Acetone and methylamine, CH3NH2, give (CH3)2C=N–CH3; acetone and dimethylamine, (CH3)2NH, give CH2=C(CH3)–N(CH3)2. Both release a molecule of water. Nucleophilic addition showed these two products side by side. This section runs the mechanism that makes them.
The first product is an imine: a compound with a carbon–nitrogen double bond, C=N. The second is an enamine: an amine nitrogen attached to one carbon of a C=C (“ene” for the alkene, “amine” for the nitrogen).
Both come from the same three steps. Only the last one differs, and one count decides it: how many hydrogens the nitrogen started with.
Step 1: the amine adds and a carbinolamine forms
An amine is a nucleophile, and the carbonyl carbon is an electrophile. So the nitrogen lone pair attacks the carbonyl carbon. As the new C–N bond forms, the C=O pi electrons move onto oxygen.
The result is a tetrahedral intermediate with a negative oxygen and a positive nitrogen on the same carbon. A molecule that carries both charges at once like this is called a zwitterion.
A proton then moves from the N+ to the O−, usually with a solvent molecule passing it along. That cancels both charges. The neutral product has an OH and an amine nitrogen on the same carbon, and it is called a carbinolamine.
Step 2: acid turns the OH into water, and water leaves
The carbinolamine now has to lose its OH. Hydroxide, HO−, is a poor leaving group. So an acid first puts a proton on the oxygen. The OH becomes OH2+, and water is a good leaving group.
Next the nitrogen lone pair pushes in toward carbon, forming a C=N double bond, while the C–O bond breaks and water leaves. The product is an iminium ion: a C=N double bond with a positive charge on nitrogen.
These are the same two moves as the loss of water in Acetals & hemiacetals: protonate the OH, then push a lone pair in as water leaves. There an oxygen lone pair pushes in. Here a nitrogen lone pair does, and nitrogen gives up its lone pair more readily than oxygen does. That is why imines form under much milder acid than acetals need.
Step 3: the last proton decides the product
The iminium ion carries a positive charge. To become a neutral product it has to lose a proton, and where that proton can come from decides the product.
| Amine | H on N after addition | Product |
|---|---|---|
| Primary, RNH2 | One left | Imine, C=N–R |
| Secondary, R2NH | None left | Enamine, C=C–NR2 |
Methylamine is a primary amine: its nitrogen carries one carbon and two hydrogens. It spent one hydrogen in Step 1, so its iminium ion still has an N–H. A water molecule takes that proton, the N–H electrons become a lone pair on nitrogen, and the neutral imine (CH3)2C=N–CH3 results.
Dimethylamine is a secondary amine: two carbons and one hydrogen on nitrogen. It spent that hydrogen in Step 1, so its iminium ion has no N–H at all. The nearest removable proton sits on the α carbon, the carbon right next to the C=N carbon (the same position that is α to a C=O in Aldehydes & ketones).
When water takes that α proton, three pairs of electrons shift at once. Water’s lone pair forms the new O–H bond. The C–H electrons form a new C=C bond. The C=N pi electrons move onto nitrogen as a lone pair. The double bond has moved from C=N to C=C, and the product is the enamine CH2=C(CH3)–N(CH3)2.
Why the pH has to be about 4.5
This reaction is the standard example of a rate that is fastest in the middle of the pH range. Acid is required, because the OH of the carbinolamine is a poor leaving group until it is protonated. But too much acid protonates the amine too. Methylamine becomes CH3NH3+, an ammonium ion, which has no lone pair and so cannot attack.
- Too basic: the OH is rarely protonated, so water leaves only slowly and the carbinolamine builds up.
- Too acidic: nearly all of the amine is ammonium, so the first step stalls.
- About pH 4–5: enough acid to protonate the OH, not enough to switch off the amine.
It is worth being precise about which step is slow on each side. On the basic side of the peak, the amine’s addition is fast and the slow step is the loss of water from the carbinolamine. That is exactly the step a proton helps, so adding acid speeds the reaction up. On the acidic side the balance flips. So much of the amine is now ammonium that the addition becomes the slow step, and more acid only slows it further. The peak sits where the two steps trade places.
Driving it forward, and back
Every step is reversible, and water is one of the products. So chemists remove water as it forms, which pulls the equilibrium toward the imine or enamine. A Dean–Stark trap does it by boiling the water off with the solvent and catching it in a side arm. Molecular sieves do it by soaking the water up into tiny pores.
Run the same steps backward and you get the reverse reaction. Add water and acid to an imine or an enamine, and it hydrolyzes: it splits back into the carbonyl compound and the amine.
The enamine is a nucleophile at carbon
This is what enamines are for. Push the nitrogen lone pair of CH2=C(CH3)–N(CH3)2 toward the neighboring carbon, and the C=C pi electrons have to move to the end CH2. That gives a second resonance structure with C=N+ and a negative end carbon.
That end carbon is the one that was α to the original C=O. The resonance structure says it carries a real share of the nitrogen’s electron pair, so it is nucleophilic. An enamine therefore gives you a nucleophilic carbon, reached with nothing stronger than an amine and mild acid.
The other way to make an α carbon nucleophilic is the enolate. A strong base pulls an α hydrogen off a ketone, and the anion left behind shares its negative charge between that carbon and the oxygen. Enolates are the subject of Enolate Chemistry, several chapters on. The table below is a preview.
The strong base usually named for this is LDA, lithium diisopropylamide, (i-Pr)2N− Li+. It is strong enough to turn nearly all of a ketone into its enolate, which is then called a lithium enolate.
| Enolate | Enamine | |
|---|---|---|
| Made with | A strong base, often LDA | A secondary amine and mild acid |
| Charge | Anionic | Neutral |
| Reactivity | Higher; can attack twice | Lower, and cleaner for it |
| Afterwards | Nothing: the product is already the ketone | Hydrolyze back to the ketone |
The Stork enamine synthesis
Take cyclohexanone and the cyclic secondary amine pyrrolidine. With mild acid and water removal they give the pyrrolidine enamine of cyclohexanone. Add allyl bromide, CH2=CH–CH2Br. The enamine’s nucleophilic carbon, C2 of the ring, attacks the CH2 that carries the bromine, and bromide leaves in an SN2 step. Finally, aqueous acid hydrolyzes the product. What comes out is 2-allylcyclohexanone and pyrrolidine.
That three-stage sequence is the Stork enamine synthesis: form the enamine, alkylate or acylate (attach an R–C=O group to) the α carbon, then hydrolyze. Net result, an α-substituted ketone.
Acylation works the same way. An acid chloride, R–CO–Cl, is a C=O carbon that carries a chlorine (Carboxylic Acids & Derivatives, later). The enamine carbon attacks its C=O carbon, chloride leaves, and hydrolysis gives a 1,3-diketone. It works well because an acid chloride is far more electrophilic than any alkyl halide.
The sequence stops cleanly after one alkylation, and the reason is structural rather than a matter of reactivity. Alkylation turns the enamine into an iminium salt. That cation has no C=C next to the nitrogen any more, so the nitrogen lone pair has no carbon to make nucleophilic. No strong base is present to pull it back to an enamine. A little proton exchange between the salt and enamine that has not yet reacted can give some dialkylated product, but it is a minor side reaction at most, not a competing pathway.
A lithium enolate has no such stop: its product can be deprotonated and alkylated again (over-alkylation), one of the problems Enolate Chemistry takes up.
Enamines in conjugate addition
Take but-3-en-2-one, CH2=CH–CO–CH3, a ketone whose C=O is conjugated with a C=C. Number the four atoms of the conjugated system from the oxygen: O is 1, the carbonyl carbon is 2, and the far CH2 is 4. Aldehydes & ketones showed that a resonance structure puts positive charge on carbon 4 (the β carbon there) as well as on carbon 2. These position numbers count along the conjugated system; they are not the IUPAC locants in the name, where C1 is the CH3.
So a nucleophile has two targets. Attack at carbon 2, the C=O carbon itself, is 1,2-addition. Attack at carbon 4, the far end of the C=C, is 1,4-addition, also called conjugate addition. The names count positions the same way as for dienes in 1,2- vs 1,4-addition. When the nucleophile is an enolate or an enamine, 1,4-addition is called a Michael addition.
Enamines are reliable Michael donors: they add 1,4 cleanly. The pyrrolidine enamine of cyclohexanone and but-3-en-2-one give, after hydrolysis, a new C–C bond from C2 of the ring to carbon 4 of the enone. Enolates vary more, and Enolate Chemistry sorts out which add 1,4.
Imines elsewhere
Reductive amination. Reduce an imine and the C=N becomes a C–N single bond, giving an amine. The usual reagent is NaBH3CN, a milder cousin of the NaBH4 in Nucleophilic addition. It reduces the protonated imine (the iminium ion) faster than it reduces the starting C=O. This is the most reliable way to make a secondary or tertiary amine. It avoids the over-alkylation that SN2 on an amine suffers, where each new amine is itself a nucleophile and reacts again. Amines builds on it.
Oximes and hydrazones. Hydroxylamine, H2N–OH, and hydrazine, H2N–NH2, are primary amines of a sort, so they give imines too. The imine from hydroxylamine is an oxime, C=N–OH. The one from a hydrazine is a hydrazone, C=N–NH2. Many are crystalline, and their melting points were once how a carbonyl compound got identified.
Wolff–Kishner reduction. Heat a hydrazone with a strong base, such as KOH in a high-boiling solvent, and the two nitrogens leave as N2 gas. A CH2 is left where the C=O was. This is the standard way to delete a carbonyl group entirely.
Clemmensen reduction. Zinc amalgam, Zn(Hg), with HCl does the same C=O to CH2 change in acid. Choose Wolff–Kishner when the molecule cannot survive acid and Clemmensen when it cannot survive base. Both matter for the same reason: a Friedel–Crafts acylation, which puts a C=O group onto a benzene ring (Aromatic Chemistry), followed by one of them is how an unrearranged alkyl group gets onto a ring.
What carries forward
Count the N–H hydrogens: one left gives an imine, none left gives an enamine. Run it near pH 4.5 and remove the water. Then remember what the enamine is for: a neutral nucleophilic carbon that alkylates once and hydrolyzes back, the gentler of the two ways to put a new group on an α carbon.