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.
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).
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.
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
- 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.
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.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.