In 1856, an eighteen-year-old student named William Henry Perkin was working in a makeshift laboratory at his home in London during the Easter holidays. He was a student of August Wilhelm von Hofmann at the Royal College of Chemistry, and he had set himself an ambitious target: to make quinine, the medicine used against malaria, in the laboratory instead of extracting it from cinchona bark.
His plan failed. Working with aniline — a nitrogen-containing compound obtained from coal tar — and oxidising it with potassium dichromate, Perkin got only a dark, unpromising sludge. Most chemists would have washed it away. Perkin instead treated the residue with alcohol, and the solution turned an intense purple. When he dipped a piece of silk into it, the silk took up the colour and kept it after washing.
That accidental purple was mauveine, the first synthetic dye made in a laboratory. Within a few years Perkin had patented it, built a factory, and started an industry that would eventually supply the world with dyes, and later with medicines, pigments and photographic chemicals. All of it grew out of the chemistry of one family of compounds — the amines.
Q1. Aniline is almost insoluble in water, yet it dissolves readily in dilute hydrochloric acid. What does the acid do to the molecule that plain water cannot? Q2. Trimethylamine has three electron-releasing methyl groups pushing electron density onto its nitrogen atom, yet in water it is a weaker base than dimethylamine, which has only two. What second factor must also be deciding basic strength? Q3. n-Butylamine and N,N-dimethylethanamine have exactly the same molar mass, but n-butylamine boils about 40 K higher. Why should merely rearranging the same atoms change the boiling point so much? Q4. Bromine water converts aniline straight into a tribromo product. A chemist who wants only one bromine atom on the ring first converts aniline into acetanilide. Why should temporarily “covering up” the –NH₂ group help? Q5. Perkin’s mauve came from aniline, and many dyes that followed are brilliant orange, red or yellow. What is it about molecules built from amines that lets them absorb visible light and appear so strongly coloured?
As you work through this chapter, you will gather exactly the ideas needed to reason out each of these questions for yourself.
13.1 Classification of Amines
Amines are organic compounds that contain nitrogen and show basic character. Structurally, every amine can be thought of as a derivative of ammonia, \(\mathrm{NH_3}\), in which one or more hydrogen atoms have been replaced by alkyl or aryl groups. The nitrogen atom still carries a lone pair of electrons, and as later sections will show, that lone pair is responsible for almost everything an amine does.
Where Amines Occur
Amines are not laboratory curiosities. They are present in a large number of natural compounds — in proteins, in several vitamins, in hormones, and in plant products such as nicotine. The nitrogen atom of an amine group appears again and again in the molecules that keep living systems working, which is one reason this family of compounds deserves careful study.
The word amine comes from ammonia, and ammonia itself traces back to sal ammoniac, the ancient name for ammonium chloride collected near the temple of Ammon. The plant product nicotine is named after Jean Nicot, the French diplomat who introduced tobacco to the French court.
Several molecules that carry messages in the human body — dopamine, serotonin and adrenaline among them — are amines, and a very large fraction of modern medicines contain an amine group somewhere in the structure. Aniline, the arylamine used industrially for dyes, is toxic and is absorbed through the skin, so it is always handled with gloves in a well-ventilated space.
Primary, Secondary and Tertiary Amines
Amines are classified as primary (1°), secondary (2°) and tertiary (3°) amines. The classification is obtained in a simple way: replace one, two or three hydrogen atoms of the \(\mathrm{NH_3}\) molecule by alkyl or aryl groups, as set out in Table 13.1.
If one hydrogen atom of ammonia is replaced, a primary amine is obtained. If two are replaced, the amine is secondary, and if all three are replaced, it is tertiary. The relationship is easiest to follow as a branching scheme, shown in FC 13.1.
A primary (1°) amine has one alkyl or aryl group on nitrogen, a secondary (2°) amine has two, and a tertiary (3°) amine has three, the remaining valencies of nitrogen being satisfied by hydrogen atoms.
Notice that the counting is done on the nitrogen atom only. The functional group of a primary amine is the amino group, \(-\mathrm{NH_2}\); of a secondary amine, the imino group, \(>\mathrm{NH}\); and of a tertiary amine, a tertiary nitrogen carrying three carbon atoms and no hydrogen.
Table 13.1 Types of amines
| Primary amine, (1°) |
Amino |
\(-\mathrm{NH_2}\) |
\(\mathrm{C_2H_5\!-\!NH_2}\) |
Ethylamine |
| Secondary amine, (2°) |
Imino |
\(>\mathrm{NH}\) |
\(\mathrm{(CH_3)_2NH}\) |
Dimethylamine |
| Tertiary amine, (3°) |
Tertiary nitrogen |
\(\geq\!\mathrm{N}\) |
\(\mathrm{(CH_3)_3N}\) |
Trimethylamine |
How amines are classified vs how alkyl halides and alcohols are classified
| What decides 1°, 2°, 3° |
Nature of the carbon atom to which the functional group is attached |
Number of alkyl or aryl groups directly attached to the nitrogen atom |
| Atom counted |
Carbon |
Nitrogen |
| Worked instance |
Isopropyl alcohol is a 2° alcohol (its –OH sits on a carbon joined to two other carbons) |
Isopropylamine is a 1° amine (only one alkyl group is on nitrogen) |
Simple and Mixed Amines
Secondary and tertiary amines are further classified as simple (symmetrical) amines and mixed (unsymmetrical) amines. When all the alkyl or aryl groups on nitrogen are the same, the compound is a simple amine. If these groups are different from one another, it is a mixed amine.
A simple or symmetrical amine has identical groups attached to nitrogen; a mixed or unsymmetrical amine has two or more different groups attached to nitrogen.
Dimethylamine, \(\mathrm{(CH_3)_2NH}\), is therefore a simple secondary amine, while ethylmethylamine, \(\mathrm{C_2H_5\!-\!NH\!-\!CH_3}\), with two different groups on nitrogen, is a mixed secondary amine. A primary amine never needs this label, since it carries only one group on nitrogen.
Aliphatic and Aromatic Amines
Amines are also divided into two major classes, namely aliphatic and aromatic amines, on the basis of the nature of the groups attached to the nitrogen atom.
In an aliphatic amine, the groups on nitrogen are alkyl groups; ethylamine and trimethylamine are examples. In an aromatic amine, an aromatic ring is joined directly to the nitrogen atom, as in aniline, \(\mathrm{C_6H_5\!-\!NH_2}\). The word directly matters: if a \(-\mathrm{CH_2}-\) group stands between the ring and the nitrogen, as in benzylamine, \(\mathrm{C_6H_5\!-\!CH_2\!-\!NH_2}\), then nitrogen is bonded to an alkyl group and the compound behaves as an aliphatic amine. This distinction will become important in Section 13.5, where the ring is found to change the basic strength of an amine sharply.
Classify each of the following as 1°, 2° or 3°; as simple or mixed where applicable; and as aliphatic or aromatic.
(i) \(\mathrm{CH_3CH_2CH_2\!-\!NH_2}\) (ii) \(\mathrm{C_6H_5\!-\!NH\!-\!CH_3}\) (iii) \(\mathrm{(CH_3CH_2)_3N}\)
Nitrogen carries a single group, the propyl group, and two hydrogen atoms. It is a primary (1°) amine. The group on nitrogen is an alkyl group, so it is aliphatic. Simple/mixed does not apply to a primary amine.
Nitrogen carries two groups — a phenyl group and a methyl group — and one hydrogen atom, so it is a secondary (2°) amine. The two groups are different, making it a mixed amine, and since the benzene ring is bonded directly to nitrogen, it is aromatic.
Nitrogen carries three ethyl groups and no hydrogen, so it is a tertiary (3°) amine. All three groups are identical, so it is a simple (symmetrical) amine, and all are alkyl groups, so it is aliphatic.
Classify each of the following amines as 1°, 2° or 3°; as simple or mixed where the label applies; and as aliphatic or aromatic:
(i) \(\mathrm{CH_3CH_2\!-\!NH\!-\!CH_2CH_3}\) (ii) \(\mathrm{(CH_3)_3C\!-\!NH_2}\) (iii) \(\mathrm{C_6H_5\!-\!CH_2\!-\!NH_2}\)
Write some examples of nitrogen-containing organic compounds. What are the types of amines?
Classify the following amines as simple/mixed; 1°, 2°, 3°; and aliphatic or aromatic: \(\mathrm{(C_2H_5)_2NH}\), \(\mathrm{(CH_3)_3N}\), \(\mathrm{C_2H_5\!-\!NH\!-\!CH_3}\), \(\mathrm{C_6H_5\!-\!NH_2}\), \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!NH_2}\), \(\mathrm{C_6H_5\!-\!NH\!-\!C_6H_5}\), \(\mathrm{(CH_3)_3C\!-\!NH_2}\), \(\mathrm{C_6H_5\!-\!N(CH_3)_2}\), \(\mathrm{(C_6H_5)_2N\!-\!C_6H_5}\)
13.2 Nomenclature of Amines
An amine can be named in two accepted ways — by its common name, which is still widely used in laboratories and in industry, and by its IUPAC name, which is systematic and can be built for any structure. A student needs both, because question papers and reagent bottles freely mix the two.
13.2.1 Common names
Common names of aliphatic amines are given by writing the name of the alkyl group followed by the suffix -amine, that is, as ‘alkyl amine’. Ethylamine and methylamine are named in exactly this way.
In the case of mixed amines, the names of the alkyl groups are written in alphabetical order. Thus an amine carrying an ethyl group and a methyl group on nitrogen is ethylmethylamine, with ethyl placed first because it comes first alphabetically.
If two or three identical alkyl groups are attached to the nitrogen atom, the prefix ‘di-’ or ‘tri-’ is added before the name of the alkyl group. This gives dimethylamine for \(\mathrm{(CH_3)_2NH}\) and trimethylamine for \(\mathrm{(CH_3)_3N}\).
The parent arylamine, \(\mathrm{C_6H_5\!-\!NH_2}\), is named as aniline. Other aromatic amines are then named as derivatives of aniline, as shown in Table 13.2. For example, an aniline molecule carrying a methyl group at the para position is called p-toluidine, and one carrying a bromine atom at the para position is p-bromoaniline.
13.2.2 IUPAC names
In the IUPAC system, primary amines are named by replacing the ending ‘e’ of the parent alkane by the suffix -amine, giving an alkanamine. Methane therefore becomes methanamine and ethane becomes ethanamine.
A locant indicating the position of the amino group is added before the suffix amine. The chain is numbered so that the carbon bearing the \(-\mathrm{NH_2}\) group gets the lowest possible number, which is how \(\mathrm{CH_3\!-\!CH(NH_2)\!-\!CH_3}\) becomes propan-2-amine rather than propan-3-amine.
When two or more amino groups are present, the prefixes ‘di-’, ‘tri-’ etc. are used with the proper locant. In this case the ending ‘e’ of the parent alkane is retained, which is why \(\mathrm{H_2N\!-\!CH_2\!-\!CH_2\!-\!NH_2}\) is named ethane-1,2-diamine and not ethan-1,2-diamine. The ‘e’ is dropped only when the suffix that follows begins with a vowel, and the prefix ‘di-’ places a consonant there instead.
Naming secondary and tertiary amines
Secondary and tertiary amines are named as N-substituted derivatives of primary amines. The largest alkyl group attached to nitrogen is taken as the parent alkane, and the other alkyl groups are treated as N-substituents. The italic capital letter N is the locant here; it tells the reader that the group is attached to nitrogen and not to a carbon of the chain.
So \(\mathrm{CH_3\!-\!CH_2\!-\!NH\!-\!CH_3}\) has ethyl as its larger group, making the parent ethanamine, and the methyl group on nitrogen is written as the prefix N-methyl, giving N-methylethanamine. The full sequence for building such a name is set out in FC 13.2.
Naming arylamines
While naming arylamines, the ending ‘e’ of the arene is replaced by ‘amine’. Benzene therefore gives benzenamine for \(\mathrm{C_6H_5\!-\!NH_2}\). The common name aniline is also accepted by IUPAC, so both aniline and benzenamine are correct, and the derivatives may be named either way — 4-bromoaniline and 4-bromobenzenamine refer to the same compound (see Table 13.2).
Common naming vs IUPAC naming of amines
| Basis of the name |
Alkyl group name + suffix -amine |
Parent alkane with final ‘e’ replaced by -amine |
| Position of the \(-\mathrm{NH_2}\) group |
Not shown by a number; shown by prefixes such as iso-, n-, sec- |
Shown by a locant, e.g. propan-2-amine |
| Two identical groups on N |
Prefix di- or tri- before the alkyl group name |
Named as N-substituents, e.g. N,N-dimethyl |
| Parent aromatic amine |
Aniline (or phenylamine) |
Aniline or benzenamine |
| Example: \(\mathrm{CH_3\!-\!CH_2\!-\!NH\!-\!CH_3}\) |
Ethylmethylamine |
N-methylethanamine |
Table 13.2 : Common and IUPAC names of some alkyl and arylamines
| a. Primary amines : |
|
|
| \(\mathrm{CH_3\!-\!NH_2}\) |
Methylamine |
Methanamine |
| \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!NH_2}\) |
n-Propylamine |
Propan-1-amine |
| \(\mathrm{CH_3\!-\!CH(NH_2)\!-\!CH_3}\) |
Isopropylamine |
Propan-2-amine |
| \(\mathrm{H_2N\!-\!CH_2\!-\!CH_2\!-\!NH_2}\) |
Ethylenediamine |
Ethane-1, 2-diamine |
| \(\mathrm{CH_2{=}CH\!-\!CH_2\!-\!NH_2}\) |
Allylamine |
Prop-2-en-1-amine |
| \(\mathrm{C_6H_5\!-\!NH_2}\) |
Aniline/ Phenylamine |
Aniline or Benzenamine |
| \(\mathrm{H_2N\!-\!(CH_2)_6\!-\!NH_2}\) |
Hexamethylenediamine |
Hexane-1, 6-diamine |
| 4-methyl-substituted aniline, \(\mathrm{CH_3\!-\!C_6H_4\!-\!NH_2}\) |
p-Toluidine |
4-Methylaniline |
| Cyclohexane ring bearing \(-\mathrm{NH_2}\) |
Cyclohexylamine |
Cyclohexanamine |
| \(\mathrm{C_6H_5\!-\!CH_2\!-\!NH_2}\) |
Benzylamine |
Phenylmethanamine |
| 4-bromo-substituted aniline, \(\mathrm{Br\!-\!C_6H_4\!-\!NH_2}\) |
p-Bromoaniline |
4-Bromoaniline or 4-Bromobenzenamine |
| b. Secondary Amines |
|
|
| \(\mathrm{CH_3\!-\!NH\!-\!CH_3}\) |
Dimethylamine |
N-methylmethanamine |
| \(\mathrm{CH_3\!-\!CH_2\!-\!NH\!-\!CH_3}\) |
Ethylmethylamine |
N-methylethanamine |
| \(\mathrm{C_6H_5\!-\!NH\!-\!CH_3}\) |
Methylphenylamine |
N-methylaniline or N-methylbenzenamine |
| \(\mathrm{C_6H_5\!-\!NH\!-\!C_6H_5}\) |
Diphenylamine |
N-Phenylbenzenamine |
| c. Tertiary Amines |
|
|
| \(\mathrm{(CH_3)_3N}\) |
Trimethylamine |
N, N-Dimethylmethanamine |
| \(\mathrm{C_2H_5\!-\!N(CH_3)_2}\) |
Ethyldimethylamine |
N, N-Dimethylethanamine |
| \(\mathrm{CH_3\!-\!N(C_2H_5)\!-\!C_3H_7}\) |
Ethylmethyl-n-propylamine |
N-Ethyl-N-methyl propan-1-amine |
| \(\mathrm{(CH_3)_2CH\!-\!N(CH_3)\!-\!C_2H_5}\) |
Ethylmethylisopropylamine |
N-Ethyl-N-methyl propan-2-amine |
| \(\mathrm{C_6H_5\!-\!N(CH_3)_2}\) |
N, N-Dimethylaniline |
N, N-Dimethylbenzenamine |
Write the IUPAC names of the following amines.
(i) \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!NH_2}\) (ii) \(\mathrm{(CH_3)_2CH\!-\!NH\!-\!CH_3}\) (iii) \(\mathrm{CH_3\!-\!CH_2\!-\!N(CH_3)\!-\!CH_2\!-\!CH_3}\)
The nitrogen carries only one group, so this is a primary amine. The parent chain has four carbon atoms, giving butane; drop the final ‘e’ and add -amine to get butanamine. The \(-\mathrm{NH_2}\) group is on carbon 1, so the locant is 1. The name is butan-1-amine.
Two groups sit on nitrogen — an isopropyl group and a methyl group — so this is a secondary amine. The larger group is the three-carbon isopropyl group, so the parent is propanamine, with the amino group on carbon 2, that is propan-2-amine. The methyl group on nitrogen is written as N-methyl. The name is N-methylpropan-2-amine.
Three groups sit on nitrogen — two ethyl groups and one methyl group — so this is a tertiary amine. The largest group is ethyl, so the parent is ethanamine. The remaining groups on nitrogen, an ethyl and a methyl, are named as N-substituents in alphabetical order. The name is N-ethyl-N-methylethanamine.
Write the IUPAC name of each compound:
(i) \(\mathrm{CH_3\!-\!CH(NH_2)\!-\!CH_2\!-\!CH_3}\) (ii) \(\mathrm{C_6H_5\!-\!NH\!-\!CH_2\!-\!CH_3}\) (iii) \(\mathrm{H_2N\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!NH_2}\)
Draw possible structures of all the isomers of \(\mathrm{C_4H_{11}N}\). Write their common as well as IUPAC names.
13.3 Preparation of Amines
There is no single universal route to an amine. The method a chemist chooses depends on the starting material available and, very importantly, on how many carbon atoms the product must contain compared with the starting compound. Six standard methods are described below, and it is worth watching the carbon count in each one.
13.3.1 By ammonolysis of alkyl halides
When an alkyl halide is heated with an alcoholic solution of excess ammonia, it undergoes a nucleophilic substitution reaction in which the halogen atom is replaced by an amino (\(-\mathrm{NH_2}\)) group to form a primary amine. Ammonia is the nucleophile here: its lone pair attacks the carbon atom carrying the halogen, and the halide ion leaves.
This process of breaking of the C–X bond by ammonia is known as ammonolysis. The reaction is also known as alkylation of ammonia, because seen from ammonia’s side, an alkyl group has been added to nitrogen. The reaction is carried out in a sealed tube at 373 K.
\[\mathrm{R\!-\!X} + \mathrm{NH_3}\text{(alc.)} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!NH_2}\]
Here the ammonia is used in excess, and the product \(\mathrm{R\!-\!NH_2}\) is a 1° amine.
Ammonolysis is the breaking of the carbon–halogen bond of an alkyl halide by ammonia, in which the halogen atom is replaced by an amino group.
Why the reaction does not stop at the primary amine
It may be noted that the primary amine obtained in the 1st step is a stronger nucleophile than ammonia. The reason is the alkyl group: it releases electron density towards nitrogen, making the lone pair on the amine nitrogen more available than the lone pair on ammonia.
Hence the primary amine further reacts with alkyl halide to form secondary and tertiary amines, and finally a quaternary ammonium salt if \(\mathrm{NH_3}\) is not used in large excess. The product is therefore a mixture, and separating the members of that mixture is troublesome. This is exactly why ammonolysis, though simple, is not a clean laboratory method for making a pure primary amine.
The order of reactivity of alkyl halides with ammonia is \(\mathrm{R\!-\!I} > \mathrm{R\!-\!Br} > \mathrm{R\!-\!Cl}\). This follows the strength of the carbon–halogen bond: the C–I bond is the weakest and breaks most easily, so iodides react fastest.
A limitation with tertiary alkyl halides
Ammonolysis also fails when the halide is tertiary. When tert-butyl bromide is treated with alcoholic \(\mathrm{NH_3}\), isobutylene is formed instead of an amine. This is the result of an elimination reaction being preferred over nucleophilic substitution, and it proceeds through the stable tertiary butyl carbocation intermediate, as drawn in Fig. 13.3.
Write chemical equations for 1. reaction of alc. \(\mathrm{NH_3}\) with \(\mathrm{C_2H_5I}\). 2. Ammonolysis of benzyl chloride followed by the reaction with two moles of \(\mathrm{CH_3\!-\!I}\). Also: why is ammonolysis of alkyl halide not a suitable method for the preparation of primary amine?
13.3.2 Reduction of nitrocompounds
Aliphatic and aromatic nitrocompounds can be reduced to primary amines by using a metal–acid mixture (Sn/HCl or Fe/HCl or Zn/HCl) or catalytic hydrogenation (\(\mathrm{H_2}\)/Ni or Pt or Pd) or \(\mathrm{LiAlH_4}\) in ether.
\[\mathrm{R\!-\!NO_2} + 6[\mathrm{H}] \xrightarrow{\ \mathrm{Sn/HCl}\ } \mathrm{R\!-\!NH_2} + 2\,\mathrm{H_2O}\]
The symbol \([\mathrm{H}]\) stands for the nascent hydrogen supplied by the metal–acid mixture at the moment of its formation; six such hydrogen atoms are needed to convert one nitro group into an amino group, with two molecules of water released. The carbon skeleton is untouched, so the amine has the same number of carbon atoms as the nitro compound.
This is the standard industrial and laboratory route to aniline, since nitrobenzene is easily made by nitrating benzene and is cheap.
13.3.3 Reduction of alkyl cyanide (alkanenitriles)
How is alkyl halide converted into alkyl cyanide?
Primary amines can be obtained by the reduction of alkyl cyanide with sodium and ethanol. This is known as Mendius reduction. The reaction can also be brought about by lithium aluminium hydride.
\[\mathrm{R\!-\!C{\equiv}N} + 4[\mathrm{H}] \xrightarrow{\ \mathrm{Na/C_2H_5OH}\ \text{or}\ \mathrm{LiAlH_4}\ } \mathrm{R\!-\!CH_2\!-\!NH_2}\]
The product is a 1° amine. Look carefully at the carbon count: the carbon of the \(-\mathrm{C{\equiv}N}\) group stays in the molecule and becomes the carbon of the \(-\mathrm{CH_2\!-\!NH_2}\) group. The amine therefore has one carbon atom more than the alkyl halide from which the cyanide was made. Such a conversion is called a step-up reaction.
\(\mathrm{LiAlH_4}\) (lithium aluminium hydride) (A powerful reducing agent that delivers hydride to polar multiple bonds → used in dry ether, since it reacts violently with water and even with moist air → reduces nitriles and amides straight through to primary amines, and also reduces nitro compounds.)
Write a reaction to convert methyl bromide into ethyl amine. Also, comment on the number of carbon atoms in the starting compound and the product.
Methyl bromide can be converted into ethyl amine in a two-stage reaction sequence as shown below.
\[\mathrm{CH_3\!-\!Br} + \mathrm{KCN} \rightarrow \mathrm{CH_3\!-\!CN} + \mathrm{KBr}\]
\[\mathrm{CH_3\!-\!CN} \xrightarrow[\text{(reduction)}]{\ \mathrm{Na/C_2H_5OH}\ } \mathrm{CH_3\!-\!CH_2\!-\!NH_2}\]
The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which the number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.
Identify ‘A’ and ‘B’ in the following conversions. (i) \(\mathrm{CH_3\!-\!I} \xrightarrow{\ \mathrm{KCN}\ } A \xrightarrow{\ \mathrm{Na/C_2H_5OH}\ } B\) (ii) \(\mathrm{CH_3\!-\!Br} \xrightarrow{\ \mathrm{AgNO_2}\ } A \xrightarrow{\ \mathrm{Sn/HCl}\ } B\) (iii) \(\mathrm{C_2H_5\!-\!I} \xrightarrow{\ \mathrm{AgCN}\ } A \xrightarrow{\ \mathrm{Na/C_2H_5OH}\ } B\)
13.3.4 By reduction of amides
Primary amines having the same number of carbon atoms can be obtained by the reduction of amides by \(\mathrm{LiAlH_4}\) in ether or by \(\mathrm{Na/C_2H_5OH}\).
\[\mathrm{CH_3\!-\!CO\!-\!NH_2} + 4[\mathrm{H}] \xrightarrow{\ \mathrm{LiAlH_4},\ \text{ether},\ \mathrm{H_3O^+}\ \text{or}\ \mathrm{Na/C_2H_5OH}\ } \mathrm{CH_3\!-\!CH_2\!-\!NH_2} + \mathrm{H_2O}\]
Here acetamide gives ethylamine. The carbonyl oxygen is removed as water and the C=O group becomes a \(-\mathrm{CH_2}-\) group, while the nitrogen atom stays attached to the same carbon. Both acetamide and ethylamine contain two carbon atoms, so the carbon count is unchanged.
13.3.5 Gabriel phthalimide synthesis
This method is used for the synthesis of primary amine. It involves the following three stages.
Formation of potassium salt of phthalimide from phthalimide on reaction with alcoholic potassium hydroxide.
Formation of N-alkyl phthalimide from the potassium salt by reaction with alkyl halide.
Alkaline hydrolysis of N-alkyl phthalimide to form the corresponding primary amine.
The three stages are drawn out in Fig. 13.4, and the same sequence is summarised as a flow scheme in FC 13.3.
The great advantage of this route is purity. Because the nitrogen is locked between two carbonyl groups, it can pick up only one alkyl group, and the amine is released only in the final hydrolysis step. The product is therefore a pure primary amine, free from the secondary and tertiary amines that spoil an ammonolysis mixture.
Aromatic amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. The carbon–halogen bond of an aryl halide has partial double-bond character and the ring’s electron cloud repels an approaching nucleophile, so aniline cannot be made this way.
The method is named after Siegmund Gabriel, the German chemist who worked it out in the late nineteenth century as a way of obtaining primary amines uncontaminated by their higher homologues.
Which alkyl halide would you use in the Gabriel phthalimide synthesis to prepare butan-1-amine? Write the structure of the N-alkylphthalimide formed in the second stage, and state why the same sequence cannot be used to prepare aniline.
13.3.6 By Hofmann degradation (Hofmann rearrangement / Hofmann bromamide degradation / Hofmann hypobromite degradation)
This is a good laboratory method for the conversion of an amide into a primary amine containing one carbon less. The reaction is brought about by warming the amide with bromine and concentrated aqueous KOH solution.
\[\mathrm{R\!-\!CO\!-\!NH_2} + \mathrm{Br_2} + 4\,\mathrm{KOH}\text{(aq)} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!NH_2} + 2\,\mathrm{KBr} + \mathrm{K_2CO_3} + 2\,\mathrm{H_2O}\]
The amide is the starting material and the product is a 1° amine. For example, acetamide gives methylamine:
\[\mathrm{CH_3\!-\!CO\!-\!NH_2} + \mathrm{Br_2} + 4\,\mathrm{KOH}\text{(aq)} \xrightarrow{\ \Delta\ } \mathrm{CH_3\!-\!NH_2} + 2\,\mathrm{KBr} + \mathrm{K_2CO_3} + 2\,\mathrm{H_2O}\]
The overall result is removal of the \(-\mathrm{CO}-\) group from the amide. As the product contains one carbon atom less than the original amide, it is a step down reaction. The carbon that is lost leaves the molecule as the carbonate ion, which is why \(\mathrm{K_2CO_3}\) appears among the products.
The six preparations of amines and how they change the carbon count
| Ammonolysis |
Alkyl halide, R–X |
alc. \(\mathrm{NH_3}\) (excess), 373 K, sealed tube |
1° amine (with 2°, 3° and quaternary salt as by-products) |
Same |
| Reduction of nitrocompounds |
\(\mathrm{R\!-\!NO_2}\) |
Sn/HCl, Fe/HCl, Zn/HCl; or \(\mathrm{H_2}\)/Ni, Pt, Pd; or \(\mathrm{LiAlH_4}\) |
1° amine |
Same |
| Reduction of alkyl cyanide (Mendius) |
\(\mathrm{R\!-\!C{\equiv}N}\) |
\(\mathrm{Na/C_2H_5OH}\) or \(\mathrm{LiAlH_4}\) |
\(\mathrm{R\!-\!CH_2\!-\!NH_2}\) |
One carbon more (step up, counting from R–X) |
| Reduction of amides |
\(\mathrm{R\!-\!CO\!-\!NH_2}\) |
\(\mathrm{LiAlH_4}\)/ether or \(\mathrm{Na/C_2H_5OH}\) |
1° amine |
Same |
| Gabriel phthalimide synthesis |
Phthalimide + R–X |
alc. KOH; then R–X; then NaOH(aq) |
Pure 1° amine (aliphatic only) |
Same as R–X |
| Hofmann degradation |
\(\mathrm{R\!-\!CO\!-\!NH_2}\) |
\(\mathrm{Br_2}\) + 4 KOH(aq), warm |
1° amine |
One carbon less (step down) |
Propanamide is warmed with bromine and aqueous potassium hydroxide. Name the amine formed, write the balanced equation, and state whether the conversion is a step-up or a step-down reaction.
Warming an amide with bromine and aqueous KOH is Hofmann degradation, which converts the amide into a primary amine with one carbon atom less. Propanamide, \(\mathrm{CH_3CH_2\!-\!CO\!-\!NH_2}\), has three carbon atoms, so the amine formed has two carbon atoms — it is ethanamine (ethylamine).
\[\mathrm{CH_3CH_2\!-\!CO\!-\!NH_2} + \mathrm{Br_2} + 4\,\mathrm{KOH}\text{(aq)} \xrightarrow{\ \Delta\ } \mathrm{CH_3CH_2\!-\!NH_2} + 2\,\mathrm{KBr} + \mathrm{K_2CO_3} + 2\,\mathrm{H_2O}\]
Since the product has one carbon atom fewer than the starting amide, the conversion is a step-down reaction.
Diamines made by these routes are the backbone of the synthetic fibre industry. Hexamethylenediamine, \(\mathrm{H_2N\!-\!(CH_2)_6\!-\!NH_2}\), obtained by treating a dihalide with a cyanide and then reducing the dinitrile, is condensed with adipic acid to give nylon-6,6 — the polymer of ropes, parachute fabric and gear wheels. Before reading on, predict how many moles of alkyl halide one mole of a primary amine would consume if it were allowed to react with excess alkyl halide.
Write the chemical equations for the following conversions: i. Methyl chloride to ethylamine. ii. Benzamide to aniline. iii. 1,4-Dichlorobutane to hexane-1,6-diamine. iv. Benzamide to benzylamine.
13.4 Physical Properties of Amines
13.4.1 Intermolecular forces, boiling points and solubility
The N–H bond in amines is polar because the electronegativities of nitrogen (3.0) and hydrogen (2.1) are different. Nitrogen pulls the shared electron pair towards itself, so the nitrogen end of the bond carries a small negative charge (\(\delta^-\)) and the hydrogen end a small positive charge (\(\delta^+\)).
Due to the polar nature of the N–H bond, primary and secondary amines have intermolecular hydrogen bonding. A hydrogen bond forms when the \(\delta^+\) hydrogen of one molecule is attracted to the lone pair on the nitrogen of a neighbouring molecule, and although each such link is far weaker than a covalent bond, a large number of them together hold the molecules firmly.
The intermolecular hydrogen bonding is to a greater extent in primary amines than in secondary amines, because primary amines have two hydrogen atoms bonded to nitrogen for hydrogen bond formation. The network this produces is shown in Fig. 13.1.
Why tertiary amines are different
Tertiary amines do not have intermolecular hydrogen bonding as there is no hydrogen atom on the nitrogen of a tertiary amine. All three valencies of nitrogen are used up by alkyl groups, so there is no \(\delta^+\) hydrogen available to bridge two molecules.
But due to polar N–C bonds, tertiary amines are polar molecules, and have intermolecular dipole–dipole attractive forces. These are real but distinctly weaker than hydrogen bonds. Thus intermolecular forces of attraction are strongest in primary amines and weakest in tertiary amines.
Intermolecular forces in 1°, 2° and 3° amines
| Hydrogen atoms on nitrogen |
Two |
One |
None |
| Intermolecular hydrogen bonding |
Present, to the greatest extent |
Present, to a smaller extent |
Absent |
| Other attractive force |
Dipole–dipole |
Dipole–dipole |
Dipole–dipole only |
| Overall strength of intermolecular forces |
Strongest |
Intermediate |
Weakest |
The observed order of boiling points of isomeric amines is: primary amine > secondary amine > tertiary amine (see Table 13.3, serial numbers 1, 2, 3). It can be explained on the basis of the intermolecular forces in them. Boiling means pulling molecules apart from one another, so the stronger the forces holding the molecules together, the more heat energy is needed and the higher the boiling point.
Q3. n-Butylamine and N,N-dimethylethanamine have exactly the same molar mass, but n-butylamine boils about 40 K higher. Why should merely rearranging the same atoms change the boiling point so much?
A. Because rearranging the atoms changes the forces between the molecules. In n-butylamine, all four carbon atoms are in a chain and nitrogen still carries two hydrogen atoms, so its molecules are held together by extensive intermolecular hydrogen bonding. In N,N-dimethylethanamine, two of those hydrogen atoms have been moved onto carbon as methyl groups; the nitrogen has no hydrogen left, hydrogen bonding is impossible, and only the weaker dipole–dipole attraction remains. Table 13.3 shows the result: 350.8 K against 310.5 K.
Physical state, odour and colour
The lower aliphatic amines are gases with a fishy odour, middle members are liquids, and higher members are solids under ordinary temperature and pressure. The trend follows molar mass: as the molecules get larger, the total attraction between them grows and the substance becomes harder to vaporise.
Aniline and other arylamines are usually colourless liquids but get coloured as they are easily oxidised by air. A freshly distilled sample of aniline is nearly colourless, while a bottle that has stood open turns yellow and then brown.
The fishy smell of stale fish comes largely from volatile lower amines such as trimethylamine, released as fish tissue breaks down. Squeezing lemon juice over fish works because the acid converts these basic amines into their non-volatile salts, which cannot reach the nose. The same chemistry explains why a fishy-smelling laboratory bench is best wiped with dilute acid rather than with water alone.
Solubility in water
Due to their ability to form hydrogen bonds with water molecules, lower aliphatic amines are soluble in water, as shown in Fig. 13.2. Water can act both ways here — its \(\delta^+\) hydrogen bonds to the lone pair on nitrogen, and the \(\delta^+\) hydrogen on nitrogen bonds to the oxygen of water.
Solubility of amines decreases with increase in molar mass of amines due to increase in the size of the hydrophobic alkyl group. The alkyl portion cannot hydrogen bond with water at all, so as it grows longer it dominates the molecule and drags the solubility down. Aromatic amines and higher aliphatic amines are insoluble in water.
Since N–H bonds in amines are less polar than the O–H bond in alcohol, water solubilities of alcohols, amines and alkanes of comparable molar mass are in the decreasing order: alcohols > amines > alkanes. Alkanes are at the bottom because they have no polar bond at all and cannot hydrogen bond with water.
The order of boiling points of alkanes, amines, alcohols and carboxylic acid of comparable molar mass is as follows:
Alkanes < Amines < Alcohols < Carboxylic acid (Table 13.3, serial numbers 4, 5, 6, 7)
Carboxylic acids stand at the top because two of their molecules pair up through two hydrogen bonds at once, so a boiling acid must break twice as many links per pair as an alcohol.
Table 13.3 Boiling points of alkane, alcohol and amines of similar molar masses
| 1 |
\(\mathrm{n\text{-}C_4H_9NH_2}\) |
73 |
350.8 |
| 2 |
\(\mathrm{(C_2H_5)_2NH}\) |
73 |
329.3 |
| 3 |
\(\mathrm{C_2H_5N(CH_3)_2}\) |
73 |
310.5 |
| 4 |
\(\mathrm{C_2H_5COOH}\) |
74 |
414.4 |
| 5 |
\(\mathrm{n\text{-}C_4H_9OH}\) |
74 |
390.3 |
| 6 |
\(\mathrm{(CH_3)_3C\!-\!NH_2}\) |
73 |
318.15 |
| 7 |
\(\mathrm{C_2H_5CH(CH_3)_2}\) |
72 |
300.8 |
Serial numbers 1 and 6 are worth a second look. Both are primary amines of molar mass 73, yet tert-butylamine boils 32.65 K lower than n-butylamine. Its compact, branched shape reduces the surface over which neighbouring molecules can touch, so the total attraction between molecules is smaller even though hydrogen bonding is possible in both.
the bar chart of boiling points of compounds of nearly equal molar mass in Fig. 13.5 (All seven compounds have molar masses between 72 and 74, yet their boiling points span more than 110 K. Which compound sits highest and which sits lowest, and what does each of them have — or lack — in the way of intermolecular forces? Reading only the three amines, what does the fall from bar 1 to bar 3 tell you about the number of N–H hydrogen atoms? Comparing bars 1 and 6, what does the chart suggest about the effect of branching on boiling point?)
Ethylamine and ethanol have nearly the same molar mass. Predict which has the higher boiling point and which is more soluble in water, and give the reason in each case.
Both molecules can form intermolecular hydrogen bonds — ethylamine through its N–H bonds and ethanol through its O–H bond. The deciding factor is bond polarity. Oxygen (3.5) is more electronegative than nitrogen (3.0), so the O–H bond of ethanol is more polar than the N–H bond of ethylamine, and the hydrogen bonds formed by ethanol are stronger.
Stronger intermolecular attraction means more energy is needed to separate the molecules, so ethanol has the higher boiling point. Stronger hydrogen bonding with water molecules also means better interaction with the solvent, so ethanol is more soluble in water. This is consistent with the general order alcohols > amines > alkanes for both properties.
(i) Which of ethylamine and butylamine is more soluble in water? Justify your answer in terms of molecular structure. (ii) Aniline contains an \(-\mathrm{NH_2}\) group and yet is insoluble in water. Explain.
Arrange the following: a. In decreasing order of the b.p.: \(\mathrm{C_2H_5\!-\!OH}\), \(\mathrm{C_2H_5\!-\!NH_2}\), \(\mathrm{(CH_3)_2NH}\) b. In increasing order of solubility in water: \(\mathrm{C_2H_5\!-\!NH_2}\), \(\mathrm{C_3H_7\!-\!NH_2}\), \(\mathrm{C_6H_5\!-\!NH_2}\)
13.5 Basicity of Amines
The basic nature of amines is due to the presence of a lone pair of electrons on the nitrogen atom. That lone pair is not involved in any bond, so it is free to be offered to a species that is short of electrons — and offering an electron pair is exactly what a base does.
In terms of Lewis theory, amines are bases because they can share a lone pair of electrons on the ‘N’ atom with an electron deficient species. For example, trimethylamine shares its lone pair of electrons with the electron deficient boron trifluoride.
\[\mathrm{Me_3N{:}} + \mathrm{BF_3} \rightarrow \mathrm{Me_3\overset{+}{N}\!-\!\overset{-}{B}F_3}\]
Boron in \(\mathrm{BF_3}\) has only six electrons around it, so it accepts the nitrogen lone pair and completes its octet. Nitrogen, having given a share of its own pair, acquires a positive charge and boron a negative charge.
A Lewis base is a species that donates a lone pair of electrons to an electron deficient species; an amine acts as a Lewis base through the lone pair on its nitrogen atom.
Measuring how strong a base is
Basic strength of amines is expressed quantitatively as \(K_b\) or \(\mathrm{p}K_b\) value. In terms of Lowry–Bronsted theory, the basic nature of amines is explained by writing the following equilibrium.
\[\mathrm{{\geq}N{:}} + \mathrm{H_2O} \rightleftharpoons \mathrm{{\geq}\overset{+}{N}\!-\!H} + \mathrm{OH^-} \qquad \ldots (13.1)\]
On the left is the amine; on the right is its conjugate acid, formed when the nitrogen lone pair picks up a proton from water, and a hydroxide ion.
In this equilibrium the amine accepts \(\mathrm{H^+}\), hence an amine is a Lowry–Bronsted base. For a stronger base, this equilibrium shifts towards the right, thereby the \(K_b\) value is larger and \(\mathrm{p}K_b\) value is smaller, and vice versa (refer to Chapter 3). Table 13.4 gives \(\mathrm{p}K_b\) values of some amines.
For a base in aqueous solution at a given temperature, the larger the \(K_b\) and the smaller the \(\mathrm{p}K_b\), the further the equilibrium of Eq. (13.1) lies to the right and the stronger the base.
\(K_b\) and \(\mathrm{p}K_b\) (\(K_b\) is the base ionisation constant of the equilibrium in Eq. (13.1), expressed in \(\text{mol dm}^{-3}\) for this reaction; \(\mathrm{p}K_b = -\log K_b\) and is a pure number with no unit. A smaller \(\mathrm{p}K_b\) means a stronger base — the sign of the trend reverses when you move from \(K_b\) to \(\mathrm{p}K_b\), and this is where most errors are made.)
The proton-transfer definition of acids and bases used here was proposed independently in the 1920s by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry, which is why it carries both their names.
Because amines are bases, many medicines built around an amine group are sold not as the free amine but as its hydrochloride salt. The salt is ionic, dissolves far better in body fluids, and is a stable crystalline solid that survives storage — which is why the label on a tablet strip so often ends in “…amine hydrochloride”.
Table 13.4 : \(\mathrm{p}K_b\) Values of some amines in aqueous medium
| Primary alkanamines : |
|
|
| Methanamine |
\(\mathrm{CH_3\!-\!NH_2}\) |
3.38 |
| Ethanamine |
\(\mathrm{CH_3\!-\!CH_2\!-\!NH_2}\) |
3.29 |
| Propan-2-amine |
\(\mathrm{(CH_3)_2\!-\!CH\!-\!NH_2}\) |
3.40 |
| Phenylmethanamine |
\(\mathrm{C_6H_5\!-\!CH_2\!-\!NH_2}\) |
4.70 |
| Secondary alkanamines : |
|
|
| N-Methylmethanamine |
\(\mathrm{(CH_3)_2NH}\) |
3.27 |
| N-Ethylethanamine |
\(\mathrm{(CH_3CH_2)_2NH}\) |
3.00 |
| Tertiary alkanamines : |
|
|
| N, N-Dimethylmethanamine |
\(\mathrm{(CH_3)_3N}\) |
4.22 |
| N, N-Diethylethanamine |
\(\mathrm{(CH_3CH_2)_3N}\) |
3.25 |
| Ammonia |
\(\mathrm{NH_3}\) |
4.75 |
| Arylamines : |
|
|
| Benzenamine (aniline) |
\(\mathrm{C_6H_5\!-\!NH_2}\) |
9.38 |
| N-Methylaniline |
\(\mathrm{C_6H_5\!-\!NHCH_3}\) |
9.30 |
| N, N-Dimethylaniline |
\(\mathrm{C_6H_5\!-\!N(CH_3)_2}\) |
8.92 |
13.5.1 Basic strength of aliphatic amines
The trend in the observed \(\mathrm{p}K_b\) values (see Table 13.4) and basic strength of 1°, 2°, 3° amines and \(\mathrm{NH_3}\) can be represented as shown below.
Order of \(\mathrm{p}K_b\) values:
\[\mathrm{NH_3} > \mathrm{R\!-\!NH_2} > \mathrm{R_2NH} < \mathrm{R_3N}\]
Order of basic strength:
\[\mathrm{NH_3} < \mathrm{R\!-\!NH_2} < \mathrm{R_2NH} > \mathrm{R_3N} \qquad \ldots (13.2)\]
Thus as per the observed \(\mathrm{p}K_b\) values of the aliphatic amines, secondary amines are the strongest bases. Basic strength increases as we move from \(\mathrm{NH_3}\) to \(\mathrm{R\!-\!NH_2}\) and from \(\mathrm{R\!-\!NH_2}\) to \(\mathrm{R_2NH}\), but basic strength decreases as we move from \(\mathrm{R_2NH}\) to \(\mathrm{R_3N}\) (Table 13.4).
The basic strength and the corresponding \(\mathrm{p}K_b\) value depend upon the position of the equilibrium shown in Eq. (13.1). The greater the stabilization of the conjugate acid, the more the equilibrium will lie on the right side, and the stronger will be the base and smaller will be its \(\mathrm{p}K_b\) value.
Basicity of amines is related to the structural effects which influence stabilization of various species. Greater the stabilization of the protonated amine, that is, the conjugate acid, greater is the basicity of the amine.
What is meant by the +I effect? Which of the following species is better stabilized and by which effect: \(\mathrm{CH_3\!-\!\overset{+}{C}H_2}\) and \(\mathrm{CH_3\!-\!\overset{+}{C}H\!-\!CH_3}\)?
a. Influence of +I effect
An alkyl group exerts an electron releasing inductive effect (+I) which stabilizes positive charge on the atom bonded to it. The alkyl group pushes electron density towards the positively charged nitrogen, spreading the charge over a larger volume and thereby lowering the energy of the ion.
As we move from the conjugate acid of ammonia (\(\mathrm{\overset{+}{N}H_4}\)) to that of a tertiary amine (\(\mathrm{R_3\overset{+}{N}H}\)), the number of alkyl groups (R) bonded to nitrogen goes on increasing steadily. This results in increasing stabilization of the conjugate acids and thereby an increasing order of basic strength is expected.
Order of stabilization:
\[\mathrm{\overset{+}{N}H_4} < \mathrm{R\!-\!\overset{+}{N}H_3} < \mathrm{R_2\overset{+}{N}H_2} < \mathrm{R_3\overset{+}{N}\!-\!H}\]
Expected order of basic strength:
\[\mathrm{NH_3} < \mathrm{R\!-\!NH_2} < \mathrm{R_2NH} < \mathrm{R_3N}\]
The expected order of basic strength on the basis of the +I effect differs from the observed order (Eq. 13.2). It is seen that the observed increasing basic strength from ammonia to 1° amine and from 1° amine to 2° amine is explained on the basis of increased stabilization of conjugate acids by the +I effect of an increased number of alkyl (R) groups.
However, decreased basic strength of the 3° amine implies that the conjugate acid of the 3° amine is less stabilized even though the +I effect of three alkyl groups in \(\mathrm{R_3\overset{+}{N}H}\) is expected to be large. This is suggestive of the existence of another influencing factor in stabilization of conjugate acids of amines.
b. Influence of solvation
Solvation by water also stabilizes the conjugate acids of aliphatic amines and ammonia, and the way it does so is drawn in Fig. 13.6.
The solvent water stabilizes the conjugate acid by hydrogen bonding through the ‘H’ bonded to the ‘\(\mathrm{N^+}\)’. Each such hydrogen bond releases energy and lowers the energy of the ion, so the more of them an ion can form, the more stable it becomes.
The number of ‘H’ atoms bonded to the ‘\(\mathrm{N^+}\)’ decreases from 4 in \(\mathrm{\overset{+}{N}H_4}\) to 1 in \(\mathrm{R_3\overset{+}{N}H}\). As a result \(\mathrm{\overset{+}{N}H_4}\) is best stabilized by solvation while the stabilization by solvation is very poor in \(\mathrm{R_3\overset{+}{N}H}\).
c. Combined influence
The combined influence of the +I effect and solvation on stabilization of conjugate acids of aliphatic amines decides the observed basic strength and \(\mathrm{p}K_b\) value. These two influencing factors operate in opposite directions, as Fig. 13.7 shows.
The net result is that as we move from \(\mathrm{NH_3}\) to \(\mathrm{RNH_2}\) to \(\mathrm{R_2NH}\), the basic strength increases due to better stabilization of the corresponding conjugate acids. Over this stretch the gain from the +I effect outweighs the loss from poorer solvation.
But a 3° amine is a weaker base than a 2° amine because the stabilization of the conjugate acid of the 3° amine by solvation is very poor. By the time three bulky alkyl groups are attached, only one N–H hydrogen is left for hydrogen bonding with water, and the loss in solvation energy more than cancels the extra electron push.
basic strength along the series \(\mathrm{NH_3} \rightarrow\) 1° \(\rightarrow\) 2° \(\rightarrow\) 3° amine (In aqueous medium: rises from \(\mathrm{NH_3}\) to 1° to 2° because the +I effect of added alkyl groups stabilizes the conjugate acid, then falls at 3° because only one N–H hydrogen is left for stabilization by solvation. Without a solvent to hydrogen bond — that is, in the gas phase — only the +I effect operates, and basic strength would rise steadily all the way to the 3° amine.)
Q2. Trimethylamine has three electron-releasing methyl groups pushing electron density onto its nitrogen atom, yet in water it is a weaker base than dimethylamine, which has only two. What second factor must also be deciding basic strength?
A. Solvation by water. Basic strength depends on how well the conjugate acid is stabilized, and that stabilization has two sources. The +I effect of the alkyl groups does favour trimethylamine. But the conjugate acid \(\mathrm{(CH_3)_3\overset{+}{N}H}\) has only one hydrogen atom on nitrogen and can form only one hydrogen bond with water, whereas \(\mathrm{(CH_3)_2\overset{+}{N}H_2}\) has two. The poorer solvation of the tertiary conjugate acid outweighs the stronger electron push, so dimethylamine (\(\mathrm{p}K_b\) 3.27) is the stronger base in water and trimethylamine (\(\mathrm{p}K_b\) 4.22) the weaker.
Refer to \(\mathrm{p}K_b\) values from Table 13.4 and answer: which compound from the following pairs is the stronger base? i. \(\mathrm{CH_3\!-\!NH_2}\) and \(\mathrm{(CH_3)_2NH}\) ii. \(\mathrm{(C_2H_5)_2NH}\) and \(\mathrm{(C_2H_5)_3N}\) iii. \(\mathrm{NH_3}\) and \(\mathrm{(CH_3)_2CH\!-\!NH_2}\)
13.5.2 Basicity of arylamines
Refer to Table 13.4 and answer: are the \(\mathrm{p}K_b\) values of aniline, N-methylaniline and N,N-dimethylaniline larger or smaller than those of \(\mathrm{NH_3}\) and \(\mathrm{CH_3NH_2}\)? Which one of the two, aniline or \(\mathrm{CH_3NH_2}\), is a stronger base?
From the \(\mathrm{p}K_b\) values we understand that arylamines in general are weaker bases than ammonia and aliphatic amines. The gap is not small: aniline has a \(\mathrm{p}K_b\) of 9.38 against 3.38 for methanamine, which means aniline is weaker by a factor of about a million.
The strength of arylamines is explained in accordance with Lowry–Bronsted theory by writing the following equilibrium (Eq. 13.3) for aniline, similar to Eq. (13.1).
\[\mathrm{C_6H_5\!-\!NH_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_6H_5\!-\!\overset{+}{N}H_3} + \mathrm{OH^-} \qquad \ldots (13.3)\]
Here aniline is the base and the anilinium ion is the conjugate acid. Both the species, base and conjugate acid, are resonance stabilized but to different extents — and that difference decides where the equilibrium settles.
In arylamines, the \(-\mathrm{NH_2}\) group is attached directly to an aromatic ring. The lone pair of electrons on nitrogen is conjugated to the aromatic ring and is less available for protonation. In effect, the ring borrows the lone pair; a lone pair that is spread into the ring cannot simultaneously be handed to a proton.
Aniline is resonance stabilized by the five resonance structures drawn in Fig. 13.8.
On the other hand, the anilinium ion obtained by accepting a proton does not have a lone pair of electrons on nitrogen. Hence it can be stabilized by only two resonance structures, shown in Fig. 13.9, and is therefore less stabilized than aniline.
As a result the equilibrium (13.3) is shifted towards the left side. The reactant aniline is stabilized more than the product anilinium ion, so the reaction has less tendency to proceed forward. This makes aniline (and also other arylamines) weaker bases than aliphatic amines and ammonia.
Using the \(\mathrm{p}K_b\) values in Table 13.4, arrange the following in increasing order of basic strength and justify the order: aniline, ammonia, ethanamine, N-ethylethanamine.
The \(\mathrm{p}K_b\) values from Table 13.4 are aniline 9.38, ammonia 4.75, ethanamine 3.29 and N-ethylethanamine 3.00. A smaller \(\mathrm{p}K_b\) means a stronger base, so arranging from the largest \(\mathrm{p}K_b\) to the smallest gives the increasing order of basic strength:
aniline < ammonia < ethanamine < N-ethylethanamine
Aniline is weakest because the nitrogen lone pair is delocalised into the benzene ring and is less available for protonation, and because its conjugate acid loses that resonance stabilization. Ammonia comes next, having no alkyl group to stabilize its conjugate acid. Ethanamine, a 1° amine, is stronger than ammonia because one ethyl group stabilizes the conjugate acid by the +I effect, and N-ethylethanamine, a 2° amine with two such groups, is the strongest of the four.
Aniline and phenylmethanamine (benzylamine) both contain a benzene ring and an \(-\mathrm{NH_2}\) group. Using the \(\mathrm{p}K_b\) values in Table 13.4, state which is the stronger base and explain the difference in terms of where the ring is attached relative to the nitrogen atom.
Arrange the following amines in decreasing order of their basic strength: \(\mathrm{NH_3}\), \(\mathrm{CH_3\!-\!NH_2}\), \(\mathrm{(CH_3)_2NH}\), \(\mathrm{C_6H_5NH_2}\).
13.6 Chemical Properties of Amines
Almost every reaction of an amine can be traced back to the same feature — the lone pair on nitrogen. It makes the amine basic, it makes the amine a nucleophile, and in arylamines it activates the ring. The reactions below are simply that one property showing itself in different company.
13.6.1 Laboratory test for amines
a. Test for amines as the ‘base’
All amines 1°, 2° and 3° are basic compounds. An aqueous solution of a water soluble amine turns red litmus blue, because the amine takes a proton from water and leaves hydroxide ions behind, exactly as written in Eq. (13.1).
The ‘basic’ nature of amines is detected in the laboratory by reaction with an aqueous solution of the strong mineral acid HCl.
\[\mathrm{{\geq}N{:}} + \mathrm{HCl}\text{(aq)} \rightarrow \mathrm{{\geq}\overset{+}{N}\!-\!H}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
The product is a substituted ammonium chloride. Being ionic, it is far more soluble in water than the amine from which it came. Adding excess aqueous NaOH takes the proton back off and returns the free amine:
\[\mathrm{{\geq}\overset{+}{N}\!-\!H}\text{(aq)} + \mathrm{Cl^-}\text{(aq)} + \mathrm{NaOH}\text{(aq, excess)} \rightarrow \mathrm{{\geq}N{:}} + \mathrm{NaCl}\text{(aq)} + \mathrm{H_2O}\]
A water insoluble amine dissolves in aqueous HCl due to the formation of a water soluble substituted ammonium chloride, which on reaction with excess aqueous NaOH regenerates the original insoluble amine. This pair of steps is a standard way of separating an amine from a mixture: dissolve it away in acid, then release it again with alkali.
Q1. Aniline is almost insoluble in water, yet it dissolves readily in dilute hydrochloric acid. What does the acid do to the molecule that plain water cannot?
A. The acid protonates it. Water can only hydrogen bond to aniline, and the large hydrophobic benzene ring easily defeats that weak interaction, so aniline stays undissolved. Hydrochloric acid instead donates a proton to the nitrogen lone pair, converting the neutral molecule into the anilinium ion of a salt, anilinium chloride. An ionic salt interacts strongly with water and dissolves. Adding excess NaOH removes the proton again and the oily, insoluble aniline separates out once more.
b. Diazotization reaction / Orange dye test
In a sample of aromatic primary amine, 1–2 mL of conc. HCl is added. The aqueous solution of \(\mathrm{NaNO_2}\) is added with cooling. This solution is transferred to a test tube containing a solution of \(\beta\)-naphthol in NaOH. Formation of an orange dye indicates the presence of an aromatic primary amino group.
It may be noted that the temperature of all the solutions and reaction mixtures is maintained near 0 °C throughout the reaction. The diazonium salt formed in the first step falls apart above this temperature, so the cooling is not a formality — without it the test simply fails. The steps are collected in FC 13.4.
The reactions involved in this test will be discussed in Section 13.7.2.
13.6.2 Alkylation of amines : Hofmann’s exhaustive alkylation
When a primary amine is heated with excess of a primary alkyl halide, it gives a mixture of secondary amine and tertiary amine along with a tetraalkylammonium halide (also refer to Sec. 13.3.1).
\[\mathrm{R\!-\!NH_2} \xrightarrow[-\mathrm{HX}]{\ \mathrm{R\!-\!X}\ } \mathrm{R_2NH} \xrightarrow[-\mathrm{HX}]{\ \mathrm{R\!-\!X}\ } \mathrm{R_3N} \xrightarrow[-\mathrm{HX}]{\ \mathrm{R\!-\!X}\ } \mathrm{R_4\overset{+}{N}X^-}\]
Here the successive products are the 1° amine, the 2° amine, the 3° amine and finally the tetraalkylammonium halide. If excess of alkyl halide is used, the tetraalkylammonium halide is obtained as the major product. The reaction is known as exhaustive alkylation of amines.
Exhaustive alkylation is the reaction of an amine with excess alkyl halide until every hydrogen on nitrogen has been replaced and the quaternary ammonium salt becomes the major product.
The tetraalkylammonium halides are called quaternary ammonium salts, which are crystalline solids. They are the derivatives of ammonium salts in which all the four hydrogen atoms attached to nitrogen in \(\mathrm{\overset{+}{N}H_4}\) are replaced by four alkyl groups (same or different).
Primary, secondary and tertiary amines consume three, two and one moles of alkyl halide respectively to get converted into the quaternary ammonium salt. The count is simply the number of hydrogen atoms still left on nitrogen.
The reaction is carried out in the presence of the mild base \(\mathrm{NaHCO_3}\), to neutralize the large quantity of HX formed. If the alkyl halide is methyl iodide, the reaction is called exhaustive methylation of amines.
For example, when methylamine is heated with excess methyl iodide, it gives tetramethylammonium iodide.
\[\mathrm{CH_3\!-\!NH_2} + \mathrm{CH_3\!-\!I} \xrightarrow{\ \Delta\ } \mathrm{(CH_3)_2NH} + \mathrm{HI}\]
\[\mathrm{(CH_3)_2NH} + \mathrm{CH_3\!-\!I} \xrightarrow{\ \Delta\ } \mathrm{(CH_3)_3N} + \mathrm{HI}\]
\[\mathrm{(CH_3)_3N} + \mathrm{CH_3\!-\!I} \xrightarrow{\ \Delta\ } \mathrm{(CH_3)_4\overset{+}{N}I^-}\]
How many moles of methyl iodide are required to convert each of the following into its quaternary ammonium iodide? (i) ethanamine (ii) N-methylethanamine (iii) N,N-dimethylethanamine
Each mole of methyl iodide replaces one hydrogen atom on nitrogen, so count the hydrogen atoms present on nitrogen in each amine.
Ethanamine, \(\mathrm{CH_3CH_2\!-\!NH_2}\), is a primary amine with two hydrogen atoms on nitrogen, and a third mole is needed to convert the tertiary amine into the quaternary salt. It therefore requires 3 moles.
N-Methylethanamine, \(\mathrm{CH_3CH_2\!-\!NH\!-\!CH_3}\), is a secondary amine with one hydrogen atom on nitrogen, and one more mole converts the tertiary amine into the salt. It requires 2 moles.
N,N-Dimethylethanamine, \(\mathrm{CH_3CH_2\!-\!N(CH_3)_2}\), is a tertiary amine with no hydrogen on nitrogen; a single mole of methyl iodide adds the fourth group directly. It requires 1 mole.
This matches the general statement that primary, secondary and tertiary amines consume three, two and one moles of alkyl halide respectively.
Complete the following: \(\mathrm{C_2H_5\!-\!NH_2} + \mathrm{C_2H_5\!-\!I}\ \text{(excess)} \xrightarrow{\ \Delta\ } ?\) \(\mathrm{(C_2H_5)_2NH} + \mathrm{CH_3\!-\!I}\ \text{(excess)} \xrightarrow{\ \Delta\ } ?\) \(\mathrm{C_6H_5\!-\!NH_2} + \mathrm{CH_3\!-\!I}\ \text{(excess)} \xrightarrow{\ \Delta\ } ?\)
13.6.3 Hofmann Elimination
When a tetraalkylammonium halide is heated with moist silver oxide, it gives a quaternary ammonium hydroxide, which is a deliquescent crystalline solid, and strongly basic like NaOH or KOH. The halide ion of the salt is exchanged for a hydroxide ion, and the silver halide formed is insoluble, which drives the exchange to completion.
Quaternary ammonium hydroxides on strong heating undergo \(\beta\)-elimination to give an alkene; the reaction is called Hofmann elimination. The hydroxide ion removes a hydrogen atom from the carbon adjacent to the positively charged nitrogen — the \(\beta\)-carbon — while the nitrogen leaves as a tertiary amine, and the new double bond forms between the \(\alpha\)- and \(\beta\)-carbons.
The least substituted alkene is obtained as the major product (in contrast to Saytzeff elimination). This is worth remembering as a contrast, since in Saytzeff elimination the more substituted alkene is the major product.
In Hofmann elimination, a quaternary ammonium hydroxide on strong heating loses a \(\beta\)-hydrogen and a tertiary amine, and the alkene carrying the fewest alkyl groups on its doubly bonded carbon atoms is formed as the major product.
For example, N,N,N-triethylpropylammonium iodide is first converted to the corresponding hydroxide:
\[\mathrm{CH_3CH_2CH_2\!-\!\overset{+}{N}(CH_2CH_3)_3\ I^-} + \mathrm{AgOH} \xrightarrow{\ \text{moist}\ \mathrm{Ag_2O},\ \Delta\ } \mathrm{CH_3CH_2CH_2\!-\!\overset{+}{N}(CH_2CH_3)_3\ OH^-}\]
The hydroxide, N,N,N-triethylpropanammonium hydroxide, then loses water on strong heating:
\[\mathrm{CH_3CH_2CH_2\!-\!\overset{+}{N}(CH_2CH_3)_3\ OH^-} \xrightarrow[-\mathrm{H_2O}]{\ \Delta\ } \mathrm{CH_2{=}CH_2} + \mathrm{CH_3CH_2CH_2\!-\!N(CH_2CH_3)_2}\]
The products are ethene and N,N-diethylpropan-1-amine. The \(\alpha\)- and \(\beta\)-positions involved are marked in Fig. 13.10, and the whole route from a simple amine to the final alkene is summarised in FC 13.5.
Quaternary ammonium salts are not only laboratory intermediates. Acetylcholine, \(\mathrm{CH_3\!-\!CO\!-\!O\!-\!CH_2\!-\!CH_2\!-\!\overset{+}{N}(CH_3)_3}\), is a quaternary ammonium salt that occurs in the nervous system and functions as a neurotransmitter, carrying signals across the gap between a nerve and the muscle it controls. Quaternary ammonium salts are also present in cationic detergents, where the permanent positive charge helps the molecule cling to negatively charged surfaces such as hair and fabric.
Complete the following reaction: \(\mathrm{CH_3\!-\!CH_2\!-\!\overset{+}{N}(CH_3)_3\ I^-} \xrightarrow{\ \text{Moist}\ \mathrm{Ag_2O}\ } ? \xrightarrow{\ \Delta\ }\)
13.6.4 Acylation of amines
What is an acyl group? How are alcohols acylated?
Aliphatic and aromatic primary and secondary amines undergo acylation reaction. These amines contain replaceable hydrogen atoms (positively polarised H) on the nitrogen atom. These hydrogen atoms are replaced by acyl groups such as the acetyl group.
On reaction of amines with acetyl chloride or acetic anhydride, an acetyl derivative of the amine is obtained. It is also called an amide. The amide is less basic than the amine, because the nitrogen lone pair is now drawn towards the neighbouring carbonyl group and is no longer freely available.
Acylation is a nucleophilic substitution reaction: the nitrogen lone pair attacks the carbonyl carbon of the acyl chloride, and the chloride ion is then expelled. The reaction is carried out in the presence of a strong base like pyridine, which neutralizes the acid produced during the reaction. Without pyridine, the HCl released would protonate the amine and stop it acting as a nucleophile.
For example, ethanamine reacts with ethanoyl chloride:
\[\mathrm{C_2H_5\!-\!NH_2} + \mathrm{CH_3\!-\!COCl} \xrightarrow{\ \text{pyridine}\ } \mathrm{C_2H_5\!-\!NH\!-\!CO\!-\!CH_3} + \mathrm{HCl}\]
The product is N-ethylethanamide. The reaction passes through the intermediate shown in Fig. 13.11, in which nitrogen has attacked the carbonyl carbon before chloride departs.
Secondary amines behave in the same way, since they too have one replaceable hydrogen on nitrogen. N-Methylaniline gives N-methyl-N-phenylethanamide:
\[\mathrm{C_6H_5\!-\!N(CH_3)\!-\!H} + \mathrm{CH_3\!-\!COCl} \xrightarrow{\ \text{pyridine}\ } \mathrm{C_6H_5\!-\!N(CH_3)\!-\!CO\!-\!CH_3} + \mathrm{HCl}\]
Benzoyl chloride also gives a similar reaction with amines, replacing the hydrogen on nitrogen by a benzoyl group instead of an acetyl group.
Tertiary amines have no hydrogen on nitrogen at all, and so cannot be acylated — a fact that is regularly used to tell them apart from primary and secondary amines.
Complete the following: \(\mathrm{CH_3\!-\!NH_2} + \mathrm{Ph\!-\!CO\!-\!Cl} \rightarrow ?\) \(\mathrm{(CH_3)_3N} + \mathrm{Ph\!-\!CO\!-\!Cl} \rightarrow ?\)
13.6.5 Carbylamine reaction
Aliphatic or aromatic primary amines on heating with chloroform give foul (offensive) smelling products called alkyl/aryl isocyanides or carbylamines. This reaction is a test for primary amines. Secondary and tertiary amines do not give this test.
\[\mathrm{R\!-\!NH_2} + \mathrm{CHCl_3} + 3\,\mathrm{KOH} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!NC} + 3\,\mathrm{KCl} + 3\,\mathrm{H_2O}\]
Here \(\mathrm{R\!-\!NH_2}\) is a 1° amine and \(\mathrm{R\!-\!NC}\) is the alkyl isocyanide. Notice that the reaction needs both hydrogen atoms of the \(-\mathrm{NH_2}\) group, which is precisely why only primary amines respond, and why the smell alone is enough to identify them in a test tube.
The carbylamine test is always performed in a fume cupboard and on the smallest possible scale. Isocyanides are not merely unpleasant — they are toxic, and their odour clings to glassware and clothing. This is also why the test, though a certain one for primary amines, is rarely used outside a teaching laboratory.
Write the carbylamine reaction using aniline as the starting material.
13.6.6 Reaction with nitrous acid
Primary, secondary and tertiary amines react differently with nitrous acid. Reactions of only primary amines will be considered here.
What is the formula of nitrous acid? Can nitrous acid be stored in a bottle?
Nitrous acid is an unstable compound. Hence it is prepared in situ by adding aqueous sodium nitrite to hydrochloric acid already mixed with the substrate, that is, the amine. Preparing it in situ means making it in the reaction vessel itself, at the moment it is needed, so that it is consumed before it can decompose.
nitrous acid, \(\mathrm{HNO_2}\) (Generated in the flask from \(\mathrm{NaNO_2} + \mathrm{HCl}\), never stored → needs cold conditions, 273–278 K → converts a primary amine into a diazonium salt; with aliphatic amines that salt at once falls apart to an alcohol, while with aromatic amines it survives long enough to be used.)
a. Aliphatic primary amines
Aliphatic primary amines on reaction with nitrous acid form aliphatic diazonium salts as very unstable intermediates, which decompose immediately by reaction with the solvent water. The corresponding alcohol is formed as the product of the reaction, and nitrogen gas is liberated.
\[\mathrm{R\!-\!NH_2} + \mathrm{HNO_2} \xrightarrow[(\mathrm{NaNO_2}\,+\,\mathrm{HCl})]{\ 273\text{–}278\ \text{K}\ } \left[\mathrm{R\!-\!\overset{+}{N}_2\ Cl^-}\right] \xrightarrow{\ \mathrm{H_2O}\ } \mathrm{R\!-\!OH} + \mathrm{N_2}\uparrow + \mathrm{HCl}\]
The square brackets mark the alkyl diazonium chloride as a species too short-lived to be isolated. The brisk effervescence of nitrogen gas is itself a useful observation in the laboratory.
b. Aromatic primary amines
Aromatic primary amines react with nitrous acid to form diazonium salts which have reasonable stability at 273 K.
\[\mathrm{C_6H_5\!-\!NH_2} + \mathrm{HNO_2} \xrightarrow[(\mathrm{NaNO_2}\,+\,\mathrm{HCl})]{\ 273\text{–}278\ \text{K}\ } \mathrm{C_6H_5\!-\!\overset{+}{N}{\equiv}N\ Cl^-}\]
The product is benzene diazonium chloride. Aryl diazonium salts are resonance stabilized and useful as versatile intermediates to obtain a variety of products. The positive charge is not confined to the nitrogen atoms but is spread into the ring, as shown by the resonance structures in Fig. 13.12, and this delocalisation is the reason an aryl diazonium salt survives while an alkyl one does not.
How 1°, 2° and 3° amines respond to the standard tests
| Dilute HCl |
Dissolves, forms substituted ammonium chloride |
Dissolves |
Dissolves |
| Acetyl chloride / acetic anhydride (pyridine) |
Acylated, gives an amide |
Acylated, gives an amide |
No reaction — no replaceable H on N |
| \(\mathrm{CHCl_3}\) + KOH (carbylamine test) |
Foul-smelling isocyanide formed |
No reaction |
No reaction |
| \(\mathrm{HNO_2}\) at 273–278 K (aliphatic) |
Alcohol formed, \(\mathrm{N_2}\) evolved |
— (not considered here) |
— (not considered here) |
An amine of molecular formula \(\mathrm{C_4H_{11}N}\) gives a foul-smelling compound when warmed with chloroform and alcoholic KOH, and on treatment with \(\mathrm{NaNO_2}\) and HCl at 273–278 K it gives an alcohol with brisk evolution of a colourless gas. What can be concluded about the amine?
The foul-smelling product with chloroform and KOH is an isocyanide, and the carbylamine reaction is given only by primary amines. So the compound must be a primary amine.
The second observation confirms this and adds more. Reaction with nitrous acid, generated in the flask from \(\mathrm{NaNO_2}\) and HCl at 273–278 K, gives an alcohol together with nitrogen gas. This is the behaviour of an aliphatic primary amine, whose diazonium salt is too unstable to survive and decomposes at once in water. An aromatic primary amine would instead have given a diazonium salt stable at that temperature.
The amine is therefore an aliphatic primary amine of formula \(\mathrm{C_4H_{11}N}\) — that is, a butylamine, with the amino group on a four-carbon chain — and the gas evolved is \(\mathrm{N_2}\).
Complete the following reactions and name the organic product in each case:
(i) \(\mathrm{C_6H_5\!-\!NH_2} + \mathrm{CH_3\!-\!COCl} \xrightarrow{\ \text{pyridine}\ } ?\)
(ii) \(\mathrm{CH_3CH_2\!-\!NH_2} + \mathrm{HNO_2} \xrightarrow{\ 273\text{–}278\ \text{K}\ } ?\)
13.7 Reactions of Arene Diazonium Salts
Aryl diazonium salts show two types of reactions. In the first type the diazo group leaves the molecule altogether and something else takes its place; in the second type the diazo group is retained and becomes part of a larger molecule. The whole of the synthetic usefulness of these salts comes from this pair of possibilities, summarised in FC 13.6.
13.7.1 Reactions involving displacement of diazo group
The diazonium group (\(-\mathrm{N_2^+}\)) is a very good leaving group due to the positive charge on the nitrogen atom bonded to the aromatic ring. It departs as neutral nitrogen gas — a stable, unreactive molecule that escapes from the mixture and never comes back, which is why these substitutions go essentially to completion.
As a result, the arene diazonium salts undergo nucleophilic substitution reaction with a variety of nucleophiles. Table 13.5 shows reactions of diazonium salts involving displacement of the diazo group.
Table 13.5 Reactions of arene diazonium salts
| Sandmeyer reaction (good yield) |
\(\mathrm{Ar\!-\!\overset{+}{N}_2X^-}\) |
\(\mathrm{CuCl/HCl}\) |
\(\mathrm{Ar\!-\!Cl} + \mathrm{N_2}\) |
|
\(\mathrm{Ar\!-\!\overset{+}{N}_2X^-}\) |
\(\mathrm{CuBr/HBr}\) |
\(\mathrm{Ar\!-\!Br} + \mathrm{N_2}\) |
|
\(\mathrm{Ar\!-\!\overset{+}{N}_2X^-}\) |
\(\mathrm{CuCN/KCN}\) |
\(\mathrm{Ar\!-\!CN} + \mathrm{N_2}\) |
| Gatterman reaction |
\(\mathrm{Ar\!-\!\overset{+}{N}_2X^-}\) |
Cu powder, HCl |
\(\mathrm{Ar\!-\!Cl} + \mathrm{N_2}\) |
|
\(\mathrm{Ar\!-\!\overset{+}{N}_2X^-}\) |
Cu powder, HBr |
\(\mathrm{Ar\!-\!Br} + \mathrm{N_2}\) |
| Iodoarene formation |
\(\mathrm{Ar\!-\!\overset{+}{N}_2Cl^-}\) |
KI |
\(\mathrm{Ar\!-\!I} + \mathrm{N_2}\) |
| Mild Reduction |
\(\mathrm{Ar\!-\!\overset{+}{N}_2Cl^-}\) |
\(\mathrm{H_3PO_2}\), \(\mathrm{H_2O}\) |
\(\mathrm{Ar\!-\!H} + \mathrm{N_2} + \mathrm{H_3PO_3} + \mathrm{HCl}\) |
|
\(\mathrm{Ar\!-\!\overset{+}{N}_2Cl^-}\) |
\(\mathrm{CH_3\!-\!CH_2\!-\!OH}\) |
\(\mathrm{Ar\!-\!H} + \mathrm{N_2} + \mathrm{CH_3CHO} + \mathrm{HCl}\) |
| Phenol formation |
\(\mathrm{Ar\!-\!\overset{+}{N}_2Cl^-}\) |
\(\mathrm{H_2O}\), 283 K |
\(\mathrm{Ar\!-\!OH} + \mathrm{N_2} + \mathrm{HCl}\) |
Read down the table and one point stands out: each of these products would be hard or impossible to make by direct substitution on benzene itself. The diazonium salt acts as a doorway, letting a chemist put a chlorine, bromine, cyanide, iodine, hydroxyl group — or nothing at all — exactly where the amino group used to stand.
Sandmeyer reaction vs Gatterman reaction
| Reagent |
Cuprous halide with the corresponding halogen acid, e.g. \(\mathrm{CuCl/HCl}\) |
Copper powder with the corresponding halogen acid, e.g. Cu powder/HCl |
| Halogens introduced |
Cl, Br and also CN (with \(\mathrm{CuCN/KCN}\)) |
Cl and Br |
| Product |
\(\mathrm{Ar\!-\!Cl}\), \(\mathrm{Ar\!-\!Br}\) or \(\mathrm{Ar\!-\!CN}\), with \(\mathrm{N_2}\) |
\(\mathrm{Ar\!-\!Cl}\) or \(\mathrm{Ar\!-\!Br}\), with \(\mathrm{N_2}\) |
| Yield |
Good yield |
Generally lower than the Sandmeyer route |
cuprous (copper(I)) salts — \(\mathrm{CuCl}\), \(\mathrm{CuBr}\), \(\mathrm{CuCN}\) (Supply the halide or cyanide nucleophile and help the diazonium group leave → used with the matching acid, HCl with CuCl and HBr with CuBr → convert an arene diazonium salt into the aryl halide or aryl cyanide in good yield. Replacing the cuprous salt by copper powder gives the Gatterman version of the same conversion.)
Diazonium salts were first prepared in the 1850s by the German chemist Johann Peter Griess, who was working on coal-tar chemistry. The substitution using cuprous salts was worked out some three decades later by the Swiss chemist Traugott Sandmeyer, and both names have stayed attached to the chemistry ever since.
Reaction with fluoroboric acid
An arene diazonium salt on reaction with fluoroboric acid gives a precipitate of diazonium fluoroborate, which on heating decomposes to yield a fluoroarene.
\[\mathrm{Ar\!-\!\overset{+}{N}_2Cl^-} \xrightarrow{\ \mathrm{HBF_4}\ } \mathrm{Ar\!-\!\overset{+}{N}_2BF_4^-} \xrightarrow{\ \Delta\ } \mathrm{Ar\!-\!F} + \mathrm{N_2}\uparrow + \mathrm{BF_3}\]
On the other hand, when the same fluoroborate is heated with aqueous sodium nitrite in the presence of copper, it gives a nitroarene.
\[\mathrm{Ar\!-\!\overset{+}{N}_2BF_4^-} \xrightarrow[\text{aq. }\mathrm{NaNO_2}]{\ \Delta,\ \mathrm{Cu}\ } \mathrm{Ar\!-\!NO_2} + \mathrm{N_2}\uparrow + \mathrm{BF_3}\]
The fluoroborate is deliberately made because it is insoluble and can be filtered off dry — a diazonium salt that can be handled as a solid is far easier to decompose in a controlled way.
Starting from aniline, write the reactions needed to prepare (a) iodobenzene and (b) phenol.
Both conversions go through the same intermediate, benzene diazonium chloride, which is prepared by treating aniline with sodium nitrite and hydrochloric acid at 273–278 K.
\[\mathrm{C_6H_5\!-\!NH_2} + \mathrm{HNO_2} \xrightarrow[(\mathrm{NaNO_2}\,+\,\mathrm{HCl})]{\ 273\text{–}278\ \text{K}\ } \mathrm{C_6H_5\!-\!\overset{+}{N}{\equiv}N\ Cl^-}\]
- Warming this salt with potassium iodide replaces the diazo group by iodine:
\[\mathrm{C_6H_5\!-\!\overset{+}{N}_2Cl^-} \xrightarrow{\ \mathrm{KI}\ } \mathrm{C_6H_5\!-\!I} + \mathrm{N_2}\]
- Warming the salt with water at 283 K replaces the diazo group by a hydroxyl group:
\[\mathrm{C_6H_5\!-\!\overset{+}{N}_2Cl^-} \xrightarrow{\ \mathrm{H_2O},\ 283\ \text{K}\ } \mathrm{C_6H_5\!-\!OH} + \mathrm{N_2} + \mathrm{HCl}\]
In both cases the driving force is the escape of nitrogen gas.
Write the product formed when benzene diazonium chloride is treated with (i) copper powder in the presence of HBr, and (ii) fluoroboric acid, followed by heating the precipitate obtained. Name each product and state the gas evolved.
Complete the following reactions: \(\mathrm{Ar\!-\!NH_2} \xrightarrow{\ ?\ } \mathrm{Ar\!-\!\overset{+}{N}_2Cl^-} \xrightarrow[\text{cuprous chloride}]{\ \mathrm{HCl}\ } ?\) \(\mathrm{Ar\!-\!\overset{+}{N}_2Cl^-} \xrightarrow{\ \mathrm{H_3PO_2}\ } ?\)
13.7.2 Reactions involving retention of diazo group: (Coupling reactions)
Arene diazonium salts, when treated with certain reactive aromatic compounds such as phenols or aromatic amines, give azo compounds. These have an extended conjugated system of double bonds in which two aromatic rings are joined through the azo group \(-\mathrm{N}{=}\mathrm{N}-\). This reaction is called azo coupling.
Azo compounds are brightly coloured and are used as dyes. Here the diazo group is not lost at all; it becomes the bridge that links the two rings.
Azo coupling is the reaction in which an arene diazonium ion attaches itself to a reactive aromatic ring, producing an azo compound in which two aromatic rings are joined through the \(-\mathrm{N}{=}\mathrm{N}-\) group.
This is an example of an electrophilic aromatic substitution reaction. Here the electrophiles are positively charged diazonium ions. Only rings that are strongly activated — those carrying an \(-\mathrm{OH}\) or an \(-\mathrm{NH_2}\) group — are electron-rich enough to be attacked by so mild an electrophile, which is why coupling works with phenols and aromatic amines but not with benzene itself.
Substitution usually occurs para to the ring activating group, since that position is the most electron-rich and is also the least crowded.
Coupling with phenol and with aniline
For example, benzenediazonium chloride reacts with phenol in mild alkaline medium to give p-hydroxyazobenzene (an orange dye), as shown in Fig. 13.13.
\[\mathrm{C_6H_5\!-\!\overset{+}{N}{\equiv}N\ Cl^-} + \mathrm{C_6H_5\!-\!OH} \xrightarrow{\ \mathrm{OH^-}\ } \mathrm{C_6H_5\!-\!N{=}N\!-\!C_6H_4\!-\!OH} + \mathrm{HCl}\]
Azo coupling with \(\beta\)-naphthol in NaOH is used as a confirmatory test for primary aromatic amines. This is the chemistry behind the orange dye test described in Section 13.6.1.
Benzenediazonium chloride reacts with aniline in mild alkaline medium to give p-aminoazobenzene (a yellow dye), drawn in Fig. 13.14.
\[\mathrm{C_6H_5\!-\!\overset{+}{N}{\equiv}N\ Cl^-} + \mathrm{C_6H_5\!-\!NH_2} \xrightarrow{\ \mathrm{OH^-}\ } \mathrm{C_6H_5\!-\!N{=}N\!-\!C_6H_4\!-\!NH_2} + \mathrm{HCl}\]
Q5. Perkin’s mauve came from aniline, and many dyes that followed are brilliant orange, red or yellow. What is it about molecules built from amines that lets them absorb visible light and appear so strongly coloured?
A. The extended conjugation. In an azo compound, two aromatic rings are joined through the \(-\mathrm{N}{=}\mathrm{N}-\) group, so the alternating double bonds of one ring, the azo linkage and the second ring form a single continuous conjugated system. Electrons in such a system are held loosely and need only a small amount of energy to be excited — an amount that corresponds to visible light rather than ultraviolet. The molecule therefore absorbs part of the visible spectrum, and we see the remaining light as an intense colour. Groups such as \(-\mathrm{OH}\) and \(-\mathrm{NH_2}\) on the ring extend the conjugation further and deepen the shade, which is exactly why p-hydroxyazobenzene is orange and p-aminoazobenzene yellow.
The acid–base indicator methyl orange is an azo dye, \(\mathrm{(CH_3)_2N\!-\!C_6H_4\!-\!N{=}N\!-\!C_6H_4\!-\!SO_3^-Na^+}\), shown in Fig. 13.15. Adding acid protonates it and alters the conjugated system, which is why the colour changes from yellow to red as a titration passes its end point. Azo dyes of the same family colour a large fraction of the world’s textiles, and because they are made from aromatic amines, factories using them must treat their effluent carefully before it reaches a river.
An aromatic primary amine is diazotised and the cold solution is added to \(\beta\)-naphthol dissolved in NaOH; an orange dye separates out. Explain which species acts as the electrophile, why the reaction is described as a coupling, and why the product is coloured.
The electrophile is the arene diazonium ion, \(\mathrm{Ar\!-\!\overset{+}{N}{\equiv}N}\), formed from the amine by treatment with nitrous acid at 273–278 K. It carries a positive charge on nitrogen and therefore seeks an electron-rich centre.
\(\beta\)-Naphthol in NaOH is strongly activated — in alkali it exists as the naphthoxide ion, whose ring is very electron-rich — so it can be attacked by this mild electrophile. The reaction is an electrophilic aromatic substitution occurring at the position para to the activating group.
It is called a coupling reaction because the diazo group is retained rather than displaced: the two aromatic systems become joined through the \(-\mathrm{N}{=}\mathrm{N}-\) bridge, giving a single molecule with an extended conjugated system of double bonds.
That extended conjugation is also the reason for the colour. Electrons spread over the whole conjugated framework absorb light in the visible region, so the product appears as an intense orange dye. Since only a primary aromatic amine can give a diazonium salt stable enough to survive until coupling, the appearance of the orange colour confirms its presence.
13.8 Reaction with arenesulfonyl chloride
(Hinsberg’s test) : Benzenesulfonyl chloride (\(\mathrm{C_6H_5SO_2Cl}\)) is known as Hinsberg’s reagent. It is valuable because the three classes of amine behave in three visibly different ways towards it, so a single reagent can sort out a primary, a secondary and a tertiary amine.
Hinsberg’s reagent is benzenesulfonyl chloride, \(\mathrm{C_6H_5SO_2Cl}\), used to distinguish primary, secondary and tertiary amines by the different products they give and by the solubility of those products in alkali.
The test is named after Oscar Hinsberg, the German chemist who introduced it in the late nineteenth century as a way of separating a mixture of amines as well as identifying them.
a. Reaction of a primary amine
Ethyl amine (a primary amine) reacts with benzenesulfonyl chloride to form N-ethyl benzenesulfonyl amide.
\[\mathrm{C_6H_5\!-\!SO_2\!-\!Cl} + \mathrm{H\!-\!NH\!-\!C_2H_5} \rightarrow \mathrm{C_6H_5\!-\!SO_2\!-\!NH\!-\!C_2H_5} + \mathrm{HCl}\]
The product, N-ethylbenzene sulfonamide, is soluble in alkali. A hydrogen atom on nitrogen has been replaced by the sulfonyl group, and one hydrogen still remains on nitrogen.
The hydrogen attached to nitrogen in the sulfonamide of ethanamine (a primary amine) is strongly acidic. Hence it is soluble in alkali. Alkali removes that acidic hydrogen and converts the sulfonamide into its sodium salt, which is ionic and passes into solution:
\[\mathrm{C_6H_5\!-\!SO_2\!-\!NH\!-\!C_2H_5} + \mathrm{NaOH} \rightarrow \mathrm{C_6H_5\!-\!SO_2\!-\!\overset{-}{N}(Na^+)\!-\!C_2H_5} + \mathrm{H_2O}\]
b. Reaction of a secondary amine
Diethyl amine reacts with benzenesulfonyl chloride to give N,N-diethyl benzene sulfonamide.
\[\mathrm{C_6H_5\!-\!SO_2\!-\!Cl} + \mathrm{H\!-\!N(C_2H_5)_2} \rightarrow \mathrm{C_6H_5\!-\!SO_2\!-\!N(C_2H_5)_2} + \mathrm{HCl}\]
N,N-Diethylbenzenesulfonamide does not contain any H-atom attached to the nitrogen atom. Hence it is not acidic and does not dissolve in alkali. A secondary amine therefore does react with the reagent, but its product settles out as an alkali-insoluble solid.
c. Behaviour of a tertiary amine
Do tertiary amines have ‘H’ bonded to ‘N’? Why do tertiary amines not react with benzene sulfonyl chloride?
A tertiary amine has all three valencies of nitrogen taken up by alkyl or aryl groups, so there is no hydrogen on nitrogen to be replaced by the sulfonyl group. It therefore gives no sulfonamide at all and simply remains as an undissolved layer or is recovered unchanged.
Behaviour of 1°, 2° and 3° amines towards Hinsberg’s reagent
| Hydrogen atoms on N available for substitution |
Two |
One |
None |
| Reaction with \(\mathrm{C_6H_5SO_2Cl}\) |
Reacts, gives N-alkylbenzenesulfonamide |
Reacts, gives N,N-dialkylbenzenesulfonamide |
Does not react |
| H left on N in the product |
One |
None |
— |
| Acidic character of the product |
Acidic, because of the electron-withdrawing \(\mathrm{SO_2}\) group |
Not acidic |
— |
| Behaviour with aqueous NaOH |
Dissolves |
Does not dissolve |
Amine remains unreacted |
benzenesulfonyl chloride, \(\mathrm{C_6H_5SO_2Cl}\) (Hinsberg’s reagent) (Replaces a hydrogen atom on the nitrogen of an amine by the \(\mathrm{C_6H_5SO_2}-\) group, releasing HCl → the mixture is then shaken with aqueous NaOH to read the result → tells a 1° amine (alkali-soluble product) from a 2° amine (alkali-insoluble product) from a 3° amine (no reaction), and can also be used to separate the three from a mixture.)
An amine of molecular formula \(\mathrm{C_3H_9N}\) reacts with benzenesulfonyl chloride, and the product formed dissolves completely in aqueous sodium hydroxide. To which class does the amine belong? Justify your answer and write the equation, taking propan-1-amine as the example.
The product dissolves in aqueous NaOH, which means it still carries a hydrogen atom on nitrogen — one that has been made acidic by the neighbouring \(\mathrm{SO_2}\) group. Only a primary amine can give such a product, since it starts with two hydrogen atoms on nitrogen and retains one after sulfonylation. A secondary amine would give a product with no N–H left, insoluble in alkali, and a tertiary amine would not react at all.
\[\mathrm{C_6H_5\!-\!SO_2\!-\!Cl} + \mathrm{H\!-\!NH\!-\!C_3H_7} \rightarrow \mathrm{C_6H_5\!-\!SO_2\!-\!NH\!-\!C_3H_7} + \mathrm{HCl}\]
The product, N-propylbenzenesulfonamide, dissolves in NaOH by forming its sodium salt.
(i) A liquid amine is shaken with benzenesulfonyl chloride and is later recovered unchanged. To which class does it belong, and why does it not react? (ii) Write the reaction of aniline with benzenesulfonyl chloride, name the product, and state whether it will dissolve in aqueous NaOH.
How will you distinguish between methylamine, dimethylamine and trimethylamine by Hinsberg’s test?
13.9 Electrophilic aromatic substitution in aromatic amines
The amino group is ortho and para directing and a powerful ring activating group. As a result, aromatic amines readily undergo electrophilic substitution reactions.
The reason lies in the lone pair on nitrogen. As Section 13.5.2 showed, that lone pair is delocalised into the ring, and the resonance structures place a negative charge at the ortho and para positions. Those positions therefore become electron-rich and attract an incoming electrophile strongly. What weakens aniline as a base is exactly what makes it so reactive towards electrophiles.
The \(-\mathrm{NH_2}\) group activates the benzene ring towards electrophilic substitution and directs the incoming group to the ortho and para positions, because the nitrogen lone pair is delocalised into the ring and raises the electron density at those positions.
a. Bromination
Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline, as shown in Fig. 13.17.
\[\mathrm{C_6H_5NH_2} + 3\,\mathrm{Br_2} \xrightarrow{\ \mathrm{Br_2/H_2O}\ } \mathrm{C_6H_2Br_3NH_2} + 3\,\mathrm{HBr}\]
No catalyst is needed and no heating is required — an ordinary benzene ring would need a Lewis acid catalyst for bromination. This is a measure of how strongly the amino group activates the ring, and it also creates a practical problem: all three available ortho and para positions are attacked, so a single bromine atom cannot be introduced this way.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
The \(\mathrm{NH_2}\)- group in aniline is highly ring activating and o-/p- directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with \(\mathrm{Br_2}\) it gives 2,4,6-tribromoaniline.
To get a monobromo product, it is necessary to decrease the ring activating effect of the \(-\mathrm{NH_2}\) group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent ring activation decreases, as Fig. 13.18 shows.
Hence, acetanilide on bromination gives a monobromo product, p-bromoacetanilide. After monobromination the original \(-\mathrm{NH_2}\) group is regenerated. The protection of the \(-\mathrm{NH_2}\) group in the form of an acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.
Q4. Bromine water converts aniline straight into a tribromo product. A chemist who wants only one bromine atom on the ring first converts aniline into acetanilide. Why should temporarily “covering up” the –NH₂ group help?
A. Because the problem is that the \(-\mathrm{NH_2}\) group activates the ring too strongly. Its lone pair is fed into the ring, all three ortho and para positions become highly reactive, and bromination cannot be stopped at one substitution. Acetylation converts the amine into acetanilide, where the same lone pair is now shared with the carbonyl group of the acetyl substituent. Less of it reaches the ring, the activation is reduced, and bromination stops after a single substitution at the para position. The acetyl group is then removed by acid-catalysed hydrolysis, regenerating the \(-\mathrm{NH_2}\) group and leaving p-bromoaniline. The acetyl group has acted as a temporary protecting group — added, used, and taken off again.
b. Nitration
Direct nitration of aniline yields a mixture of ortho, meta and para nitroanilines, in the proportions shown in Fig. 13.19.
In acidic medium the \(-\mathrm{NH_2}\) group is protonated to the \(-\mathrm{\overset{+}{N}H_3}\) group, which is meta-directing and deactivating. Hence a considerable amount of m-nitroaniline is obtained.
\(-\mathrm{NH_2}\) and \(-\mathrm{\overset{+}{N}H_3}\) as substituents on a benzene ring
| Lone pair on nitrogen |
Present, delocalised into the ring |
Used up in bonding to \(\mathrm{H^+}\) |
| Effect on ring electron density |
Increases it |
Decreases it |
| Effect on electrophilic substitution |
Strongly activating |
Deactivating |
| Position directed to |
Ortho and para |
Meta |
However, to get p-nitroaniline as the major product, the \(-\mathrm{NH_2}\) group is first protected by acetylation, nitration is carried out, and then the amide is hydrolysed. The sequence is set out in FC 13.9.
Protection does two jobs at once here. It stops the nitrogen from being protonated in the acidic mixture, so the meta-directing anilinium ion never forms, and it moderates the ring activation so that substitution is clean and mainly at the para position.
Search the \(\mathrm{p}K_a\) or \(\mathrm{p}K_b\) values of ortho, meta and para nitroaniline on the internet and arrange them in increasing order of their basic strength.
c. Sulfonation
Aniline reacts with concentrated sulfuric acid to form anilinium hydrogen sulfate, which on heating with sulfuric acid at 453–473 K produces p-aminobenzene sulfonic acid (sulfanilic acid) as the major product.
\[\mathrm{C_6H_5NH_2} \xrightarrow{\ \mathrm{H_2SO_4}\ } \mathrm{C_6H_5\!-\!\overset{+}{N}H_3\ HSO_4^-} \xrightarrow{\ 453\text{–}473\ \text{K}\ } \mathrm{H_2N\!-\!C_6H_4\!-\!SO_3H}\]
Sulfanilic acid exists as a salt, called a dipolar ion or zwitter ion. It is produced by the reaction between an acidic group and a basic group present in the same molecule. The \(-\mathrm{SO_3H}\) group at one end gives up its proton and the \(-\mathrm{NH_2}\) group at the other end accepts it, so a single molecule ends up carrying both a positive and a negative charge, as drawn in Fig. 13.20.
A zwitter ion (dipolar ion) is a neutral molecule that carries a positive charge on one group and a negative charge on another group within the same molecule.
Because it is effectively an ionic compound, sulfanilic acid is a high-melting solid, insoluble in ordinary organic solvents but soluble in aqueous alkali — behaviour that a simple covalent molecule of the same size would never show.
Sulfanilic acid is the parent of the sulfonamide family. Sulfanilamide, closely related to it, was the first widely used antibacterial drug of the 1930s, and the “sulfa drugs” that followed saved very large numbers of lives before penicillin became available. The same compound is also a standard starting material for azo dyes, since its amino group can be diazotised and coupled.
(i) Sulfanilic acid melts at a much higher temperature than would be expected for a molecule of its size and does not dissolve in common organic solvents. Explain this in terms of its structure. (ii) Arrange benzene, aniline and the anilinium ion in decreasing order of ease of electrophilic substitution, giving a reason for the order.
Can aniline react with a Lewis acid? Why does aniline not undergo Friedel–Craft’s reaction using aluminium chloride?
13.10 Summary
What amines are
- Amines are nitrogen-containing organic compounds with basic character, formed by replacing one, two or three hydrogen atoms of \(\mathrm{NH_3}\) by alkyl or aryl groups. They occur widely in nature — in proteins, vitamins, hormones and plant products such as nicotine.
- Amines are classified as primary (1°), secondary (2°) and tertiary (3°) according to the number of groups attached to nitrogen. Secondary and tertiary amines are further called simple (symmetrical) when all groups on nitrogen are the same, and mixed (unsymmetrical) when they differ.
- On the basis of the groups attached to nitrogen, amines are also divided into aliphatic amines and aromatic amines, the latter having a ring joined directly to nitrogen.
Naming an amine
- In common names, the alkyl group name is followed by the suffix -amine, written as one word; groups in mixed amines are named in alphabetical order, and identical groups take di- or tri-. The parent arylamine \(\mathrm{C_6H_5NH_2}\) is aniline, and other arylamines are named as its derivatives.
- In IUPAC names, the final ‘e’ of the parent alkane is replaced by -amine, with a locant for the position of the amino group. Two or more amino groups keep the ‘e’ and take di-, tri- with locants, while secondary and tertiary amines are named as N-substituted derivatives of the largest alkanamine.
Preparation
- Ammonolysis of alkyl halides: \(\mathrm{R\!-\!X} + \mathrm{NH_3}\text{(alc.)} \xrightarrow{\Delta} \mathrm{R\!-\!NH_2}\), in a sealed tube at 373 K with excess ammonia. The 1° amine formed is a stronger nucleophile than ammonia, so 2°, 3° amines and a quaternary ammonium salt also appear. Reactivity order: \(\mathrm{R\!-\!I} > \mathrm{R\!-\!Br} > \mathrm{R\!-\!Cl}\).
- Reduction of nitrocompounds: \(\mathrm{R\!-\!NO_2} + 6[\mathrm{H}] \xrightarrow{\mathrm{Sn/HCl}} \mathrm{R\!-\!NH_2} + 2\,\mathrm{H_2O}\), also with Fe/HCl, Zn/HCl, \(\mathrm{H_2}\)/Ni, Pt or Pd, or \(\mathrm{LiAlH_4}\) in ether.
- Reduction of alkyl cyanide (Mendius reduction): \(\mathrm{R\!-\!C{\equiv}N} + 4[\mathrm{H}] \xrightarrow{\mathrm{Na/C_2H_5OH}\ \text{or}\ \mathrm{LiAlH_4}} \mathrm{R\!-\!CH_2\!-\!NH_2}\). The product has one carbon more, so this is a step-up conversion.
- Reduction of amides: \(\mathrm{LiAlH_4}\) in ether or \(\mathrm{Na/C_2H_5OH}\) converts an amide into a 1° amine with the same number of carbon atoms.
- Gabriel phthalimide synthesis: phthalimide → potassium phthalimide (alc. KOH) → N-alkylphthalimide (R–X) → 1° amine (alkaline hydrolysis). Aromatic amines cannot be made this way, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
- Hofmann degradation: \(\mathrm{R\!-\!CO\!-\!NH_2} + \mathrm{Br_2} + 4\,\mathrm{KOH}\text{(aq)} \xrightarrow{\Delta} \mathrm{R\!-\!NH_2} + 2\,\mathrm{KBr} + \mathrm{K_2CO_3} + 2\,\mathrm{H_2O}\). The \(-\mathrm{CO}-\) group is removed, so the amine has one carbon less — a step-down reaction.
Physical properties
- The N–H bond is polar because nitrogen (3.0) and hydrogen (2.1) differ in electronegativity, so 1° and 2° amines form intermolecular hydrogen bonds, most extensively in 1° amines. Tertiary amines have no N–H hydrogen and are held only by dipole–dipole forces.
- Boiling points of isomeric amines therefore fall in the order 1° > 2° > 3°, and across families the order is alkanes < amines < alcohols < carboxylic acids.
- Lower aliphatic amines are gases with a fishy odour, middle members are liquids and higher members solids. Aniline and other arylamines are colourless liquids that darken on standing because they are easily oxidised by air.
- Lower aliphatic amines are soluble in water through hydrogen bonding; solubility falls as the hydrophobic alkyl group grows, and aromatic and higher aliphatic amines are insoluble. Solubility order: alcohols > amines > alkanes.
Basicity
- An amine is a Lewis base because its lone pair can be shared with an electron-deficient species, and a Lowry–Bronsted base because it accepts \(\mathrm{H^+}\): \(\mathrm{{\geq}N{:}} + \mathrm{H_2O} \rightleftharpoons \mathrm{{\geq}\overset{+}{N}\!-\!H} + \mathrm{OH^-}\).
- Basic strength is measured by \(K_b\) or \(\mathrm{p}K_b\); a stronger base has a larger \(K_b\) and a smaller \(\mathrm{p}K_b\). The stronger the stabilization of the conjugate acid, the stronger the base.
- The +I effect of alkyl groups predicts \(\mathrm{NH_3} < \mathrm{RNH_2} < \mathrm{R_2NH} < \mathrm{R_3N}\), but solvation of the conjugate acid falls as N–H hydrogens are replaced. The two factors oppose each other, and the observed order in water is \(\mathrm{NH_3} < \mathrm{RNH_2} < \mathrm{R_2NH} > \mathrm{R_3N}\), making secondary amines the strongest bases.
- Arylamines are much weaker bases than ammonia and aliphatic amines. Aniline is stabilized by five resonance structures, so its lone pair is delocalised into the ring and less available for protonation, while the anilinium ion has only two resonance structures.
Chemical properties
- All amines are basic: they turn red litmus blue and dissolve in aqueous HCl as substituted ammonium chlorides, from which excess NaOH regenerates the free amine.
- The orange dye test — conc. HCl, aqueous \(\mathrm{NaNO_2}\) with cooling, then \(\beta\)-naphthol in NaOH near 0 °C — confirms an aromatic primary amino group.
- Exhaustive alkylation with excess alkyl halide converts 1°, 2° and 3° amines into quaternary ammonium salts, consuming three, two and one moles of halide respectively, with \(\mathrm{NaHCO_3}\) present to neutralise the HX formed.
- Hofmann elimination: a tetraalkylammonium halide with moist \(\mathrm{Ag_2O}\) gives the quaternary ammonium hydroxide, which on strong heating undergoes \(\beta\)-elimination to give the least substituted alkene, in contrast to Saytzeff elimination.
- Acylation: 1° and 2° amines react with acetyl chloride or acetic anhydride in the presence of pyridine to give amides; the amide is less basic than the amine, and 3° amines do not react.
- Carbylamine reaction: \(\mathrm{R\!-\!NH_2} + \mathrm{CHCl_3} + 3\,\mathrm{KOH} \xrightarrow{\Delta} \mathrm{R\!-\!NC} + 3\,\mathrm{KCl} + 3\,\mathrm{H_2O}\) — a test given by primary amines only.
- With nitrous acid (made in situ from \(\mathrm{NaNO_2}\) and HCl at 273–278 K), aliphatic 1° amines give an alcohol with brisk evolution of \(\mathrm{N_2}\), while aromatic 1° amines give arene diazonium salts that are reasonably stable at 273 K and are resonance stabilized.
Reactions of arene diazonium salts
- Displacement of the diazo group: Sandmeyer reaction (CuCl/HCl, CuBr/HBr, CuCN/KCN) gives Ar–Cl, Ar–Br, Ar–CN; Gatterman reaction (Cu powder with HCl or HBr) gives Ar–Cl or Ar–Br; KI gives Ar–I; \(\mathrm{H_3PO_2}\) or ethanol gives Ar–H; water at 283 K gives Ar–OH. Nitrogen gas is released in every case.
- With fluoroboric acid, the diazonium fluoroborate precipitates and on heating gives the fluoroarene, while heating with aqueous \(\mathrm{NaNO_2}\) and copper gives the nitroarene.
- Retention of the diazo group (azo coupling): the diazonium ion acts as an electrophile towards a phenol or an aromatic amine, substituting para to the activating group, to give brightly coloured azo compounds used as dyes — p-hydroxyazobenzene (orange) with phenol and p-aminoazobenzene (yellow) with aniline.
Hinsberg’s test
- With benzenesulfonyl chloride, a 1° amine gives a sulfonamide whose remaining N–H hydrogen is strongly acidic, so it dissolves in alkali; a 2° amine gives a sulfonamide with no N–H, insoluble in alkali; a 3° amine, having no replaceable hydrogen on nitrogen, does not react.
Electrophilic aromatic substitution in aromatic amines
- The \(-\mathrm{NH_2}\) group is a powerful ring-activating and ortho–para directing group, so arylamines undergo electrophilic substitution readily.
- Bromination: aniline with bromine water at room temperature gives a white precipitate of 2,4,6-tribromoaniline. To obtain the monobromo product, the \(-\mathrm{NH_2}\) group is first protected by acetylation; bromination of acetanilide gives p-bromoacetanilide, and acid hydrolysis then restores the amino group to give p-bromoaniline.
- Nitration: direct nitration at 288 K gives p-nitroaniline (51%), m-nitroaniline (47%) and o-nitroaniline (2%), the large meta fraction arising because in acid the group is protonated to \(-\mathrm{\overset{+}{N}H_3}\), which is meta-directing and deactivating. Acetylation, nitration and hydrolysis give p-nitroaniline as the major product.
- Sulfonation: aniline with concentrated \(\mathrm{H_2SO_4}\) gives anilinium hydrogen sulfate, which on heating at 453–473 K gives p-aminobenzene sulfonic acid (sulfanilic acid). Sulfanilic acid exists as a dipolar ion or zwitter ion, formed by reaction between the acidic and basic groups of the same molecule.
13.11 Objective Questions
- The hybridisation of nitrogen in primary amine is …………
- sp
- \(\mathrm{sp^2}\)
- \(\mathrm{sp^3}\)
- \(\mathrm{sp^3d}\)
- Isobutylamine is an example of …………
- 2° amine
- 3° amine
- 1° amine
- quaternary ammonium salt
- Which one of the following compounds has the highest boiling point?
- n-Butylamine
- sec-Butylamine
- isobutylamine
- tert-Butylamine
- Which of the following has the highest basic strength?
- Trimethylamine
- Methylamine
- Ammonia
- Dimethylamine
- Which type of amine does produce \(\mathrm{N_2}\) when treated with \(\mathrm{HNO_2}\)?
- Primary amine
- Secondary amine
- Tertiary amine
- Both primary and secondary amines
- Carbylamine test is given by
- Primary amine
- Secondary amine
- Tertiary amine
- Both secondary and tertiary amines
- Which one of the following compounds does not react with acetyl chloride?
- \(\mathrm{CH_3\!-\!CH_2\!-\!NH_2}\)
- \(\mathrm{(CH_3\!-\!CH_2)_2NH}\)
- \(\mathrm{(CH_3\!-\!CH_2)_3N}\)
- \(\mathrm{C_6H_5\!-\!NH_2}\)
- Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
- Ethylamine
- Triethylamine
- Trimethylamine
- Diethylamine
- Identify ‘B’ in the following reactions: \(\mathrm{CH_3\!-\!C{\equiv}N} \xrightarrow{\mathrm{Na/C_2H_5OH}} A \xrightarrow{\mathrm{NaNO_2/dil\ HCl}} B\).
- \(\mathrm{CH_3\!-\!CH_2\!-\!NH_2}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!NO_2}\)
- \(\mathrm{CH_3\!-\!CH_2\overset{+}{N}_2Cl^-}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!OH}\)
- Which one of the following compounds contains azo linkage?
- Hydrazine
- p-Hydroxyazobenzene
- N-Nitrosodiethylamine
- Ethylenediamine
13.12 Check Your Concepts
Isopropylamine is a primary amine, while isopropyl alcohol is a secondary alcohol. Explain why the same carbon skeleton leads to two different classifications.
Classify each of the following as 1°, 2° or 3°, as simple or mixed where the label applies, and as aliphatic or aromatic: (i) \(\mathrm{C_6H_5\!-\!NH\!-\!C_6H_5}\) (ii) \(\mathrm{C_6H_5\!-\!CH_2\!-\!NH_2}\) (iii) \(\mathrm{(CH_3CH_2)_2N\!-\!CH_3}\)
Write the structures of all isomeric amines of molecular formula \(\mathrm{C_3H_9N}\) and give the IUPAC name of each.
Assertion: Ammonolysis of an alkyl halide is rarely used to prepare a pure primary amine. Reason: The primary amine formed in the first step is a stronger nucleophile than ammonia. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Why does tert-butyl bromide give isobutylene rather than tert-butylamine when treated with alcoholic ammonia?
Aryl halides do not react with the potassium salt of phthalimide. What consequence does this have for the preparation of aniline, and which alternative method would you choose instead?
Classify each of the following conversions as step-up, step-down or same-carbon, giving a reason: (i) acetamide to methylamine (ii) methyl bromide to ethylamine (iii) acetamide to ethylamine.
Arrange in increasing order of boiling point and justify your order: butan-1-amine, N,N-dimethylethanamine, N-methylpropan-1-amine.
Aniline has an \(-\mathrm{NH_2}\) group capable of hydrogen bonding, yet it is insoluble in water while ethylamine dissolves freely. Explain.
Explain, in terms of the +I effect and of solvation, why a tertiary amine is a weaker base than a secondary amine in aqueous solution but a stronger base than it in the gaseous phase.
Assertion: Aniline is a weaker base than ammonia. Reason: The anilinium ion is stabilised by more resonance structures than aniline. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Match the following:
| a. \(\mathrm{C_6H_5SO_2Cl}\) |
i. Hofmann elimination |
| b. \(\mathrm{CHCl_3}\) + alc. KOH |
ii. Sandmeyer reaction |
| c. moist \(\mathrm{Ag_2O}\) |
iii. Hinsberg’s test |
| d. \(\mathrm{CuCl/HCl}\) |
iv. Carbylamine test |
What is observed when (i) ethylamine, (ii) diethylamine and (iii) triethylamine are each warmed with chloroform and alcoholic KOH? Write the equation wherever a reaction occurs.
Predict the major organic product: \(\mathrm{CH_3CH_2CH_2\!-\!\overset{+}{N}(CH_3)_3\ I^-} \xrightarrow{\text{moist Ag}_2\mathrm{O}} ? \xrightarrow{\Delta} ?\) and state why the alkene you have chosen is the major product.
Why is the acylation of an amine carried out in the presence of pyridine? What would happen if pyridine were omitted?
An aliphatic primary amine and an aromatic primary amine are each treated with \(\mathrm{NaNO_2}\) and dilute HCl at 273–278 K. Describe what happens in each case and explain the difference in terms of the stability of the intermediate formed.
Complete and name the organic product: (i) \(\mathrm{C_6H_5\overset{+}{N}_2Cl^-} + \mathrm{KI} \rightarrow\) (ii) \(\mathrm{C_6H_5\overset{+}{N}_2Cl^-} + \mathrm{H_2O}\) at 283 K \(\rightarrow\) (iii) \(\mathrm{C_6H_5\overset{+}{N}_2Cl^-} \xrightarrow{\mathrm{CuCN/KCN}}\)
Why can fluorobenzene not be prepared by direct halogenation of benzene, and how does the diazonium route overcome the difficulty?
Why is azo coupling described as an electrophilic aromatic substitution? Identify the electrophile and explain why the ring attacked must be strongly activated.
A sulfonamide obtained from Hinsberg’s reagent dissolves in aqueous NaOH. What does this tell you about the amine used, and what makes the N–H hydrogen of that sulfonamide acidic?
Direct nitration of aniline gives a large proportion of the meta product although \(-\mathrm{NH_2}\) is an ortho–para director. Explain this apparent contradiction and outline the strategy used to obtain p-nitroaniline as the major product.
Why does aniline fail to undergo the Friedel–Crafts reaction in the presence of anhydrous aluminium chloride?
Sulfanilic acid is a solid with a very high melting point and is insoluble in common organic solvents. Explain these properties using the structure of its zwitter ion.
Arrange in decreasing order of ease of electrophilic substitution and justify: benzene, aniline, acetanilide, anilinium ion.
Write the reaction of p-toluenesulfonyl chloride with diethylamine.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Why are primary aliphatic amines stronger bases than ammonia?
Predict the product of the following reaction: Nitrobenzene \(\xrightarrow{\mathrm{Sn/Conc.\ HCl}}\) ?
Write the IUPAC name of benzylamine.
Arrange the following amines in an increasing order of boiling points: n-propylamine, ethylmethyl amine, trimethylamine.
Write the balanced chemical equation for the action of dil \(\mathrm{H_2SO_4}\) on diethylamine.
Arrange the following amines in the increasing order of their \(\mathrm{p}K_b\) values: Aniline, Cyclohexylamine, 4-Nitroaniline.
Identify A and B in the following reactions: \(\mathrm{C_6H_5CH_2Br} \xrightarrow{\text{alco. KCN}} A \xrightarrow{\mathrm{Na/ethanol}} B\)
Explain the basic nature of amines with a suitable example.
What is diazotisation? Write the diazotisation reaction of aniline.
Write the reaction to convert acetic acid into methylamine.
Write a short note on coupling reactions.
Explain Gabriel phthalimide synthesis.
Explain the carbylamine reaction with suitable examples.
Write reactions to convert (i) methanamine into ethanamine (ii) aniline into p-bromoaniline.
Complete the following reactions:
- \(\mathrm{C_6H_5\overset{+}{N}_2Cl^-} + \mathrm{C_2H_5OH} \rightarrow\)
- \(\mathrm{C_6H_5NH_2} + \mathrm{Br_2}\text{(aq)} \rightarrow\) ?
Explain ammonolysis of alkyl halides.
Write the reaction to convert ethylamine into methylamine.
Write the IUPAC names of the following amines:
- \(\mathrm{CH_3\!-\!CH_2\!-\!N(CH_3)\!-\!CH_2\!-\!CH_2\!-\!CH_3}\)
- \(\mathrm{CH_3\!-\!C(CH_3)_2\!-\!CH_2\!-\!CH_2\!-\!NH_2}\)
- \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!NH\!-\!CH_2\!-\!CH_3}\)
What are amines? How are they classified?
Write the IUPAC names of the following amines:
- \(\mathrm{H_2N\!-\!(CH_2)_6\!-\!NH_2}\)
- the dimethyl-substituted aniline shown in Fig. 13.21
- the diaminobenzene shown in Fig. 13.21
- Write reactions to prepare ethanamine from
- Acetonitrile
- Nitroethane
- Propionamide
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent.
Write reactions to bring about the following conversions:
- Aniline into p-nitroaniline
- Aniline into sulphanilic acid
13.13 Practice With Numericals
The \(\mathrm{p}K_b\) of N-ethylethanamine is 3.00 and that of ammonia is 4.75. Calculate the \(K_b\) of each and find how many times stronger a base the amine is than ammonia.
The \(K_b\) of a certain aliphatic primary amine is \(5.0\times10^{-4}\). Calculate its \(\mathrm{p}K_b\) and state, using Table 13.4, whether it is a stronger or weaker base than methanamine.
How many moles of methyl iodide are required to convert (i) 1 mole of ethanamine, (ii) 1 mole of N-methylethanamine and (iii) 1 mole of N,N-dimethylethanamine into the corresponding quaternary ammonium iodide? Write the formula of each product.
5.9 g of acetamide is warmed with bromine and excess aqueous KOH. Calculate the mass of methylamine that would be obtained if the reaction goes to completion. (Molar masses: acetamide 59 g mol\(^{-1}\), methylamine 31 g mol\(^{-1}\))
9.3 g of aniline is completely diazotised with \(\mathrm{NaNO_2}\) and HCl at 273 K, and the diazonium salt formed is then warmed with water at 283 K. Calculate (i) the mass of phenol obtained and (ii) the volume of nitrogen gas liberated at STP. (Molar masses: aniline 93 g mol\(^{-1}\), phenol 94 g mol\(^{-1}\); molar volume at STP = 22.4 L mol\(^{-1}\))
Direct nitration of aniline at 288 K gives p-nitroaniline (51%), m-nitroaniline (47%) and o-nitroaniline (2%). If 18.6 g of aniline is nitrated and is completely converted into this mixture, calculate the mass of each isomer formed. (Molar masses: aniline 93 g mol\(^{-1}\), nitroaniline 138 g mol\(^{-1}\))
0.1 mole of propan-1-amine is warmed with chloroform and potassium hydroxide in the carbylamine reaction. Calculate the number of moles of KOH consumed and the mass of KCl formed. (Molar mass of KCl = 74.5 g mol\(^{-1}\))
An amine of molar mass 59 g mol\(^{-1}\) contains 23.7% nitrogen by mass. Determine the number of nitrogen atoms per molecule, write the molecular formula of the amine, and draw all its isomeric structures. (Atomic mass of N = 14)
4.65 g of aniline is treated with an excess of bromine water. Calculate the mass of 2,4,6-tribromoaniline precipitated and the number of moles of HBr released. (Molar masses: aniline 93 g mol\(^{-1}\), 2,4,6-tribromoaniline 330 g mol\(^{-1}\))
Ethanenitrile is reduced with \(\mathrm{Na/C_2H_5OH}\). Starting from 8.2 g of ethanenitrile, calculate the mass of the amine formed, assuming complete conversion, and state whether the conversion is step-up or step-down. (Molar masses: ethanenitrile 41 g mol\(^{-1}\), ethanamine 45 g mol\(^{-1}\))
How many moles of methylbromide are required to convert ethanamine to N,N-dimethyl ethanamine?