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

Question

Isabella Acosta · Accepted Answer

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.

Using #"NaH"# would deprotonate acetylene (similar to #"NaNH"_2#), which can then act as a nucleophile in an #"S"_N2# reaction and form methylacetylene.
Hydroboration adds #"OH"# onto the less-substituted carbon (anti-Markovnikov) instead of the more-substituted carbon (Markovnikov).
Keto-enol tautomerization occurs to stabilize the terminal enol into an aldehyde.
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.
Adding #"H"_2# on #"Pd/C"# reduces the #"C"="N"# bond to a #"C"-"N"# bond, similar to how it works on alkenes.

Ella Allen · Answer

undefined (a) Methyl bromide can be converted to dimethylamine by reacting it with excess methylamine (CH3NH2) in the presence of a strong base, such as sodium hydroxide (NaOH). This reaction forms dimethylamine and sodium bromide as products.

(b) Propanamine can be prepared from methyl chloride by a series of reactions. First, methyl chloride is reacted with excess ammonia (NH3) in the presence of a catalyst, such as iron (Fe), to form methylamine (CH3NH2). Then, methylamine undergoes a nucleophilic substitution reaction with chloroethane (C2H5Cl) to form diethylamine (C2H5)2NH. Finally, diethylamine is subjected to Hofmann degradation by treatment with excess sodium hypobromite (NaOBr) or sodium hydroxide (NaOH) to yield propanamine (CH3CH2CH2NH2).