. The molecule is chiral and exists as a pair of enantiomers that can be resolved. The cis isomer would be meso (internal mirror plane).