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JEE Advanced levelSingle correct optionDifficulty 3 of 3g1170

Free-radical bromination of 2,3-dimethylbutane, $(CH X_{3}) X_{2} CH - CH (CH X_{3}) X_{2}$ , with $Br X_{2}$ and light is very selective. The major monobromo product is:

A2,3-dibromo-2,3-dimethylbutane
B2-bromo-2,3-dimethylbutane and 1-bromo-2,3-dimethylbutane in equal amounts
C2-bromo-2,3-dimethylbutane
D1-bromo-2,3-dimethylbutane

Answer: C

Bromine radical is highly selective and abstracts the hydrogen that gives the most stable radical. Abstraction of a tertiary C-H gives the tertiary radical

(CH X_{3}) X_{2} C^{∙} - CH (CH X_{3}) X_{2}

(six alpha hydrogens), far more stable than the primary radical. The product is therefore 2-bromo-2,3-dimethylbutane.

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