Question

An alkyl halide with molecular formula C₆H₁₃Br on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH, CHO and (CH₃)₂ CHCHO. Find the alkyl halide.

Answer 1

1. C₆H₁₃Br is 3 – Bromo – 4 methylpentanc.

\[\begin{array}{cc}\ce{Br}\\
|\\\ce{CH3 - CH - C - CH2 - CH3}\\
|\phantom{......}|\phantom{.......}\\
\ce{CH3}\phantom{...}\ce{H}\phantom{.......}
\end{array}\]

2. 3 – Bromo -4 methylpentane on dehydrogenation give two isomers X and Y as follows:

Therefore C₆H₁₃ Br is 3 – Bromo – 4 – methy ipentane.