m-Aminophenol on treatment with NaOH and CO_2 | vedclass

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Question

(A) The reaction of $m$-aminophenol with $NaOH$ and $CO_2$ is a Kolbe-Schmitt reaction. In $m$-aminophenol,the $-OH$ group is a strong ortho/para-dire

Vedclass · Accepted Answer

$4-$amino$-2-$hydroxybenzoic acid

Vedclass · Answer

(A) The reaction of $m$-aminophenol with $NaOH$ and $CO_2$ is a Kolbe-Schmitt reaction. In $m$-aminophenol,the $-OH$ group is a strong ortho/para-directing group,and the $-NH_2$ group is also an ortho/para-directing group. The $-OH$ group is more activating than the $-NH_2$ group. Electrophilic substitution by $CO_2$ occurs at the position ortho to the $-OH$ group. In $m$-aminophenol,the positions ortho to the $-OH$ group are positions $2$ and $6$. Position $2$ is between the $-OH$ and $-NH_2$ groups,which is sterically hindered. Position $6$ is ortho to the $-OH$ group and para to the $-NH_2$ group. Therefore,the electrophilic attack occurs at the $6-$position (relative to $-OH$ at $1$),which corresponds to the $4-$position relative to the $-NH_2$ group. The major product is $4$-amino-$2$-hydroxybenzoic acid.

List-$I$ (Reactants)	List-$II$ (Products)
$A$. Phenol,$Zn / \Delta$	$I$. Salicylaldehyde
$B$. Phenol,$CHCl_3, NaOH, HCl$	$II$. Salicylic acid
$C$. Phenol,$CO_2, NaOH, HCl$	$III$. Benzene
$D$. Phenol,Conc. $HNO_3$	$IV$. Picric acid

$m$-Aminophenol on treatment with $NaOH$ and $CO_2$ gives which of the following as major product?

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Match List-$I$ with List-$II$.

List-$I$ (Reactants) List-$II$ (Products)

$A$. Phenol,$Zn / \Delta$ $I$. Salicylaldehyde

$B$. Phenol,$CHCl_3, NaOH, HCl$ $II$. Salicylic acid

$C$. Phenol,$CO_2, NaOH, HCl$ $III$. Benzene

$D$. Phenol,Conc. $HNO_3$ $IV$. Picric acid

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