Animated Solution for Chemistry - Organic Chemistry: $S_{N}1$ reaction is feasible in

Decoding $S_{N}1$ Feasibility

When we talk about nucleophilic substitution reactions, the $S_{N}1$ mechanism is a fascinating two-step dance. The first step—and the most crucial one—is the departure of the leaving group to form a carbocation intermediate. Because this step is the slowest, it acts as the rate-determining step.

The golden rule of $S_{N}1$ reactions is simple: The more stable the carbocation intermediate, the faster and more feasible the reaction. To determine which of our given molecules will most readily undergo an $S_{N}1$ reaction, we must play the role of a chemical detective and evaluate the stability of the carbocation each one forms.

Let's break down the options one by one and see what happens when the chloride ion ($C{l}^{-}$) packs its bags and leaves.

Option (a): tert-Butyl Chloride When the $C-Cl$ bond breaks heterolytically in tert-butyl chloride, it leaves behind a tertiary ($3^{\circ }$) carbocation, $(C{H}_3{)}_3{C}^{+}$. This is the heavyweight champion of carbocations. It is immensely stabilized by two factors: the electron-donating inductive effect ($+I$ effect) of three methyl groups, and the hyperconjugation provided by nine $\alpha$-hydrogens. This dual stabilization makes it highly stable and very eager to form.

Option (b): Allyl Chloride Cleaving the $C-Cl$ bond in allyl chloride yields an allyl carbocation, $C{H}_2=CH-C{H}_2^{+}$. While this is a primary ($1^{\circ }$) carbocation, it possesses a secret weapon: resonance. The positive charge is delocalized over two carbon atoms. However, despite this resonance stabilization, a purely tertiary carbocation like the one in option (a) is generally more stable than a primary allyl carbocation.

Option (c): Chlorobenzene If chlorobenzene were to undergo an $S_{N}1$ reaction, it would have to form a phenyl cation, $C_6{H}_5^{+}$. This is a chemical nightmare. The positive charge would reside on an $s{p}^2$ hybridized carbon atom, which is highly electronegative and strongly resists bearing a positive charge. Furthermore, the empty orbital is perpendicular to the $\pi$-electron cloud of the benzene ring, meaning it gets absolutely zero stabilization from resonance. Consequently, this carbocation is highly unstable, making the $S_{N}1$ pathway practically impossible.

Option (d): 1-Chloro-2-phenylethane Here, the departure of the chloride ion forms a primary carbocation: $C_6{H}_5-C{H}_2-C{H}_2^{+}$. Not only is it a primary carbocation (which is inherently unstable), but it is also situated adjacent to a phenyl ring. Because the positive charge is separated from the ring by a $C{H}_2$ group, it cannot be stabilized by resonance. Instead, the $s{p}^2$ carbons of the phenyl ring exert a strong electron-withdrawing inductive effect ($-I$ effect), which pulls electron density away from an already electron-deficient carbon, further destabilizing it.

By comparing the stability of the resulting intermediates, the hierarchy is clear:

Because tert-butyl chloride forms the exceptionally stable tertiary carbocation, it provides the lowest activation energy barrier for the rate-determining step. Therefore, the $S_{N}1$ reaction is most feasible in option (a).