Question

Homolytic fission of the following alkanes forms free radicals
CH₃ – CH₃, CH₃ – CH₂ – CH₃, (CH₃)₂ CH – CH₃, CH₃ – CH₂
– CH (CH₃)₂. Increasing order of stability of the radicals is
[NEET Kar. 2013]
⦁ ⦁
(a) (CH₃)₃ C < (CH₃)₂ C – CH₂CH₃ <
⦁ ⦁
CH₃ – CH – CH₃ < CH₃ – CH₂
⦁ ⦁ ⦁
(b) (CH₃)₂C – CH₂CH₃ < CH₃ – CH – CH₃ < CH₃ – CH₂
⦁
< (CH₃)₃C
⦁ ⦁
(c) CH₃ – CH₂ < CH₃ – CH – CH₃ <
⦁ ⦁
(CH₃)₂C – CH₂ – CH₃ < (CH₃)₃C
⦁ ⦁ ⦁
(d) CH₃ – CH₂ < CH₃ – CH – CH₃ < (CH₃)₃ C < (CH₃)₂
⦁
C – CH₂CH₃

Answer 1

Answer:

(a) (CH₃)₃ C < (CH₃)₂ C – CH₂CH₃ <  

CH₃ – CH – CH₃ < CH₃ – CH₂  

(b) (CH₃)₂C – CH₂CH₃ < CH₃ – CH – CH₃ < CH₃ – CH₂  

< (CH₃)₃C  

(c) CH₃ – CH₂ < CH₃ – CH – CH₃ <  

(CH₃)₂C – CH₂ – CH₃ < (CH₃)₃C

(d) CH₃ – CH₂ < CH₃ – CH – CH₃ < (CH₃)₃ C < (CH₃)₂

⦁

C – CH₂CH₃

Explanation: