Compare the stability of the following free radicals:
$(1)$ $CH_3 - \dot{CH} - CH_3$
$(2)$ $C_6H_5 - \dot{CH_2}$
$(3)$ $\dot{CH_2} - CH(CH_3)_2$
$(4)$ $\dot{CH_2} - CH_3$

  • A
    $II > I > III > IV$
  • B
    $II > I > IV > III$
  • C
    $I > II > III > IV$
  • D
    $IV > III > I > II$

Explore More

Similar Questions

What type of reaction is oxidative rancidity?

$p$ will be

Difficult
View Solution

Which of the following carbanions is the least stable?

Arrange the following carbocations in decreasing order of stability:
$I: CH_3-CH_2^+$
$II: (CH_3)_2CH^+$
$III: Ph-CH_2^+$

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.

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