Which is most reactive towards soda-lime deca | vedclass

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(B) The decarboxylation of carboxylic acids with soda-lime $(NaOH + CaO)$ proceeds through the formation of a carbanion intermediate. The rate of deca

Vedclass · Accepted Answer

$p-O_2N-C_6H_4-CH_2-COOH$

Vedclass · Answer

(B) The decarboxylation of carboxylic acids with soda-lime $(NaOH + CaO)$ proceeds through the formation of a carbanion intermediate. The rate of decarboxylation is directly proportional to the stability of the carbanion formed after the loss of $CO_2$. Electron-withdrawing groups (EWGs) stabilize the carbanion,thereby increasing the rate of decarboxylation,while electron-donating groups (EDGs) destabilize it. Comparing the substituents: $1$. $-NO_2$ is a strong electron-withdrawing group $(EWG)$ due to both $-I$ and $-M$ effects. $2$. $-OCH_3$ is an electron-donating group $(EDG)$ due to its $+M$ effect. Between the nitro-substituted compounds,the $-NO_2$ group at the ortho or para position exerts a stronger electron-withdrawing effect via resonance ($-M$ effect) compared to the meta position. However,in the case of $p-O_2N-C_6H_4-CH_2-COOH$,the $-NO_2$ group is at the para position relative to the $-CH_2COOH$ group. The electron-withdrawing effect of the $-NO_2$ group stabilizes the negative charge on the benzylic carbon formed during the transition state. Thus,$p-O_2N-C_6H_4-CH_2-COOH$ is the most reactive.

Which is most reactive towards soda-lime decarboxylation?

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