# How will you convert methyl bromide to dimethylamine?

Sep 10, 2016

$2 {H}_{3} C - B r + N {H}_{4} B r + 3 N {\left(C {H}_{2} C {H}_{3}\right)}_{3} \rightarrow {\left({H}_{3} C\right)}_{2} N H + 3 \left[H B r \cdot N {\left(C {H}_{2} C H\right)}_{3}\right]$

#### Explanation:

Possibly there would be a laboratory synthesis along these lines. I assume you are answering a problem, and not doing an actual synthesis. The industrial process takes ammonia, methanol, and heat and pressure:

$N {H}_{3} + 2 C {H}_{3} O H \rightarrow H N {\left(C {H}_{3}\right)}_{2} + 2 {H}_{2} O$

But this is not a reaction that you could reproduce in a laboratory.

In the first reaction, each $N - C$ bond is formed with the elimination of triethylamine hydrobromide. Hence we need 2 equiv of $E {t}_{3} N$ for the $2 \times C - N$ bond formation, and a third equiv to give the ammonia, the free base. The trithylamine operates as a non-nucleophilic base here to sop up the hydrogen bromide produced. These days, we have non-nucleophilic bases such as $\text{DBU}$, and $\text{DABCO}$; these are more basic than the amine, and are a lot less pfaff to use.