Explain the mechanism of the following reaction: $CH_3Cl + OH^-_{(aq)} \to HO-CH_3 + Cl^-$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The reaction follows an $S_N2$ mechanism.
$1$. Kinetics: The rate of this reaction depends on the concentration of both the substrate $(CH_3Cl)$ and the nucleophile $(OH^-)$. Thus,it is a second-order (bimolecular) reaction: $\text{Rate} = k[CH_3Cl][OH^-]$.
$2$. Mechanism: It is a single-step concerted process. The nucleophile $(OH^-)$ attacks the electrophilic carbon from the side opposite to the leaving group $(Cl^-)$.
$3$. Transition State: As the $C-OH$ bond begins to form,the $C-Cl$ bond begins to break. This leads to a pentacoordinate transition state where the carbon is partially bonded to both the nucleophile and the leaving group.
$4$. Stereochemistry: The process results in the inversion of configuration (Walden inversion) at the carbon atom.

Explore More

Similar Questions

By passing excess $Cl_{2(g)}$ in boiling toluene, which one of the following compounds is exclusively formed?

Which one of the following is most reactive towards nucleophilic substitution reaction?

Consider the following reaction. Which of the following products is not expected to be formed?
$CH_3CH_2CH(Br)CH_3 + HC \equiv C^- Na^+ \rightarrow ?$

The major product obtained in the following reaction is $C_2H_5ONa + (CH_3)_3C-Cl \rightarrow$

The chlorine atom of the following compound that reacts most readily with $AgNO_3$ to give a precipitate is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo