Imines and enamines — Carbonyl & Enolate Breadth

An amine is a nucleophile and a carbonyl is an electrophile, so the first step is not in doubt: the nitrogen attacks the carbonyl carbon and you get a tetrahedral intermediate with an OH and an NR₂ on the same carbon — a carbinolamine.

What happens next depends entirely on how many hydrogens the nitrogen brought, and that one question separates the two products of this section.

Two amines, two endpoints

Amine	H on N after addition	Product
Primary, RNH₂	One left	Imine, C=N–R
Secondary, R₂NH	None left	Enamine, C=C–NR₂

In both cases the carbinolamine loses water under acid catalysis to give a cation with a C=N⁺. A primary amine still has a hydrogen on that nitrogen, so losing it gives the neutral imine. A secondary amine has none, so the only proton available is on the α carbon — and removing it gives the enamine, with the double bond between the two carbons instead.

Count the hydrogens on the nitrogen and the answer is forced. A tertiary amine, with no N–H at all, cannot lose water in either direction and so gives no stable product — which is why tertiary amines are used as bases in these reactions rather than as reagents.

Why the pH has to be about 4.5

Both amines give the same iminium cation, and everything up to that point is identical. What separates the two products is where the last proton can come from — the nitrogen, if it still has one, and otherwise the α carbon.This is why the answer is a count rather than a mechanism. You do not need to run the steps: look at how many hydrogens the nitrogen brought and subtract the one it spends reaching the cation. Two becomes one, so a primary amine gives an imine; one becomes none, so a secondary amine has to take the alpha proton instead. A tertiary amine brings none, which is why it can add and still give nothing, and why tertiary amines appear in these reactions as bases.

This reaction is the standard example of a rate that is fastest in the middle. Acid is required, because the OH of the carbinolamine is a terrible leaving group and has to be protonated first. But too much acid protonates the amine, and an ammonium ion has no lone pair, so there is no nucleophile left.

Too basic: the OH never leaves, and the carbinolamine sits there.
Too acidic: the amine is fully protonated, and nothing attacks in the first place.
About pH 4–5: enough acid to protonate the OH, not enough to switch off the amine.

Every step is reversible, so imine and enamine formation are run with the water removed — a Dean–Stark trap, or molecular sieves — to pull the equilibrium across. Add water back with acid and you get the carbonyl compound and the amine returned.

The enamine is a nucleophile at carbon

This is what the section is really for. Draw the enamine's second resonance structure: the nitrogen lone pair pushes into the C=C, giving an iminium cation with a negative α carbon.

That makes an enamine a nucleophilic carbon — the same role an enolate plays, reached without a strong base. The comparison is worth holding:

	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	Protonate	Hydrolyze back to the ketone

The Stork enamine synthesis is the sequence built on that: form the enamine, alkylate or acylate the α carbon, then hydrolyze. Net result, an α-substituted ketone — and because the enamine is neutral and less reactive than an enolate, it usually stops after one alkylation rather than going twice.

An enamine is also a good Michael donor, adding 1,4 to an α,β-unsaturated carbonyl. That is the classic use, because conjugate addition of a full enolate often competes with 1,2-addition, and the milder nucleophile does not.

Imines elsewhere

Imines are worth recognizing beyond this chapter. Reducing one with NaBH₃CN gives an amine, which is reductive amination — the most reliable way to make a secondary or tertiary amine without the over-alkylation problem that direct S_N2 on ammonia has. And an imine formed with hydroxylamine or a hydrazine gives the crystalline oximes and hydrazones that were once how a carbonyl compound got identified.

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, which is the gentler half of the two ways to functionalize an α carbon.