CBSE Class 12-science Answered

Free-radical bromination of this alkane would produce 4 products: 1-bromo-3-methylbutane, 2-bromo-3-methylbutane, 3-bromo-3-methylbutane, and 1-bromo-2-methylbutane.

Bromine is much more regioselective than chlorine. The relative rates of reaction for tertiary:secondary:primary is 1640:82:1. This means that in the free-radical bromination of 2-methylbutane the major product would be 3-bromo-3-methylbutane because of the tertiary nature of that particular carbon. It would be about 90% 3-bromo-3-methylbutane and 9% 2-bromo-3-methylbutane with the remaining 1% being a mixture of 1-bromo-3-methylbutane and 1-bromo-2-methylbutane.