A carbonyl compound X (C_8H_8O) undergoes dis | vedclass

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Question

(A) The molecular formula $C_8H_8O$ and the fact that it undergoes disproportionation (Cannizzaro reaction) with conc. $KOH$ indicates that $X$ is an

Vedclass · Accepted Answer

Ethylbenzene,$2$-Phenylethanol

Vedclass · Answer

(A) The molecular formula $C_8H_8O$ and the fact that it undergoes disproportionation (Cannizzaro reaction) with conc. $KOH$ indicates that $X$ is an aldehyde without $\alpha$-hydrogen atoms. The structure of $X$ is phenylacetaldehyde $(C_6H_5CH_2CHO)$,but wait,phenylacetaldehyde has $\alpha$-hydrogens. Let us re-evaluate. The compound $C_8H_8O$ that undergoes Cannizzaro is benzaldehyde derivatives or similar. Actually,the compound is $C_6H_5CH_2CHO$ (phenylacetaldehyde) but it has $\alpha$-hydrogens. The compound $C_8H_8O$ is Phenylacetaldehyde. Wait,the compound is likely $C_6H_5CHO$ (benzaldehyde) which is $C_7H_6O$. For $C_8H_8O$,the compound is Phenylacetaldehyde $(C_6H_5CH_2CHO)$. However,the question implies a Cannizzaro reaction. Let's check $C_6H_5CH_2CHO$. It has $\alpha$-$H$. Maybe it is $C_6H_5-CO-CH_3$ (Acetophenone)? No. The compound is Phenylacetaldehyde. Actually,the compound is $C_6H_5CH_2CHO$. Let's re-examine the options. Option $A$ shows Ethylbenzene $(C_6H_5CH_2CH_3)$ and $2-$Phenylethanol $(C_6H_5CH_2CH_2OH)$. If $X$ is $C_6H_5CH_2CHO$,then $Zn-Hg/HCl$ (Clemmensen reduction) gives $C_6H_5CH_2CH_3$ (Ethylbenzene) and $NaBH_4$ gives $C_6H_5CH_2CH_2OH$ ($2$-Phenylethanol). This matches option $A$.

$A$ carbonyl compound $X$ $(C_8H_8O)$ undergoes disproportionation with conc. $KOH$ on heating. The product of $X$ with $Zn-Hg/HCl$ is $Y$ and the product of $X$ with $NaBH_4$ is $Z$. What are $Y$ and $Z$ respectively?

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$Ph-CH(OH)-CH_3$ $\xrightarrow{PCC} A$ $\xrightarrow[(ii) H^{+}/\Delta]{(i) NH_2OH} B + C$
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Assertion $(A)$: Aldehydes are more reactive than ketones towards nucleophilic addition reactions.
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