Homolytic fission of the following alkanes forms free radicals CH3-CH3, CH3-CH2-CH3, (CH3)2CH-CH3, CH3-CH2-CH(CH3)2. Increasing order of stability of the radicals is
- (CH3)2 C-CH2CH3 < CH3- CH-CH3 < CH3- CH2 < (CH3)3 C
- CH3- CH2 < CH3- CH-CH3 < (CH3)2 C-CH2-CH3 < (CH3)3 C
- CH3- CH2 < CH3- CH-CH3 < (CH3)3 C < (CH3)2 C-CH2CH3
- (CH3)3 C < (CH3)2 C-CH2CH3 < CH3- CH-CH3 < CH3- CH2
Answer
(2) CH3- CH2 < CH3- CH-CH3 < (CH3)2 C-CH2-CH3 < (CH3)3 C
Solution
Stability of alkyl free radicals: 3° > 2° > 1°, due to hyperconjugation (and +I effect of alkyl groups).
CH3- CH2 (1°) < CH3- CH-CH3 (2°) < 3° radicals
Between the 3° radicals: (CH3)3 C has 9 α-hydrogens, (CH3)2 C-CH2-CH3 has 8 α-hydrogens.
⇒ (CH3)3 C is the most stable.
∴ CH3- CH2 < CH3- CH-CH3 < (CH3)2 C-CH2-CH3 < (CH3)3 C
Answer: (2)