Question

Which of the following haloalkanes reacts with aqueous \[\ce{KOH}\] most easily? Explain giving reason.

Options

• 1-Bromobutane

• 2-Bromobutane

• 2-Bromo-2-methylpropane

• 2-Chlorobutane

Answer 1

2-Bromo-2-methylpropane

Explanation:

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

will react the aqueous KOH most easily because the corbocation formed during hte reaction is tertiary and most stable.

\[\begin{array}{cc}
\phantom{}\ce{CH3}\phantom{.......}\\
\phantom{}|\phantom{..........}\\
\phantom{}\ce{(CH3)3CBr ->[Step I]}\phantom{}\ce{C+ + Br- ->[Step I][OH] (CH3)3COH}\phantom{}\\
\phantom{}/\phantom{...}\phantom{}\backslash\phantom{.........}\\
\phantom{}\ce{\underset{(Stable 3° carbocation)}{\phantom{..}CH3\phantom{...}CH3}}\phantom{..........}
\end{array}\]