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 two primary halides of molecular formula C₄H₉Br can be the following:

\[\ce{\underset{n-butyl bromide}{CH3CH2CH2CH2Br}}\] and

\[\begin{array}{cc}
\phantom{...........}\ce{CH3}\\
\phantom{........}|\\
\ce{\underset{isobutyl bromide}{CH3 - CH - CH2Br}}
\end{array}\]

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Since compound ‘a’ reacts with sodium metal to produce compound ‘d’ (molecular formula C₈H₁₈), which is different from the compound obtained when n-butyl bromide reacts with sodium metal, compound ‘a’ must be isobutyl bromide, and compound ‘d’ should be 2, 5-dimethylhexane.

\[\ce{2CH3CH2CH2CH2Br + 2Na ->[Wurtz reaction] \underset{n-octane}{CH3CH2CH2CH2CH2CH2CH2CH3}}\]

\[\begin{array}{cc}
\phantom{....}\ce{CH3}\phantom{.....................}\ce{CH3}\phantom{................}\ce{CH3}\phantom{........}\\
\phantom{,...}|\phantom{.........................}|\phantom{....................}|\phantom{...........}\\
\ce{2CH3 - CH - CH2Br ->[Wurtz reaction] \underset{2, 5-Dimethylhexane (d)}{CH3 - CH - CH2 - CH2 - CH - CH3}}
\end{array}\]

If compound 'a' is isobutyl bromide, compound 'b' obtained by reacting compound 'a' with alcoholic KOH must be 2-methyl-1-propene.

\[\begin{array}{cc}
\phantom{.........}\ce{CH3}\phantom{...........................}\ce{CH3}\phantom{.}\\
\phantom{......}|\phantom{...............................}|\phantom{.}\\
\ce{\underset{Isobutyl Bromide (a)}{CH3 - CH - CH2Br} ->[KOH (alc.),heat][Dehydrohalogenation] \underset{2-Methyl-1-Propene (b)}{CH3 - C = CH2}}
\end{array}\]

Compound ‘b’ reacts with HBr to give compound ‘c’ according to Markovnikov rule. Hence, compound ‘c’ is tertiary-butyl bromide, an isomer of compound ‘a’ (isobutyl bromide).

\[\begin{array}{cc}
\phantom{......}\ce{CH3}\phantom{............................}\ce{CH3}\phantom{....}\\
\phantom{..}|\phantom{................................}|\phantom{..}\\
\ce{\underset{2-Methyl-1-Propene (b)}{CH3 - C = CH2} ->[HBr][Markovnikov's rule] CH3 - C - CH3}\\
\phantom{.................................}|\\
\phantom{..................................}\ce{\underset{(Isomer of compound A)}{\underset{Tertiary-butyl bromide (c)}{Br}}}
\end{array}\]

Thus, ‘a’ is isobutyl bromide, ‘b’ is 2-methyl-1-propene, ‘c’ is tertiary-butyl bromide and ‘d’ is 2, 5-dimethylhexane.