Properties of Amines

ChemistryAminesFor JEE aspirants

Every reaction in the properties of amines starts at the nitrogen lone pair. It takes a proton (basicity), attacks carbon (alkylation, acylation, reactions with nitrous acid) and pushes electrons into a benzene ring (ring substitution in aniline). This page covers the physical and chemical properties of amines: basic strength in water and in the gas phase, reactions with acids, acyl halides and $HN O_{2}$ , Hofmann elimination and the ring reactions asked in JEE Main and NEET.

Key Formulas - Quick Reference

Basicity: $RN H_{2} + H_{2} O RN H_{3}^{+} + O H^{-}$ ; smaller $pKb$ means a stronger base
In water: $(C H_{3})_{2} NH$ 3.27 > $C H_{3} N H_{2}$ 3.38 > $(C H_{3})_{3} N$ 4.22 > $N H_{3}$ 4.75 ( $pKb$ )
Without solvent (gas phase): 3° > 2° > 1° > $N H_{3}$ , from the $+ I$ effect alone
Aromatic: aniline $pKb$ 9.38, benzylamine 4.70, cyclohexylamine 3.32; $- N O_{2}$ weakens, $- OC H_{3}$ strengthens
Acylation: $RN H_{2} + C H_{3} COCl C H_{3} CONHR + HCl$ ; 3° amines have no N-H and do not react
Nitrous acid: 1° aliphatic gives $N_{2}$ + alcohol; 1° aromatic gives $Ar N_{2}^{+} C l^{-}$ ; 2° gives $R_{2} N - N = O$
Carbylamine (1° only): $RN H_{2} + CHC l_{3} + 3KOH RNC + 3KCl + 3 H_{2} O$
Hofmann elimination: $R_{4} N^{+} O H^{-} heat 3° amine + less substituted alkene + H_{2} O$
Aniline + $3B r_{2}$ (water) gives 2,4,6-tribromoaniline; acetylation first gives only the para product
Oxidation: 3° amine + $H_{2} O_{2}$ gives the amine oxide $R_{3} N^{+} - O^{-}$ , which gives a Cope elimination on heating

1. Physical Properties

Lower aliphatic amines are gases with a fishy smell; higher ones are liquids or solids. Methylamine and ethylamine smell of ammonia.
Pure aryl amines are colourless, but they darken on standing because they are oxidised by air.
Boiling points: 1° and 2° amines form intermolecular hydrogen bonds through their N-H bonds, so for isomeric amines the order is 1° > 2° > 3°. Amines boil below alcohols of similar mass, because N is less electronegative than O.
Solubility: lower amines dissolve in water (they hydrogen-bond to it), and solubility falls as the alkyl group grows. All amines dissolve in organic solvents such as alcohol, ether and benzene.

Figure 1: For isomeric amines the boiling point order is

1^{\circ} > 2^{\circ} > 3^{\circ}

, because only N-H bonds can donate a hydrogen bond.

2. Basic Character of Amines

2.1 Amines as Bases

The lone pair on nitrogen can accept a proton, so an amine is a Bronsted base and a Lewis base. In water:

RN H_{2} + H_{2} O ⇌ RN H_{3}^{+} + O H^{-}

K_{b} = \frac{[ RN H _{3}^{+} ] [ O H ^{-} ]}{[ RN H _{2} ]}, p K_{b} = - lo g K_{b}

A larger $K_{b}$ , and so a smaller $p K_{b}$ , means a stronger base. Anything that makes the lone pair more available, or that stabilises the cation formed, increases basic strength.

2.2 Aliphatic Amines: Two Effects Pull Opposite Ways

The inductive effect. Alkyl groups release electrons ( $+ I$ ), so they raise the electron density on nitrogen. On this count alone the order should be 3° > 2° > 1° > $N H_{3}$ , and that is exactly what is found in the gas phase or in a non-polar solvent such as chlorobenzene, where the butylamines follow $BuN H_{2}$ < $B u_{2} NH$ < $B u_{3} N$ .

Solvation of the cation. In water the ammonium ion formed is stabilised by hydrogen bonds. The more N-H bonds the cation has, the more water molecules hold it, and the more stable it is. This effect favours the primary amine.

Figure 2: Basicity in water is a tug of war between the

+ I

effect (favours

3^{\circ}

) and solvation of the cation (favours

1^{\circ}

); the

2^{\circ}

amine wins both ways.

The two effects together explain the odd orders measured in water:

Amine	$pKb$ (water)	Amine	$pKb$ (water)
$C H_{3} N H_{2}$	3.38	$C_{2} H_{5} N H_{2}$	3.29
$(C H_{3})_{2} NH$	3.27	$(C_{2} H_{5})_{2} NH$	3.00
$(C H_{3})_{3} N$	4.22	$(C_{2} H_{5})_{3} N$	3.25
$N H_{3}$	4.75	$C_{6} H_{5} C H_{2} N H_{2}$	4.70

Exam Trick

In water the 2° amine always wins. For methyl groups the order is 2° > 1° > 3° > $N H_{3}$ ; for ethyl and larger groups it is 2° > 3° > 1° > $N H_{3}$ . Take away the solvent (gas phase or chlorobenzene) and the order becomes the simple inductive one, 3° > 2° > 1° > $N H_{3}$ .

2.3 Aromatic Amines Are Much Weaker Bases

In aniline the nitrogen lone pair is delocalised into the ring, so it is not fully available to a proton. Protonation would destroy that resonance stabilisation, which costs energy. Aniline ( $pKb$ 9.38) is therefore about a million times weaker a base than cyclohexylamine ( $pKb$ 3.32), and even weaker than ammonia.

Figure 3: The lone pair that makes an amine basic is tied up in the ring, so aryl amines are weak bases; benzylamine, whose N is off the ring, stays strong.

Every extra phenyl group weakens the base further: diphenylamine ( $pKb$ 13.21) is very weak and triphenylamine is not basic at all.
Benzylamine is not an aryl amine: its nitrogen sits on a $C H_{2}$ group, so it behaves like an aliphatic amine ( $pKb$ 4.70).
The nitrogen of aniline is bonded to an $sp^{2}$ carbon, which is more electronegative than an $sp^{3}$ carbon and also pulls electron density away.

2.4 Substituents on the Ring

Electron-withdrawing groups make an aryl amine weaker; electron-releasing groups make it stronger. The values below are $p K_{a}$ of the anilinium ion, so a larger number means a stronger base.

Substituent	ortho	meta	para
none (aniline)	4.62	4.62	4.62
$- N O_{2}$	−0.28	2.45	0.95
$- OH$	4.72	4.17	5.30
$- OC H_{3}$	4.49	4.20	5.29
$- C H_{3}$	4.38	4.67	5.10

$- N O_{2}$ withdraws electrons by both $- I$ and $- R$ ; from the ortho and para positions its $- R$ effect reaches the $N H_{2}$ group directly, so those isomers are the weakest bases. o-Nitroaniline is so weak that its salts are hydrolysed in water, and 2,4,6-trinitroaniline (picramide) behaves like an amide.
$- OH$ and $- OC H_{3}$ release electrons by resonance from the ortho and para positions but withdraw by induction from the meta position, so the meta isomers are weaker bases than aniline itself.
Whatever the group, the ortho isomer is always weaker than expected. This ortho effect mixes steric crowding, hydrogen bonding and direct interaction with the $N H_{2}$ group.

JEE Advanced

Steric inhibition of resonance. In 2,6-dinitro-N,N-dimethylaniline the two bulky $- N O_{2}$ groups force the $- N (C H_{3})_{2}$ group to twist out of the plane of the ring. The lone pair can no longer overlap with the ring $π$ system, so it stays on nitrogen and the compound is a strong base, unlike other aryl amines. The same twist explains why N,N-dimethyl-2,6-disubstituted anilines are more basic than aniline.

Figure 4: Check resonance first, then the solvent, then the size of the alkyl groups.

3. Reaction with Acids: Salt Formation

Amines neutralise acids to give crystalline salts in which nitrogen is quadricovalent and unielectrovalent. The salts dissolve in water but not in ether, and a stronger base such as NaOH sets the amine free again. This is how an amine is separated from a neutral organic mixture.

C H_{3} N H_{2} + HCl C H_{3} N H_{3}^{+} C l^{-}

C H_{3} N H_{3}^{+} C l^{-} + NaOH C H_{3} N H_{2} + NaCl + H_{2} O

C_{6} H_{5} NH (C H_{3}) + H_{2} S O_{4} [C_{6} H_{5} N H_{2} (C H_{3})]^{+} HS O_{4}^{-}

Because the lone pair can also be donated to a metal ion, amines act as ligands: ethylamine dissolves in $CuS O_{4}$ and in $AgN O_{3}$ solution by forming complexes such as $[Cu (C H_{3} C H_{2} N H_{2})_{4}]^{2 +}$ .

4. Alkylation

Amines are nucleophiles, so they attack alkyl halides and move up one class each time: 1° gives 2°, then 3°, then the quaternary salt. This is the ammonolysis reaction used to prepare amines (see Preparation of Amines), and it is also the first step of exhaustive methylation before a Hofmann elimination.

RN H_{2} + R^{'} X RNH R^{'} + HX

5. Acylation and Benzoylation

Primary and secondary amines have an N-H bond, so they react with acid chlorides, anhydrides and esters to give substituted amides. The reaction is run with a base such as pyridine or NaOH to remove the HCl formed. Tertiary amines have no N-H and do not give amides.

C_{6} H_{5} N H_{2} + C H_{3} COCl pyridine acetanilide C_{6} H_{5} NHCOC H_{3} + HCl

C_{6} H_{5} N H_{2} + (C H_{3} CO)_{2} O C_{6} H_{5} NHCOC H_{3} + C H_{3} COOH

(C_{2} H_{5})_{2} NH + (C H_{3} CO)_{2} O C H_{3} CON (C_{2} H_{5})_{2} + C H_{3} COOH

Benzoylation with benzoyl chloride in aqueous NaOH is the Schotten-Baumann reaction:

C_{6} H_{5} N H_{2} + C_{6} H_{5} COCl + NaOH N-phenylbenzamide C_{6} H_{5} NHCO C_{6} H_{5} + NaCl + H_{2} O

Amides are far less basic than amines, because the lone pair is delocalised onto the C=O group. This is why acylation is used to protect the

N H_{2}

group of aniline before halogenation or nitration (Section 9), and why the resulting amide is only moderately activating and directs ortho and para.

6. Reaction with Other Electrophiles

(i) Aldehydes and ketones. A primary amine condenses with a carbonyl compound to give an imine, also called a Schiff base. A secondary amine gives an enamine instead, because there is no second N-H to lose.

C_{6} H_{5} CHO + C_{6} H_{5} N H_{2} benzylidene aniline C_{6} H_{5} CH = N C_{6} H_{5} + H_{2} O

(ii) Phosgene gives a symmetrically disubstituted urea:

2RN H_{2} + COC l_{2} RNHCONHR + 2HCl

(iii) Isocyanates and isothiocyanates give unsymmetrical ureas and thioureas:

R^{'} NCO + RN H_{2} a substituted urea R^{'} NHCONHR

R^{'} NCS + RN H_{2} a thiourea R^{'} NHCSNHR

(iv) Carbon disulphide. A primary amine gives a dithiocarbamic acid, which $HgC l_{2}$ converts to an isothiocyanate with the smell of mustard oil. This is the mustard oil test for primary amines (see Analysis of Amines).

RN H_{2} + C S_{2} RNHCSSH

RNHCSSH + HgC l_{2} RNCS + HgS + 2HCl

7. Reaction with Nitrous Acid

Nitrous acid is made in the flask from $NaN O_{2}$ and a mineral acid at 273-278 K. Every class of amine behaves differently, which makes this the most useful single test in amine chemistry (Figure 5).

RN H_{2} + HN O_{2} 1° aliphatic: nitrogen is evolved ROH + N_{2} + H_{2} O

C_{6} H_{5} N H_{2} + HN O_{2} + HCl 273-278 K 1° aromatic: diazonium salt C_{6} H_{5} N_{2}^{+} C l^{-} + 2 H_{2} O

(C H_{3})_{2} NH + HN O_{2} 2°: yellow oily N-nitrosamine (C H_{3})_{2} N - N = O + H_{2} O

(C H_{3})_{3} N + HN O_{2} 3° aliphatic: a soluble salt [(C H_{3})_{3} NH]^{+} N O_{2}^{-}

C_{6} H_{5} N (C H_{3})_{2} + HN O_{2} 281 K 3° aromatic: green p-nitroso compound ON C_{6} H_{4} N (C H_{3})_{2} + H_{2} O

Figure 5: Nitrous acid is the classic class test:

N_{2}

gas (

1^{\circ}

aliphatic), a stable salt (

1^{\circ}

aromatic), a yellow oil (

2^{\circ}

) or ring attack (

3^{\circ}

aromatic).

The diazonium ion from a primary aliphatic amine loses nitrogen at once and leaves a carbocation, which gives alcohols, alkenes and halides, often after a hydride shift. The reaction is useless for synthesis, but the nitrogen given off can be measured to estimate amino groups (Van Slyke method). Diazonium salts of aromatic amines are stable in the cold and are the starting point of a whole family of syntheses, covered in Diazonium Salt.

Exam Trick

Watch what you see. Brisk effervescence of a colourless gas means a 1° aliphatic amine; a clear solution that couples to a dye means a 1° aromatic amine; a yellow oil means a 2° amine; a green solid means a 3° aromatic amine; and nothing visible means a 3° aliphatic amine.

8. Quaternary Ammonium Salts and Hofmann Elimination

Exhaustive alkylation of any amine ends at a quaternary ammonium salt. Moist silver oxide exchanges the halide for hydroxide, giving a quaternary ammonium hydroxide, which is as strong a base as NaOH:

2 R_{4} N^{+} X^{-} + A g_{2} O + H_{2} O 2 R_{4} N^{+} O H^{-} + 2AgX

On heating, the hydroxide ion removes a $β$ -hydrogen and the amine leaves: an E2 elimination, and but-1-ene is the major product. The alkene formed is the less substituted one, the opposite of the Saytzeff rule. This is Hofmann's rule, and the reason is that the bulky $- N (C H_{3})_{3}^{+}$ group makes the base attack the least hindered $β$ -carbon, whose hydrogens are also the most acidic.

[(C H_{3})_{3} NCH (C H_{3}) C H_{2} C H_{3}]^{+} O H^{-} heat (C H_{3})_{3} N + C H_{2} = CHC H_{2} C H_{3} + H_{2} O

When the groups on nitrogen differ, the one with the largest number of $β$ -hydrogens gives the major alkene. A quaternary hydroxide with no $β$ -hydrogen cannot eliminate; tetramethylammonium hydroxide gives methanol instead.

(C H_{3})_{4} N^{+} O H^{-} heat (C H_{3})_{3} N + C H_{3} OH

Figure 6: Both eliminations give the less substituted alkene, but Hofmann uses an external base and anti-periplanar geometry, while Cope uses the oxide's own oxygen and is syn.

Exam Trick

Hofmann gives the ugly alkene. Elimination from a quaternary ammonium hydroxide (and the Cope elimination of an amine oxide) gives the least substituted alkene, while dehydrohalogenation of an alkyl halide with alcoholic KOH follows Saytzeff and gives the most substituted one. Exhaustive methylation followed by elimination, repeated until no nitrogen is left, is the classic way of finding the carbon skeleton of an unknown amine.

9. Ring Reactions of Aromatic Amines

The groups $- N H_{2}$ , $- NHR$ and $- N R_{2}$ release electrons into the ring by resonance, so they are strongly activating and direct ortho and para. The ring becomes so reactive that substitution is hard to stop at one group (Figure 7).

9.1 Halogenation

C_{6} H_{5} N H_{2} + 3B r_{2} bromine water 2,4,6-tribromoaniline, white ppt C_{6} H_{2} B r_{3} N H_{2} + 3HBr

To introduce a single halogen, the amino group is first acetylated. Acetanilide is only moderately activating, so bromination gives mainly the para product, and hydrolysis returns the amine:

C_{6} H_{5} N H_{2} (C H_{3} CO)_{2} O C_{6} H_{5} NHCOC H_{3} B r_{2} / C H_{3} COOH Br C_{6} H_{4} NHCOC H_{3} H_{3} O^{+} Br C_{6} H_{4} N H_{2}

Figure 7: To stop at one substituent, protect

- N H_{2}

as the amide first; the acetyl group is removed again by hydrolysis.

9.2 Nitration

Direct nitration is messy. In the strongly acidic mixture most of the aniline is present as the anilinium ion, which is meta-directing, so the product contains about 51% para, 47% meta and 2% ortho nitroaniline, along with oxidation products. Nitration of acetanilide instead gives mainly p-nitroacetanilide, and hydrolysis then gives p-nitroaniline.

C_{6} H_{5} NHCOC H_{3} HN O_{3} / H_{2} S O_{4} N O_{2} C_{6} H_{4} NHCOC H_{3} O H^{-}, heat N O_{2} C_{6} H_{4} N H_{2}

9.3 Sulphonation

Aniline and conc. $H_{2} S O_{4}$ first give anilinium hydrogensulphate, which on heating at 453-473 K rearranges to sulphanilic acid. Sulphanilic acid exists as an internal salt (a zwitterion): the $- S O_{3} H$ group is a strong enough acid to protonate the weakly basic $- N H_{2}$ group.

C_{6} H_{5} N H_{2} + H_{2} S O_{4} anilinium hydrogensulphate C_{6} H_{5} N H_{3}^{+} HS O_{4}^{-}

C_{6} H_{5} N H_{3}^{+} HS O_{4}^{-} 453-473 K sulphanilic acid (zwitterion) H_{3} N^{+} C_{6} H_{4} S O_{3}^{-} + H_{2} O

The zwitterion explains the high melting point of sulphanilic acid, its insolubility in organic solvents and in dilute HCl, and its solubility in aqueous NaOH.
p-Aminobenzoic acid does not form a zwitterion: $- COOH$ is too weak an acid to protonate the weakly basic aryl $- N H_{2}$ group. Glycine does, because its aliphatic $- N H_{2}$ group is basic enough.

Figure 8: Strong acid removes the lone pair: the anilinium ion directs meta, and in sulphanilic acid the

- S O_{3} H

group protonates the ring's own

- N H_{2}

9.4 Friedel-Crafts Reactions Fail

Aniline does not undergo Friedel-Crafts alkylation or acylation. Its lone pair forms a salt with the Lewis acid, $C_{6} H_{5} N H_{2}$ + $AlC l_{3}$ giving the salt $C_{6} H_{5} N H_{2} \cdot AlC l_{3}$ , and the positive nitrogen then deactivates the ring strongly and directs meta. Diazotisation, the most useful reaction of aniline, is covered in Diazonium Salt.

10. Rearrangements of N-Substituted Anilines

A group attached to the nitrogen of an aniline can migrate to the ring, usually to the para position, when the compound is treated with acid. The pattern is the same in each case: the N-substituted aniline is protonated, the N-X bond breaks, and X returns at the para position.

Name	Starting compound	Reagent	Product
Fischer-Hepp	N-nitroso-N-methylaniline $C_{6} H_{5} N (NO) C H_{3}$	HCl	p-nitroso-N-methylaniline
Bamberger	phenylhydroxylamine $C_{6} H_{5} NHOH$	dil. $H_{2} S O_{4}$	4-aminophenol
Orton	N-chloroacetanilide $C_{6} H_{5} N (Cl) COC H_{3}$	HCl	p-chloroacetanilide
Sulphamic acid	phenylsulphamic acid $C_{6} H_{5} NHS O_{3} H$	heat	sulphanilic acid

C_{6} H_{5} N (NO) C H_{3} HCl Fischer-Hepp ON C_{6} H_{4} NHC H_{3}

C_{6} H_{5} NHOH dil. H_{2} S O_{4} Bamberger HO C_{6} H_{4} N H_{2}

C_{6} H_{5} NHS O_{3} H heat H_{2} N C_{6} H_{4} S O_{3} H

11. Oxidation of Amines and the Cope Elimination

Amines are easily oxidised, and what forms depends on the class:

1° amines with $H_{2} O_{2}$ give the hydroxylamine, which is oxidised further to a nitroso compound and finally to an oxime. A primary amine on a tertiary carbon gives a nitro compound with $KMn O_{4}$ .
2° amines give N,N-dialkylhydroxylamines.
3° amines are oxidised cleanly to amine oxides by $H_{2} O_{2}$ or a peroxy acid.
Aniline is oxidised by $K_{2} C r_{2} O_{7}$ / $H_{2} S O_{4}$ to p-benzoquinone, and slow aerial oxidation gives the coloured products that darken a sample on standing. With dichromate on cotton it gives the dye aniline black.

RC H_{2} N H_{2} H_{2} O_{2} RC H_{2} NHOH H_{2} O_{2}, - H_{2} O RCH = NOH

R_{2} NH + H_{2} O_{2} R_{2} NOH + H_{2} O

R_{3} N + H_{2} O_{2} a tertiary amine oxide R_{3} N^{+} - O^{-} + H_{2} O

Heating an amine oxide above about 425 K gives an alkene: the Cope elimination. The oxygen of the oxide takes a $β$ -hydrogen from the same side of the molecule through a five-membered cyclic transition state, so the elimination is syn, unlike the anti-periplanar E2 of the Hofmann elimination (Figure 6). As in the Hofmann elimination, the less substituted alkene is the major product.

(C H_{3})_{2} N (O) C H_{2} C H_{2} R 425 K RCH = C H_{2} + (C H_{3})_{2} NOH

Figure 9: Almost every property traces back to the nitrogen lone pair: where it is, and whether it is free.

12. Solved Examples

Solved Example 1

4-Methylbutanolide (a

γ

-lactone) is treated with methylamine. The product is
(A) a cyclic imine
(B) a cyclic hemiaminal
(C) 4-hydroxy-N-methylpentanamide
(D) no reaction

Solution:

Answer: (C). A lactone is an ester, and an amine is a better nucleophile than an alcohol. Methylamine attacks the C=O carbon, the tetrahedral intermediate collapses, and the ring C-O bond breaks. The product is the open-chain hydroxy amide $C H_{3} NHCO - C H_{2} C H_{2} - CH (OH) C H_{3}$ . Amides are far more stable than esters, so the ring does not close again.

Solved Example 2

Piperidine is treated with excess

C H_{3} I

, then with moist

A g_{2} O

, and the product is heated; the sequence is repeated. The final organic product is
(A) penta-1,3-diene
(B) penta-1,4-diene
(C) N,N-dimethylpiperidinium iodide
(D) cyclohexene

Solution:

Answer: (A). Piperidine is a cyclic 2° amine. Exhaustive methylation and Hofmann elimination open the ring to an unsaturated amine; a second round of methylation and elimination removes nitrogen completely as trimethylamine and leaves a pentadiene, which is isolated as the conjugated penta-1,3-diene. Option (C) is only the first intermediate.

C_{5} H_{10} NH 2 C H_{3} I C_{5} H_{10} N (C H_{3})_{2}^{+} I^{-}

Two rounds of $A g_{2} O$ / $H_{2} O$ and heating then give penta-1,3-diene and $(C H_{3})_{3} N$ .

Solved Example 3

The sequence that is NOT correct is
(A) dimethylamine > methylamine > trimethylamine in water
(B) diethylamine > triethylamine > ethylamine in water
(C) methylamine > pyridine > aniline
(D) aniline > pyrrole > pyridine

Solution:

Answer: (D). The correct order is pyridine > aniline > pyrrole. Pyridine ( $pKb$ 8.8) has its lone pair in an $sp^{2}$ orbital in the ring plane, not in the $π$ system, so it is still basic; aniline ( $pKb$ 9.38) shares its lone pair with the ring; and in pyrrole the lone pair is part of the aromatic sextet, so protonation destroys aromaticity and the compound is practically non-basic. (A), (B) and (C) match the measured $pKb$ values.

Solved Example 4

An aqueous solution of ethylamine gives a reddish-brown precipitate with ferric chloride. Explain.

Solution:

Ethylamine is a base, so in water it produces an appreciable concentration of $O H^{-}$ ions. These precipitate $F e^{3 +}$ as ferric hydroxide.

C_{2} H_{5} N H_{2} + H_{2} O ⇌ C_{2} H_{5} N H_{3}^{+} + O H^{-}

F e^{3 +} + 3O H^{-} Fe (OH)_{3}

Solved Example 5

p-Methoxyaniline is a stronger base than aniline, but p-nitroaniline is weaker. Why?

Solution:

The $- OC H_{3}$ group releases electrons by resonance from the para position, so it raises the electron density on nitrogen and makes the lone pair more available: $pKa$ of the conjugate acid rises from 4.62 to 5.29. The $- N O_{2}$ group withdraws electrons by both $- I$ and $- R$ ; from the para position its $- R$ effect drains the nitrogen lone pair into the ring, so the base is much weaker ( $pKa$ 0.95).

Solved Example 6

Which is the better nucleophile, aniline or the anilinium ion, and why?

Solution:

Aniline. In the anilinium ion $C_{6} H_{5} N H_{3}^{+}$ the nitrogen has used its lone pair to hold a proton and carries a positive charge, so it has no pair to donate. Aniline still has its lone pair, even though it is partly delocalised into the ring. This is why electrophilic substitution of aniline is carried out in a medium that is not strongly acidic.

Solved Example 7

Which of these will NOT give an amide with ethanoyl chloride?
(A) piperidine
(B) quinuclidine (a bicyclic 3° amine)
(C)

RC H_{2} N H_{2}

(D) none of these

Solution:

Answer: (B). Acylation replaces a hydrogen on nitrogen. A 3° amine such as quinuclidine has no N-H, so it cannot form an amide; it only forms an unstable acylammonium salt. Piperidine (2°) and any 1° amine give amides.

Solved Example 8

The action of nitrous acid on ethylamine gives mainly
(A) ethyl alcohol
(B) acetamide
(C) ethane
(D) ethanoic acid

Solution:

Answer: (A). A 1° aliphatic amine forms a diazonium ion that loses $N_{2}$ at once. The carbocation left behind reacts mostly with water, giving the alcohol, along with some alkene and halide.

C_{2} H_{5} N H_{2} + HN O_{2} C_{2} H_{5} OH + N_{2} + H_{2} O

Solved Example 9

Why is a Lewis acid not used when an electrophilic substitution is carried out on aniline?

Solution:

Aniline is a base, so its lone pair attacks the Lewis acid and forms a coordinate bond, as in $C_{6} H_{5} N H_{2} \cdot AlC l_{3}$ . The nitrogen then carries a positive charge, which strongly deactivates the ring and directs meta, so the intended reaction fails. This is why aniline cannot be used in the Friedel-Crafts reaction.

Solved Example 10

During the nitration of benzene with a mixture of conc.

HN O_{3}

and conc.

H_{2} S O_{4}

, nitric acid acts as a base. Explain.

Solution:

Both are strong acids, but $H_{2} S O_{4}$ is the stronger one, so it gives up a proton and $HN O_{3}$ accepts it. The protonated nitric acid then loses water to give the nitronium ion, the actual electrophile.

HN O_{3} + 2 H_{2} S O_{4} ⇌ N O_{2}^{+} + H_{3} O^{+} + 2HS O_{4}^{-}

Solved Example 11

Which of these reactions involves migration of an alkyl or aryl group?
(A) Baeyer-Villiger oxidation
(B) Hofmann bromamide reaction
(C) Beckmann rearrangement
(D) all of these

Solution:

Answer: (D). In the Baeyer-Villiger oxidation a group migrates from carbon to oxygen, in the Hofmann bromamide reaction from carbon to nitrogen, and in the Beckmann rearrangement the group anti to the oxime $- OH$ migrates from carbon to nitrogen. In each case the group keeps its configuration.

Solved Example 12

Which reaction is described as passing through a nitrene intermediate?
(A) Baeyer-Villiger oxidation
(B) Hofmann elimination
(C) Hofmann bromamide rearrangement
(D) carbylamine reaction

Solution:

Answer: (C). Older texts describe an acyl nitrene in the Hofmann bromamide reaction, although the migration and the loss of $B r^{-}$ are now known to be concerted. The carbylamine reaction goes through a carbene, dichlorocarbene, not a nitrene.

Solved Example 13

Why does aniline acquire a pink to brown colour on standing?

Solution:

The ring of aniline is very electron rich, so oxygen of the air slowly oxidises it. The coloured oxidation products build up even in a stoppered bottle, so aniline is stored in dark bottles and is distilled before use.

Solved Example 14

Arrange ethylamine, 2-aminoethanol and 3-amino-1-propanol in order of decreasing basicity.

Solution:

$C H_{3} C H_{2} N H_{2}$ > $HO (C H_{2})_{3} N H_{2}$ > $HO (C H_{2})_{2} N H_{2}$ .

The $- OH$ group withdraws electrons by induction, which lowers the electron density on nitrogen. The effect weakens with distance, so it is strongest in 2-aminoethanol, where $- OH$ is on the carbon next to the $- N H_{2}$ group, and is absent in ethylamine.

Solved Example 15

Suggest a structure for a chiral compound (A),

C_{8} H_{11} N

, that dissolves in dilute HCl and releases

N_{2}

with

HN O_{2}

Solution:

$C_{8} H_{11} N$ has 4 degrees of unsaturation, which is one benzene ring (a saturated amine of 8 carbons would be $C_{8} H_{19} N$ ). Dissolving in HCl shows it is an amine, and evolving $N_{2}$ with nitrous acid shows it is a primary aliphatic amine. That leaves two carbons and $N H_{2}$ to build a stereocentre, so (A) is 1-phenylethylamine, $C_{6} H_{5} CH (N H_{2}) C H_{3}$ .

Solved Example 16

An optically active amine (A) is exhaustively methylated and then subjected to Hofmann elimination, giving an alkene (B). Ozonolysis of (B) gives equimolar amounts of methanal and butanal. Deduce (A) and (B), and give a structural isomer of (A).

Solution:

Ozonolysis cleaves C=C into two carbonyl compounds, so joining $HCHO$ and $C H_{3} C H_{2} C H_{2} CHO$ at their carbonyl carbons gives $C H_{2} = CHC H_{2} C H_{2} C H_{3}$ , pent-1-ene: that is (B).

Hofmann elimination gives the less substituted alkene, and (A) is optically active, so the nitrogen must be on C-2: (A) is pentan-2-amine, $C H_{3} CH (N H_{2}) C H_{2} C H_{2} C H_{3}$ , whose C-2 carries four different groups.

A structural isomer is pentan-1-amine, $C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} N H_{2}$ , which is not optically active.

Solved Example 17

C H_{3} C H_{2} N H_{2}

is soluble in
(A) dilute HCl
(B)

CuS O_{4}

solution
(C)

AgN O_{3}

solution
(D) all of these

Solution:

Answer: (D). With dilute HCl it forms the soluble salt $C H_{3} C H_{2} N H_{3}^{+} C l^{-}$ . With $C u^{2 +}$ and $A g^{+}$ it acts as a ligand: the lone pair forms coordinate bonds and soluble complex ions are produced.

Solved Example 18

Which amine forms an N-nitroso derivative with

NaN O_{2}

and HCl?
(A) 4-methylaniline
(B) aniline
(C) N,N-dimethylaniline
(D) N-methylaniline

Solution:

Answer: (D). N-Nitrosation needs an N-H on a secondary nitrogen. N-Methylaniline is a 2° amine and gives the yellow N-nitroso compound. (A) and (B) are 1° aromatic amines and give diazonium salts; (C) is a 3° aromatic amine and is nitrosated on the ring instead.

C_{6} H_{5} NH (C H_{3}) + HN O_{2} C_{6} H_{5} N (NO) C H_{3} + H_{2} O

Solved Example 19

Identify
(A) in this oxidation:
(A) with

K_{2} C r_{2} O_{7}

H_{2} S O_{4}

gives p-benzoquinone.
(A) benzene-1,4-diamine
(B) 4-aminophenol
(C) benzene-1,4-diol
(D) all of these

Solution:

Answer: (D). Any para-disubstituted benzene carrying two groups that can be oxidised to C=O, whether $- N H_{2}$ or $- OH$ , gives p-benzoquinone. Aniline itself also gives it, through 4-aminophenol.

Solved Example 20

The strongest base among the following is
(A) guanidine,

(H_{2} N)_{2} C = NH

(B) urea,

(H_{2} N)_{2} C = O

Solution:

Answer: (A). Guanidine takes a proton to give the guanidinium ion, in which the positive charge is shared equally by three nitrogen atoms in a symmetrical resonance hybrid. That stabilisation makes guanidine one of the strongest neutral organic bases ( $pKb$ about 0.4). In urea and acetamide the lone pair is pulled away by the C=O group, and aniline loses its pair to the ring.

Practice Questions

Why do tertiary amines not undergo acylation?Answer: they have no hydrogen on nitrogen to be replaced by the acyl group
Give the order of basic strength: (a) $C H_{3} N H_{2}$ , $(C H_{3})_{2} NH$ , $(C H_{3})_{3} N$ in an aprotic solvent; (b) the ethyl series in an aprotic solvent; (c) the methyl series in water; (d) the ethyl series in water.Answer: (a) and (b): 3° > 2° > 1° (inductive effect only); (c) $(C H_{3})_{2} NH$ > $C H_{3} N H_{2}$ > $(C H_{3})_{3} N$ ; (d) $(C_{2} H_{5})_{2} NH$ > $(C_{2} H_{5})_{3} N$ > $C_{2} H_{5} N H_{2}$
Explain the mechanism: 1-(aminomethyl)cyclopentan-1-ol with $HN O_{2}$ gives cyclohexanone.Answer: $HN O_{2}$ makes the diazonium ion; loss of $N_{2}$ gives a 1° carbocation, and a ring C-C bond migrates to it (ring expansion). The resulting cyclohexyl cation carries $- OH$ , which loses $H^{+}$ to give cyclohexanone (Tiffeneau-Demjanov rearrangement)
Arrange in decreasing order of basic strength: $RN H_{2}$ , $RCON H_{2}$ and succinimide.Answer: $RN H_{2}$ > $RCON H_{2}$ > succinimide; the lone pair is in an $sp^{3}$ orbital in the amine, delocalised onto one C=O in the amide and onto two C=O groups in the imide
Which is the stronger acid, o-chlorophenol or o-fluorophenol, and why?Answer: o-chlorophenol; chlorine is less electronegative but its larger, more polarisable orbitals delocalise the negative charge of the phenoxide better, and intramolecular hydrogen bonding in o-fluorophenol stabilises the acid rather than its anion
Sulphanilic acid is insoluble in organic solvents. Explain.Answer: it exists as a zwitterion, $H_{3} N^{+} C_{6} H_{4} S O_{3}^{-}$ , so it is effectively an ionic salt; it has a high melting point, dissolves in aqueous NaOH and does not dissolve in ether or benzene
Unlike other aromatic amines, 2,6-dinitro-N,N-dimethylaniline is strongly basic. Why?Answer: the two bulky ortho $- N O_{2}$ groups twist the $- N (C H_{3})_{2}$ group out of the ring plane, so the lone pair cannot overlap with the ring (steric inhibition of resonance) and stays available
In $HO - C H_{2} C H_{2} C H_{2} - N H_{2}$ , which site acts as an acid and which as a base?Answer: the $- OH$ group is the acid (O-H is more acidic than N-H) and the $- N H_{2}$ group is the base (it is more basic than $- OH$ )
3-Chloroaniline is treated with $NaN H_{2}$ to give B, and B with $NaOC H_{3}$ gives C. Identify B and C.Answer: B = a benzyne (dehydrobenzene) intermediate from elimination of HCl; C = 3-methoxyaniline, formed by addition of methoxide to the benzyne
One mole of a bromo compound (A) with $N H_{3}$ gives (B); (B) with $C H_{3} I$ gives (C); (B) and (C) with $HN O_{2}$ give (D) and (E); (D) on oxidation and decarboxylation gives 2-methoxy-2-methylpropane. Identify (A) to (E).Answer: A = $(C H_{3})_{2} C (OC H_{3}) C H_{2} C H_{2} Br$ ; B = the 1° amine; C = its N-methyl derivative; D = $(C H_{3})_{2} C (OC H_{3}) C H_{2} C H_{2} OH$ ; E = the N-nitrosamine of (C)
What happens when cyclopentanone reacts with (a) $C H_{3} C H_{2} N H_{2}$ and (b) $(C H_{3} C H_{2})_{2} NH$ ?Answer: (a) a 1° amine gives the imine (Schiff base) cyclopentylidene-ethylamine; (b) a 2° amine has no second N-H, so it gives the enamine, 1-(N,N-diethylamino)cyclopentene
Cyclohexylamine is a stronger base than aniline. Why?Answer: the lone pair of cyclohexylamine is on an $sp^{3}$ nitrogen and is fully available, while in aniline it is delocalised into the ring and protonation destroys that resonance
Why is aniline acetylated before it is brominated?Answer: the free $- N H_{2}$ group activates the ring so strongly that bromine water gives 2,4,6-tribromoaniline; acetylation lowers the activation, so only one bromine enters, mainly at the para position
Dimethylamine is a stronger base than methylamine, but trimethylamine is weaker than both. Why?Answer: two methyl groups add $+ I$ effect while the cation still keeps two N-H bonds for solvation; in trimethylamine the cation has only one N-H and the three bulky groups also hinder solvation and protonation
An aromatic compound (A), $C_{7} H_{5} N O_{2} C l_{2}$ , on reduction with Sn/HCl gives (B), which with $NaN O_{2}$ /HCl gives (C). (B) gives no dye with $β$ -naphthol, and (C) gives a red colour with ceric ammonium nitrate and on oxidation gives an acid (D) of equivalent weight 191. Decarboxylation of (D) gives (E), which gives a single mononitro derivative (F). Identify (A) to (F).Answer: A = 2,5-dichloro-1-(nitromethyl)benzene; B = the 1° aliphatic amine (so no azo dye); C = the alcohol; D = 2,5-dichlorobenzoic acid (equivalent weight 191); E = 1,4-dichlorobenzene; F = 1,4-dichloro-2-nitrobenzene
Two isomeric amines (A) and (B), $C_{4} H_{11} N$ , lose $N_{2}$ with $HN O_{2}$ to give alcohols (C) and (D), $C_{4} H_{10} O$ . (C) reacts at once with Lucas reagent and resists oxidation; (D) does not react in the cold but is easily oxidised. Exhaustive methylation of either gives no but-1-ene. Identify (A) to (D).Answer: C = 2-methylpropan-2-ol and A = 2-methylpropan-2-amine; D = 2-methylpropan-1-ol and B = 2-methylpropan-1-amine
Account for the fact that glycine and sulphanilic acid exist as dipolar ions but p-aminobenzoic acid does not.Answer: in glycine the aliphatic $- N H_{2}$ is basic enough to take the proton of $- COOH$ ; in sulphanilic acid $- S O_{3} H$ is a strong enough acid to protonate even the weakly basic aryl $- N H_{2}$ ; $- COOH$ is too weak an acid to protonate an aryl $- N H_{2}$ , so p-aminobenzoic acid stays neutral
Activation of the benzene ring by $- N H_{2}$ in aniline can be reduced by treating it with (a) dilute HCl (b) ethyl alcohol (c) acetic acid (d) acetyl chloride.Answer: (d) acetyl chloride, which acetylates the nitrogen; the amide lone pair is delocalised onto C=O, so the ring is only moderately activated

Common Mistakes to Avoid

Watch out

Using the gas-phase order 3° > 2° > 1° for aqueous solutions. In water solvation of the cation matters, and the 2° amine is the strongest base.
Explaining basicity with the inductive effect alone. Solvation and, for aryl amines, resonance usually decide the answer.
Expecting a 3° amine to give an amide with an acid chloride, or to give a carbylamine or N-nitroso product. All of these need an N-H bond.
Writing a stable diazonium salt from a primary aliphatic amine. Only aryl diazonium salts survive, and only below about 278 K.
Giving the Saytzeff product for a Hofmann elimination. A quaternary ammonium hydroxide gives the least substituted alkene.
Proposing a Friedel-Crafts reaction on aniline. The Lewis acid bonds to the lone pair and deactivates the ring.
Saying that direct nitration of aniline gives only ortho and para products. In acid the anilinium ion forms, so about 47% of the product is the meta isomer.
Treating benzylamine as an aryl amine. Its nitrogen is not on the ring, so it is as basic as an aliphatic amine.
Writing p-aminobenzoic acid as a zwitterion. Only glycine and sulphanilic acid form internal salts.

Frequently Asked Questions

Why are aliphatic amines stronger bases than ammonia?

Alkyl groups release electrons by the inductive effect, so they raise the electron density on nitrogen and make the lone pair more available to a proton. They also stabilise the ammonium ion formed. Methylamine has a pKb of 3.38 against 4.75 for ammonia, so it is about twenty times stronger.

Why is aniline a much weaker base than cyclohexylamine?

In aniline the nitrogen lone pair is delocalised into the benzene ring, so it is not fully available, and protonation destroys that resonance stabilisation. Nitrogen is also bonded to an sp2 carbon, which pulls electrons away. Aniline has a pKb of 9.38 against 3.32 for cyclohexylamine.

Why is the order of basicity of amines different in water and in the gas phase?

In the gas phase only the inductive effect operates, so the order is tertiary, secondary, primary, ammonia. In water the ammonium ion is stabilised by hydrogen bonds, and a cation with more N-H bonds is better solvated. The two effects together make the secondary amine the strongest base in water.

Why can aniline not undergo the Friedel-Crafts reaction?

Aniline is basic, so its lone pair bonds to the Lewis acid catalyst such as aluminium chloride. The nitrogen then carries a positive charge, which strongly deactivates the ring and directs incoming groups to the meta position, so the intended alkylation or acylation does not happen.

Why is aniline acetylated before nitration or bromination?

The free amino group activates the ring so strongly that bromine water gives 2,4,6-tribromoaniline, and nitration in acid gives a mixture with about half the meta isomer. Acetylation ties up the lone pair in an amide, which activates only moderately, so a single group enters, mainly at the para position.

What is Hofmann elimination and what does Hofmann's rule say?

Heating a quaternary ammonium hydroxide gives an alkene, a tertiary amine and water by an E2 elimination, and but-1-ene is the major product. Hofmann's rule says the least substituted alkene is the major product, because the bulky ammonium group makes the base remove a hydrogen from the least hindered beta carbon.

How do the different classes of amine behave with nitrous acid?

A primary aliphatic amine gives nitrogen gas and a mixture of alcohols, alkenes and halides. A primary aromatic amine gives a diazonium salt, stable between 273 and 278 K. A secondary amine gives a yellow oily nitrosamine. A tertiary aliphatic amine gives a salt, while a tertiary aromatic amine is nitrosated on the ring.

Which properties of amines are asked most in NEET?

NEET concentrates on the NCERT points: the basic strength order of amines in water, why aniline is less basic than ethylamine, the reaction with nitrous acid, acylation and the carbylamine test, and why aniline gives 2,4,6-tribromoaniline with bromine water while acetylation controls it.

What does JEE Main ask about the properties of amines?

JEE Main sets comparisons and product questions: orders of basicity including substituted anilines and heterocycles, the effect of the solvent, Hofmann versus Saytzeff products in eliminations, the Cope elimination, distinguishing amine classes with nitrous acid, and rearrangements such as Fischer-Hepp and Bamberger.

Previous year questions on Properties of Amines

12 questions from past papers, each with a step-by-step solution.

Show all 12 questions

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