The correct order of relative stability for t | vedclass

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

Question

(B) The stability of free radicals is determined by the resonance and inductive effects of the attached groups.$I$ is an ethyl radical $(CH_3CH_2^{\bu

Vedclass · Accepted Answer

$I < III < II$

Vedclass · Answer

(B) The stability of free radicals is determined by the resonance and inductive effects of the attached groups.$I$ is an ethyl radical $(CH_3CH_2^{\bullet})$, which is stabilized by hyperconjugation.$II$ is a radical with an amino group $(N(CH_3)_2)$ attached to the radical carbon. The lone pair on the nitrogen atom provides resonance stabilization through the $+M$ effect, which is very effective for radical stabilization.$III$ has both an amino group $(N(CH_3)_2)$ and an ester group $(COOEt)$ attached to the radical carbon. While the amino group provides strong resonance stabilization, the ester group is electron-withdrawing ($-I$ and $-M$ effects), which destabilizes the radical compared to $II$.Therefore, the stability order is $III < II < I$ is incorrect based on the provided options; the correct order is $I < II < III$ is also not quite right as $II$ is more stable than $I$ due to resonance. Comparing $II$ and $III$, $II$ is more stable than $III$ because the electron-withdrawing ester group in $III$ reduces the electron density on the radical carbon. Thus, the order is $I < III < II$.

The correct order of relative stability for the given free radicals is:

Similar Questions

Among the given species,which carbocations are resonance stabilized?

Which of the following is the most stable carbanion?

Find the correct stability order for the following carbocations:

List the reactive intermediates produced by the fission of a covalent bond and describe their main characteristics.

What are free radicals,carbanions,and carbocations? How are they formed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module