Chemistry - What are the products of the reaction between methanamine and nitrous acid?

Solution 1:

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to S_N2 reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.

Solution 2:

Both the teacher and the friend are partially right, but keep in mind that reaction products depend on reaction conditions, not just reactants.

According to Chemical Mutagens Environmental Effects on Biological Systems (2012):

In the presence of Cl-,

1. $\ce{CH3+}$, is formed.

2a. $\ce{CH3+ + H2O -> CH3OH +H+}$

2b. $\ce{CH3+ + NO2- -> CH3NO2}$ and $\ce{CH3+ + NO2- -> CH3ONO}$

2c. $\ce{CH3+ + Cl- -> CH3Cl}$

2d. $\ce{CH3+}$ reacts with methylamine to form dimethyl amine

3. Methanol formed above can react with $\ce{CH3+}$ to form dimethyl ether

4. Methanol can also react with nitrous acid to form methyl nitrite

The above book is relying upon Austin, A. T., The Action of Nitrous Acid on Aliphatic Amines Nature 188, 1086-1088 (1960) as reference 472 to support the above information.

Austin explains that, under the reaction conditions:

0.5 mole methylamine hydrochloride
1.5 moles sodium nitrite
550 ml water
slowly acidified with 0.4 to 0.6 equivalents of 2N sulphuric acid
room temperature

products are as follows:

methanol: 6-25%
methyl nitrite: 45-35%
methyl chloride: 13%
nitromethane : 6%
methylnitrolic acid: 10-12%

Austin then states:

The multiplicity of products formed in the above reaction is rationally explained if one assumes that the electron-deficient entity $\ce{CH3+}$ is formed - a suggestion originally put forth by Whitmore [reference 7] for this type of reaction.

While Austin discusses the possibility of dimethyl ether and dimethyl amine also forming, he does not indicate that they were observed experimentally.