Question

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

Answer 1

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^@`, `2^@`, and `3^@` amines, as well as a `4^@` 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^@ "C"` and leave dimethylamine (whose boiling point is `7^@ "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 `"NaH"` would deprotonate acetylene (similar to `"NaNH"_2`), which can then act as a nucleophile in an `"S"_N2` reaction and form methylacetylene.
2. Hydroboration adds `"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 (`"R"-"C"="NH"`, in this case). What happens is that the oxygen gets protonated by the two protons transferred from `"NH"_3`, and leaves as `"OH"_2^(+)` when the tetrahedral intermediate collapses to form the imine.
5. Adding `"H"_2` on `"Pd/C"` reduces the `"C"="N"` bond to a `"C"-"N"` bond, similar to how it works on alkenes.