Question

An alkyl halide with molecular formula C₆H₁₃Br on dehydrohalogenation gave two isomeric alkenes X and Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂CHCHO. The alkyl halide is
(a) 2-bromohexane
(b) 2, 2-dimethyl-1-bromobutane
(c) 4-bromo-2-methylpentane
(d) 3-bromo-2-methylpentane

Answer 1

Answer:

d) 3-bromo-2methylpentane

Explanation:

The reactions for the same is attached

APPROACH to SOLVE:

The actual clue in solving this still lies in the fact that one of the products of oxidising is a ketone

Which means that on dehydrohalogenation, one of the alkenes must have be secondary and all others must be primary, i.e only double bond is of form -C=< whereas the other is of form -C=C-

And to get such abproduct, the halogen present before dehydrohalogenation should be on a carbon atom adjacent to a 2° carbon atom which gives us our answer of D

Thanks for the great sum

Hope this answer helped you

Answer 2

Hello Friend..!!

The answer of ur question is..!!

Option.D

.

Thank you