I came across a question asking for the major product on chlorination of 2-methylbutane. The answer in the back of the book says it will be 2-chloro-3-methylbutane. However, I thought the major product would be 2-chloro-2-methylbutane. The later is formed through a tertiary free radical with eight hyper-conjugative structures, whereas the former forms through a secondary free radical with only four hyper-conjugative structures. How is this possible?
Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.
According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:
22 % 2-chloro-2-methylbutane
33 % 2-chloro-3-methylbutane
30 % 1-chloro-2-methylbutane
15 % 1-chloro-3-methylbutane
The statistical factor is responsible for this.
If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be
1 hydrogen atom (8 %) → 2-chloro-2-methylbutane
2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane
6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane
3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane
In terms per hydrogen atom, the experimental yield is
22 % / 1 = 22 % → 2-chloro-2-methylbutane
33 % / 2 = 16.5 % → 2-chloro-3-methylbutane
30 % / 6 = 5 % → 1-chloro-2-methylbutane
15 % / 3 = 5 % → 1-chloro-3-methylbutane
Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.