Which of the following shows tautomerism?

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(D) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom on a carbon atom adjacent to a carbonyl group $(-C=O)$.$A$. $(CH_3)_3C-CH

Vedclass · Accepted Answer

Phenylacetaldehyde $(C_6H_5CH_2CHO)$

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(D) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom on a carbon atom adjacent to a carbonyl group $(-C=O)$.$A$. $(CH_3)_3C-CHO$: The carbon adjacent to the carbonyl group is a tertiary carbon with no $\alpha$-hydrogen. Thus,it does not show tautomerism.$B$. $4,4-$dimethylcyclohexa$-2,5-$dien$-1-$one: The $\alpha$-carbons (positions $2$ and $6$) are part of double bonds and have no $\alpha$-hydrogen. Thus,it does not show tautomerism.$C$. $7,7-$dimethylbicyclo[$2.2$.$1$]heptan$-2-$one: The $\alpha$-carbons (positions $1$ and $3$) have no $\alpha$-hydrogen available for tautomerization. Thus,it does not show tautomerism.$D$. Phenylacetaldehyde $(C_6H_5CH_2CHO)$: The $\alpha$-carbon (the carbon in the $-CH_2-$ group) has two $\alpha$-hydrogen atoms. Therefore,it can undergo keto-enol tautomerism to form $C_6H_5CH=CHOH$. Hence,it shows tautomerism.

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