# Hofmann elimination

An unknown amine is treated with an excess of methyl iodide. Two equivalents of methyl iodide react with the amine. The amine is treated with silver oxide and water, and then heated to $120\ ^\circ \mathrm{C}$. The resulting products are trimethylamine and ethylene. The unknown amine is

1. $\ce{CH3CH2NHCH3}$

2. $\ce{CH3CH2NH2}$

3. $\ce{CH2=CHNH2}$

4. $\ce{CH2=CHNHCH3}$

I tried to get the answer using concept of Hofmann elimination.

The unknown amine has to be a secondary amine because after adding 2 equivalents of $\ce{CH3I}$, a quaternary ammonium iodide will be formed which on treatment with most silver hydroxide and heating thereafter well give tertiary amine, alkene and water. This is what happens when one of the alkyl group is other than methyl.

I have never seen this reaction anywhere working with amine containing a alkyl group having double bond, so I eliminated the fourth option and went for the first option and the answer was correct.

My question is that why the fourth option is not correct? What is the corresponding mechanism?

• If the reaction worked, it would produce acetylene (ethyne/$\ce{HC#CH}$) instead of ethylene. – DHMO Jan 12 '17 at 15:22
• @DHMO can you please provide the mechanism – Resorcinol Jan 12 '17 at 15:34
• The mechanism is here but it doesn't matter. The point is that the carbon region would have one additional bond. So if you start with a double bond, you would end up having a triple bond. – DHMO Jan 12 '17 at 15:37

## 1 Answer

The reaction has three steps.

I shall use N-methylethenamine, $\ce{CH2=CH-NH-CH3}$ (the fourth option), as an example.

Step $1$:

$$\ce{CH2=CH-NH-CH3 + 2CH3I -> [CH2=CH-N(CH3)3]+ + I-}$$

This step is an $\mathrm{S_N2}$ step.

The carbon in the methyl iodide is the electrophile while the nitrogen in the N-methylethenamine is the nucleophile.

As the comments have pointed out, this step is not viable. The following tautomerism happens:

$$\ce{CH2=CH-NH-CH3 <=> CH3-CH=N-CH3}$$

Also, the electrophile methyl iodide would attack on the carbon-carbon double bond instead of the nitrogen.

Step $2$:

$$\ce{Ag2O + I- -> AgI + AgO-}$$ $$\ce{AgO- + H2O -> AgOH + OH-}$$

Step $3$:

$$\ce{[CH2=CH-N(CH3)3]+ + OH- ->[\Delta] CH#CH + CH3-N(CH3)-CH3 + H2O}$$

This resembles an $\mathrm{E1}$ step.

According to this forum post, there should be no problem with the first step. The methyl iodide would not destroy the double bond.

There is also no hydrogen available to perform an addition reaction with the iodide ion to the N-methylethenamine to destroy the double bond.

I believe that the second step is also viable, also because of the lack of a good electrophile (silver doesn't count because silver can't react with carbon-carbon double bonds).

The third step should be slower because eliminiation reactions that result in alkynes are slower than elimination reactions that result in alkenes in general.

Therefore, the product would be ethyne $\ce{HC#CH}$ instead of ethylene $\ce{HC=CH}$.

• It would not work at all as enamines tautomerise to imines. – Mithoron Jan 12 '17 at 21:49
• Yeah the supposed enamine product doesn't exist. Furthermore even if you could isolate the enamine it is more likely to react with MeI on carbon not nitrogen. – orthocresol Jan 13 '17 at 20:00
• @orthocresol do your remarks contradict with the link to the forum post? – DHMO Jan 14 '17 at 1:54
• No. That post assumes that the amine and the alkene are isolated. When they are conjugated like in an enamine the reactivity is different. – orthocresol Jan 14 '17 at 8:45