# How can a methyl halide react by intramolecular SN2?

Out of the following, $\ce{CH3Br}$ doesn't react by which mechanism?

1. SN2
2. SN1
3. E2
4. Intramolecular SN2
5. E1

My textbook gives 2, 3 and 5 as the correct answers, which means it does react by 1 and 4.

But how can a methyl halide react by intramolecular SN2, without a nucleophile present in the molecule? There's just a leaving group ($\ce{-Br}$).

## 1 Answer

I agree. There is no way that a one-carbon and one-heteroatom compound could react by intramolecular substitution. All intramolecular nucleophilic substitutions will require some ring-like transition state. The smallest ring you can build is a three-membered ring — but bromomethane does not have three atoms that could build up that ring (hydrogen does not participate).