Ph-CH_2-CN xrightarrow[(1) EtONa]{(2) CH_3-CO | vedclass

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Question

(B) Step $1$: The base $EtONa$ abstracts an acidic $\alpha$-hydrogen from $Ph-CH_2-CN$ to form a carbanion,$Ph-CH^{\ominus}-CN$.Step $2$: This carbani

Vedclass · Accepted Answer

$Ph-CH_2-CO-CH_3$

Vedclass · Answer

(B) Step $1$: The base $EtONa$ abstracts an acidic $\alpha$-hydrogen from $Ph-CH_2-CN$ to form a carbanion,$Ph-CH^{\ominus}-CN$.Step $2$: This carbanion acts as a nucleophile and attacks the electrophilic carbonyl carbon of acetyl chloride $(CH_3-COCl)$,resulting in the formation of $Ph-CH(CN)-CO-CH_3$.Step $3$: Acidic hydrolysis $(H_3O^{\oplus})$ converts the cyano group $(-CN)$ into a carboxylic acid group $(-COOH)$,yielding $Ph-CH(COOH)-CO-CH_3$,which is a $\beta$-keto acid.Step $4$: Upon heating $(\Delta)$,the $\beta$-keto acid undergoes decarboxylation (loss of $CO_2$) to form the final product,$Ph-CH_2-CO-CH_3$ (phenylacetone).

$Ph-CH_2-CN \xrightarrow[(1) EtONa]{(2) CH_3-COCl, (3) H_3O^{\oplus}/\Delta} (P)$; Product $(P)$ of the reaction will be

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