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

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(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

$C_{6}H_{5}COCl$

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(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 $C_{6}H_{5}COCl$.

List-$I$	List-$II$
$1.$ Ammoniacal $AgNO_3$	$a.$ Primary amine
$2.$ $HIO_4$	$b.$ Aldehyde
$3.$ Alkaline $KMnO_4$	$c.$ Vicinal-$OH$
$4.$ Chloroform $NaOH$	$d.$ Double bond

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

Similar Questions

Match List-$I$ with List-$II$.
List-$I$ List-$II$
$1.$ Ammoniacal $AgNO_3$ $a.$ Primary amine
$2.$ $HIO_4$ $b.$ Aldehyde
$3.$ Alkaline $KMnO_4$ $c.$ Vicinal-$OH$
$4.$ Chloroform $NaOH$ $d.$ Double bond

Identify compound $A$ from the following reaction:
$A + C_2H_5MgBr$ $\xrightarrow[ether]{dry} B$ $\xrightarrow{H_3O^{+}} 3-\text{methylpentan}-3-\text{ol}$

Iodoform test is not given by

The reaction of $1-(\text{aminomethyl})cyclohexan-1-ol$ with $NaNO_2/HCl$ gives $A$. Identify $A$.

Which of the following has the lowest enol content?

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