JEE Main 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.