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

I'm finding this particular reaction very problematic. Here's why:

My teacher tells me that methanamine, when treated with nitrous acid (created by reacting sodium nitrite and hydrochloric acid) converts to an the unstable (at room temperature) intermediate: methane diazonium chloride, which decomposes in the (aqueous) solution to yield methanol, hydrogen chloride and dinitrogen.

A friend of mine (who happens to be a fairly trustworthy senior), tells me that the actual products are methoxymethane and dinitrogen, since methanamine is an exception and doesn't give the same kind of products the other primary aliphatic amines do (he read it up a long time ago, but doesn't remember where).

In hope of clearing up this mess, I resorted to my trusty copy of Organic Chemistry (Morrison and Boyd, 7th edition), according to which, the products of a reaction between a primary aliphatic amine and nitrous acid should be: dinitrogen and a mixture of alcohols and alkenes.

enter image description here

It doesn't mention anything about methanamine being an exception, and "a mixture of alcohols and alkenes" is a little ambiguous. Also I didn't discard this idea because maybe - through some method I don't even know - the methyl bits from the amines might couple and somehow result in an alkene. I wouldn't know since I'm relatively new to amine + nitrous acid reactions, so I won't take the risk.

I couldn't find any trust-worthy reference on this issue for this reaction. I need help resolving this ambiguity.


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

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

  2. 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.

  • 5
    $\begingroup$ Nearly all chemistry that is taught in undergrad etc. was performed decades ago. $\endgroup$ – orthocresol Oct 7 '16 at 14:37
  • $\begingroup$ I would need to see their kinetic or mechanistic data, but I am extremely suspicious of a methyl cation in aqueous solution. This seems extremely implausible. This isn't FVP, it's at 0 degrees C. $\endgroup$ – Zhe Oct 7 '16 at 17:44
  • 2
    $\begingroup$ According to the terminology I learnt at uni, it should be a carbenium ion (lack of a bond) not a carbonium (an additional bond). $\endgroup$ – Jan Oct 8 '16 at 0:13
  • 2
    $\begingroup$ Unfortunately I can’t access the mechanism either. But to be perfectly honest, I see no reason why the reaction must happen under $\mathrm{S_N1}$ conditions. Until proven otherwise, I will assume a mixture of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ because that seems most reasonable. (CC@Zhe; I approve of bringing Occam’s razor into the discussion!) $\endgroup$ – Jan Oct 8 '16 at 0:21
  • 1
    $\begingroup$ I added information from the original A.T. Austin reference that says the mechanism is via CH3+ , but I agree it isn't proof. The reference is very good for answering the OP's question about what the actual products are, however. $\endgroup$ – DavePhD Oct 11 '16 at 16:50

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 SN2 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.