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

Concept: Styrene undergoes anti-Markovnikov addition of HBr in presence of peroxide (Kharasch effect), followed by functional group transformations leading to carboxylic acids.

Step 1: Formation of product. \[ C_6H_5CH=CH_2 \xrightarrow{HBr/peroxide} C_6H_5CH_2CH_2Br \]

Step 2: Identify X. Bromide undergoes nucleophilic substitution: \[ C_6H_5CH_2CH_2Br \xrightarrow{KCN} nitrile \] Thus X = KCN is correct for cyanide formation.

Step 3: Identify Y. Nitrile hydrolysis: \[ H_3O^+ \rightarrow COOH \] Thus formation of: \[ C_6H_5CH_2CH_2CO_2H \] So set I is valid.

Step 4: Check III. Grignard formation and carbonation: \[ Mg/dry\ ether \rightarrow RMgX \] \[ CO_2 \rightarrow RCOOH \] Also gives same acid. Thus III valid.

Step 5: Check IV. Leads to one extra carbon acid: \[ C_6H_5CH_2CH_2CH_2CO_2H \] So incorrect.

Step 6: Final answer. Valid sets: \[ I,\;III \] Hence: \[ \boxed{C} \]