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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.
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Solution
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.
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