# (a) How will you convert methyl bromide to dimethylamine? (b) How will you prepare propanamine from methyl chloride using suitable reactions?

Apr 11, 2016

Here's what I have come up with.

a) While one might expect that it can be accomplished in one step by reacting with methylamine, it would more likely produce a mixture of ${1}^{\circ}$, ${2}^{\circ}$, and ${3}^{\circ}$ amines, as well as a ${4}^{\circ}$ ammonium salt. A similar thing can happen if you try to react ammonia with methyl bromide twice...

Furthermore, it is difficult to control this reaction so that you get to a specific step and stop. So, that isn't the best idea.

However, I actually can't think of any other ideas for this. Since this is probably a theoretical exercise, I guess it's OK to give the following synthesis:

To do this, you may have to increase the pH to be between $10.64$ and $10.72$ so that methylamine is unprotonated, while dimethylamine is protonated. Then you could separate them via extraction. But it would be a pain to get the pH to lie between a 0.08 interval...

You could, however, have a pH above $10.72$ to get both neutral, and then boil methylamine off at $- {6}^{\circ} \text{C}$ and leave dimethylamine (whose boiling point is ${7}^{\circ} \text{C}$). Still hard, but easier.

b) This also asks us to consider reacting amines with alkyl halides. Here's a way to do it without resorting to that.

1. Using $\text{NaH}$ would deprotonate acetylene (similar to ${\text{NaNH}}_{2}$), which can then act as a nucleophile in an ${\text{S}}_{N} 2$ reaction and form methylacetylene.
2. Hydroboration adds $\text{OH}$ onto the less-substituted carbon (anti-Markovnikov) instead of the more-substituted carbon (Markovnikov).
3. Keto-enol tautomerization occurs to stabilize the terminal enol into an aldehyde.
4. Adding ammonia in trace acid (pH near $4.5$ for optimal activity) forms an imine ($\text{R"-"C"="NH}$, in this case). What happens is that the oxygen gets protonated by the two protons transferred from ${\text{NH}}_{3}$, and leaves as ${\text{OH}}_{2}^{+}$ when the tetrahedral intermediate collapses to form the imine.
5. Adding ${\text{H}}_{2}$ on $\text{Pd/C}$ reduces the $\text{C"="N}$ bond to a $\text{C"-"N}$ bond, similar to how it works on alkenes.