Animated Solution for Chemistry - Organic Chemistry: Increasing rate of $S_{N}1$ reaction in the following compounds is

When tackling an $S_{N}1$ reaction problem, the very first thing you must visualize is the formation of the carbocation intermediate. The rate of an $S_{N}1$ reaction is entirely dictated by the stability of this intermediate. The more stable the carbocation, the faster it forms, and consequently, the faster the overall reaction proceeds.

To determine stability, we must look at the electronic effects of the substituents attached to the benzene ring. Electron-donating groups (EDG) pump electron density into the ring, stabilizing the positive charge. Conversely, electron-withdrawing groups (EWG) pull electron density away, destabilizing the carbocation.

Compound A

Let's start with Compound A. When the iodine leaving group departs, it leaves behind a $2^{\circ }$-benzyl carbocation. Notice that there are no additional substituents on the benzene ring. This makes Compound A our perfect baseline. We will compare the stability of all other carbocations against this neutral standard.

Compound B

Now, focus on Compound B. Here, we have a methoxy ($-OC{H}_3$) group located at the meta position relative to the carbocation center. This is where many students fall into a trap!

Because the lone pairs on the oxygen cannot delocalize into the benzylic carbon, the methoxy group can only exert its inductive effect. Oxygen is highly electronegative, so it exerts a strong $-I$ effect, pulling electron density away from the ring. This actively destabilizes the positive charge, making Compound B less stable than our neutral baseline, Compound A.

Compound C

Moving on to Compound C, we find a methyl ($-C{H}_3$) group at the para position. Alkyl groups are classic electron donors. They push electron density into the ring through two mechanisms: the $+I$ inductive effect and, more importantly, hyperconjugation ($+H$).

The three alpha-hydrogens of the methyl group provide significant stabilization to the benzylic carbocation. Because it has these stabilizing forces, Compound C is more stable than our baseline, Compound A.

Compound D

Finally, let's visualize Compound D. The methoxy ($-OC{H}_3$) group is now situated at the para position. This changes everything!

From the para position, the lone pairs on the oxygen atom can fully participate in resonance. The oxygen donates its electrons directly into the pi system of the ring, exerting a powerful $+R$ effect. This resonance stabilization is incredibly strong—much stronger than the hyperconjugation seen in Compound C. As a result, the carbocation formed from Compound D is highly stabilized, making it the most stable of the four.

By comparing the electronic effects, we can easily rank the stabilities of the carbocation intermediates:

1. Compound B is the least stable due to the destabilizing $-I$ effect. 2. Compound A is next, serving as our neutral baseline. 3. Compound C follows, stabilized by $+I$ and hyperconjugation. 4. Compound D is the most stable, crowned by the powerful $+R$ resonance effect.

Therefore, the increasing order of the rate of the $S_{N}1$ reaction perfectly mirrors this stability trend: (B) $<$ (A) $<$ (C) $<$ (D).