m-chlorobenzaldehyde on treatment with 50 % K | vedclass

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(B) $m$-chlorobenzaldehyde does not have any $\alpha$-hydrogen atom. Therefore,it undergoes the Cannizzaro reaction in the presence of a concentrated

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(B) $m$-chlorobenzaldehyde does not have any $\alpha$-hydrogen atom. Therefore,it undergoes the Cannizzaro reaction in the presence of a concentrated base like $50 \% KOH$. In this reaction,one molecule of the aldehyde is oxidized to the corresponding carboxylate ion ($m$-chlorobenzoate) and another molecule is reduced to the corresponding alcohol ($m$-chlorobenzyl alcohol). The reaction is as follows:$2 	ext{ } m	ext{-Cl-C}_6	ext{H}_4	ext{CHO} + KOH$ $ightarrow m	ext{-Cl-C}_6	ext{H}_4	ext{COO}^- 	ext{K}^+ + m	ext{-Cl-C}_6	ext{H}_4	ext{CH}_2	ext{OH}$

List-$I$	List-$II$
$A$. Grignard reagent	$1$. $H_2 / Pd-BaSO_4$
$B$. Clemmensen reduction	$2$. $N_2H_4 / KOH / (CH_2OH)_2$
$C$. Rosenmund reduction	$3$. $CH_3MgX$
$D$. Wolff-Kishner reduction	$4$. $Zn-Hg / \text{conc. } HCl$
	$5$. $H_2 / Ni$

$m$-chlorobenzaldehyde on treatment with $50 \% KOH$ solution yields

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Match the following lists:
List-$I$ List-$II$
$A$. Grignard reagent $1$. $H_2 / Pd-BaSO_4$
$B$. Clemmensen reduction $2$. $N_2H_4 / KOH / (CH_2OH)_2$
$C$. Rosenmund reduction $3$. $CH_3MgX$
$D$. Wolff-Kishner reduction $4$. $Zn-Hg / \text{conc. } HCl$
$5$. $H_2 / Ni$

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