An alkene Y ( $C X_{6} H X_{12}$ ) gives propanone and propanal on reductive ozonolysis. Y is hydrogenated completely and the alkane formed is monochlorinated by free radicals. How many different monochloro products (structural isomers only, ignoring stereoisomers) are possible?

Answer: C

Propanone + propanal fit

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

(2-methylpent-2-ene). Hydrogenation gives 2-methylpentane,

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

. Its hydrogen sets: two equivalent

CH X_{3}

on C2, the C2-H, C3-H2, C4-H2 and the C5

CH X_{3}

= 5 types, hence 5 structural monochloro isomers.