Question

Identify correct set(s) of X, Y, Z in the following reaction sequence. \[ Styrene \xrightarrow{(C_6H_5CO)_2O_2, HBr} Product \xrightarrow{(ii)X} \xrightarrow{(iii)Y} Z \]

• A. I, II, III
• B. II, III only
• C. I, III only
• D. IV only

Hint:

Nitrile hydrolysis and Grignard carboxylation both produce carboxylic acids with same carbon count only if chain length is preserved.

Answer 1

The Correct Option is C

Step 1: Add HBr to styrene under peroxide.
With benzoyl peroxide, HBr adds to styrene by the anti-Markovnikov (Kharasch) route. The bromine goes to the terminal carbon: $C_6H_5CH=CH_2 \rightarrow C_6H_5CH_2CH_2Br$.
Step 2: Use X to make a nitrile (route I).
Treating the primary bromide with $KCN$ substitutes bromine by cyanide: $C_6H_5CH_2CH_2Br \xrightarrow{KCN} C_6H_5CH_2CH_2CN$. So $X=KCN$ is valid.
Step 3: Use Y to hydrolyse the nitrile.
Acidic hydrolysis $H_3O^+$ converts the nitrile to the carboxylic acid: $C_6H_5CH_2CH_2CN \rightarrow C_6H_5CH_2CH_2COOH$. This gives the target acid Z, so route I is correct.
Step 4: Check the Grignard route (route III).
The same bromide can form a Grignard reagent $RMgBr$ with $Mg$/dry ether, and carbonation with $CO_2$ followed by workup gives the same acid $C_6H_5CH_2CH_2COOH$. So route III is also valid.
Step 5: Reject the wrong route.
A route that adds an extra carbon to give $C_6H_5CH_2CH_2CH_2CO_2H$ (the higher homologue) does not match the required product, so that set (IV) is incorrect.
Step 6: Combine valid routes.
The valid sets are I and III, matching option (3).
\[ \boxed{\text{Sets I and III (Option 3)}} \]