Question

Which compound would undergo dehydrohalogenation with strong base to give the alkene shown below as the only alkene product?

CH₃ – CH₂CH = CH – CH₃

Options

• 1-Chloropentane

• 2-Chloropentane

• 3-Chloropentane

• 1-Chloro-methylbutane

Answer 1

3-Chloropentane

Explanation:

\[\begin{array}{cc}
\ce{CH3 - CH2 - CH - CH2 - CH3 ->[Strong Base] \underset{Pent = 2 = ene}{CH3 - CH} = CH - CH2 - CH3}\\
|\phantom{...........................................}\\
\ce{\underset{3-Chloropentane}{Cl}}\phantom{...........................................}
\end{array}\]

In case of 3 chloropentane, pentane-2-ene will be the only product obtained by its reaction with strong base.

\[\begin{array}{cc}
\ce{CH3 - CH2 - CH2 - CH - CH3 -> \underset{Zaitsev + product (Major)}{CH3 = - CH  - CH2 - CH3} - \underset{Hoffman product (Minor)}{CH2 = CH - CH2 - CH2 - CH3}}\\
|\phantom{......................................................}\\
\ce{\underset{2-Chloropentane}{Cl}}\phantom{.....................................................}
\end{array}\]

In this case, two products may be obtained.