Question

In the presence of peroxide, styrene reacts with \(HBr\) to give \(X\). When \(X\) reacts with magnesium in dry ether followed by \(CO_2\) and hydrolysis gave \(Y\). Treatment of \(Y\) with \(PCl_5\) and then next with \(H_2, Pd-BaSO_4\) gave \(Z\). What is \(Z\)?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In presence of peroxide, \(HBr\) adds to alkenes by anti-Markovnikov rule. Grignard reagent with \(CO_2\) gives carboxylic acid, and acid chloride with \(H_2/Pd-BaSO_4\) gives aldehyde by Rosenmund reduction.

Answer 1

The Correct Option is A

Step 1: Reaction of styrene with \(HBr\) in presence of peroxide.
Styrene is
\[ C_6H_5CH=CH_2 \] In the presence of peroxide, \(HBr\) adds by anti-Markovnikov rule.
Therefore, bromine attaches to the terminal carbon.
\[ C_6H_5CH=CH_2 \xrightarrow{HBr/ROOR} C_6H_5CH_2CH_2Br \] So,
\[ X=C_6H_5CH_2CH_2Br \]

Step 2: Formation of Grignard reagent.
When \(X\) reacts with magnesium in dry ether, it forms a Grignard reagent.
\[ C_6H_5CH_2CH_2Br \xrightarrow{Mg/dry\ ether} C_6H_5CH_2CH_2MgBr \]

Step 3: Reaction with \(CO_2\) followed by hydrolysis.
Grignard reagent reacts with \(CO_2\), followed by hydrolysis, to form a carboxylic acid with one extra carbon atom.
\[ C_6H_5CH_2CH_2MgBr \xrightarrow{CO_2/H_3O^+} C_6H_5CH_2CH_2COOH \] So,
\[ Y=C_6H_5CH_2CH_2COOH \]

Step 4: Reaction of \(Y\) with \(PCl_5\).
Carboxylic acid reacts with \(PCl_5\) to form acid chloride.
\[ C_6H_5CH_2CH_2COOH \xrightarrow{PCl_5} C_6H_5CH_2CH_2COCl \]

Step 5: Rosenmund reduction.
Acid chloride on treatment with \(H_2/Pd-BaSO_4\) undergoes Rosenmund reduction to form aldehyde.
\[ C_6H_5CH_2CH_2COCl \xrightarrow{H_2/Pd-BaSO_4} C_6H_5CH_2CH_2CHO \] So,
\[ Z=C_6H_5CH_2CH_2CHO \]

Step 6: Final conclusion.
Hence, the product \(Z\) is
\[ \boxed{C_6H_5CH_2CH_2CHO} \]