Question

Consider all the structural isomers with molecular formula C₃H₅Br are separately treated with KOH(aq) to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is_______.

Hint:

To be optically active, a molecule usually needs a chiral center. In a 3-carbon chain with a double bond, there aren't enough atoms to satisfy the "4 different groups" requirement on a single carbon.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
First, identify all structural isomers of $C_3H_5Br$ (degree of unsaturation = 1, so one double bond or one ring). Then, replace the Br with OH (substitution) and check the resulting products for chiral centers.
Step 2: Detailed Explanation:
Structural isomers of $C_3H_5Br$: 1. 3-bromoprop-1-ene ($CH_2=CH-CH_2Br$) $\rightarrow$ $CH_2=CH-CH_2OH$ (Achiral). 2. 2-bromoprop-1-ene ($CH_2=C(Br)CH_3$) $\rightarrow$ $CH_2=C(OH)CH_3$ (Enol, achiral). 3. 1-bromoprop-1-ene ($CHBr=CH-CH_3$) $\rightarrow$ $CH(OH)=CH-CH_3$ (Enol, achiral). 4. Bromocyclopropane (Cyclic) $\rightarrow$ Cyclopropanol (Achiral). Wait, let's re-examine for chirality in the alcohol products: None of the open-chain unsaturated alcohols from $C_3H_5Br$ substitution contain a chiral carbon. However, if we consider substitution on a saturated carbon in a way that creates a chiral center: The only way to have a chiral center in a 3-carbon system with an -OH is to have 4 different groups on a carbon. In $CH_3-CH(OH)-...$ we need a third carbon. In $C_3H_5OH$, there are no chiral centers.
Step 3: Re-evaluating the Isomers:
Actually, if we look at the cyclic isomers: - 1-bromo-1-methyl... (Not possible with 3 carbons). - If the question implies the number of isomers of the reactant that are chiral, or products, let's check: All standard products of $C_3H_5Br + KOH(aq)$ (substitution) are achiral. However, in many JEE-style problems of this type, the answer is often 0.
Step 4: Final Answer:
The number of products exhibiting optical isomerism is 0.