Chem 17 Lecture 19: Imines, Iminium Ions, Reductive Amination & Enamines

Imine formation mechanism and pH dependence, electrophilicity of imines vs iminium ions, reductive amination with NaBH3CN, enamine formation. Based on Lecture 19 slides + annotated notes (Clayden Ch. 11).

Imine

A C=N compound made from an aldehyde or ketone + a primary amine (or NH3) under mildly acidic conditions (pH ~5), with loss of H2O. A reversible, acid-catalyzed condensation.

Hemiaminal

The tetrahedral intermediate in imine formation: one carbon carrying both –OH and –NHR. It is the nitrogen analog of a hemiacetal.

Iminium ion

The positively charged C=N(+) species formed when the protonated hemiaminal loses H2O. It is the key electrophilic intermediate in imine formation, enamine formation, and reductive amination.

Zwitterion

A neutral molecule that contains separate opposite formal charges, e.g. the first adduct when an amine attacks a carbonyl (O– and N+ on the same molecule).

Overall mechanism of imine formation (3 stages)

(1) Addition: amine lone pair attacks C=O (no acid needed) → zwitterion → PT → hemiaminal. (2) Dehydration: protonate –OH, expel H2O → iminium ion. (3) Deprotonation: lose the N–H proton → neutral imine. Acid is regenerated (catalytic).

Why does imine formation require a primary amine (or NH3)?

The iminium must still carry an N–H so it can be deprotonated to the neutral C=N. A secondary amine gives an iminium with no N–H (→ enamine instead); a tertiary amine cannot form a neutral product at all.

Why is imine formation very slow below pH ~4?

Too much acid protonates the amine to R–NH3+, which has no lone pair and is not a nucleophile. The addition step (Stage 1) stalls.

Why is imine formation very slow above pH ~6?

Not enough acid to protonate the hemiaminal –OH. Hydroxide is a poor leaving group, so dehydration to the iminium (Stage 2) stalls.

Optimal pH for imine formation and why

About pH 5. It balances two opposing needs: enough free (unprotonated) amine to add to the carbonyl, and enough acid to turn –OH into the good leaving group H2O.

Rate-determining step at low pH vs high pH (imine formation)

Below pH 4: addition (amine attack) is rate-determining. Above pH 6: dehydration (loss of H2O from the protonated hemiaminal) is rate-determining. Near pH 5 the two rates are comparable.

Role of H3O+ in imine formation

Catalyst: it is consumed when it protonates the hemiaminal –OH (making H2O a good leaving group) and regenerated in the final deprotonation of the iminium N–H.

PT (abbreviation)

Proton transfer. From now on you may write "PT" for proton transfers within a tetrahedral intermediate (e.g. moving H+ from N+ to O– in the zwitterion).

Rank the electrophilicity: aldehyde/ketone, imine, iminium ion

Iminium > aldehyde/ketone > imine. The positive charge on the iminium strongly lowers its LUMO (π*). A neutral imine is less electrophilic than a carbonyl because N is less electronegative than O.

Why is a neutral imine a weaker electrophile than a ketone?

Nitrogen is less electronegative than oxygen, so C=N is less polarized and its π* LUMO is higher in energy than that of C=O.

Why is an iminium ion a stronger electrophile than a ketone?

The full positive charge on N dramatically lowers the π* LUMO. Experimentally, the charge effect wins over the electronegativity difference (N vs O).

Reductive amination

One-pot conversion of an aldehyde/ketone + amine (NH3, 1°, or 2°) into an amine: the iminium forms at pH ~5, then NaBH3CN delivers hydride to the iminium carbon. Irreversible overall.

Why can't LiAlH4 be used for reductive amination?

AlH4– reacts violently with water and protic solvents, so it is incompatible with the aqueous pH ~5 conditions needed to form the iminium.

Why is NaBH4 a poor choice for reductive amination?

BH4– tolerates water but is too strong a hydride nucleophile: it reduces the starting aldehyde/ketone (→ alcohol) as well as the iminium, so you get a mixture.

Why does NaBH3CN work selectively in reductive amination?

The CN group is electron-withdrawing (inductive effect), which makes B–H a weaker hydride donor. The weaker nucleophile reacts only with the most electrophilic species present, the iminium, and leaves the aldehyde/ketone alone.

Which step makes reductive amination irreversible?

Hydride addition to the iminium. Forming the C–H bond cannot be reversed, so the whole reversible imine equilibrium is pulled forward to the amine product.

Reductive amination: product by nitrogen source

NH3 → primary amine. Primary amine → secondary amine. Secondary amine → tertiary amine. Each round adds one alkyl group (from the carbonyl carbon) to the nitrogen.

Why is aqueous NH4Cl the preferred NH3 source for reductive amination?

Ammonium chloride is slightly acidic and holds the pH near 5, the optimum for iminium formation.

Over-alkylation in reductive amination

The amine product is itself a nucleophile. With excess aldehyde/ketone and NaBH3CN it reacts again: 1° → 2° → 3° amine. Example: NH4Cl + 3 equiv PhCHO + NaBH3CN gives tribenzylamine.

Enamine

An alkene with a nitrogen directly on one alkene carbon: C=C–NR2. Formed from an aldehyde/ketone + a secondary amine (acid catalysis, pH ~5) via the iminium, which is deprotonated at the α-carbon.

Why does a secondary amine give an enamine instead of an imine?

The iminium from a 2° amine has no N–H to lose. The next most acidic proton is on the α-carbon (its C–H is next to the C=N+), so a weak base removes that proton and the C=C forms.

Two possible fates of an iminium made from a secondary amine

(1) It is attacked by a nucleophile (e.g. hydride from NaBH3CN → tertiary amine), or (2) it is deprotonated at the α-carbon by a weak base (e.g. H2O) → enamine.

Are imine and enamine formation reversible?

Yes. Both are acid-catalyzed condensations (lose H2O), analogous to acetal formation. Driven forward by removing water; reversed (hydrolysis) by adding water and mild acid. Neither proceeds under basic conditions.

Acetal vs imine formation: the key mechanistic difference

Same logic in both: add Nu, protonate OH, lose H2O, form a cation. Acetal: oxocarbenium ion, then a second ROH adds. Imine: iminium ion, then a proton is lost from N. Alcohol addition needs acid catalysis; amine addition does not (amines are better nucleophiles).

Imine hydrolysis (reverse reaction) conditions and steps

Mildly acidic water (pH ~5). Protonate N → iminium; H2O adds → PT → hemiaminal; protonate N and expel the amine; deprotonate O → aldehyde/ketone. You should be able to write the mechanism in both directions.

How to find the carbonyl and amine "parents" of an imine or iminium in a complex molecule

Locate the C=N (or C=N+). The carbon came from a C=O (aldehyde or ketone); the nitrogen came from an amine (NH3, RNH2, or R2NH). Mentally replace C=N with C=O and an N–H to disconnect it.

Why is imine formation called the "most subtle" mechanism of the course?

Because of its delicate dependence on acid concentration: the first stage is inhibited by acid, the second stage requires acid, so only a narrow pH window (~5) works.