Question

Primary alkyl halide C₄H₉Br (a) 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) is reacted with sodium metal it gives compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Answer 1

The given reaction sequence is:

When (A) is treated with Na metal, the Wurtz reaction takes place.

It is given that (A) is a primary alkyl halide.

∴ The possible structures of C₄H₉Br (A) are:

(I)
\[\begin{array}{cc}
\ce{H3C - CH - CH2Br}\\
|\phantom{.......}\\
\ce{CH3}\phantom{...}\\
\end{array}\]

(II) \[\ce{CH3 - CH2 - CH2 - CH2 - Br}\]

It is established that (A) reacts with sodium metal to produce the alkane C₈H₁₈, which differs from the alkane formed from the Wurtz reaction of n-butylbromide. Therefore, (A) ≠ II.

Thus,

\[\begin{array}{cc}
\phantom{......................................................................................}\ce{Br}\phantom{}\\
\phantom{.....................................................................................}|\phantom{}\\
\phantom{}\ce{CH3 - CH - CH2 - Br ->[alc.KOH] CH3 - C = CH2 ->[HBr] CH3 - C - CH3}\phantom{}\\
\phantom{.............}|\phantom{..................................................}|\phantom{.................................}|\phantom{.............}\\
\phantom{............................}\ce{\underset{(A)}{CH3}}\phantom{............................................}\ce{\underset{(B)}{CH3}}\phantom{........................}\ce{\underset{\underset{(isomeric with (A))}{(C)}}{CH3}}\phantom{....................}
\end{array}\]

\[\begin{array}{cc}
\ce{H3C - CH - CH2 - CH2 - CH - CH3}\\
|\phantom{...............................}|\phantom{...}\\
\phantom{.}\ce{\underset{(D)}{\ce{CH3}\phantom{.........................}\ce{CH3}}}\\
\end{array}\]