Question

Compound ‘X’ with molecular formula \( C_4H_9Br \) reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of compound ‘X’. When an optically active isomer ‘Y’ of the compound ‘X’ was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound ‘Y’ and aqueous KOH both.

22(a). Write down the structural formula of both ‘X’ and ‘Y’.

Hint:

To identify $S_N1$ vs $S_N2$ from rate laws: If the rate depends on one reactant, it's $S_N1$ ($3^\circ$ halide); if two, it's $S_N2$ ($1^\circ$ or $2^\circ$ halide).

Answer 1

Step 1: Understanding the Concept:
Reactions where the rate depends only on the concentration of the alkyl halide follow the \( S_N1 \) mechanism (Unimolecular).
Reactions where the rate depends on both the alkyl halide and the nucleophile follow the \( S_N2 \) mechanism (Bimolecular).
Step 2: Detailed Explanation:
1. Compound 'X': The rate depends only on [X], so it is a \( 3^\circ \) alkyl halide. The only \( 3^\circ \) isomer of \( C_4H_9Br \) is 2-Bromo-2-methylpropane (tert-butyl bromide).
Structural formula: \( (CH_3)_3C-Br \).
2. Compound 'Y': The rate depends on both [Y] and [KOH], so it follows \( S_N2 \). It is stated to be optically active. Among isomers of \( C_4H_9Br \), only 2-Bromobutane has a chiral center (C2 is attached to \( -H, -CH_3, -C_2H_5, -Br \)).
Structural formula: \( CH_3-\text{CH}(Br)-CH_2-CH_3 \).
Step 3: Final Answer:
The structural formula of X is \( (CH_3)_3CBr \) and Y is \( CH_3CH(Br)CH_2CH_3 \).