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

The given molecular formula of compound ‘X’ is \( C_4H_9Br \), and it reacts with aqueous KOH to give an alcohol. Since the rate of reaction depends only on the concentration of compound ‘X’, this indicates an SN1 mechanism, which involves the formation of a carbocation intermediate. This suggests that compound ‘X’ is a primary alkyl halide or a structure that can form a stable carbocation. Compound 'Y' is optically active, meaning it has chirality and is likely to be a chiral isomer of compound ‘X’. Since the rate of reaction with aqueous KOH depends on both the concentration of compound ‘Y’ and KOH, this indicates an SN2 mechanism, where the nucleophile (OH⁻) attacks the carbon bearing the leaving group (Br) directly in a single step.

Structural formula of compound ‘X’:
Compound ‘X’ is 1-Bromobutane (CH₃-CH₂-CH₂-CH₂Br), which is a primary alkyl halide that undergoes an SN1 reaction.

Structural formula of compound ‘Y’:
Compound ‘Y’ is 2-Bromobutane (CH₃-CH(Br)-CH₂-CH₃), which is a secondary alkyl halide that undergoes an SN2 reaction.