Write hydrocarbon radicals that can be formed as intermediates during monochlorinationof 2-methylpropane? Which of them is more stable? Give reasons

During the monochlorination of 2-methylpropane, three different types of hydrocarbon radicals can be formed as intermediates. These are:

1. Primary (1°) radical: This is formed when a hydrogen atom is removed from the end carbon atom of 2-methylpropane. The formula for this radical is CH3CH2CH(CH3)•.

2. Secondary (2°) radical: This is formed when a hydrogen atom is removed from the second carbon atom of 2-methylpropane. The formula for this radical is CH3CH(CH3)CH2•.

3. Tertiary (3°) radical: This is formed when a hydrogen atom is removed from the central carbon atom of 2-methylpropane. The formula for this radical is CH3C(CH3)2•.

Among these, the tertiary (3°) radical is the most stable. This is because of the phenomenon known as hyperconjugation. In hyperconjugation, the adjacent carbon-hydrogen bonds can donate electron density to the radical center, stabilizing it. The tertiary radical has the most adjacent carbon-hydrogen bonds (nine in total), and therefore, it is the most stable.