The reaction of 50% aq KOH on an equimolar mi | vedclass

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(B) The reaction involves a cross-Cannizzaro reaction between $4-$methylbenzaldehyde (which has no $\alpha-$hydrogen) and formaldehyde (which also has

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(B) The reaction involves a cross-Cannizzaro reaction between $4-$methylbenzaldehyde (which has no $\alpha-$hydrogen) and formaldehyde (which also has no $\alpha-$hydrogen). In a cross-Cannizzaro reaction,the more reactive aldehyde (formaldehyde) is oxidized to the corresponding acid salt (formate),and the less reactive aldehyde ($4-$methylbenzaldehyde) is reduced to the corresponding alcohol ($4-$methylbenzyl alcohol). Reaction: $CH_3-C_6H_4-CHO + HCHO + KOH \rightarrow CH_3-C_6H_4-CH_2OH + HCOOK$. Upon acidification,the formate salt $HCOOK$ converts to $HCOOH$ and the alcohol remains as $4-$methylbenzyl alcohol. Thus,the products are $4-$methylbenzyl alcohol and formic acid.

The reaction of $50\%$ aq $KOH$ on an equimolar mixture of $4-$methylbenzaldehyde and formaldehyde followed by acidification gives

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