CH_3-C(=O)-O-CH_2-CH_3 + H-bullet^- to (where | vedclass

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Question

(C) The reaction is a nucleophilic acyl substitution where the nucleophile $H-\bullet^-$ (where $\bullet = ^{18}O$) attacks the carbonyl carbon of the

Vedclass · Accepted Answer

$CH_3-C(=O)-\bullet^-$

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(C) The reaction is a nucleophilic acyl substitution where the nucleophile $H-\bullet^-$ (where $\bullet = ^{18}O$) attacks the carbonyl carbon of the ester $CH_3-C(=O)-O-CH_2-CH_3$.$1$. The nucleophile $H-\bullet^-$ attacks the electrophilic carbonyl carbon,leading to the formation of a tetrahedral intermediate.$2$. The tetrahedral intermediate collapses,expelling the ethoxide ion $(EtO^-)$ as a leaving group.$3$. This results in the formation of $CH_3-C(=O)-\bullet-H$ (an ester with $^{18}O$ in the carbonyl oxygen position).$4$. The ethoxide ion $(EtO^-)$ then acts as a base and abstracts the proton from the newly formed ester,resulting in the formation of $CH_3-C(=O)-\bullet^-$ and ethanol $(EtOH)$.Thus,one of the products is $CH_3-C(=O)-\bullet^-$.

$CH_3-C(=O)-O-CH_2-CH_3 + H-\bullet^- \to$ (where $\bullet = ^{18}O$) One of the products of the reaction is

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