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Stoichiometry

Primary alkyl...

Question

Primary alkyl halide (a) $C_{4} H_{9} B r$ reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) was reacted with sodium metal it gave a compound (d) $C_{8} H_{18}$ that was different than the compound when n-butyl bromide was reacted with sodium. Give the structural of (a) and write the equation for all the reactions.

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Solution

As $A$ is an isomer of $C$ and $B$ do not form $n - o c t a n e (C_{8} H_{18})$ as it is form by $n - b u t y l b r o m i d e$ . Hence $A$ must be $1^{o} - a l k y l$ $B r o m i d e$ .

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