When carboxylic acid reacts with organolithiu | vedclass

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Question

(B) The reaction of a carboxylic acid with organolithium reagents involves multiple steps due to the acidity of the protons present.$1$. The first equ

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The reaction of a carboxylic acid with organolithium reagents involves multiple steps due to the acidity of the protons present.$1$. The first equivalent of $CH_3Li$ reacts with the carboxylic acid proton $(-COOH)$ to form a carboxylate salt $(R-COO^-Li^+)$ and methane $(CH_4)$.$2$. The second equivalent of $CH_3Li$ reacts with the hydroxyl proton $(-OH)$ of the $HOCH_2-$ group to form an alkoxide salt $(Li^+O^-CH_2-)$.$3$. The third equivalent of $CH_3Li$ attacks the carbonyl carbon of the carboxylate group to form a gem-diol dianion intermediate.$4$. Upon workup with $NH_4Cl/H_2O$,the intermediate is protonated to give the final ketone product.Since the substrate contains both a carboxylic acid group and a primary alcohol group,it requires $3$ equivalents of $CH_3Li$ to fully deprotonate both acidic sites and then attack the carbonyl group. Thus,$(x) = 3$.

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