Heat butan-2-amine (also called 2-aminobutane) with a strong base, and no alkene forms. Turn its NH2 group into an N+(CH3)3 group first, then heat it with hydroxide, and it gives mainly but-1-ene. This two-stage route from an amine to an alkene is the Hofmann elimination.
Two questions follow. Why must the nitrogen be changed before it can leave? And why is the main alkene the less substituted one, the opposite of what E2 leads you to expect from a small base?
Why the amine has to be converted first
In an E2, the leaving group takes the electron pair of its bond to carbon with it. The carbon it leaves from is the α carbon, and a β carbon is any carbon bonded to that one. For butan-2-amine, the group that would leave is H2N−, the amide ion. This is the conjugate base of ammonia; it has nothing to do with the carbonyl amides of Esters & amides.
Ammonia’s N–H has a pKa near 38, so the amide ion is one of the strongest bases in the course. As Leaving groups showed, a strong base is a poor leaving group. The amide ion is so poor that the E2 never gets started.
The fix is to change what leaves. Treat the amine with excess methyl iodide, CH3I, and the nitrogen picks up methyl groups until it carries four carbon groups and a positive charge. The product is a quaternary ammonium salt, with iodide as its counterion. Adding methyl groups until no more can go on is called exhaustive methylation. It is the runaway alkylation from Amine reactions, now done on purpose.
Each methyl group goes on in an SN2 step: the nitrogen lone pair attacks the methyl carbon, and iodide leaves. The ammonium ion that forms still has an N–H, and a base removes that hydrogen as a proton. Potassium carbonate, K2CO3, is usually added for this job, though a second amine molecule can do it too. The neutral amine left behind carries one more methyl group, and it reacts again.
So the methyl groups never replace the hydrogens directly. Each N–H hydrogen costs one CH3I and leaves as a proton, and one last CH3I makes the quaternary salt. The count is the number of N–H hydrogens plus one. Butan-2-amine has two, so it takes three equivalents and ends as the sec-butyltrimethylammonium ion (systematic name N,N,N-trimethylbutan-2-aminium).
In that salt, the group that would leave is trimethylamine, N(CH3)3, a neutral molecule. The nitrogen is positive before it leaves and neutral afterward, so it carries no negative charge away. Trimethylamine is a far weaker base than the amide ion, and that makes it a workable leaving group.
Methyl is the right group to add because a methyl group has no β carbon. It therefore has no β hydrogen, and it cannot be lost as an alkene itself. The only C–N bond that can break in the elimination is the one to the original carbon chain.
Next comes silver oxide, Ag2O, in water. It swaps the iodide for hydroxide: silver iodide, AgI, falls out of solution as a solid, and hydroxide takes its place as the counterion. That supplies the base the E2 needs and removes iodide, a nucleophile that would otherwise compete.
Heating this quaternary ammonium hydroxide then drives an ordinary E2. Hydroxide takes a β hydrogen, the C–H bonding pair becomes the new π bond, and the C–N bond breaks, all in one step. Nothing about the mechanism is new; only the leaving group is.
The Hofmann rule: the less substituted alkene
Butan-2-amine shows which alkene wins. The nitrogen sits on C2, so the β carbons are C1 and C3. C1 is a CH3 with three hydrogens. C3 is a CH2 with two hydrogens, and it also carries C4. Losing a hydrogen from C1 gives but-1-ene, a monosubstituted alkene. Losing one from C3 gives but-2-ene, which is disubstituted and more stable.
Zaitsev’s rule would predict but-2-ene. The Hofmann elimination gives mainly but-1-ene.
This is the Hofmann rule: a Hofmann elimination gives mainly the less substituted alkene, the one with fewer carbon groups on its C=C. E2 gave this alkene a name, the Hofmann product, and made it with a bulky base such as tert-butoxide.
The reason starts with the E2 geometry. The C–H bond that breaks and the C–N bond that breaks must be anti-periplanar: in one plane, pointing in opposite directions, 180° apart. In an open chain, the molecule turns about the C–C bond until they line up. The figure below looks down the C1–C2 bond and the C3–C2 bond to show what that turn costs at each β carbon.
For C1, it costs nothing. C1 carries only hydrogens, so whichever one is anti to the nitrogen, the two beside it are small. For C3, it costs a lot. Once a C3 hydrogen is anti to the nitrogen, C3’s other two bonds each sit 60° from the nitrogen, a gauche position. One of them holds C4, so that methyl group is pushed right up against the very large N+(CH3)3 group.
That crowded arrangement is higher in energy, so the route to but-2-ene is slower. Hydroxide also reaches the exposed hydrogens on C1 more easily than the shielded ones on C3. Both are steric effects, and together they are the bulky-base argument from E2 with the bulk moved from the base to the leaving group.
A second effect points the same way. The positive nitrogen pulls electron density away from the β carbons, which makes their hydrogens more acidic than usual. The C–H bond therefore breaks further ahead of the C–N bond than in a typical E2, and the β carbon picks up some negative charge in the transition state. Chemists call this transition state E1cb-like. In a true E1cb, such as the aldol dehydration in Aldol reactions, the proton leaves in one step and the leaving group in a second. The Hofmann elimination is still one step; it only leans that way.
Neither effect makes but-1-ene the more stable alkene. They make it form faster. An E2 cannot run backward, so the Hofmann elimination is under kinetic control, as Kinetic vs thermodynamic control explained: but-1-ene is the kinetic product. But-2-ene is more stable, yet heating the mixture for longer never turns one alkene into the other.
The two effects do not always agree. Take the salt from 1-phenylpropan-2-amine, PhCH2–CH(NH2)–CH3. One β carbon is the CH3. The other is the CH2 bonded directly to the benzene ring, a benzylic carbon. A hydrogen there is much more acidic than an ordinary one, because the negative charge that builds on that carbon spreads into the ring.
Here acidity beats crowding. The main product is 1-phenylprop-1-ene, PhCH=CH–CH3, whose C=C is conjugated with the ring. The less substituted 3-phenylprop-1-ene, PhCH2–CH=CH2, is the minor product. A C=O next to a β carbon has the same effect.
So the full rule is that hydroxide removes the most acidic β hydrogen it can reach. In most substrates, the easiest hydrogen to reach is also the most acidic one. When a benzene ring or a C=O makes one hydrogen far more acidic, acidity wins.
| Ordinary E2, small base | Hofmann elimination | |
|---|---|---|
| Leaving group | a halide, small | N(CH3)3, bulky |
| Where the bulk is | nowhere | on the leaving group |
| β hydrogen taken | the one that gives the more substituted C=C | usually the least crowded one; a far more acidic one (benzylic, or next to a C=O) wins |
| Product | Zaitsev alkene | Hofmann alkene |
Worked example: 2-methylpentan-3-amine
Exhaustive methylation of 2-methylpentan-3-amine, then Ag2O, water and heat. What is the major alkene?
Count the CH3I. The NH2 is on C3. It is a primary amine, with two N–H hydrogens, so it takes three equivalents and becomes the trimethylammonium salt on C3.
Find the β carbons. They are C2 and C4. C2 carries one hydrogen and two methyl groups, C1 and the branch. C4 carries two hydrogens and one methyl group, C5.
Compare the two alkenes. A hydrogen from C2 gives 2-methylpent-2-ene, which is trisubstituted. A hydrogen from C4 gives 4-methylpent-2-ene, which is disubstituted. The major product is 4-methylpent-2-ene. Neither choice is a terminal alkene, but the less substituted one still wins.
The Cope elimination: similar regiochemistry, the other geometry
A tertiary amine can also eliminate by a second route. Take N,N-dimethylbutan-2-amine, butan-2-amine with two methyl groups already on its nitrogen. Oxidize it with hydrogen peroxide, H2O2, or with mCPBA, the peroxyacid from Oxidative cleavage & dihydroxylation. The nitrogen lone pair forms a bond to an oxygen atom, and the product is an amine oxide, R3N+–O−, with a positive nitrogen bonded to a negative oxygen.
Heat the amine oxide to 100–150 °C and it eliminates with no added base. Its own oxygen removes a β hydrogen. The O–H bond forms, the C–H bond breaks, the π bond forms and the C–N bond breaks, all in one step, through a flat ring of five atoms: O, H, Cβ, Cα and N. This is the Cope elimination (not the Cope rearrangement named in The Diels–Alder reaction). It gives the alkene and N,N-dimethylhydroxylamine, (CH3)2N–OH.
The oxygen is bonded to the nitrogen, so it can only reach a hydrogen on the nitrogen’s side of the Cα–Cβ bond. That makes the Cope elimination syn-periplanar: the H and the N leave from the same side, 0° apart. The Hofmann elimination, like the E2 reactions in this course, is normally anti-periplanar, 180° apart.
Its regiochemistry usually matches the Hofmann elimination’s: it tends to give the less substituted alkene, so N,N-dimethylbutan-2-amine oxide gives mainly but-1-ene.
The conditions are its advantage. A Cope elimination needs no strong base and no silver, only heat. A molecule that also carries an ester, which hot hydroxide would hydrolyze, survives it.
Counting methyl groups: working out structures before spectroscopy
Before instrument methods such as IR, mass spectrometry and NMR (NMR Spectroscopy) existed, chemists used exhaustive methylation to work out the structure of an unknown amine. The amount of CH3I used in the first methylation gave its class, by the N–H + 1 rule above:
- three equivalents → a primary amine, RNH2
- two equivalents → a secondary amine, R2NH
- one equivalent → a tertiary amine, R3N
The alkenes that came off then showed the carbon skeleton around the nitrogen. A cyclic amine gave the most information. Piperidine, for example, is a six-membered ring with one nitrogen, and two C–N bonds hold that nitrogen in the ring. The first elimination breaks only one of them, so it opens the ring but leaves the nitrogen attached to the chain.
A second round of methylation and elimination breaks the other C–N bond and frees the nitrogen as trimethylamine. The whole degradation uses three CH3I, two in the first round and one in the second.
Two rounds alone do not prove a ring. Diethylamine, an open-chain secondary amine, also needs two rounds, and each one splits off a molecule of ethene. The sign of a ring is what the first round gives. In an open-chain amine, the first round splits off an alkene that contains no nitrogen. In a ring, it splits nothing off: all the carbons stay in one molecule with the nitrogen. That is how chemists knew the nitrogen had been in a ring.
The two measurements answer different questions. The CH3I count gives the class of the amine, and the products of the first round show whether the nitrogen sat in a ring. The method used up the sample, and it was only as reliable as the identification of each fragment. Even so, chemists worked out the structures of many alkaloids, the nitrogen compounds of plants, this way. Today NMR spectroscopy answers the same questions in minutes, on a few milligrams, without destroying anything.
What carries forward
E2 treated Zaitsev’s rule as a preference and showed two ways to override it: geometry, when only one β hydrogen can be anti-periplanar, and a bulky base. The Hofmann elimination adds a third: a bulky leaving group. Before you predict the more substituted alkene, check the size of the base and of the leaving group.