P xrightarrow{H_2 / Pd-BaSO_4} Q xrightarrow{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. The reaction $P \stackrel{H_{2} / Pd-BaSO_{4}}{\longrightarrow} Q$ is the Rosenmund reduction,which converts an acid chloride $(RCOCl)$ to an

Vedclass · Accepted Answer

$C_{6}H_{5}COCl$

Vedclass · Answer

(A) $1$. The reaction $P \stackrel{H_{2} / Pd-BaSO_{4}}{\longrightarrow} Q$ is the Rosenmund reduction,which converts an acid chloride $(RCOCl)$ to an aldehyde $(RCHO)$. Thus,$P$ is $C_{6}H_{5}COCl$ (benzoyl chloride) and $Q$ is $C_{6}H_{5}CHO$ (benzaldehyde).
$2$. The reaction of $Q$ (benzaldehyde) with conc. $NaOH$ is the Cannizzaro reaction,which produces a carboxylic acid salt and an alcohol. Upon acidification,we get $R$ ($C_{6}H_{5}COOH$,benzoic acid) and $S$ ($C_{6}H_{5}CH_{2}OH$,benzyl alcohol).
$3$. Benzoic acid $(R)$ and benzyl alcohol $(S)$ undergo esterification to form benzyl benzoate $(C_{6}H_{5}COOCH_{2}C_{6}H_{5})$.
$4$. Therefore,$P$ is $C_{6}H_{5}COCl$.

$P \xrightarrow{H_2 / Pd-BaSO_4} Q \xrightarrow{(i) \text{ conc. } NaOH} R + S$. $R$ and $S$ form benzyl benzoate when treated with each other. Hence,$P$ is

Similar Questions

In the scheme given below,the total number of intra molecular aldol condensation products formed from '$Y$' is

Which of the following is reduced by $Zn-Hg/HCl$ to give the corresponding hydrocarbon?

The correct order of rate of reaction towards nucleophilic addition reaction for the following compounds is:

When an aldehyde is treated with an excess of monohydric alcohol,what is formed?

The reaction of $CH_3CH_2CHO$ with $PCl_5$ yields a product. Identify the type of dihalide formed.

Vedclass Test Series

Exam Paper Generator

Online Exam Module