Source: Organic Chemistry, The Ohio State University
Difficulty: Intermediate to Advanced | Prerequisites: Part 1 of these notes (structure, basicity), SN2 mechanisms, carbonyl chemistry basics (reduction with LiAlH4, NaBH3CN), acid-base workup.
This is Part 2 of 3 on Chapter 22. Part 1 covered structure, nomenclature, properties and basicity. This part covers every major synthetic route to amines: reductions of nitriles, amides and oximes, SN2 alkylation and its problems, azide and Gabriel syntheses, reductive amination, and two named rearrangements (Hoffmann and Curtius). It also covers how to make arylamines from nitro groups and how to protect an arylamine for Friedel-Crafts chemistry. Knowing which method gives you which class of amine (1°, 2°, or 3°) is the central organising idea.
There are many ways to make amines, and each method has a specific scope. LiAlH4 reduction of nitriles and oximes gives only primary amines. LiAlH4 reduction of amides can give 1°, 2° or 3° amines depending on substitution. SN2 alkylation of amines is straightforward but suffers from over-alkylation; azide synthesis and Gabriel synthesis solve that problem for primary amines. Reductive amination (ketone/aldehyde + amine + NaBH3CN) is the most versatile method, producing 1°, 2° or 3° amines. The Hoffmann and Curtius rearrangements convert amides or acyl chlorides into amines with loss of one carbon (as CO2). Arylamines are made by reducing nitrobenzene derivatives.
Tags: nitrile reduction, amide reduction, reductive amination, Gabriel synthesis, azide synthesis, Hoffmann rearrangement, Curtius rearrangement, LiAlH4, NaBH3CN, imine, oxime, isocyanate, phthalimide
Reductive amination
A two-step, one-pot process: a carbonyl compound reacts with an amine (or ammonia) to form an imine (or iminium ion), which is then reduced in situ by NaBH3CN. The most versatile method for making amines. In simple terms, you are replacing a C=O with a C-NH bond.
Gabriel synthesis
A method for making primary amines without over-alkylation. Potassium phthalimide acts as a nitrogen nucleophile in an SN2 reaction. The two flanking carbonyls prevent the nitrogen from reacting a second time. Hydrolysis with KOH/H2O then frees the primary amine.
Azide synthesis
Sodium azide (NaN3) displaces a halide by SN2, and the resulting alkyl azide is then reduced (LiAlH4) to a primary amine. The azide ion is a good nucleophile but a poor base, so it works well even with secondary halides.
Hoffmann rearrangement
A reaction that converts a primary amide (RCONH2) into a primary amine (RNH2) with loss of one carbon as CO2. Conditions: NaOH, Br2, H2O. Proceeds through an isocyanate intermediate.
Curtius rearrangement
Similar outcome to the Hoffmann but starting from an acyl azide (RC(O)N3). Heating drives off N2 to form an isocyanate, which is hydrolysed to the amine plus CO2.
Isocyanate
The intermediate R-N=C=O formed in both the Hoffmann and Curtius rearrangements. Hydrolysis of the isocyanate gives the amine.
Imine
A C=N-H (or C=N-R) double bond formed by condensation of an amine with a carbonyl compound. The key intermediate in reductive amination.
Oxime
A C=N-OH compound formed from a ketone or aldehyde plus hydroxylamine (NH2OH). Reduction with LiAlH4 gives a primary amine.
LDA (lithium diisopropylamide)
A very strong, bulky, non-nucleophilic base. Made from diisopropylamine and n-BuLi. Relevant here because it illustrates the deprotonation of amines by organolithium reagents.
Reagents: 1) LiAlH4, 2) H2O
Scope: Produces primary amines only. The triple bond is fully reduced to a single bond, and you gain a CH2 between the original R group and the NH2.
Key points:
LiAlH4 reduces both the C≡N triple bond and any other reducible groups present (carbonyls, epoxides).
If the molecule contains both a nitrile and a ketone, both will be reduced.
Example: A cyclohexane bearing a -CN group becomes cyclohexylmethylamine (the -CN becomes -CH2NH2).
Example: A molecule with both -OH and -CN: the hydroxyl is untouched, but the nitrile becomes -NH2.
Example: A cyclohexanone with a -CN group: both the ketone (→ -OH) and the nitrile (→ -CH2NH2) are reduced.
Limitation: You can only make primary amines this way. There is no option to introduce secondary or tertiary amine functionality directly from a nitrile.
Reagents: 1) LiAlH4, 2) H2O
Scope: Can produce 1°, 2° or 3° amines, depending on the substitution of the starting amide.
An unsubstituted amide (RCONH2) gives a primary amine (RCH2NH2).
A monosubstituted amide (RCONHR') gives a secondary amine (RCH2NHR').
A disubstituted amide (RCONR'2) gives a tertiary amine (RCH2NR'2).
The carbonyl C=O is fully reduced to a CH2 in each case. This is one of the cleanest routes to amines because you control the degree of substitution through the starting amide.
Amines are nucleophiles, so they can attack alkyl halides by SN2.
The reaction: R-NH2 + R'-CH2-X → R-NH-CH2R' + HX. A second equivalent of starting amine can then deprotonate the ammonium salt product.
The problem: over-alkylation. The product (a secondary amine) is also a nucleophile, and it can react with another molecule of alkyl halide to give a tertiary amine. That tertiary amine can react again to give a quaternary ammonium salt. It is very difficult to stop the reaction at mono-alkylation.
Additional limitation: SN2 requirements still apply, so the alkyl halide must be primary or methyl (not tertiary, which would undergo elimination instead).
Two solutions to the over-alkylation problem:
Azide synthesis (Section below)
Gabriel synthesis (Section below)
Step 1: SN2 displacement of a halide with sodium azide (NaN3) in ethanol. RCH2-Br + NaN3 → RCH2-N3.
Step 2: Reduction of the azide with LiAlH4, then aqueous workup. RCH2-N3 → RCH2-NH2.
Why it works: The azide ion (N3-) is an excellent nucleophile but a very poor base, so it works well in SN2 reactions. It even handles secondary halides reasonably well.
Result: A primary amine, with no risk of over-alkylation (the azide has no hydrogen to lose, and the azide intermediate is not nucleophilic at nitrogen).
Comparison with cyanide route: RCH2-Br + NaCN → RCH2-CN, then LiAlH4/H2O → RCH2-CH2-NH2. Note the cyanide route adds one extra carbon, while the azide route does not.
The idea: Use a nitrogen nucleophile that cannot react twice.
Step 1: Deprotonate phthalimide with KOH in ethanol. The resulting potassium phthalimide anion has a nitrogen flanked by two carbonyl groups, making it nucleophilic at nitrogen but unable to undergo a second alkylation (the nitrogen's remaining lone pair is tied up in resonance with the carbonyls).
Step 2: SN2 reaction with an alkyl halide (R-CH2-X). The nitrogen attacks the halide to form N-alkyl phthalimide.
Step 3: Hydrolysis with KOH/H2O cleaves both amide bonds, releasing the free primary amine (R-CH2-NH2) and phthalic acid (as its dipotassium salt).
Key points:
Produces only primary amines.
Very similar in logic to a Williamson ether synthesis (SN2 with an alkoxide), except the nucleophile is a nitrogen anion.
The carbonyl imide "masks" the nitrogen from over-alkylation.
Subject to the same SN2 limitations as any other SN2: works best with primary and methyl halides.
Step 1: Form the oxime by reacting a ketone or aldehyde with hydroxylamine (NH2OH) in water. The oxime (C=N-OH) is favoured at equilibrium.
Step 2: Reduce the oxime with LiAlH4, then aqueous workup.
Result: A primary amine.
This is a two-step route from a carbonyl to a primary amine. It is less commonly used than reductive amination (below) but appears in synthesis problems.
This is the most versatile and most frequently tested method for preparing amines.
The process (two steps, one pot):
Step 1 (imine formation): A carbonyl compound (aldehyde or ketone) reacts with ammonia or an amine to form an imine (C=N-R). No need to isolate the imine.
Step 2 (hydride reduction): NaBH3CN (sodium cyanoborohydride) in methanol reduces the imine (or iminium ion) to an amine.
Scope:
Ammonia + aldehyde/ketone → primary amine (1°)
Primary amine + aldehyde/ketone → secondary amine (2°). The intermediate is an imine (C=N-R), which is reduced.
Secondary amine + aldehyde/ketone → tertiary amine (3°). The intermediate is an iminium ion (C=N+R2), which is reduced.
Why NaBH3CN and not NaBH4? NaBH3CN is a milder, more selective reducing agent. It reduces imines and iminium ions but does not reduce the starting carbonyl, so the imine has time to form before reduction occurs. NaBH4 would reduce the ketone/aldehyde directly.
Example: Cyclohexanone + dimethylamine + NaBH3CN/CH3OH → N,N-dimethylcyclohexylamine (a 3° amine). The intermediate iminium ion is reduced in situ.
Example: 4-methylcyclohexanone + NH3 + NaBH3CN/CH3OH → 4-methylcyclohexylamine (a 1° amine).
This reaction is powerful because you can dial in the degree of substitution at nitrogen by choosing the right amine partner.
Starting material: Primary amide only (RCONH2).
Reagents: NaOH, Br2, H2O.
Product: Primary amine (RNH2) + CO2. The amine has one fewer carbon than the amide.
Mechanism overview:
Base (OH-) deprotonates the amide NH.
Br2 brominates the nitrogen.
A second deprotonation gives an N-bromo anion.
The R group migrates from carbon to nitrogen (a 1,2-shift), expelling bromide. This forms an isocyanate (R-N=C=O).
The isocyanate hydrolyses in water: it first forms a carbamic acid (unstable), which decomposes to the amine + CO2.
Key constraint: Works only on primary amides (RCONH2). If the amide nitrogen has any substituents, the rearrangement cannot proceed because you need the N-H for bromination and the subsequent deprotonation step.
Starting material: Acyl azide, RC(=O)-N3, made from an acyl chloride (RCOCl) and sodium azide (NaN3).
Conditions: Heat.
Mechanism overview:
The acyl chloride reacts with NaN3 to form the acyl azide.
Heating causes loss of N2 gas and migration of R from carbon to nitrogen (a 1,2-shift, same type as in the Hoffmann), producing an isocyanate (R-N=C=O).
Hydrolysis of the isocyanate with water gives the primary amine (RNH2) + CO2 + N2.
Comparison with Hoffmann: Both rearrangements give the same type of product (a primary amine one carbon shorter than the carbonyl precursor) and both go through an isocyanate intermediate. The Hoffmann starts from a primary amide with Br2/NaOH; the Curtius starts from an acyl chloride via the azide.
The standard route to arylamines is reduction of a nitrobenzene (Ar-NO2 → Ar-NH2).
Methods (all give the same product):
H2 with Pt catalyst in ethanol (catalytic hydrogenation)
Fe, HCl (step 1), then NaOH, H2O (step 2, to neutralise the protonated amine product)
SnCl2, H3O+ (step 1), then NaOH, H2O (step 2). This is considered a "gentle" method.
LiAlH4 (step 1), then H2O (step 2)
The second step (base workup) is necessary in the acidic methods because the amine product gets protonated under acidic conditions. You need to free the neutral amine.
The problem: Arylamines are Lewis bases. In Friedel-Crafts alkylation (using AlCl3 as Lewis acid catalyst), the nitrogen lone pair coordinates with AlCl3, deactivating both the catalyst and the ring. The reaction fails.
The solution: acetylation as a protecting group.
Acetylate the amine: Ar-NH2 + acetic anhydride → Ar-NHC(O)CH3 (an acetamide). The electron-withdrawing acetyl group pulls electron density away from nitrogen, making the lone pair much less available for coordination with AlCl3.
Run the Friedel-Crafts reaction on the protected arylamine. It now works because the nitrogen does not tie up the Lewis acid.
Remove the protecting group: hydrolysis with NaOH/H2O cleaves the amide bond, restoring the free Ar-NH2.
This protect-react-deprotect sequence is a standard strategy in organic synthesis whenever a functional group is incompatible with the reaction conditions.
Students often forget that direct SN2 alkylation of amines leads to over-alkylation. On an exam, if someone proposes "just react the amine with an alkyl halide" to get a specific secondary amine, the answer is usually wrong unless they account for the mixture of products.
Confusing the Hoffmann rearrangement with the Hoffmann elimination. They are different reactions with different substrates and outcomes. The rearrangement converts a primary amide to an amine; the elimination converts a quaternary ammonium salt to an alkene (covered in Part 3).
Thinking LiAlH4 is selective. LiAlH4 reduces almost every reducible functional group (nitriles, amides, esters, ketones, aldehydes, epoxides). If a molecule has two reducible groups, both will be reduced.
Forgetting that the Hoffmann rearrangement works only on primary amides. If the amide nitrogen has substituents, the reaction does not proceed.
⚠️ "Choose a method to prepare [specific amine]" is one of the most common exam question formats for this chapter. You must know which methods give 1° only, which give 1°/2°/3°.
⚠️ Reductive amination is the workhorse reaction. Be ready to propose the correct carbonyl partner and amine partner for any target amine.
⚠️ The Gabriel synthesis mechanism (phthalimide anion → SN2 → hydrolysis) is a classic multi-step mechanism question.
⚠️ Comparing the Hoffmann and Curtius rearrangements: know that both go through an isocyanate, both lose CO2, and both shorten the carbon chain by one.
⚠️ Arylamine protection (acetylation before Friedel-Crafts, then deprotection) is tested both as a standalone question and within multi-step synthesis problems.
True or false: Reduction of a nitrile with LiAlH4 can produce a secondary amine. Answer: False. Nitrile reduction gives only primary amines.
Fill in the blank: The reagent used for the mild reduction step in reductive amination is ___. Answer: NaBH3CN (sodium cyanoborohydride).
True or false: The Gabriel synthesis can produce tertiary amines. Answer: False. It produces only primary amines.
True or false: Both the Hoffmann and Curtius rearrangements proceed through an isocyanate intermediate. Answer: True.
Fill in the blank: To protect an arylamine from interfering with a Friedel-Crafts reaction, you convert it to an ___ using acetic anhydride. Answer: Acetamide (an amide).
Q: Propose a synthesis of benzylamine (PhCH2NH2) from benzyl bromide (PhCH2Br) that avoids over-alkylation.
A: Two good options. (1) Azide synthesis: treat PhCH2Br with NaN3 in EtOH, then reduce the azide with LiAlH4/H2O. (2) Gabriel synthesis: react potassium phthalimide with PhCH2Br (SN2), then hydrolyse with KOH/H2O.
Q: What amine product(s) would you expect from the LiAlH4 reduction of N,N-dimethylbenzamide (PhC(O)N(CH3)2)?
A: A tertiary amine: PhCH2N(CH3)2. The carbonyl is reduced to CH2, and both methyl groups on nitrogen are retained.
Q: Design a reductive amination to produce N-methylcyclohexylamine from cyclohexanone.
A: React cyclohexanone with methylamine (CH3NH2) in the presence of NaBH3CN in methanol. An imine forms, which is reduced in situ to give the secondary amine.
Q: Why does the Hoffmann rearrangement not work on N-methylbenzamide?
A: The Hoffmann rearrangement requires a primary amide (RCONH2). N-methylbenzamide is a secondary amide (PhC(O)NHCH3). The mechanism needs an N-H that can be brominated and deprotonated twice; with only one N-H, the second deprotonation cannot occur.
Q: List four different reducing conditions that convert nitrobenzene to aniline.
A: (1) H2/Pt in EtOH. (2) Fe/HCl, then NaOH/H2O. (3) SnCl2/H3O+, then NaOH/H2O. (4) LiAlH4, then H2O.
Reduction reactions here (LiAlH4, NaBH3CN, catalytic hydrogenation) all connect back to the carbonyl chemistry chapters. LiAlH4 is the same reagent used to reduce esters, carboxylic acids, and aldehydes/ketones.
SN2 alkylation of amines is a direct extension of the SN2 chapter. The over-alkylation problem is a practical lesson in why nucleophile selectivity matters.
The Hoffmann and Curtius rearrangements involve 1,2-shifts, connecting to the carbocation rearrangement chemistry from earlier chapters. The difference is that here the migration terminus is nitrogen rather than a carbocation centre.
preparation of amines, synthesis of amines, nitrile reduction, amide reduction, LiAlH4 reduction, reductive amination, NaBH3CN, sodium cyanoborohydride, imine, iminium ion, oxime reduction, hydroxylamine, Gabriel synthesis, phthalimide, azide synthesis, NaN3, sodium azide, SN2 alkylation, over-alkylation, Hoffmann rearrangement, Curtius rearrangement, isocyanate, carbamic acid, acyl azide, nitro reduction, aniline synthesis, Fe HCl, SnCl2, arylamine protection, acetylation, acetic anhydride, Friedel-Crafts protection, organic chemistry chapter 22