Question

here is the reaction you drew for protonation of methanol with an acid.
now predict what would happen if the acid does not contain a good nucleophile (hint: what else could serve as a nucleophile?)

Explanation:

First, methanol ($CH_3OH$) gets protonated by the acid to form the positively charged oxonium ion $CH_3OH_2^+$, as shown in the initial reaction. Since the acid lacks a good nucleophile, another methanol molecule (which has lone electron pairs on its oxygen atom, acting as a nucleophile) will attack the protonated methanol. This leads to the loss of a water molecule and formation of dimethyl ether ($(CH_3)_2O$).

Answer:

1. Protonation of methanol first occurs:

$CH_3OH + H^+
ightarrow CH_3OH_2^+$

1. A second methanol molecule (as nucleophile) attacks the oxonium ion, leading to elimination of $H_2O$ and formation of dimethyl ether:

$CH_3OH_2^+ + CH_3OH
ightarrow (CH_3)_2O + H_3O^+$
The final major product is dimethyl ether ($(CH_3)_2O$).