Warm bromoethane with ammonia and you do get ethylamine, CH3CH2NH2. You also get diethylamine, triethylamine and tetraethylammonium bromide, all in the same flask. The obvious way to make an amine does not stop where you want it to, and every method on this page is a way around that one failure.
The problem: alkylation cannot be stopped
The first step is the SN2 alkylation from Amine reactions: ammonia’s lone pair attacks the carbon of bromoethane, and bromide leaves. The ethylamine this makes is a primary amine, one whose nitrogen carries one carbon.
The trouble is that ethylamine is a better nucleophile than ammonia. Its ethyl group pushes electron density onto the nitrogen, so it attacks the next molecule of bromoethane faster than the remaining ammonia does. The secondary amine it makes, diethylamine, is faster again.
So the flask ends up holding primary, secondary and tertiary amine plus the quaternary ammonium salt. The relative rates set the proportions, not the amount of halide you weighed out.
The speed-up does not last all the way. A tertiary amine is a poorer nucleophile than the secondary amine that made it, because its three alkyl groups crowd the SN2 transition state. So the speed-up stops at the secondary amine: the tertiary amine reacts more slowly than the secondary amine did.
A large excess of ammonia does help. Most of the halide then meets an ammonia molecule rather than an amine, and the mixture leans toward the primary amine. It still never gives a single product.
The runaway is useful in one case. If a quaternary ammonium salt is the target, excess CH3I left to react to completion gives a single product, because the nitrogen has nowhere further to go.
Every other route on this page escapes in one of three ways. Gabriel alkylates a nitrogen that can react only once. Azide and cyanide use a nucleophile that is not an amine. Reductive amination makes the C–N bond by reducing a C=N, and amide reduction makes it by an acylation that stops after one step.
Gabriel synthesis: alkylate something that can only react once
Phthalimide is a benzene ring fused to a five-membered ring that holds C=O, N–H and C=O. A nitrogen bonded to two carbonyl groups like this is called an imide. Both carbonyls pull electron density away from the N–H. When the proton leaves, the negative charge spreads onto both oxygens. So this N–H is unusually acidic, with a pKa of about 8.3, and KOH removes it easily.
The anion is a good nucleophile, and it does one SN2 on the alkyl halide. The product, an N-alkylphthalimide, has no N–H left to remove. Its lone pair is pulled into both carbonyls, the way an amide’s lone pair is pulled into one (Esters & amides). So it cannot attack a second molecule of halide.
Heating with hydrazine, H2N–NH2, then frees a clean primary amine. The rest of the molecule leaves as phthalhydrazide. Hot aqueous acid or base also releases the amine.
The catch is that the alkylation is still an SN2. The halide should be methyl or primary and unhindered. Secondary halides give poorer yields, with elimination alongside, and tertiary halides give only alkene.
Because the nitrogen can pick up only one carbon, Gabriel always gives a primary amine. It cannot make a secondary amine, whatever halide you use.
Azide displacement: use a nucleophile that is not an amine
Treat 1-bromobutane with sodium azide, NaN3, and the azide ion does the SN2. The azide ion is three nitrogens in a straight line with one negative charge overall. It is small, and it is a strong nucleophile but a very weak base (Substrate & solvent effects).
The product, 1-azidobutane, is an alkyl azide. It is not an amine, and it does not attack a second molecule of alkyl halide, so there is nothing to over-alkylate. LiAlH4, or H2 over Pd, then reduces it to butan-1-amine, and the other two nitrogens leave as N2 gas. Like Gabriel, the route gives a primary amine and needs an SN2 substrate, so a tertiary halide fails. But azide is small and a very weak base, so it also works well on secondary halides, where Gabriel struggles.
Cyanide, then reduce: the route that adds a carbon
1-Bromopropane and NaCN give butanenitrile, CH3CH2CH2C≡N. LiAlH4 then reduces it to butan-1-amine, CH3CH2CH2CH2NH2. Nitriles covers both steps.
What matters here is the count. The nitrile carbon becomes the CH2 next to the nitrogen, so the amine has one more carbon than the halide had. Gabriel and azide hand back the carbon skeleton you put in; this route lengthens it by one.
Reductive amination: build the C–N bond by reduction instead
Stir cyclohexanone with methylamine and NaBH3CN in mildly acidic solution, and the product is N-methylcyclohexylamine.
Two reactions run in the same flask. First the amine adds to the C=O and gives a carbinolamine, a carbon that carries both an OH and the nitrogen. The carbinolamine loses water to give an imine, a compound with a C=N bond (Imines and enamines). In the mildly acidic solution the imine picks up a proton and becomes an iminium ion, C=N+. Then NaBH3CN delivers a hydride to the carbon of the C=N+, which turns the C=N into a C–N single bond.
A secondary amine works the same way. It condenses with the carbonyl compound to an iminium ion that has no N–H, and reduction gives a tertiary amine.
This route escapes the runaway for a structural reason. The C–N bond forms when hydride adds to a C=N, not by displacement, so there is no alkyl halide for the product amine to attack. What you put in, and how much of it, decides what you get.
The choice of reducing agent is what makes this work. NaBH3CN and NaBH(OAc)3 are attenuated hydride sources, meaning weaker hydride donors than NaBH4. They survive the mildly acidic solution that imine formation needs. There they reduce the iminium ion much faster than the ketone or aldehyde, so the ketone lasts long enough to condense. NaBH4 would simply reduce the ketone to the alcohol. Hydrogen gas over a metal catalyst, such as Pd or Ni, can also reduce the C=N.
Two limits remain, and they are not the SN2 ones. The first comes from where the nitrogen lands: always on the old carbonyl carbon, which gains a hydrogen from the hydride. So this route cannot put a nitrogen on a carbon that carries no hydrogen. tert-Butylamine is not a reductive amination product, and neither is an aryl amine.
The second is that over-alkylation is not gone entirely. The first condensation is fast. A second one, by the more crowded secondary amine, is slower and can form only a charged iminium ion. Once a C=N is reduced to a C–N bond, that step does not reverse. Even so, dialkylation is a known side reaction. It is worst with ammonia as the nitrogen source, or with an unhindered aldehyde such as formaldehyde.
Stereochemistry separates the two kinds of route. An SN2 route, Gabriel or azide, attacks the carbon from the side opposite the leaving group, so a stereocenter at that carbon is inverted. Reductive amination adds hydride to a flat C=N, from the front or the back. When nothing else in the molecule is chiral, both faces react equally, and a new stereocenter comes out as a racemic mixture. Where the stereochemistry matters, that difference alone decides the route.
Amide reduction: acylate once, then take the oxygen away
Cyclohexylamine and acetyl chloride, CH3COCl, give an amide, N-cyclohexylacetamide. LiAlH4 then reduces its C=O to CH2, which gives N-ethylcyclohexylamine. The oxygen leaves, and no carbon is lost (Esters & amides).
The acylation stops after one step. The amide nitrogen has its lone pair tied up in the carbonyl, so it is not nucleophilic, and one acylation gives one product. That makes this the cleanest general route to a secondary or tertiary amine: RCOCl + R′NH2 → RCONHR′ → RCH2NHR′.
Nitro reduction: the route to an aromatic amine
Nitrobenzene, C6H5NO2, with H2 over Pd or Pt gives aniline, C6H5NH2. Fe/HCl, Sn/HCl and SnCl2 do the same job. The acidic methods give the anilinium salt first, aniline with an extra proton on its nitrogen (C6H5NH3+), and a base such as NaOH then frees the amine.
None of the earlier routes can make an aryl amine, because each one leaves the nitrogen on an sp3 carbon: SN2 cannot happen at a ring carbon, and a ring carbon cannot be a C=O. Nitro reduction does not make the C–N bond at all. Nitration already put the nitrogen on the ring (Electrophilic aromatic substitution), and the reduction only changes its oxidation level. The carbon count does not change.
The nitrogen stays exactly where the nitro group was. Nitrobenzene brominates at the meta position, because the nitro group is a meta director. Reducing that product gives 3-bromoaniline, with the NH2 on the carbon that held the NO2. So the amine ends up wherever nitration put the NO2, and while it is still a nitro group it steers later substituents meta (Ortho/meta/para directing effects).
Losing a carbon on purpose
Butanamide (four carbons) with Br2 and NaOH gives propan-1-amine (three carbons) and CO2. This is the Hofmann rearrangement. It turns a primary amide, one with a CONH2 group, into a primary amine with one fewer carbon.
The key step is a migration: a group moves, together with its bonding pair of electrons, from one atom to the next. Here the propyl group moves from the carbonyl carbon to the nitrogen.
Three quick steps set it up. Hydroxide removes one N–H proton. The nitrogen anion attacks Br2, which gives an N-bromoamide, R–CO–NHBr. Hydroxide then removes the second N–H proton, which leaves an N-bromoamide anion.
That anion does the key step. The propyl group migrates to the nitrogen at the same moment bromide leaves. Because this happens in one step, no nitrene (a neutral nitrogen with only six electrons around it) is ever free. The product is an isocyanate, R–N=C=O, whose N=C=O unit is a straight line.
Water adds to the isocyanate’s carbon and gives a carbamic acid, R–NH–COOH: a nitrogen that keeps its N–H and carries a COOH group. A carbamic acid loses CO2 on its own, and the amine is left. The carbon that leaves as CO2 is the original carbonyl carbon.
The migrating group keeps its configuration. If the carbon that moves is a stereocenter, it arrives on the nitrogen with retention.
The Curtius rearrangement reaches the same isocyanate another way. An acid chloride and NaN3 give an acyl azide, R–CO–N3. On heating, the R group migrates to the nitrogen as N2 leaves, in one step, and the isocyanate forms. The same carbonyl carbon is lost.
Choosing a route
| Route | Gives | Carbon count |
|---|---|---|
| Direct alkylation | a mixture of 1°, 2°, 3°, 4° | unchanged |
| Gabriel | 1° only | unchanged |
| Azide, then reduce | 1° only | unchanged |
| Cyanide, then LiAlH4 | 1° only | +1 |
| Reductive amination | 1°, 2° or 3°, your choice | carbonyl’s carbons + amine’s carbons |
| Amide, then LiAlH4 | 1°, 2° or 3° | unchanged |
| Hofmann or Curtius | 1° only | −1 |
| Nitro, then reduce | 1° aryl amines | unchanged |
Step 1: count carbons. The target has four and the starting material has three, so the count has to go up by one. That rules out Gabriel, azide, amide reduction and reductive amination at once. It rules out Hofmann and Curtius twice over, since they go the other way.
Step 2: see what is left. Only the cyanide route adds exactly one carbon.
Step 3: check the substrate. Cyanide displacement is an SN2, so the halide should be methyl or primary. 1-Bromopropane is primary, so the route works.
Step 4: write it. CH3CH2CH2Br + NaCN → CH3CH2CH2C≡N, butanenitrile. Then LiAlH4 and an aqueous workup → CH3CH2CH2CH2NH2.
Step 5: check the answer. Four carbons, a primary amine, and the nitrogen on the end of the chain. The nitrile carbon became the CH2 that carries the nitrogen, which is always where the extra carbon ends up.
The target is a cyclohexane ring with an NH–CH3 group on one carbon, a secondary amine. The carbon count sets no limit here, so ask the other question: how substituted is the nitrogen? It is secondary, so every route that runs an SN2 with a stand-in for the amine nitrogen is out before you start.
Attempt A: Gabriel on bromocyclohexane, then methylate. This fails twice over. Bromocyclohexane is a secondary halide, so the phthalimide alkylation gives a poor yield, with cyclohexene from E2 alongside. And even the cyclohexylamine it does give must then be methylated, which is Attempt B’s runaway.
Attempt B: cyclohexylamine + CH3I. The connectivity is right, but nothing controls it. The secondary amine product is at least as nucleophilic as the starting amine, so it takes a second methyl and then a third. The result is a mixture that ends at the quaternary salt.
Attempt C: reductive amination. There are two ways round, and one is better. Cyclohexanone + CH3NH2 + NaBH3CN puts the nitrogen on the old carbonyl carbon and gives the target cleanly. Cyclohexylamine + formaldehyde + NaBH3CN delivers the methyl instead, but it does not stop there. Formaldehyde is so reactive that the secondary amine condenses with a second molecule of it, and the product goes on to N,N-dimethylcyclohexylamine. Use the ketone.
Attempt D: acylate, then reduce. This route is also clean. It is the one to use when a reductive amination is not an option. But watch the count. Cyclohexylamine + CH3COCl gives the amide, and LiAlH4 reduces its C=O to CH2. That gives N-ethylcyclohexylamine, the ethyl amine, because the acetyl group carried two carbons. To reach N-methyl, acylate with a formate ester such as ethyl formate, HCOOCH2CH3, whose single carbon becomes the methyl.