Question

Observe the following reaction sequence \[ \text{Styrene} \xrightarrow[\left(\mathrm{C_6H_5CO}\right)_2\mathrm{O_2}]{\mathrm{HBr}} X \xrightarrow[(ii)\ \mathrm{H_3O^+}]{(i)\ \mathrm{KCN}} Y \xrightarrow[(ii)\ \mathrm{H_2O}]{(i)\ \mathrm{Br_2/red\ P}} Z \] The correct statement regarding \(Y\) and \(Z\) is

• A. \(Y\) is stronger acid than benzoic acid
• B. \(Z\) is stronger acid than \(Y\)
• C. \(Y\) can be reduced with \(\mathrm{NaBH_4}\)
• D. \(Z\) on decarboxylation gives benzyl bromide

Hint:

The \(\alpha\)-halogen atom exerts a strong \(-I\) effect, increasing the acidity of carboxylic acids. \[ \boxed{ \alpha\text{-halo acid}>\text{corresponding carboxylic acid} } \]

Answer 1

The Correct Option is B

Step 1: Formation of \(X\). Peroxide effect with HBr gives anti-Markovnikov addition. \[ \mathrm{C_6H_5CH=CH_2} \xrightarrow{\mathrm{HBr/peroxide}} \boxed{\mathrm{C_6H_5CH_2CH_2Br}} \]

Step 2: Formation of \(Y\). \[ \mathrm{C_6H_5CH_2CH_2Br} \xrightarrow{\mathrm{KCN}} \mathrm{C_6H_5CH_2CH_2CN} \xrightarrow{\mathrm{H_3O^+}} \boxed{\mathrm{C_6H_5CH_2CH_2COOH}} \] Thus, \[ Y=\text{3-phenylpropanoic acid.} \]

Step 3: Formation of \(Z\). Hell-Volhard-Zelinsky reaction introduces bromine at the \(\alpha\)-carbon. \[ \boxed{ \mathrm{C_6H_5CH_2CHBrCOOH} } \] The electron-withdrawing Br atom increases acidity. Hence, \[ \boxed{Z\text{ is a stronger acid than }Y.} \] Therefore, \[ \boxed{(B)} \] is the correct answer.