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 \(Z\)?
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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.
Step 1: Add HBr to styrene with peroxide. Styrene is $C_6H_5CH=CH_2$. With HBr in the presence of peroxide the addition follows the anti-Markovnikov rule, so bromine goes to the terminal carbon: \[ C_6H_5CH=CH_2 \xrightarrow{HBr/ROOR} C_6H_5CH_2CH_2Br \] giving $X = C_6H_5CH_2CH_2Br$. Step 2: Make the Grignard reagent. With magnesium in dry ether, $X$ becomes the Grignard reagent $C_6H_5CH_2CH_2MgBr$. Step 3: React with $CO_2$ then hydrolyse. The Grignard reagent adds to $CO_2$ and, after hydrolysis, gives a carboxylic acid with one extra carbon: \[ C_6H_5CH_2CH_2MgBr \xrightarrow{CO_2/H_3O^+} C_6H_5CH_2CH_2COOH \] so $Y = C_6H_5CH_2CH_2COOH$. Step 4: Convert acid to acid chloride. Treatment with $PCl_5$ turns the $-COOH$ into $-COCl$, giving $C_6H_5CH_2CH_2COCl$. Step 5: Apply Rosenmund reduction. The acid chloride with $H_2$ over $Pd\text{-}BaSO_4$ undergoes Rosenmund reduction to an aldehyde: \[ C_6H_5CH_2CH_2COCl \xrightarrow{H_2/Pd\text{-}BaSO_4} C_6H_5CH_2CH_2CHO \] Step 6: Identify Z. Therefore $Z = C_6H_5CH_2CH_2CHO$, which is the answer marked option 1. \[ \boxed{C_6H_5CH_2CH_2CHO} \]
