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

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.