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

1 Answer
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^@, 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.