Which of the following compounds is most reac | vedclass

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(B) Decarboxylation with sodalime $(NaOH + CaO)$ proceeds via the formation of a carbanion intermediate in the rate-determining step. The stability of

Vedclass · Accepted Answer

$O_2N-CH_2-COOH$

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(B) Decarboxylation with sodalime $(NaOH + CaO)$ proceeds via the formation of a carbanion intermediate in the rate-determining step. The stability of the carbanion determines the rate of decarboxylation. Electron-withdrawing groups (EWGs) stabilize the carbanion by dispersing the negative charge,thereby increasing the rate of decarboxylation. Among the given options,the nitro group $(-NO_2)$ in $O_2N-CH_2-COOH$ is a strong electron-withdrawing group,which significantly stabilizes the resulting carbanion $(O_2N-CH_2^-)$. Therefore,$O_2N-CH_2-COOH$ is the most reactive toward decarboxylation.

Which of the following compounds is most reactive toward decarboxylation when reacting with sodalime?

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