Question

Several hydrocarbon radicals can be formed as intermediates during monochlorination of 2-methylpropane. Which of them is the most stable?

• A. ${\left({CH}_3\right)}_2\stackrel{\bullet }{C}{CH}_3$
• B. ${\left({CH}_3\right)}_{2}CH\stackrel{\bullet }{C}{H}_2$
• C. $\left({CH}_3\right)\stackrel{\bullet }{C}H{CH}_2{CH}_3$
• D. All are equally stable

Answer 1

The correct option is A ${\left({CH}_3\right)}_2\stackrel{\bullet }{C}{CH}_3$

2-Methypropane gives 2 types of radicals.

Radical (i) is more stable because it is $3^0$ and stabilized by nine hyperconjugative structures.

Radical (ii) is less stable because it is $1^0$ and stabilized by only one hyperconjugative structure.