The product (C) of the above reaction is: | vedclass

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(C) $1$. The starting material is a carboxylic acid protected as an acetal at the other end of the chain. $2$. Treatment with $CH_3Li$ (an organolithi

Vedclass · Accepted Answer

$CH_3-C(=O)-CH_2-CH_2-C(=O)-CH_3$

Vedclass · Answer

(C) $1$. The starting material is a carboxylic acid protected as an acetal at the other end of the chain. $2$. Treatment with $CH_3Li$ (an organolithium reagent) first deprotonates the carboxylic acid to form a lithium carboxylate salt. $3$. $A$ second equivalent of $CH_3Li$ attacks the carboxylate carbon to form a gem-diol lithium salt intermediate. $4$. Upon acidic workup $(H_3O^+)$,the acetal protecting group is hydrolyzed to a ketone,and the gem-diol salt is converted into a ketone. $5$. The final product $(C)$ is $CH_3-C(=O)-CH_2-CH_2-C(=O)-CH_3$ (hexane$-2,5-$dione).

The product $(C)$ of the above reaction is:

Similar Questions

Identify the wrong statement from the following.

Increasing order of boiling points in the following compounds is:
$I$. $CH_3COOH$
$II$. $CH_3CH_2CHO$
$III$. $CH_3CH_2CH_2OH$
$IV$. $CH_3COCH_3$

What is the product of the following reaction?

In the given reaction,$[X]$ will be:

What is the ratio of $(x/y)$ in the above reactions?

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