Source: Klein, Organic Chemistry, Chapter 20 (Sections 20.6–20.7)
Tags: imine formation, Schiff base, enamine formation, carbinolamine, oxime, hydrazone, Wolff-Kishner reduction, mechanism strategies, acid-catalysed hydrolysis, primary amine, secondary amine, nitrogen nucleophile, organic chemistry
Difficulty: Intermediate Prerequisites: Parts 1–2 of these notes (Sections 20.1–20.5). Acetal formation mechanism (seven steps). Arrow-pushing fundamentals (Ch. 6). pKa and pH concepts.
Nitrogen nucleophiles react with aldehydes and ketones through mechanisms that are nearly identical to acetal formation. The first five steps are the same across acetals, imines, and enamines; only the ending differs. Grasping this shared skeleton is the single most efficient way to master three seemingly different reactions at once. The Wolff-Kishner reduction, which converts a ketone all the way to an alkane via a hydrazone intermediate, rounds out the nitrogen nucleophile toolkit.
Primary amines react with carbonyls to form imines (C=N); secondary amines form enamines (C=C next to nitrogen). Both proceed through a carbinolamine intermediate via a six-step acid-catalysed mechanism. The Wolff-Kishner reduction uses hydrazine then strong base/heat to reduce a ketone to an alkane. All three mechanisms (acetal, imine, enamine) share the same first five steps.
Imine (Schiff base)
A compound containing a C=N double bond, formed when a primary amine (RNH₂) reacts with an aldehyde or ketone under mildly acidic conditions, with loss of water.
Think of it as the nitrogen analogue of a carbonyl: the oxygen has been swapped for NR.
Carbinolamine
The tetrahedral intermediate formed after the amine attacks the carbonyl carbon, containing both an OH and an NHR (or NR₂) group on the same carbon. Analogous to the hemiacetal in acetal formation.
In simple terms, it is the "halfway" structure on the path to an imine or enamine.
Enamine
A compound in which a nitrogen lone pair is conjugated with an adjacent C=C double bond. Formed when a secondary amine (R₂NH) reacts with an aldehyde or ketone under acidic conditions, with loss of water.
Think of it as what you get when the nitrogen of the iminium ion has no proton to lose, so a proton is removed from the adjacent carbon instead, generating a C=C bond.
Oxime
The product formed when hydroxylamine (NH₂OH) reacts with a carbonyl. Structurally, C=N-OH.
In simple terms, it is an imine where the R group on nitrogen is an OH.
Hydrazone
The product formed when hydrazine (NH₂NH₂) reacts with a carbonyl. Structurally, C=N-NH₂.
In simple terms, it is an imine where the R group on nitrogen is an NH₂. Hydrazones are the starting point for the Wolff-Kishner reduction.
Wolff-Kishner reduction
A two-step procedure that converts a ketone to an alkane: first form the hydrazone ([H⁺], NH₂NH₂, lose H₂O), then treat with KOH/H₂O and heat. Nitrogen gas (N₂) is expelled, driving the reaction to completion.
Think of it as a way to completely remove a carbonyl oxygen and replace it with two hydrogens.
Overall: aldehyde or ketone + RNH₂, [H⁺], gives imine + H₂O.
The mechanism has six steps, divided into two parts:
Part 1, formation of carbinolamine (3 steps):
Proton transfer: carbonyl is protonated
Nucleophilic attack: the primary amine attacks the protonated carbonyl
Proton transfer: the tetrahedral intermediate is deprotonated to give the carbinolamine
These three steps are identical to the first three steps of acetal formation.
Part 2, conversion of carbinolamine to imine (3 steps):
Proton transfer: the OH group of the carbinolamine is protonated (converts it to a good leaving group)
Loss of leaving group: water departs, forming a C=N bond (iminium ion)
Proton transfer: a proton is removed from nitrogen to give the neutral imine
Note: the first two steps of Part 2 are also identical to acetal formation. The difference is only in the final step: acetal formation has a nucleophilic attack then a proton transfer, whereas imine formation simply has a proton transfer (removing H from N).
The rate of imine formation is greatest around pH 4.5
If pH is too high (too little acid): the carbonyl is not protonated (step 1 is slow), and the carbinolamine OH is not protonated (step 4 is slow)
If pH is too low (too much acid): the amine is fully protonated (RNH₃⁺), which is not a nucleophile, so step 2 is slow
Optimal conditions require mildly acidic pH to balance these competing effects
Several compounds of the form RNH₂ react with carbonyls through the same six-step mechanism:
Hydroxylamine (NH₂OH) → oxime (C=N-OH)
Hydrazine (NH₂NH₂) → hydrazone (C=N-NH₂)
Primary alkyl amine (RNH₂) → imine (C=N-R)
The mechanism is directly analogous in each case; only the identity of the R group on nitrogen changes.
Overall: aldehyde or ketone + R₂NH, [H⁺], gives enamine + H₂O.
The mechanism is identical to imine formation for the first five steps. The difference is only in the last step:
In imine formation, the iminium ion has a proton on nitrogen that can be removed → gives C=N (imine)
In enamine formation, the iminium ion has no proton on nitrogen (both positions are occupied by R groups) → instead, a proton is removed from the adjacent carbon → gives C=C with nitrogen attached (enamine)
Part 1, formation of carbinolamine (3 steps): same as imine formation.
Part 2, conversion of carbinolamine to enamine (3 steps):
Proton transfer: OH is protonated
Loss of leaving group: water departs (iminium ion forms)
Proton transfer: a proton is removed from the carbon adjacent to nitrogen (not from nitrogen itself)
A two-step procedure for reducing a ketone to an alkane:
Step 1: form the hydrazone by treating the ketone with hydrazine and acid catalyst (standard imine-type mechanism)
Step 2: treat the hydrazone with KOH in water at high temperature
The mechanism of Step 2 (Mechanism 20.8) involves five steps:
Proton transfer: one proton is removed from the NH₂ of the hydrazone, forming a resonance-stabilised anion
Proton transfer: the intermediate is reprotonated on carbon
Proton transfer: another proton is removed from nitrogen
Loss of leaving group: N₂ gas is expelled, generating a carbanion
Proton transfer: the carbanion is protonated by water to give the alkane product
The loss of N₂ gas is thermodynamically unfavourable in isolation, but the gas bubbles out of the reaction mixture, continuously shifting the equilibrium forward (Le Chatelier's principle). This makes yields generally very good.
Comparing the mechanisms for acetal, imine, and enamine formation:
All three share the same first five steps (through loss of the water leaving group)
Acetal formation then adds a nucleophilic attack + proton transfer (7 steps total)
Imine formation ends with a proton transfer from nitrogen (6 steps total)
Enamine formation ends with a proton transfer from an adjacent carbon (6 steps total)
The master rule for proton transfers in acid-catalysed conditions: all reagents, intermediates, and leaving groups should either be neutral (no charge) or bear one positive charge. Every proton transfer in the mechanism exists to satisfy this rule. Specifically:
The carbonyl is protonated before attack (generates a better electrophile, avoids a negative charge on the product)
Do not form two positive charges on a single intermediate
The leaving group should be neutral when it departs (protonate OH to water before it leaves; do not expel hydroxide)
A final proton transfer yields a neutral product
Acetals, imines, and enamines can all be hydrolysed back to ketones by treatment with excess water and acid (H₃O⁺).
Strategy for drawing the hydrolysis mechanism:
Write all the intermediates from the forward (formation) mechanism, but in reverse order
The first intermediate of the forward mechanism becomes the last intermediate before the product in the hydrolysis
Then draw curved arrows working forward through the reversed intermediates
Obey the master rule at every step
Imine formation: R₂C=O + RNH₂ → [H⁺] → R₂C=NR + H₂O
Enamine formation: R₂C=O + R₂NH → [H⁺] → enamine + H₂O
Oxime: R₂C=O + NH₂OH → [H⁺] → R₂C=NOH + H₂O
Hydrazone: R₂C=O + NH₂NH₂ → [H⁺] → R₂C=NNH₂ + H₂O
Wolff-Kishner: R₂C=O → hydrazone → KOH/H₂O/heat → R₂CH₂ + N₂
Shared mechanistic skeleton (first five steps):
Proton transfer → Nucleophilic attack → Proton transfer → Proton transfer → Loss of leaving group
Then: acetal adds (nucleophilic attack → proton transfer); imine adds (proton transfer from N); enamine adds (proton transfer from adjacent C).
Imine formation is central to vision: beta-carotene is metabolised to 11-cis-retinal (an aldehyde), which reacts with an amino group of the protein opsin to form rhodopsin (an imine). Absorption of light triggers cis-to-trans isomerisation, producing the nerve signal we perceive as sight. Vitamin A deficiency impairs this pathway, causing night blindness.
Imine and acetal hydrolysis is exploited in prodrug design. Progabide, a prodrug for gamma-aminobutyric acid (GABA), uses an imine moiety to cross the blood-brain barrier in an uncharged form; once inside the brain, hydrolysis releases the active neurotransmitter. Methenamine, a nitrogen analogue of an acetal, is hydrolysed to formaldehyde in the acidic environment of the urinary tract, where it acts as an antiseptic.
Students often think imine and enamine formation are entirely different reactions. They share the same first five mechanistic steps; only the final proton transfer differs.
A common error is attempting imine formation under strongly acidic conditions (pH well below 4). At very low pH the amine is protonated and no longer nucleophilic, so the reaction slows dramatically. Optimal pH is around 4.5.
Students sometimes draw the last step of enamine formation as removal of a proton from nitrogen. In enamine formation, nitrogen has no proton to lose (it bears two R groups), so the proton must come from the adjacent carbon.
When drawing hydrolysis mechanisms, students frequently introduce hydroxide as a reagent. Under acidic conditions (H₃O⁺), use water for deprotonation, not hydroxide.
⚠️ The six-step imine mechanism and the six-step enamine mechanism are extremely high-frequency exam questions. Know them cold, and know how they differ from the seven-step acetal mechanism.
⚠️ Be able to predict the product when different nitrogen nucleophiles (RNH₂, NH₂OH, NH₂NH₂, R₂NH) react with a given aldehyde or ketone.
⚠️ The Wolff-Kishner reduction (hydrazone + KOH/H₂O/heat → alkane + N₂) is a classic exam transformation. Know it alongside the Clemmensen reduction (Zn-Hg/HCl) as two methods for converting a carbonyl to a methylene.
⚠️ Hydrolysis mechanisms (reverse of formation) are frequently tested. Practice drawing them by reversing the intermediates and then adding curved arrows.
⚠️ The "master rule" for acid-catalysed mechanisms (all species neutral or +1) is a reliable check on your arrow-pushing: if you have produced a negative charge or a double positive charge, you have made an error.
True or False: A secondary amine reacts with a ketone to form an imine. ___
Fill in the blank: The intermediate common to both imine and enamine formation is called a ________.
True or False: The Wolff-Kishner reduction uses acidic conditions in both steps. ___
Fill in the blank: The optimal pH for imine formation is approximately ________.
True or False: In enamine formation, the final proton transfer removes a proton from nitrogen. ___
Answers: 1. False (a secondary amine forms an enamine; a primary amine forms an imine). 2. Carbinolamine. 3. False (step 1 is acidic; step 2 is strongly basic, KOH/H₂O/heat). 4. 4.5. 5. False (the proton is removed from the carbon adjacent to nitrogen).
Q: What product forms when cyclohexanone is treated with ethylamine (EtNH₂) and a catalytic amount of acid?
A: An imine: N-ethylcyclohexanimine (cyclohexanone's oxygen is replaced by N-Et, giving a C=N-Et group).
Q: How do the mechanisms of imine formation and enamine formation differ?
A: They are identical for the first five steps (through loss of water). In the sixth step, imine formation removes a proton from nitrogen (which bears an H in the iminium ion), while enamine formation removes a proton from the carbon adjacent to nitrogen (because nitrogen bears two R groups and has no H to lose).
Q: Propose a two-step procedure to convert a ketone into an alkane using nitrogen chemistry.
A: (1) Treat the ketone with hydrazine (NH₂NH₂) and catalytic acid to form the hydrazone. (2) Treat the hydrazone with KOH/H₂O at high temperature (Wolff-Kishner reduction) to give the alkane and N₂ gas.
Q: Why does the Wolff-Kishner reduction proceed to completion, even though carbanion formation is thermodynamically unfavourable?
A: The N₂ gas produced in the carbanion-forming step bubbles out of the reaction mixture. Its irreversible departure shifts the equilibrium forward (Le Chatelier's principle), driving the reaction to completion.
Q: Draw the hydrolysis product when an enamine is treated with H₃O⁺.
A: The original ketone (or aldehyde) is regenerated, along with the secondary amine (R₂NH). The mechanism is the reverse of enamine formation.
The Wolff-Kishner reduction is one of three methods for converting a C=O to CH₂ (alongside the Clemmensen reduction from Section 19.6 and the thioacetal/Raney Ni method from Section 20.8). Imine chemistry connects to biological processes including vision (rhodopsin), amino acid metabolism (transamination uses imine intermediates), and prodrug design. Enamine chemistry becomes synthetically powerful in Chapter 23, where enamines are used as nucleophiles in alkylation and acylation reactions.
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