Question

t – butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n – butyl chloride reacts with S_N² mechanism.

Answer 1

t – butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n – butyl chloride reacts with S_N² mechanism.

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

\[\begin{array}{cc}\phantom{........}\ce{CH3}\phantom{....................}\ce{CH3}\\
\phantom{......}|\phantom{.......................}|\\
\ce{CH3 - C - Cl ->[Ionization][Cl-/slow] CH3 - C+}\\
\phantom{......}|\phantom{.......................}|\\
\phantom{................}\ce{CH3}\phantom{...............}\ce{\underset{\text{(stable)}}{\underset{\text{Tert-butyl varbocation}}{CH3}}}\phantom{.....}
\end{array}\]

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

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

S_N¹ mechanism follows the reactivity order as 3° > 2°> 1° while S_N² mechanism follows the reactivity order as 1° > 2° > 3°. Therefore, tert-butyl chloride (3°) reacts by S_N¹ mechanism while n-butyl chloride (1°) reacts by an S_N² mechanism.