# Reactivity of benzyl halides towards nucleophilic substitution

I experienced some difficulty as I was doing question 22 of this quiz on nucleophilic substitution at saturated carbons:

I picked option B but the answer turned out to be C. My rationale for picking B was that I thought that IV was going to be more reactive than I because I has an additional electron-donating methyl substituent which would increase the energy of the transition state in the $S_{N}2$ pathway.

Evidently, II and III are in the 3rd and 4th places respectively because alkyl bromides are less reactive than alkyl iodides and that phenyl bromides are not as reactive towards nucleophilic substitution as the $\pi$ electron cloud of the neighbouring benzene ring would repel the nucleophile away.

What went wrong with my thinking?

• Consider the contribution of the methyl group of 1 towards the stability of the benzylic cation – Waylander Dec 30 '17 at 11:43
• If you’re just putting it in methanol it’s going to be SN1 not SN2. – orthocresol Dec 30 '17 at 12:39
• @orthocresol How could you tell so surely? – Tan Yong Boon Dec 30 '17 at 13:25
• SN2 needs a decent nucleophile, which methanol is not – orthocresol Dec 30 '17 at 14:36
• @orthocresol So I was on the wrong train of thought because I thought it was SN2. On a side note, would the benzylic halides react by SN1 or SN2 if the nucleophile was a good one? – Tan Yong Boon Dec 30 '17 at 14:41

The answer to this question hinges on figuring out whether the reaction occurs by the SN1 or the SN2 pathway. As noted by a comment, methanol is not a very good nucleophile, and SN2 reactions require a strong nucleophile for the nucleophilic attack to occur.

Also, it is important to realize that the likelihood of the SN1 pathway being taken largely depends on the stability of the carbocation intermediate that forms. The benzylic carbocation that forms in this case is extremely stable:

1. due to resonance and

2. due the the presence of the polar protic substance, methanol.

Thus, SN1 is highly preferred here.

In fact, due to these three reasons, the probability of the SN2 pathway occurring is almost negligible in I, II, and IV.

I believe you have already worked the rest of it out; the answer is C.

As you already noted, phenyl bromide is least reactive towards nucleophilic substitution, so any answer suggestion that does not place III last is wrong a priori. The reasoning you gave does not quite add up with me so here is my version: no $\mathrm{S_N2}$ back attack is possible through the phenyl ring since that pathway is blocked by the other ring atoms; and no $\mathrm{S_N1}$ pathway is possible since that would require a very unstable phenyl cation with the empty orbital being of the $\mathrm{sp^2}$ type.

Thankfully, this also immediately tells us that II is penultimate, since no other options exist and the question boils down to whether I or IV is more reactive.

If you were to assume an $\mathrm{S_N2}$ pathway, then indeed I would lose to IV due to increased steric inhibition: the electrophilic carbon is not easily accessible from behind.

However, both reactive carbons are benzylic and the neighbouring phenyl ring is very good at stabilising ionic charges. Considering that you also have a rather weak nucleophile — methanol — it is very likely that this reaction will proceed under $\mathrm{S_N1}$ conditions, so the rate determining step is the formation of the cation. From this point of view, I gives the better intermediate cation than IV because of the additional inductive stabilisation from the methyl group.

Therefore, the correct ordering is $\mathrm{I>IV>II>III}$ or answer C.