Define carbocation,carbanion,and free radical.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $1$. $Carbocation$: $A$ species containing a carbon atom bearing a positive charge and having only $6$ electrons in its valence shell. It is an electron-deficient species (electrophile) with $sp^2$ hybridization and a planar geometry.
$2$. $Carbanion$: $A$ species containing a carbon atom bearing a negative charge and having $8$ electrons in its valence shell. It is an electron-rich species (nucleophile) with $sp^3$ hybridization and a pyramidal geometry.
$3$. $Free \text{ } radical$: $A$ species containing a carbon atom with an unpaired electron. It is formed by homolytic fission of a covalent bond. It is neutral,electron-deficient,and highly reactive.

Explore More

Similar Questions

The compound $(X)$ will be:

Difficult
View Solution

Which is the decreasing order of stability?
$(i)$ $CH_3-CH^{+}-CH_3$
$(ii)$ $CH_3-CH^{+}-OCH_3$
$(iii)$ $CH_3-CH^{+}-COCH_3$

Arrange the following anions in decreasing order of their stability:
$CH_3^{\Theta}, NH_2^{\Theta}, OH^{\Theta}, Cl^{\Theta}$

Difficult
View Solution

Which of the following compounds forms the most stable carbocation: $(CH_3)_3C-Br$,$(C_6H_5)_3CBr$,$(C_6H_5)_2CHBr$,and $C_6H_5CH_2Br$?

Which of the following intermediates is not expected to be formed in the following reaction?

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