Represent the structure of the free radical,carbocation,and carbanion derived from methane.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The formation of reactive intermediates from methane $(CH_4)$ occurs via bond cleavage:
$1$. Homolytic cleavage: The covalent bond breaks equally,resulting in the formation of a methyl free radical $(CH_3^{\bullet})$ and a hydrogen radical $(H^{\bullet})$.
$2$. Heterolytic cleavage: The covalent bond breaks unequally,where the shared electron pair stays with one fragment.
- If the carbon atom retains the electron pair,a methyl carbanion $(CH_3^-)$ and a proton $(H^+)$ are formed.
- If the hydrogen atom retains the electron pair,a methyl carbocation $(CH_3^+)$ and a hydride ion $(H:^-)$ are formed.

Explore More

Similar Questions

Consider the following carbocations:
$I. C_6H_5CH_2^+$
$II. CH_2=CH^+$
$III. CH_3-CH^+(CH_3)$
$IV. CH_3-CH_2^+$
$V. HC \equiv C^+$
Arrange the above carbocations in the order of decreasing stability.

The following carbocation is stabilized by the interaction of the empty $p$ orbital with:

The central carbon atom of a free radical contains

Which of the following carbocations will be most stable?

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

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