# What are the products of the following reaction in methoxide in methanol?

May 23, 2016

You aren't given the reaction conditions, so you might have to assume room temperature. But for the sake of learning, we'll assume any temperature.

One issue that might arise is that the solvent is protic, so a strong nucleophile, which tends to also be a strong base (in this case it is), would likely act as a lewis base to donate a pair of electrons and acquire a proton. That would deactivate the nucleophile and disfavor ${\text{S}}_{N} 2$.

However, since the solvent is the protic form of the nucleophile, it doesn't matter whether the nucleophile is protonated or not:

${\text{CH"_3"OH" rightleftharpoons "CH"_3"CO"^(-) + "H}}^{+}$

because some solvent methanol would have to be deprotonated in order to protonate methoxide, so either way you have your nucleophile.

Thus, the type of reaction depends solely on the substrate cyclohaloalkane.

The substrate is very sterically hindered (the alkyl halide carbon is tertiary), so at room temperature, ${\text{S}}_{N} 1$ is likely.

Or, at elevated temperatures, $\text{E} 1$ is likely, because higher temperatures favor elimination over substitution.

A good rule of thumb is that if ${S}_{N} 1$ is favored, then boosting the temperature favors $E 1$, and similarly, if ${S}_{N} 2$ is favored, then boosting the temperature favors $\text{E} 2$.

Now that we've established that you get either ${\text{S}}_{N} 1$, $\text{E} 1$, or both, let's draw both mechanisms.

In the ${\text{S}}_{N} 1$ mechanism, we have to wait until the bromine leaves on its own, because the nucleophile is not strong enough to simply brute-force the reaction and displace the bromide---the steric hindrance is too much.

So, the first step is rate-limiting, i.e. it is slow.

Then, the carbocation intermediate forms, and the methoxide can either:

• Act as a nucleophile and attack the carbocation center to form a new $\text{C"-"O}$ bond, thus forming the ${\text{S}}_{N} 1$ product. This step is fast.
• Act as a base and steal a proton from the carbon adjacent to the carbocation and form a $\pi$ bond, thus forming the $\text{E} 1$ product. This step is fast.

Thus, we see that the two possible products formed are the original compound with a methoxide displacing a bromide, or the alkene of the original compound.