Ph-COOH xrightarrow{SOCl_2} (A) xrightarrow{H | vedclass

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Question

(A) Step $1$: Benzoic acid $(Ph-COOH)$ reacts with $SOCl_2$ to form benzoyl chloride $(Ph-COCl)$ as product $(A)$.$Ph-COOH + SOCl_2 \rightarrow Ph-COC

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$Ph-CHO$

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(A) Step $1$: Benzoic acid $(Ph-COOH)$ reacts with $SOCl_2$ to form benzoyl chloride $(Ph-COCl)$ as product $(A)$.$Ph-COOH + SOCl_2 \rightarrow Ph-COCl + SO_2 + HCl$Step $2$: Benzoyl chloride undergoes Rosenmund reduction in the presence of $H_2$ and $Pd-BaSO_4$ to form benzaldehyde $(Ph-CHO)$ as product $(B)$.$Ph-COCl + H_2 \xrightarrow{Pd-BaSO_4} Ph-CHO + HCl$

$Ph-COOH$ $\xrightarrow{SOCl_2} (A)$ $\xrightarrow{H_2, Pd-BaSO_4} (B)$. Product $(B)$ is:

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