The product (B) of the following reaction is: | vedclass

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(B) The given reaction is a variation of the Gattermann-Koch reaction or a form of Friedel-Crafts formylation using formyl chloride equivalent $(HN=CH

Vedclass · Accepted Answer

$4-$methylbenzaldehyde

Vedclass · Answer

(B) The given reaction is a variation of the Gattermann-Koch reaction or a form of Friedel-Crafts formylation using formyl chloride equivalent $(HN=CH-Cl)$.$1$. The reagent $HN=CH-Cl$ in the presence of $AlCl_3$ generates an electrophilic species $HN=CH^+$.$2$. This electrophile attacks the toluene ring. Since the methyl group $(-CH_3)$ is an ortho/para-directing group and para-position is sterically less hindered,the electrophile attacks the para-position to form an imine intermediate $(A)$.$3$. Subsequent hydrolysis of the imine intermediate $(A)$ with $H_3O^+$ yields the corresponding aldehyde,which is $4$-methylbenzaldehyde (p-tolualdehyde).

The product $(B)$ of the following reaction is:

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