Question

tert-Butylbromide reacts with aq. \[\ce{NaOH}\] by S_N1 mechanism while n-butylbromide reacts by S_N2 mechanism. Why?

Answer 1

In general, the S_N1 reaction proceeds through the formation of carbocation. The tert-butylbromide readily loses Br ion to form stable 3° carbocation. Therefore, it reacts with aqueous KOH by S_N1 mechanism as:

\[\begin{array}{cc}
\phantom{.........}\ce{CH3}\phantom{...................}\ce{CH3}\phantom{}\\
\phantom{.......}|\phantom{.......................}|\phantom{.}\\
\ce{CH3 - C - Br ->[ionization][Br- slow] CH3 - C^+}\\
\phantom{.......}|\phantom{.......................}|\phantom{.}\\
\phantom{...............}\ce{CH3}\phantom{.............}\ce{\underset{\underset{(stable)}{tert-butyl carbocation}}{CH3}}\phantom{}
\end{array}\]

\[\begin{array}{cc}
\phantom{....}\ce{CH3}\phantom{...........}\ce{CH3}\phantom{}\\
\phantom{...}|\phantom{...............}|\phantom{..}\\
\ce{CH3 - C^+ ->[OH][fast] CH3 - C - OH}\\
\phantom{...}|\phantom{...............}|\phantom{..}\\
\phantom{........}\ce{CH3}\phantom{.......}\ce{\underset{tert-butyl alcohol}{CH3}}\phantom{}
\end{array}\]

On the other hand, n-Butyl bromide does not undergo ionization to form n-Butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo a reaction by the S_N2 mechanism, which occurs in one step through a transition state involving the nucleophilic attack of OH^– ion from the back side with simultaneous expulsion of Br^– ion from the front side.

S_N1 mechanism follows the reactivity order as 3° > 2° > 1° while the S_N2 mechanism follows the reactivity order as 1° > 2° > 3°.

Therefore, tert-butybromide (3°) reacts by the S_N1 mechanism while n-Butylbromide (1°) reacts by the S_N2 mechanism.