Imines and Enamines

Quick answer A primary amine (RNH₂) plus an aldehyde or ketone condenses to an imine (C=N, a Schiff base) and water; a secondary amine (R₂NH) gives an enamine (C=C–N) instead. Both share one carbinolamine intermediate and run best around pH 4–5.

⚙ Mechanism · Imine Formation5 steps

Step 1 — the amine adds to the carbonyl carbon.

A neutral amine attacks and the π electrons go up onto oxygen. Nothing has been given or taken away, so the adduct carries both charges at once — negative on oxygen, positive on nitrogen.

Step 2 — a proton moves from nitrogen to oxygen.

The alkoxide takes the proton off the ammonium. Both charges are neutralised in one move, leaving the carbinolamine — an OH and an NH on the same carbon.

Step 3 — acid protonates the OH.

Same reason as always: the OH has to become water before it can leave. This is why imine formation needs acid — and why it is fastest around pH 4–5.

Step 4 — water leaves as the C=N forms.

The nitrogen lone pair pushes in to make the double bond as water departs. Too acidic and the amine is fully protonated so step 1 never happens; too basic and there is no acid for step 3. That tension is what sets the pH optimum.

Step 5 — solvent removes the N–H proton.

The neutral imine is left. A secondary amine has no second N–H to lose here, so it takes a proton from the α-carbon instead and gives an enamine.

A nitrogen nucleophile adds to a C=O carbon and then the molecule eliminates water — a condensation. The one variable that decides the product is how many hydrogens the nitrogen brings.

An imine (C=N) — from a 1° amine

An enamine (C=C–N) — from a 2° amine

Same reaction type, one variable (the amine), two different products.

1. A Primary Amine + a Carbonyl Gives an Imine (C=N)

A primary amine has two N–H bonds, so after adding to the carbonyl one N–H survives to expel water and close a C=N — an imine (Schiff base).

Acetone + a primary amine → a ketimine + H₂O.

2. A Secondary Amine Gives an Enamine (C=C–N) Instead

A secondary amine has only one N–H, spent when it bonds to carbon; with no N–H left to form a C=N, it loses a proton from the α-carbon instead, giving an enamine (C=C–N). A tertiary amine, having no N–H, gives neither.

Acetaldehyde + a secondary amine → an enamine + H₂O.

3. Both Products Pass Through the Same Carbinolamine

Both routes are identical up to losing water: nitrogen adds to give a neutral carbinolamine, acid expels its –OH as water to a resonance-stabilized iminium ion (C=N⁺), and only the last deprotonation differs (N for the imine, α-carbon for the enamine).

Start: the carbonyl (acetaldehyde).

Amine adds → carbinolamine (hemiaminal), with –OH and –NHR on one carbon.

Protonate –OH, lose water, deprotonate → the imine (via an iminium ion).

4. The Reaction Only Runs Well Near pH 4–5

Some acid is needed to protonate the –OH for dehydration, but too much acid protonates the amine to R–NH₃⁺, killing its lone pair.

The free amine (methylamine) must keep its lone pair — too much acid protonates it out of action.

So a mildly acidic pH of about 4–5 is optimal, giving the classic bell-shaped rate-vs-pH curve.

5. Every Step Is Reversible — and Other N-Nucleophiles Do the Same Thing

Every step is an equilibrium, so removing water drives the product forward and adding water hydrolyzes it back; any N–H nucleophile works — hydroxylamine gives an oxime, hydrazine a hydrazone.

An oxime (from H₂NOH)

A hydrazone (from H₂NNH₂)

6. Enamines Are Nucleophilic at the α-Carbon (Stork Synthesis)

The nitrogen lone pair pushes electron density onto the α-carbon, so an enamine alkylates or acylates a ketone there like an enolate — the Stork enamine synthesis — and hydrolysis then reveals the new carbonyl.

The enamine — nucleophilic at the terminal (α) carbon of the C=C.

7. Summary

Amine + carbonyl condenses (add, lose water) · 1° amine → imine (C=N) · 2° amine → enamine (C=C–N, loses an α-H) · shared carbinolamine → iminium → product · optimal at pH 4–5 (bell curve) · all reversible; oximes/hydrazones form the same way and enamines are α-nucleophiles (Stork).

Worked example

Problem. What forms when acetone reacts with a primary amine (CH₃NH₂), and how would a secondary amine differ?

The amine's lone pair attacks the carbonyl carbon → a carbinolamine (hemiaminal).
Acid-catalysed loss of water gives an iminium ion, which loses N–H to form a neutral imine (C=N).
A secondary amine has no N–H to lose at that stage, so it eliminates toward carbon instead, giving an enamine (C=C–N).

Answer. A 1° amine → the imine (N-isopropylidenemethylamine); a 2° amine → the corresponding enamine.

How each reagent works — the arrow pushing

Electron flow only. Follow the arrows; the structures do the talking.

RNH₂ / H⁺ — imines & enamines2 steps

Why it works · The amine nitrogen lone pair is the nucleophile; it adds to the δ+ carbonyl carbon → a carbinolamine. Acid then removes the OH as water. A primary amine loses the remaining N–H proton to give an imine (C=N); a secondary amine has no N–H, so it loses an α C–H instead to give an enamine (C=C–N). Mildly acidic pH is best — enough to dehydrate, not so much it protonates the amine.

Quiz yourself

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A secondary amine has only one N–H, and that hydrogen is consumed when nitrogen bonds to the carbon. After losing water there is no N–H left to form a C=N, so the molecule instead loses a proton from the neighboring α-carbon, forming a C=C conjugated to nitrogen — an enamine.

Both go through a neutral carbinolamine (hemiaminal), the carbon bearing both –OH and an amino group. Protonating and losing the –OH as water gives a resonance-stabilized iminium ion (C=N⁺), which then loses a proton to give the neutral imine or enamine.

Some acid is needed to protonate the carbinolamine –OH so water can leave. But strong acid protonates the amine to R–NH₃⁺, which has no lone pair and cannot add to the carbonyl. pH 4–5 balances both effects, giving a bell-shaped rate-vs-pH curve.

The nitrogen lone pair conjugates into the C=C, putting electron density on the α-carbon so it acts like an enolate (the Stork enamine synthesis — alkylation/acylation at the α-carbon). Because every step is reversible, a final acidic hydrolysis removes the nitrogen and regenerates the substituted ketone.

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