A hydrocarbon A ( $C X_{6} H X_{12}$ ) gives only propanone and propanal on reductive ozonolysis. A is then treated with $HBr$ in the presence of peroxide. The major product is

Answer: A

Joining the carbonyl carbons gives A =

(CH X_{3}) X_{2} C = CH - CH X_{2} CH X_{3}

(2-methylpent-2-ene). With peroxide,

Br \cdot

adds to the less substituted carbon (

C X_{3}

) to give the more stable tertiary radical at

C X_{2}

. The product is

(CH X_{3}) X_{2} CH - CHBr - CH X_{2} CH X_{3}

, 3-bromo-2-methylpentane.