The reaction of methyltrichloroacetate (Cl_{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction of methyltrichloroacetate $(Cl_{3}CCO_{2}Me)$ with sodium methoxide $(NaOMe)$ involves the nucleophilic attack of the methoxide ion $

Vedclass · Accepted Answer

carbene

Vedclass · Answer

(B) The reaction of methyltrichloroacetate $(Cl_{3}CCO_{2}Me)$ with sodium methoxide $(NaOMe)$ involves the nucleophilic attack of the methoxide ion $(^-OMe)$ on the carbonyl carbon of the ester.This leads to the formation of a tetrahedral intermediate, which then collapses to eliminate the trichloromethyl carbanion $(^-CCl_{3})$.The trichloromethyl carbanion $(^-CCl_{3})$ is unstable and subsequently loses a chloride ion $(Cl^-)$ to generate dichlorocarbene $(:CCl_{2})$.Therefore, the final reactive intermediate generated in this process is a carbene.

The reaction of methyltrichloroacetate $(Cl_{3}CCO_{2}Me)$ with sodium methoxide $(NaOMe)$ generates

Similar Questions

Which of the following carbocations is the most stable?

The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$

The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical,respectively,are due to

Find the correct stability order for the following carbocations:

Using curved-arrow notation,show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage:
$(a)$ $CH_{3}-SCH_{3}$
$(b)$ $CH_{3}-CN$
$(c)$ $CH_{3}-Cu$

Vedclass Test Series

Exam Paper Generator

Online Exam Module