Chapter 9 · Amines

9.1 Structure, Classification, Nomenclature and Preparation

Amines are ammonia with carbon groups in place of hydrogen. This section looks at the pyramidal nitrogen, sorts amines into primary, secondary and tertiary, names them, makes them by six routes (and asks which give a pure primary amine), and explains their boiling points and solubility.

Every year the world makes several million tonnes of aniline. Most of it becomes polyurethane foam for mattresses, car seats and insulation; the rest goes into rubber chemicals, herbicides, dyes and medicines such as paracetamol. Other amines are just as useful: hexane-1,6-diamine is one half of nylon-6,6, ethane-1,2-diamine grips metal ions as a ligand, and the drug salbutamol in an asthma inhaler is a secondary amine. Living things are full of them too. The amino group of every amino acid, the neurotransmitters adrenaline, dopamine and serotonin, and the histamine released in an allergic reaction are all amines.

An amine is what you get when one or more hydrogen atoms of ammonia are replaced by carbon groups. The nitrogen keeps its lone pair, so amines behave like a souped-up (or sometimes toned-down) ammonia: they are bases and nucleophiles. In this section you will see the shape of the amino group, learn to classify and name amines, meet six ways of making them (and learn which of them give a pure primary amine, which is what a chemist usually wants), and explain their boiling points and solubility in water. Their reactions follow in Section 9.2.

Structure and Classification

📘 Definition 9.1 — Amines and the shape of the amino group

Amines are derivatives of ammonia, $\mathrm{NH_{3}}$, in which one, two or all three hydrogen atoms have been replaced by alkyl or aryl groups. The functional group $\mathrm{{-}NH_{2}}$ is the amino group.

As in ammonia, the nitrogen atom of an amine is $sp^{3}$ hybridised. Three of its $sp^{3}$ orbitals overlap with orbitals of hydrogen or carbon; the fourth holds the lone pair. The four electron pairs point roughly to the corners of a tetrahedron, so the molecule is pyramidal, with nitrogen at the apex. Because the lone pair repels the bonding pairs more strongly than they repel one another, the bond angles are a little smaller than $109.5^{\circ}$: the C–N–C angle in trimethylamine is about $108^{\circ}$.

📘 Definition 9.2 — Classification of amines

(a) By the number of hydrogens replaced (the number of carbon atoms bonded to nitrogen):

  • Primary (1°) amine: one carbon group on N, $\mathrm{R{-}NH_{2}}$, e.g. $\mathrm{CH_{3}CH_{2}NH_{2}}$.
  • Secondary (2°) amine: two carbon groups on N, $\mathrm{R_{2}NH}$ or $\mathrm{RR'NH}$, e.g. $\mathrm{(CH_{3})_{2}NH}$.
  • Tertiary (3°) amine: three carbon groups on N, $\mathrm{R_{3}N}$, e.g. $\mathrm{(CH_{3})_{3}N}$.

A 2° or 3° amine is simple if all the groups on N are the same and mixed if they differ, e.g. $\mathrm{CH_{3}NHC_{2}H_{5}}$.

(b) By the type of carbon bonded to N: in an aliphatic amine nitrogen is bonded only to $sp^{3}$ (alkyl) carbons; in an aromatic amine or arylamine nitrogen is bonded directly to a carbon of an aromatic ring, e.g. aniline, $\mathrm{C_{6}H_{5}NH_{2}}$. Phenylmethanamine (benzylamine), $\mathrm{C_{6}H_{5}CH_{2}NH_{2}}$, is aliphatic: its nitrogen is on the $\mathrm{CH_{2}}$, not on the ring.

(c) If nitrogen forms four bonds to carbon it carries a positive charge and has no lone pair: $\mathrm{R_{4}N^{+}X^{-}}$ is a quaternary ammonium salt, e.g. tetramethylammonium chloride, $\mathrm{(CH_{3})_{4}N^{+}Cl^{-}}$. It is made from amines but is not itself an amine.

⚠️ "Primary" means something different for amines and alcohols
For an alcohol, 1°, 2° and 3° describe the carbon that carries the $\mathrm{OH}$. For an amine, they describe the nitrogen: how many carbon groups it carries. So 2-methylpropan-2-amine (tert-butylamine), $\mathrm{(CH_{3})_{3}C{-}NH_{2}}$, is a primary amine even though its nitrogen sits on a tertiary carbon, while 2-methylpropan-2-ol, $\mathrm{(CH_{3})_{3}C{-}OH}$, is a tertiary alcohol. Count the carbons on N, and ignore what lies beyond them.
✏️ Example 1 — Classify the amine
Classify each compound as a 1°, 2° or 3° amine (or a quaternary salt), and as aliphatic or aromatic: (a) $\mathrm{CH_{3}CH_{2}NH_{2}}$; (b) $\mathrm{(CH_{3})_{3}CNH_{2}}$; (c) $\mathrm{(CH_{3}CH_{2})_{2}NH}$; (d) $\mathrm{C_{6}H_{5}N(CH_{3})_{2}}$; (e) $\mathrm{C_{6}H_{5}CH_{2}NH_{2}}$; (f) $\mathrm{C_{6}H_{5}NHCH_{3}}$; (g) $\mathrm{(CH_{3})_{4}N^{+}I^{-}}$; (h) $\mathrm{H_{2}NCH_{2}CH_{2}NH_{2}}$.
Method Find the nitrogen, count the carbons bonded to it, then ask whether any of them is a ring carbon.
(a), (b) One carbon on N: both 1° aliphatic. In (b) that carbon is tertiary, but the amine is still primary.
(c), (d) (c) two ethyl groups: 2°, aliphatic, simple. (d) a phenyl and two methyls: 3°, aromatic (N on the ring), mixed.
(e), (f) (e) one carbon on N, the $\mathrm{CH_{2}}$: 1° aliphatic (an aralkyl amine). (f) a ring carbon and a methyl: 2° aromatic.
(g), (h) (g) four carbons on a positive nitrogen: a quaternary ammonium salt (tetramethylammonium iodide). (h) two $\mathrm{{-}NH_{2}}$ groups, each on one carbon: a 1° aliphatic diamine, ethane-1,2-diamine.
Interactive 1 — Amine Sorter: Count the Carbons on Nitrogen

A structure appears. Decide whether it is a 1°, 2° or 3° amine or a quaternary ammonium salt, and press your answer. The sorter then lights up, one by one, the carbons bonded to nitrogen (red), marks the nitrogen in blue, and says whether the amine is aliphatic or aromatic. Press Play to let it sort the whole set for you.

Your answer:
carbons on N—
H on N—
class—
type—
your score0 / 0

Naming Amines

Simple amines have common names that you will still hear in the laboratory, built by listing the groups on nitrogen followed by "amine": methylamine, ethylmethylamine, trimethylamine. The IUPAC system treats amines as alkanamines, in the same way that alcohols are alkanols.

📐 Rule 9.1 — IUPAC names of amines
  1. Primary amines: take the longest chain containing the carbon bonded to N, replace the final "e" of the alkane by "amine", and give the position of the amino group the lowest possible number: $\mathrm{CH_{3}NH_{2}}$ methanamine, $\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$ propan-1-amine, $\mathrm{CH_{3}CH(NH_{2})CH_{3}}$ propan-2-amine.
  2. More than one amino group: keep the "e" and use di-, tri-: $\mathrm{H_{2}NCH_{2}CH_{2}NH_{2}}$ ethane-1,2-diamine, $\mathrm{H_{2}N(CH_{2})_{6}NH_{2}}$ hexane-1,6-diamine. With a double bond: $\mathrm{CH_{2}{=}CHCH_{2}NH_{2}}$ prop-2-en-1-amine.
  3. Secondary and tertiary amines: the largest carbon group on N gives the parent name; the other groups are prefixes with the locant $N$: $\mathrm{CH_{3}NHCH_{2}CH_{3}}$ $N$-methylethanamine, $\mathrm{(CH_{3})_{2}NCH_{2}CH_{3}}$ $N,N$-dimethylethanamine.
  4. Arylamines: $\mathrm{C_{6}H_{5}NH_{2}}$ is aniline, a name accepted by IUPAC; its systematic name is benzenamine. Ring substituents are numbered from the carbon bearing $\mathrm{NH_{2}}$ (C-1): 4-bromobenzenamine (4-bromoaniline). Groups on nitrogen again take $N$: $\mathrm{C_{6}H_{5}N(CH_{3})_{2}}$ is $N,N$-dimethylbenzenamine.
AmineCommon nameIUPAC name
$\mathrm{CH_{3}CH_{2}NH_{2}}$ethylamineethanamine
$\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$$n$-propylaminepropan-1-amine
$\mathrm{CH_{3}CH(NH_{2})CH_{3}}$isopropylaminepropan-2-amine
$\mathrm{(CH_{3})_{3}CNH_{2}}$tert-butylamine2-methylpropan-2-amine
$\mathrm{CH_{3}NHCH_{2}CH_{3}}$ethylmethylamine$N$-methylethanamine
$\mathrm{(CH_{3})_{3}N}$trimethylamine$N,N$-dimethylmethanamine
$\mathrm{(C_{2}H_{5})_{2}NCH_{2}CH_{2}CH_{3}}$diethylpropylamine$N,N$-diethylpropan-1-amine
$\mathrm{C_{6}H_{5}NH_{2}}$anilineaniline or benzenamine
$\mathrm{C_{6}H_{5}NHCH_{3}}$$N$-methylaniline$N$-methylbenzenamine
$o$-$\mathrm{CH_{3}C_{6}H_{4}NH_{2}}$$o$-toluidine2-methylaniline (2-methylbenzenamine)
$\mathrm{C_{6}H_{5}CH_{2}NH_{2}}$benzylaminephenylmethanamine
✏️ Example 2 — Name the amine
Give IUPAC names: (a) $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}}$; (b) $\mathrm{(CH_{3})_{2}CHCH_{2}NH_{2}}$; (c) $\mathrm{CH_{3}CH(NH_{2})CH_{2}CH_{3}}$; (d) $\mathrm{CH_{3}CH_{2}N(CH_{3})_{2}}$; (e) $\mathrm{(CH_{3}CH_{2})_{2}NCH_{2}CH_{2}CH_{3}}$; (f) $p$-$\mathrm{BrC_{6}H_{4}NH_{2}}$; (g) $\mathrm{C_{6}H_{5}N(C_{2}H_{5})_{2}}$; (h) $\mathrm{CH_{2}{=}CHCH_{2}NH_{2}}$.
(a), (b) (a) Four-carbon chain, $\mathrm{NH_{2}}$ on C-1: butan-1-amine. (b) Longest chain through the $\mathrm{CH_{2}}$ bonded to N has three carbons, with a methyl on C-2: 2-methylpropan-1-amine.
(c), (d) (c) Number from the end nearer N: butan-2-amine. (d) Largest group on N is ethyl, two methyls are prefixes: $N,N$-dimethylethanamine.
(e), (f) (e) Propyl is the largest group: $N,N$-diethylpropan-1-amine. (f) 4-bromobenzenamine (4-bromoaniline).
(g), (h) (g) $N,N$-diethylbenzenamine ($N,N$-diethylaniline). (h) Number from the carbon bonded to N, which gets C-1; the double bond starts at C-2: prop-2-en-1-amine.
✏️ Example 3 — All the amines with formula $\mathrm{C_{4}H_{11}N}$
Write the structures and IUPAC names of all the structurally isomeric amines of formula $\mathrm{C_{4}H_{11}N}$ and classify each. Which of them can exist as a pair of optical isomers?
Check the formula $\mathrm{C_{n}H_{2n+3}N}$ with $n=4$ gives $\mathrm{C_{4}H_{11}N}$: a saturated, open-chain amine (no rings or double bonds). Sort by the number of carbons on N.
1° (four carbons in one group) The four butyl skeletons: $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}}$ butan-1-amine; $\mathrm{CH_{3}CH_{2}CH(NH_{2})CH_{3}}$ butan-2-amine; $\mathrm{(CH_{3})_{2}CHCH_{2}NH_{2}}$ 2-methylpropan-1-amine; $\mathrm{(CH_{3})_{3}CNH_{2}}$ 2-methylpropan-2-amine.
2° (split 3 + 1 or 2 + 2) $\mathrm{CH_{3}CH_{2}CH_{2}NHCH_{3}}$ $N$-methylpropan-1-amine; $\mathrm{(CH_{3})_{2}CHNHCH_{3}}$ $N$-methylpropan-2-amine; $\mathrm{(CH_{3}CH_{2})_{2}NH}$ $N$-ethylethanamine.
3° (split 2 + 1 + 1) $\mathrm{CH_{3}CH_{2}N(CH_{3})_{2}}$ $N,N$-dimethylethanamine. (A 1 + 1 + 1 split has only three carbons.)
Answer 8 isomers: four 1°, three 2°, one 3°. Only butan-2-amine has a carbon with four different groups ($\mathrm{H}$, $\mathrm{NH_{2}}$, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$), so it exists as two enantiomers: 9 stereoisomers in all.
Check A computer enumeration of every open-chain skeleton of four carbons and one nitrogen gives the same 8 (4 + 3 + 1); the same method gives 4 isomers for $\mathrm{C_{3}H_{9}N}$ and 17 for $\mathrm{C_{5}H_{13}N}$.

Preparation of Amines

There are two broad strategies. Either a nitrogen that is already bonded to carbon is reduced (a nitro group, a nitrile or an amide), or a carbon–nitrogen bond is made by letting a nitrogen nucleophile attack a haloalkane. Two special methods, the Gabriel synthesis and the Hoffmann degradation, solve the main problem of the second strategy: getting a primary amine and nothing else.

📐 Rule 9.2 — Reduction of nitro compounds

Nitro compounds are reduced to primary amines by hydrogen gas over a finely divided metal catalyst (Ni, Pt or Pd), or by a metal in acid (Sn and HCl, or Fe and HCl):

$$\mathrm{C_{6}H_{5}NO_{2}+3H_{2}\xrightarrow{Ni,\ Pt\ or\ Pd}C_{6}H_{5}NH_{2}+2H_{2}O}$$ $$\mathrm{C_{6}H_{5}NO_{2}\xrightarrow{Fe/HCl\ or\ Sn/HCl}C_{6}H_{5}NH_{2}}$$

This is the standard route to aniline, because nitrobenzene is easily made by nitrating benzene. With metal and acid the amine is first obtained as its salt, $\mathrm{C_{6}H_{5}NH_{3}^{+}Cl^{-}}$, and the free amine is released by adding NaOH. Iron scrap and hydrochloric acid are preferred in industry: the $\mathrm{FeCl_{2}}$ formed is hydrolysed, releasing HCl, so only a small amount of acid is needed to start the reaction. Nitroalkanes are reduced in the same way to primary alkylamines.

📐 Rule 9.3 — Ammonolysis of haloalkanes

Ammonia, with its lone pair, is a nucleophile. Heated with a haloalkane in ethanol in a sealed tube at about $373\ \mathrm{K}$, it displaces the halide ion (a nucleophilic substitution) to give the salt of a primary amine. The cleavage of the C–X bond by ammonia is called ammonolysis:

$$\mathrm{R{-}X+NH_{3}\rightarrow R{-}NH_{3}^{+}X^{-}}$$ $$\mathrm{R{-}NH_{3}^{+}X^{-}+NaOH\rightarrow R{-}NH_{2}+H_{2}O+NaX}$$

But the primary amine is itself a nucleophile, and a stronger one than ammonia, so it competes for the haloalkane. Further substitution gives the secondary amine, the tertiary amine and finally the quaternary ammonium salt:

$$\mathrm{RNH_{2}\xrightarrow{RX}R_{2}NH\xrightarrow{RX}R_{3}N\xrightarrow{RX}R_{4}N^{+}X^{-}}$$

Ammonolysis therefore gives a mixture, which is its main drawback. A large excess of ammonia makes the primary amine the main product. The reactivity order of the haloalkanes is $\mathrm{RI\gt RBr\gt RCl}$. Aryl halides do not react under these conditions (the C–X bond of a haloarene has partial double-bond character), so aniline cannot be made this way.

📐 Rule 9.4 — Reduction of nitriles and amides

Nitriles are reduced to primary amines by lithium aluminium hydride, $\mathrm{LiAlH_{4}}$, by catalytic hydrogenation ($\mathrm{H_{2}/Ni}$) or by sodium amalgam and ethanol; amides are reduced by $\mathrm{LiAlH_{4}}$ followed by water:

$$\mathrm{R{-}C{\equiv}N\xrightarrow[\text{or }LiAlH_{4}]{H_{2}/Ni}R{-}CH_{2}{-}NH_{2}}$$ $$\mathrm{R{-}CO{-}NH_{2}\xrightarrow{(i)\,LiAlH_{4}\ \ (ii)\,H_{2}O}R{-}CH_{2}{-}NH_{2}}$$

In both, the nitrogen ends up on a $\mathrm{CH_{2}}$ group, and no carbon is gained or lost. Because a nitrile is made from a haloalkane and cyanide ion, $\mathrm{R{-}X\rightarrow R{-}CN}$, the sequence haloalkane → nitrile → amine gives an amine with one carbon more than the haloalkane: a way of stepping up the carbon chain.

📘 Definition 9.3 — Gabriel phthalimide synthesis

Phthalimide (benzene-1,2-dicarboximide) is treated with ethanolic KOH to give potassium phthalimide; its nitrogen anion is a nucleophile that attacks a haloalkane to give an $N$-alkylphthalimide. Hydrolysis with aqueous alkali then releases the primary amine:

$$\text{phthalimide}\xrightarrow{\text{KOH (ethanol)}}\text{potassium phthalimide}$$ $$\xrightarrow{\mathrm{R{-}X}}N\text{-alkylphthalimide}$$ $$\xrightarrow{\mathrm{NaOH(aq)}}\mathrm{R{-}NH_{2}}+\text{sodium phthalate}$$

Because the nitrogen of the $N$-alkylphthalimide has no hydrogen left and is held between two carbonyl groups, it cannot be alkylated again: the product is a pure 1° amine, free of 2° and 3° amines. Aromatic primary amines cannot be made by this method, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

📘 Definition 9.4 — Hoffmann bromamide degradation

A primary amide heated with bromine and an aqueous or ethanolic solution of sodium hydroxide gives a primary amine with one carbon atom fewer than the amide:

$$\mathrm{R{-}CO{-}NH_{2}+Br_{2}+4NaOH}$$ $$\mathrm{\rightarrow R{-}NH_{2}+Na_{2}CO_{3}+2NaBr+2H_{2}O}$$

The carbonyl carbon is lost as carbonate. During the reaction the alkyl or aryl group R migrates from the carbonyl carbon to the nitrogen, so R keeps its structure and ends up bonded to N. The method works for aryl groups too: benzamide gives aniline. It is a way of stepping down the carbon chain.

🧭 Problem-solving strategy — Choosing a route to an amine
  1. Is N on a benzene ring? Then reduce the nitro compound (or use Hoffmann degradation of the aryl amide). Ammonolysis and Gabriel synthesis fail with aryl halides.
  2. Need a pure 1° alkylamine from a haloalkane? Use the Gabriel synthesis, not ammonolysis.
  3. Count the carbons. Nitrile reduction from R–X: one carbon more (and N ends up on a $\mathrm{CH_{2}}$). Hoffmann degradation: one carbon fewer. Nitro, amide reduction, Gabriel, ammonolysis: no change.
  4. Is the N on a $\mathrm{CH_{2}}$? Only then can the amine come from reducing a nitrile or an amide; $\mathrm{(CH_{3})_{2}CH{-}NH_{2}}$ or $\mathrm{(CH_{3})_{3}C{-}NH_{2}}$ cannot.
✏️ Example 4 — Pick the route
Suggest a method, with reagents, for each conversion, and explain your choice: (a) 1-bromopropane to pure propan-1-amine; (b) benzene to aniline; (c) bromomethane to ethanamine; (d) ethanamide to methanamine; (e) propanamide to propan-1-amine. (f) Why can aniline not be made from chlorobenzene by the Gabriel synthesis?
(a) Gabriel synthesis: potassium phthalimide + $\mathrm{CH_{3}CH_{2}CH_{2}Br}$, then NaOH(aq). Ammonolysis would also give di- and tripropylamine.
(b) Nitration, then reduction: $\mathrm{C_{6}H_{6}\xrightarrow[\text{below }333\ K]{conc.\ HNO_{3}/H_{2}SO_{4}}C_{6}H_{5}NO_{2}}$, then $\mathrm{\xrightarrow{Fe/HCl}}$ and NaOH to free the amine.
(c) One carbon more, N on a $\mathrm{CH_{2}}$: $\mathrm{CH_{3}Br\xrightarrow{KCN}CH_{3}CN\xrightarrow{LiAlH_{4}}CH_{3}CH_{2}NH_{2}}$.
(d) One carbon fewer: Hoffmann bromamide degradation, $\mathrm{CH_{3}CONH_{2}\xrightarrow{Br_{2}/NaOH}CH_{3}NH_{2}}$.
(e) Same number of carbons, N on a $\mathrm{CH_{2}}$: reduce the amide with (i) $\mathrm{LiAlH_{4}}$, (ii) $\mathrm{H_{2}O}$: $\mathrm{CH_{3}CH_{2}CONH_{2}\rightarrow CH_{3}CH_{2}CH_{2}NH_{2}}$. (Hoffmann degradation of propanamide would give ethanamine.)
(f) The key step is nucleophilic substitution of the halide by the phthalimide anion. The C–Cl bond of chlorobenzene has partial double-bond character and the ring repels the nucleophile, so the substitution does not occur.
✏️ Example 5 — How much aniline from benzamide?
A student heats $6.05\ \mathrm{g}$ of benzamide, $\mathrm{C_{6}H_{5}CONH_{2}}$, with the theoretical amounts of bromine and sodium hydroxide. (a) What masses of $\mathrm{Br_{2}}$ and NaOH are needed? (b) What mass of aniline is formed if the yield is $80.0\%$? Use $M$: $\mathrm{C_{6}H_{5}CONH_{2}}=121.1$, $\mathrm{Br_{2}}=159.8$, $\mathrm{NaOH}=40.00$, $\mathrm{C_{6}H_{5}NH_{2}}=93.13\ \mathrm{g\,mol^{-1}}$.
Equation $\mathrm{C_{6}H_{5}CONH_{2}+Br_{2}+4NaOH}$ $\mathrm{\rightarrow C_{6}H_{5}NH_{2}+Na_{2}CO_{3}+2NaBr+2H_{2}O}$: mole ratio amide : $\mathrm{Br_{2}}$ : NaOH : amine $=1:1:4:1$.
Moles $n(\text{amide})=\dfrac{6.05}{121.1}=0.04996\ \mathrm{mol}$.
(a) $m(\mathrm{Br_{2}})=0.04996\times159.8=7.98\ \mathrm{g}$; $m(\mathrm{NaOH})=4\times0.04996\times40.00=7.99\ \mathrm{g}$.
(b) Theoretical aniline $=0.04996\times93.13=4.653\ \mathrm{g}$; at $80.0\%$: $0.800\times4.653=3.72\ \mathrm{g}$.
Check The amine has one carbon fewer: $\mathrm{C_{7}H_{7}NO}$ (benzamide) gives $\mathrm{C_{6}H_{7}N}$ (aniline), and the missing carbon leaves as $\mathrm{Na_{2}CO_{3}}$. The equation balances: C 7, H 11, N 1, O 5, Na 4, Br 2 on each side.
Interactive 2 — Amine Synthesis Planner: Which Route Gives a Pure 1° Amine?

Choose a target amine and a route. The planner works backwards to the starting material the route needs, counts its carbons (dots) against the target's, and runs the reaction: a green bar means a pure primary amine, a mixture of bars means a mixture of products, and a red cross means the route cannot give this target. The product proportions shown for ammonolysis are illustrative only: they depend on the excess of ammonia. Press Play to try every route in turn.

starting material—
reagents—
carbons—
outcome—

Physical Properties

Methanamine and ethanamine are gases that smell like ammonia with a fishy edge; primary amines with three or more carbons are liquids, and the higher ones are solids. Pure aniline and other arylamines are colourless, but they slowly turn brown on standing because air oxidises them to coloured products.

📐 Rule 9.5 — Boiling points and solubility of amines
  • Hydrogen bonding between amine molecules. 1° amines (two N–H) and 2° amines (one N–H) form intermolecular hydrogen bonds $\mathrm{N{-}H\cdots N}$; 3° amines have no N–H and cannot. For isomeric amines the boiling points are therefore 1° $\gt$ 2° $\gt$ 3°.
  • Amines against alcohols. Nitrogen is less electronegative than oxygen, so N–H is less polar than O–H and its hydrogen bonds are weaker. Amines boil lower than alcohols (and carboxylic acids) of similar molar mass, but higher than alkanes, which have no hydrogen bonds at all.
  • Solubility in water. All amines can hydrogen-bond to water through the lone pair on N (and 1° and 2° amines through N–H as well), so the lower aliphatic amines are soluble in water. Solubility falls as the hydrophobic carbon part grows; higher amines, and arylamines such as aniline, are almost insoluble in water. Amines dissolve readily in organic solvents such as ethanol, ether and benzene.
Compound ($M\approx73\text{–}74\ \mathrm{g\,mol^{-1}}$)TypeBoiling point (approx.)
butan-1-ol, $\mathrm{CH_{3}(CH_{2})_{3}OH}$alcohol$391\ \mathrm{K}$
butan-1-amine, $\mathrm{CH_{3}(CH_{2})_{3}NH_{2}}$1° amine$351\ \mathrm{K}$
$N$-ethylethanamine, $\mathrm{(C_{2}H_{5})_{2}NH}$2° amine$329\ \mathrm{K}$
$N,N$-dimethylethanamine, $\mathrm{C_{2}H_{5}N(CH_{3})_{2}}$3° amine$310\ \mathrm{K}$
pentane, $\mathrm{C_{5}H_{12}}$ ($M=72$)alkane$309\ \mathrm{K}$

Notice that the tertiary amine, with no N–H to hydrogen-bond through, boils at almost the same temperature as the alkane of similar mass.

✏️ Example 6 — Rank the boiling points
(a) Arrange in increasing order of boiling point: propan-1-ol, $N,N$-dimethylmethanamine, propan-1-amine, $N$-methylethanamine. All have molar masses of about $60\ \mathrm{g\,mol^{-1}}$. (b) $N,N$-dimethylmethanamine boils at about $276\ \mathrm{K}$ yet is very soluble in water. Explain.
Idea The three amines are isomers, $\mathrm{C_{3}H_{9}N}$ ($M=59.11$), and propan-1-ol, $\mathrm{C_{3}H_{8}O}$ ($M=60.10$), has nearly the same mass, so the differences come from hydrogen bonding.
(a) $\mathrm{(CH_{3})_{3}N}$ (3°, no N–H) $\lt$ $\mathrm{CH_{3}NHC_{2}H_{5}}$ (2°, one N–H) $\lt$ $\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$ (1°, two N–H) $\lt$ $\mathrm{CH_{3}CH_{2}CH_{2}OH}$ (stronger O–H hydrogen bonds). Approximate values: $276$, $310$, $321$ and $370\ \mathrm{K}$.
(b) Boiling depends on attractions between amine molecules: a 3° amine has no N–H, so it cannot hydrogen-bond to itself. Dissolving depends on attractions to water: the lone pair on N accepts a hydrogen bond from an H of water, $\mathrm{O{-}H\cdots N}$, so the small amine dissolves easily.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — why chiral amines are rarely separated

In JEE Main, NEET and JEE Advanced 2026 (structure of amines; stereoisomerism and enantiomers), though no syllabus names nitrogen inversion; not required by the CBSE syllabus.

A nitrogen with three different groups and a lone pair, as in $N$-ethyl-$N$-methylpropan-1-amine, has no mirror plane, so you might expect two enantiomers. In practice they cannot be separated: the pyramid turns itself inside out like an umbrella in the wind, passing through a planar form in which N is $sp^{2}$ and the lone pair is in a $p$ orbital. This nitrogen inversion needs only about $25\ \mathrm{kJ\,mol^{-1}}$ and happens billions of times a second at room temperature, so the two forms interconvert far too fast to be isolated. A quaternary ammonium ion with four different groups has no lone pair, cannot invert, and can be resolved into enantiomers.

Exercises

I
Structure, classification and names.
  • 1 Describe the hybridisation of nitrogen and the shape of trimethylamine. Why is the C–N–C angle about $108^{\circ}$ rather than $109.5^{\circ}$?
  • 2 Classify as 1°, 2° or 3°, aliphatic or aromatic: (a) $\mathrm{(CH_{3})_{2}CHNH_{2}}$; (b) $\mathrm{C_{6}H_{5}NHC_{2}H_{5}}$; (c) $\mathrm{(C_{2}H_{5})_{3}N}$; (d) cyclohexanamine, $\mathrm{C_{6}H_{11}NH_{2}}$.
  • 3 Explain why $\mathrm{(CH_{3})_{3}CNH_{2}}$ is called a primary amine while $\mathrm{(CH_{3})_{3}COH}$ is a tertiary alcohol.
  • 4 Give the IUPAC name of $\mathrm{CH_{3}NHCH(CH_{3})_{2}}$.
  • 5 Give the IUPAC name of $\mathrm{(CH_{3})_{2}NCH_{2}CH_{2}CH_{3}}$.
  • 6 Give the IUPAC name of $m$-$\mathrm{O_{2}NC_{6}H_{4}NH_{2}}$ and of $\mathrm{C_{6}H_{5}CH_{2}NHCH_{3}}$.
  • 7 Write the structures of (a) $N$-ethyl-$N$-methylbutan-2-amine and give its molecular formula; (b) 2,4,6-tribromoaniline.
  • 8 Write the structures and IUPAC names of all the amines of formula $\mathrm{C_{3}H_{9}N}$, and classify each.
  • 9 How many secondary amines have the formula $\mathrm{C_{5}H_{13}N}$? Name them.
  • 10 Is tetraethylammonium bromide an amine? Explain, and say how many carbons are bonded to its nitrogen.
1 $sp^{3}$ nitrogen; three bonding pairs and one lone pair point towards the corners of a tetrahedron, so the molecule is pyramidal. Lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion and squeezes the C–N–C angle to about $108^{\circ}$.
2 (a) 1° aliphatic (propan-2-amine); (b) 2° aromatic; (c) 3° aliphatic; (d) 1° aliphatic (the ring is saturated, not aromatic).
3 For amines the label counts the carbon groups on nitrogen (one here); for alcohols it describes the carbon carrying OH (bonded to three other carbons here).
4 $N$-methylpropan-2-amine.
5 $N,N$-dimethylpropan-1-amine.
6 3-nitrobenzenamine (3-nitroaniline); $N$-methyl-1-phenylmethanamine ($N$-methylbenzylamine).
7 (a) $\mathrm{CH_{3}CH_{2}CH(CH_{3}){-}N(CH_{3}){-}CH_{2}CH_{3}}$, $\mathrm{C_{7}H_{17}N}$. (b) A benzene ring with $\mathrm{NH_{2}}$ on C-1 and Br on C-2, C-4 and C-6.
8 Four: propan-1-amine and propan-2-amine (1°), $N$-methylethanamine (2°), $N,N$-dimethylmethanamine (3°).
9 Six. Split 4 + 1: $N$-methylbutan-1-amine, $N$-methylbutan-2-amine, $N$,2-dimethylpropan-1-amine, $N$,2-dimethylpropan-2-amine; split 3 + 2: $N$-ethylpropan-1-amine, $N$-ethylpropan-2-amine.
10 No. $\mathrm{(C_{2}H_{5})_{4}N^{+}Br^{-}}$ is a quaternary ammonium salt: four carbons on N, a positive charge and no lone pair, so it is not basic like an amine.
II
Preparation. Molar masses ($\mathrm{g\,mol^{-1}}$): $\mathrm{C_{4}H_{9}Br}$ 137.02, $\mathrm{C_{4}H_{11}N}$ 73.14.
  • 1 Write the equation for the reduction of nitrobenzene by hydrogen over palladium.
  • 2 Why is iron scrap with hydrochloric acid preferred to tin with hydrochloric acid for reducing nitrobenzene in industry?
  • 3 Bromoethane is heated with ethanolic ammonia in a sealed tube. Name all the possible organic products. How would you make ethanamine the main product?
  • 4 Write the steps of the Gabriel synthesis of phenylmethanamine from (chloromethyl)benzene, $\mathrm{C_{6}H_{5}CH_{2}Cl}$.
  • 5 Give the product of reducing propanenitrile, $\mathrm{CH_{3}CH_{2}CN}$, with $\mathrm{LiAlH_{4}}$.
  • 6 Give the product of the Hoffmann bromamide degradation of 2-phenylethanamide, $\mathrm{C_{6}H_{5}CH_{2}CONH_{2}}$.
  • 7 Convert ethanoic acid into methanamine.
  • 8 Convert bromoethane into propan-1-amine.
  • 9 Amide A, $\mathrm{C_{3}H_{7}NO}$, gives amine B with $\mathrm{Br_{2}/NaOH}$ and amine C with $\mathrm{LiAlH_{4}}$. Identify A, B and C.
  • 10 What mass of 1-bromobutane is needed to make $5.00\ \mathrm{g}$ of butan-1-amine by the Gabriel synthesis if the overall yield is $75.0\%$?
1 $\mathrm{C_{6}H_{5}NO_{2}+3H_{2}\xrightarrow{Pd}C_{6}H_{5}NH_{2}+2H_{2}O}$.
2 Iron is cheap, and the $\mathrm{FeCl_{2}}$ formed is hydrolysed, regenerating HCl; only a little acid is needed to start the reaction.
3 $\mathrm{C_{2}H_{5}NH_{2}}$, $\mathrm{(C_{2}H_{5})_{2}NH}$, $\mathrm{(C_{2}H_{5})_{3}N}$ and $\mathrm{(C_{2}H_{5})_{4}N^{+}Br^{-}}$ (as their salts until base is added). A large excess of ammonia makes ethanamine the main product.
4 Phthalimide $\xrightarrow{\text{KOH (ethanol)}}$ potassium phthalimide $\xrightarrow{\mathrm{C_{6}H_{5}CH_{2}Cl}}$ $N$-benzylphthalimide $\xrightarrow{\mathrm{NaOH(aq)}}$ $\mathrm{C_{6}H_{5}CH_{2}NH_{2}}$ + sodium phthalate.
5 Propan-1-amine, $\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$.
6 Phenylmethanamine, $\mathrm{C_{6}H_{5}CH_{2}NH_{2}}$ (one carbon fewer; the benzyl group migrates to N).
7 $\mathrm{CH_{3}COOH\xrightarrow{NH_{3}}CH_{3}COO^{-}NH_{4}^{+}}$; heat: $\mathrm{\rightarrow CH_{3}CONH_{2}}$; then $\mathrm{\xrightarrow{Br_{2}/NaOH}CH_{3}NH_{2}}$.
8 $\mathrm{CH_{3}CH_{2}Br\xrightarrow{KCN}CH_{3}CH_{2}CN}$, then $\mathrm{\xrightarrow{LiAlH_{4}}CH_{3}CH_{2}CH_{2}NH_{2}}$.
9 A: propanamide, $\mathrm{CH_{3}CH_{2}CONH_{2}}$; B: ethanamine (one carbon fewer); C: propan-1-amine.
10 $n(\text{amine})=\dfrac{5.00}{73.14}=0.06836\ \mathrm{mol}$; needed $\dfrac{0.06836}{0.750}=0.09115\ \mathrm{mol}$ of $\mathrm{C_{4}H_{9}Br}$; mass $=0.09115\times137.02=12.5\ \mathrm{g}$.
III
Physical properties.
  • 1 Arrange in increasing order of boiling point: butan-1-amine, $N,N$-dimethylethanamine, butan-1-ol, $N$-ethylethanamine.
  • 2 Why does the solubility of primary amines in water decrease from methanamine to hexan-1-amine?
  • 3 Explain why a tertiary amine boils lower than its primary isomer.
  • 4 Why does a colourless sample of aniline turn brown on storage?
  • 5 Ethanamine ($M=45.08$) boils at about $290\ \mathrm{K}$, ethanol ($M=46.07$) at about $351\ \mathrm{K}$. Explain the difference.
  • 6 Which forms the stronger hydrogen bond with water, $\mathrm{N{-}H\cdots O}$ from an amine or $\mathrm{O{-}H\cdots N}$ from water to an amine? Use your answer to explain why even 3° amines dissolve in water.
1 $N,N$-dimethylethanamine $\lt$ $N$-ethylethanamine $\lt$ butan-1-amine $\lt$ butan-1-ol (about $310$, $329$, $351$, $391\ \mathrm{K}$).
2 The $\mathrm{NH_{2}}$ group hydrogen-bonds to water, but the growing hydrocarbon chain is hydrophobic and cannot; larger amines break more water–water hydrogen bonds than they replace.
3 A 3° amine has no N–H, so its molecules cannot hydrogen-bond to one another; a 1° amine has two N–H per molecule.
4 Atmospheric oxidation forms coloured impurities.
5 Both hydrogen-bond, but O is more electronegative than N, so O–H is more polar and its hydrogen bonds are stronger; more energy is needed to separate ethanol molecules.
6 $\mathrm{O{-}H\cdots N}$ is the stronger: the very polar O–H of water bonds to the lone pair of N. Every amine has that lone pair, so even a 3° amine is a hydrogen-bond acceptor and the small ones dissolve.
IV
Applications and synthesis. $M$ ($\mathrm{g\,mol^{-1}}$): $\mathrm{C_{6}H_{5}NO_{2}}$ 123.11, $\mathrm{C_{6}H_{5}NH_{2}}$ 93.13, $\mathrm{H_{2}}$ 2.016.
  • 1 Industry. A plant makes aniline by hydrogenating nitrobenzene. What mass of nitrobenzene is needed for $1.00$ tonne of aniline if the yield is $95.0\%$?
  • 2 For the same tonne of aniline, how many moles and what mass of hydrogen are consumed (assume the hydrogen is used only for this reaction)?
  • 3 A dye works has benzoic acid but no benzene. Suggest a route to aniline, naming each reagent.
  • 4 A primary amine X, $\mathrm{C_{3}H_{9}N}$, is formed from amide Y by Hoffmann bromamide degradation, and also from 1-bromopropane by the Gabriel synthesis. Identify X and Y, and give the product of reducing Y with $\mathrm{LiAlH_{4}}$.
1 $n(\text{aniline})=\dfrac{1.00\times10^{6}}{93.13}=1.074\times10^{4}\ \mathrm{mol}$; nitrobenzene $=\dfrac{1.074\times10^{4}}{0.950}\times123.11$ $=1.39\times10^{6}\ \mathrm{g}=1.39$ tonne.
2 $3\ \mathrm{mol\ H_{2}}$ per mol aniline: $3\times1.074\times10^{4}=3.22\times10^{4}\ \mathrm{mol}$, i.e. $3.22\times10^{4}\times2.016=6.49\times10^{4}\ \mathrm{g}=64.9\ \mathrm{kg}$ (at least; more is used to make the lost 5%).
3 $\mathrm{C_{6}H_{5}COOH\xrightarrow{SOCl_{2}}C_{6}H_{5}COCl}$, then $\mathrm{\xrightarrow{NH_{3}}C_{6}H_{5}CONH_{2}}$, then $\mathrm{\xrightarrow{Br_{2}/NaOH}C_{6}H_{5}NH_{2}}$.
4 X: propan-1-amine (a 1° amine with N on $\mathrm{CH_{2}}$, from $\mathrm{CH_{3}CH_{2}CH_{2}Br}$). Y has one carbon more: butanamide, $\mathrm{CH_{3}CH_{2}CH_{2}CONH_{2}}$. $\mathrm{LiAlH_{4}}$ keeps all four carbons: butan-1-amine.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Write the reactions involved in the Gabriel phthalimide synthesis of ethanamine. Why can this method not be used to prepare aniline?

Reactions Phthalimide $\xrightarrow{\text{KOH (ethanol)}}$ potassium phthalimide $\xrightarrow{\mathrm{C_{2}H_{5}Br}}$ $N$-ethylphthalimide $\xrightarrow{\mathrm{NaOH(aq)}}$ $\mathrm{C_{2}H_{5}NH_{2}}$ + sodium phthalate. (1 mark)
Reason Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, so no $N$-arylphthalimide forms. (1 mark)

E2. NEET Benzamide is heated with bromine and aqueous sodium hydroxide. The organic product is (A) phenylmethanamine  (B) aniline  (C) benzonitrile  (D) $N$-bromobenzamide

Idea Hoffmann bromamide degradation: the amide loses its carbonyl carbon and the phenyl group migrates to nitrogen.
Answer (B) aniline, $\mathrm{C_{6}H_{5}NH_{2}}$. (Phenylmethanamine would need one carbon more; $N$-bromobenzamide is only an intermediate.)

E3. JEE Main Numerical value. The number of primary amines of formula $\mathrm{C_{4}H_{11}N}$, counting stereoisomers separately, is ______.

Structures One butyl group on $\mathrm{NH_{2}}$, and there are four butyl skeletons: butan-1-amine, butan-2-amine, 2-methylpropan-1-amine, 2-methylpropan-2-amine.
Stereo Butan-2-amine has a stereocentre (H, $\mathrm{NH_{2}}$, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$): two enantiomers.
Answer $4+1=\mathbf{5}$.

E4. JEE MainJEE Advanced An amide A, $\mathrm{C_{4}H_{9}NO}$, gives a primary amine B, $\mathrm{C_{3}H_{9}N}$, with $\mathrm{Br_{2}/NaOH}$. B is also formed when 2-bromopropane is used in the Gabriel synthesis. (a) Identify A and B. (b) What is formed when A is reduced with $\mathrm{LiAlH_{4}}$? (c) Could B be made by reducing a nitrile? Explain.

(a) Gabriel from $\mathrm{(CH_{3})_{2}CHBr}$ puts N on the middle carbon: B is propan-2-amine, $\mathrm{(CH_{3})_{2}CHNH_{2}}$. Hoffmann removes the carbonyl carbon and the isopropyl group migrates unchanged, so A is 2-methylpropanamide, $\mathrm{(CH_{3})_{2}CHCONH_{2}}$ ($\mathrm{C_{4}H_{9}NO}$ ✓).
(b) $\mathrm{LiAlH_{4}}$ turns $\mathrm{C{=}O}$ into $\mathrm{CH_{2}}$: 2-methylpropan-1-amine, $\mathrm{(CH_{3})_{2}CHCH_{2}NH_{2}}$ ($\mathrm{C_{4}H_{11}N}$).
(c) No. Nitrile reduction always gives $\mathrm{R{-}CH_{2}{-}NH_{2}}$; in B the nitrogen is on a CH carbon.

E5. Board · 3 marks Arrange $\mathrm{C_{2}H_{5}OH}$, $\mathrm{(CH_{3})_{2}NH}$ and $\mathrm{C_{2}H_{5}NH_{2}}$ in increasing order of boiling point and explain the order. All three have $M\approx45\text{–}46\ \mathrm{g\,mol^{-1}}$.

Order $\mathrm{(CH_{3})_{2}NH\lt C_{2}H_{5}NH_{2}\lt C_{2}H_{5}OH}$ (about $280$, $290$, $351\ \mathrm{K}$). (1 mark)
Amines The 1° amine has two N–H per molecule and forms more intermolecular hydrogen bonds than the 2° amine, which has one. (1 mark)
Alcohol O is more electronegative than N, so the O–H bond is more polar and the hydrogen bonds in ethanol are stronger. (1 mark)

E6. Board · Case study Read the passage and answer. Most aniline is made by reducing nitrobenzene. In an older process, nitrobenzene was stirred with iron turnings and a little hydrochloric acid; the product was steam-distilled after adding alkali. Today most plants pass nitrobenzene vapour and hydrogen over a metal catalyst. (a) Write equations for the catalytic process and for freeing aniline from anilinium chloride with NaOH. (2 marks) (b) Why does the iron process need only a little acid? (1 mark) (c) Freshly distilled aniline is colourless but a stored sample is brown. Why? (1 mark)

(a) $\mathrm{C_{6}H_{5}NO_{2}+3H_{2}\xrightarrow{Ni}C_{6}H_{5}NH_{2}+2H_{2}O}$; $\ \mathrm{C_{6}H_{5}NH_{3}^{+}Cl^{-}+NaOH}$ $\mathrm{\rightarrow C_{6}H_{5}NH_{2}+NaCl+H_{2}O}$.
(b) The $\mathrm{FeCl_{2}}$ formed is hydrolysed and releases HCl again, so the acid is regenerated.
(c) Aniline is slowly oxidised by air to coloured products.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Which way of making aniline wastes less: iron and acid, or hydrogen and a catalyst?

Why it matters. Aniline is made on a scale of millions of tonnes a year, mostly to produce polyurethanes. Green chemistry judges a process partly by its atom economy: the percentage of the mass of all the reactants that ends up in the desired product. The older iron-reduction process (the Béchamp reduction) left mountains of iron oxide; catalytic hydrogenation leaves only water.

What to do.

  1. Check that the equation for the iron process balances: $\mathrm{4C_{6}H_{5}NO_{2}+9Fe+4H_{2}O}$ $\mathrm{\rightarrow4C_{6}H_{5}NH_{2}+3Fe_{3}O_{4}}$.
  2. Calculate the atom economy, $\dfrac{\text{mass of desired product}}{\text{total mass of reactants}}\times100\%$, for this process and for $\mathrm{C_{6}H_{5}NO_{2}+3H_{2}\rightarrow C_{6}H_{5}NH_{2}+2H_{2}O}$. Use $M$: $\mathrm{C_{6}H_{5}NO_{2}}$ 123.11, Fe 55.85, $\mathrm{H_{2}O}$ 18.02, $\mathrm{H_{2}}$ 2.016, aniline 93.13.
  3. For one tonne of aniline, calculate the mass of $\mathrm{Fe_{3}O_{4}}$ ($M=231.5$) produced by the iron process at $100\%$ yield.
  4. Find out what the iron oxide by-product was later used for, and which metals are used as hydrogenation catalysts today.

Tools: A calculator or spreadsheet; the planner in Interactive 2; a search engine for industrial process descriptions.

Hand in: A half-page comparison of the two processes: balanced equations, atom economies, waste per tonne, and one advantage of each.

Going further. Atom economy ignores yield, solvents and energy. Look up the "E-factor" (kg of waste per kg of product) and discuss which measure better reflects the environmental cost of making aniline.

Instructor note: the iron equation balances (C 24, H 28, N 4, O 12, Fe 9 on each side). Atom economy: iron process, reactants $4\times123.11+9\times55.85+4\times18.02=1067.2$, product $4\times93.13=372.5$, so $\tfrac{372.5}{1067.2}=34.9\%$; hydrogenation $\tfrac{93.13}{123.11+3\times2.016}=\tfrac{93.13}{129.16}=72.1\%$. $\mathrm{Fe_{3}O_{4}}$ per tonne of aniline: $\tfrac{3}{4}\times1.074\times10^{4}\ \mathrm{mol}\times231.5\ \mathrm{g\,mol^{-1}}=1.86$ tonne. The iron oxide was sold as a pigment; nickel, copper, palladium and platinum catalysts are used for hydrogenation.

🧪 Self-check quiz

0/5 correct

Q1. Which compound is a secondary amine?

Q2. Hoffmann bromamide degradation of propanamide, $\mathrm{CH_{3}CH_{2}CONH_{2}}$, gives

Q3. The Gabriel phthalimide synthesis cannot be used to prepare

Q4. The IUPAC name of $\mathrm{(CH_{3})_{2}NCH_{2}CH_{3}}$ is

Q5. Which has the highest boiling point?

← Chapter 9 — Amines9.2 Basicity and Chemical Reactions of Amines →