Nitrogen compounds: amines

Class 11 Chemistry, unit 20, book pages 373–379

Amines are organic compounds derived from ammonia by replacing one or more hydrogen atoms with alkyl or aryl groups. This unit classifies amines as primary, secondary and tertiary, explains how the sp3 nitrogen with its lone pair gives them their shape, their hydrogen bonding and their basic and nucleophilic character, and shows how they are made from alkyl halides and by reducing nitriles, nitro compounds and amides. It ends with the main reactions of amines, including the preparation of diazonium salts and their use in azo dyes and pigments.

Learning outcomes (4)
  • Define primary and secondary amines and explain their basic properties and reactivity.
  • Identify the differences between primary and secondary amines in terms of their structure and chemical properties.
  • Describe the preparation methods of primary and secondary amines including nucleophilic reactions and reduction of nitro compounds.
  • Explain the properties and reactivity of phenylamine and azo compounds, including their uses as dyes and pigments.

20.1AminesOften examined

Amines are organic compounds that contain nitrogen linked to alkyl or aryl groups, with a lone pair of electrons on the nitrogen atom. They are derived from ammonia by replacing one or more of its hydrogen atoms by alkyl or aryl groups, so amines are derivatives of ammonia.

Amines are classified into three types according to the number of carbon atoms directly attached to nitrogen: primary, secondary and tertiary.

Classification of amines by the number of alkyl groups linked to nitrogen. The tertiary example is the triethylamine met in section 20.3.
TypeAlkyl groups on NGeneral formulaExample
Primary amineOne (methylamine)
Secondary amineTwo (dimethylamine)
Tertiary amineThree (triethylamine)

Examples of primary amines

Methylamine (CH3-NH2)
Ethylamine (CH3-CH2-NH2)
Propylamine (CH3-CH2-CH2-NH2)

Examples of secondary amines

Dimethylamine (CH3-NH-CH3)
Methylethylamine (CH3-CH2-NH-CH3)
Diethylamine (CH3-CH2-NH-CH2-CH3)

20.2Physical properties of amines

The N-H bond is polar because nitrogen and hydrogen differ in electronegativity (nitrogen is more electronegative, so N carries a partial negative charge and H a partial positive charge). This polarity lets amine molecules form hydrogen bonds with other amine molecules.

  • Because of hydrogen bonding, amines have high melting and boiling points compared with analogous alkanes, whose molecules are held only by weak dispersion forces.
  • Because of hydrogen bonding, amines also have high solubility in water, since the lone pair on N and the N-H bonds can hydrogen bond with water molecules.

Structure of amines

In amines the nitrogen atom is -hybridised and has a nearly tetrahedral arrangement. It forms three sigma bonds with three of its -hybrid orbitals, while the fourth, non-bonding -hybrid orbital carries a pair of electrons (the lone pair). Amines have the ability to donate this electron pair, which makes them a base and a nucleophile.

Structure of amines: the nitrogen atom is -hybridised, with three sigma bonds and a lone pair in the fourth orbital (shown in the accent). In (b) the three groups are different, so the amine is optically active.

An amine with three different groups on nitrogen is optically active, as the book shows with in the figure.

Worked example: hydrogen bonding in ethylamine
Draw hydrogen bonds (a) between two molecules of ethylamine and (b) between ethylamine and water. Use the drawing to explain why amines have high boiling points and dissolve in water.
Hydrogen bonds (accent): the polar N–H of one amine molecule is attracted to the lone pair on nitrogen of the next (a), and a water O–H is attracted to the lone pair on nitrogen of an amine (b).
  1. Nitrogen is more electronegative than hydrogen, so the bond is polar: N carries a partial negative charge and H a partial positive charge.
  2. In (a) the hydrogen of one molecule is attracted to the lone pair on the nitrogen of its neighbour. Many such bonds hold the liquid together, so extra energy is needed to separate the molecules: the boiling point is high.
  3. An alkane of similar size has no polar bond, so only weak dispersion forces act between its molecules, and its boiling point is much lower.
  4. In (b) the water hydrogen bonds to the nitrogen lone pair, and the amine can bond to the water oxygen. These bonds with water explain the high solubility of amines.

20.3Preparation of aminesOften examined

1. By the reaction of alkyl halides with ammonia

When an alcoholic solution of ammonia is heated with an alkyl halide in a sealed tube at , a mixture of primary, secondary and tertiary amines and a quaternary ammonium salt is obtained. The nitrogen lone pair of ammonia attacks the alkyl halide, so ammonia acts as the nucleophile, and each new amine formed can react again with more alkyl halide.

Stepwise alkylation of ammonia
  1. Ammonia reacts with ethyl iodide to give the primary amine, ethylamine.
  2. Ethylamine, still having a lone pair, reacts with another ethyl iodide to give the secondary amine, diethylamine.
  3. Diethylamine reacts again to give the tertiary amine, triethylamine.
  4. Triethylamine reacts once more and forms the quaternary ammonium salt, quaternary ethylammonium iodide.
Each alkylation step turns the product of the previous step into the next amine; the last step gives the quaternary ammonium salt (accent).

At the end of the reaction, addition of a strong alkali such as KOH liberates the free amines from their salts, but the quaternary salt is unaffected. The three types of amines are then separated by fractional distillation.

Worked example: ammonia and iodomethane: all the products
Iodomethane is heated with alcoholic ammonia in a sealed tube at . Write the equation for each stage, name each product, and say how the free amines are obtained and why excess ammonia helps.
  1. The lone pair on ammonia attacks the carbon of (ammonia is the nucleophile). The first product is the primary amine, methylamine.
  2. Methylamine still has a lone pair, so it reacts with another to give the secondary amine, dimethylamine.
  3. Again, giving the tertiary amine, trimethylamine.
  4. Trimethylamine has no hydrogen left on nitrogen, and the last attack gives the quaternary ammonium salt.
  5. The amines are present as their salts. Adding a strong alkali such as KOH liberates the free amines, but the quaternary salt is unaffected. The three amines are then separated by fractional distillation.
  6. Excess ammonia makes it more likely that a molecule meets than an amine, so over-alkylation is reduced, although the yield is low.
Methylamine (primary)
Dimethylamine (secondary)
Trimethylamine (tertiary)

2. By the reduction of nitrogen-containing functional groups

Nitrogen-containing functional groups such as nitriles, amides and nitro compounds are reduced to the amino group by various reducing agents. In each case the nitrogen is already in the molecule; reduction adds hydrogen to it (shown as ).

a. Reduction of nitriles

Reduction of alkyl or aryl nitriles gives primary amines. The reduction may be brought about by a reducing agent such as , or by sodium in ethanol. Nitriles have the general formula .

Nitrile, R-C≡N (here ethanenitrile)
b. Catalytic reduction

Nitriles react with hydrogen gas in the presence of catalysts such as , Pt or Raney nickel to give primary amines.

Benzyl cyanide
Phenylethylamine
c. Reduction of nitro compounds

Nitro compounds on catalytic or chemical reduction produce primary amines. Nitrobenzene can be reduced to aniline (phenylamine) using tin or iron in the presence of concentrated hydrochloric acid, or by catalytic hydrogenation (for example with Pd).

Nitrobenzene
Aniline (phenylamine)
4-Nitroaniline
1,4-Diaminobenzene
d. Reduction of amides

When an amide is treated with the reducing agent , the group of the amide is reduced to and a primary amine is formed. The carbon chain remains intact in this reaction.

Acetamide
Ethylamine

Hofmann degradation of amides

When an amide is treated with bromine in the presence of KOH, the amide is degraded and a primary amine is formed. This reaction is known as the Hofmann degradation reaction.

Worked example: propanamide to a primary amine, two ways
Propanamide, , can be converted into a primary amine by or by bromine and KOH. Give the amine formed in each case and compare the number of carbon atoms.
  1. With the group becomes and the carbon chain stays intact. Propanamide (three carbons) gives propylamine (three carbons).
  2. Check the balance: 3 = 3, 1 = 1, 7 + 4 = 11 = 9 + 2, 1 = 1.
  3. In the Hofmann degradation the amide is degraded: the carbonyl carbon is lost as . Propanamide gives ethylamine (two carbons), in the book's form of the reaction:
  4. Common mistake: the two reactions do not give the same amine. keeps the carbon chain; Hofmann degradation shortens it by one carbon.
Propanamide
Propylamine (LiAlH4)
Ethylamine (Hofmann)
Summary of the preparation methods in section 20.3.
Starting compoundReagentProductRemark
Alkyl halide (alcoholic), , sealed tubeMixture of 1°, 2°, 3° amines and quaternary saltSeparated by fractional distillation
Nitrile , or Na in ethanolPrimary amine
Nitrile with Rh-, Pt or NiCatalytic reduction
NitrobenzeneSn or Fe with conc. HCl, or /PdAnilinePrimary aromatic amine
Amide Chain stays intact
Amide + KOHHofmann degradation

20.4Basicity of ammonia and methylamineOften examined

Amines are bases and nucleophiles because the non-bonding pair of electrons on nitrogen can easily be donated to an electron-deficient centre. Consider the following reactions with hydrogen chloride, in which the lone pair on N bonds to \ce{H+}.

\ce{NH3 + HCl -> NH4+ + Cl-}
\ce{CH3-NH2 + HCl -> CH3-NH3+ + Cl-}

The strength of a base is expressed in terms of \mathrm{p}K_b. For ammonia and methylamine the \mathrm{p}K_b values are 4.76 and 3.38 respectively.

A smaller \mathrm{p}K_b means a stronger base.
Base\mathrm{p}K_bStrength
Ammonia \ce{NH3}4.76Weaker base
Methylamine \ce{CH3NH2}3.38Stronger base

Since \mathrm{p}K_{b}(\ce{NH3}) > \mathrm{p}K_{b}(\ce{CH3NH2}), methylamine is a stronger base than ammonia.

Why methylamine is the stronger base

In ammonia the lone pair on nitrogen is attracted by the s-orbitals of the hydrogen atoms, whereas in \ce{CH3-NH2} the sp^3 orbital of carbon pushes electrons towards nitrogen. The methyl group is an electron-donating group, so the electron density on N is higher, the lone pair is more available to accept \ce{H+}, and the ion \ce{CH3NH3+} formed is stabilised.

Comparison of basicity of primary and secondary amines

The presence of an electron-donating group increases the basicity of amines. A secondary amine is a stronger base than a primary amine, since it has two electron-donating alkyl groups.

\text{Secondary amine} > \text{primary amine}
Worked example: ranking bases and showing the effect of the methyl group
Arrange ammonia, methylamine and dimethylamine in increasing order of basicity. Draw how methylamine accepts \ce{H+} and show the effect of the methyl group.
Methylamine accepts H⁺ with its lone pair (curved arrow). The electron-donating methyl group (accent arrow) makes the nitrogen richer in electrons, so the lone pair is more available.
  1. A base here is a lone-pair donor. Ammonia has three H atoms on nitrogen and no electron-donating group, so it is the weakest of the three.
  2. Methylamine has one electron-donating \ce{CH3} group. It pushes electrons towards N, so the lone pair is more available to bond to \ce{H+}, and the ion \ce{CH3NH3+} is stabilised.
  3. Dimethylamine is a secondary amine with two donating groups, so it is stronger still.
  4. Check with the data: \mathrm{p}K_b is 4.76 for \ce{NH3} and 3.38 for \ce{CH3NH2}. The smaller \mathrm{p}K_b belongs to the stronger base, and 3.38<4.76 agrees.
  5. Common mistake: a larger \mathrm{p}K_b means a weaker base, not a stronger one.
\ce{NH3}<\ce{CH3NH2}<\ce{(CH3)2NH}
Basicity increases with the number of electron-donating alkyl groups on nitrogen.
PointAmmoniaPrimary amineSecondary amine
Groups on NNone (three H)One alkyl groupTwo alkyl groups
Electron donation to NNoneOne donating groupTwo donating groups
BasicityWeakestStronger than ammoniaStrongest of the three

20.5Reactions of aminesOften examined

1. Reaction of amine with acid halide

When an amine reacts with an acid halide, a secondary amide (an N-substituted amide) is formed. The lone pair on N attacks the carbonyl carbon and chloride leaves as HCl.

\ce{R-CO-Cl + RNH2 -> R-CO-NHR + HCl}
\ce{CH3COCl + CH3CH2NH2 -> CH3CONHCH2CH3 + HCl}\quad(\text{N-ethylacetamide})
N-Ethylacetamide

2. Reaction of amine with alkyl halide (alkylation of amine by alkyl halides)

Primary amines react with alkyl halides to form secondary and tertiary amines and an ammonium salt. This reaction is called an alkylation reaction. It is the same stepwise alkylation met in the preparation from ammonia, now starting from a primary amine.

\ce{C2H5I + C2H5NH2 -> (C2H5)2NH + HI}\quad(\text{diethylamine})
\ce{C2H5I + (C2H5)2NH -> (C2H5)3N + HI}\quad(\text{triethylamine})
\ce{(C2H5)3N + C2H5I -> [(C2H5)4N]+ I-}\quad(\text{quaternary ethylammonium iodide})

3. Reaction with nitrous acid (preparation of diazonium salts)

When primary amines react with nitrous acid in the presence of hydrochloric acid, diazonium compounds are formed. The reaction mixture must be kept below 10^\circ\mathrm{C}.

\ce{RNH2 + HNO2 + HCl -> [R-N#N]+Cl- + 2H2O}\quad(\text{diazonium salt})

Alkyl diazonium salts are generally too unstable for practical use. When the diazonium group is attached to the benzene ring, the delocalised electrons of the ring stabilise the ion to some extent. Because of this, the benzene diazonium ion is far more stable than the aliphatic diazonium salt.

\ce{C6H5NH2 + HNO2 ->[HCl] [C6H5-N#N]+Cl- + 2H2O}\quad(\text{benzene diazonium chloride})
Benzene diazonium ion (chloride salt)
Why the benzene diazonium ion can be used and the aliphatic one cannot.
PropertyAlkyl diazonium saltBenzene diazonium chloride
StabilityGenerally too unstable for practical useFar more stable, because the delocalised ring electrons stabilise the ion
FormNot isolated in practiceColourless crystalline solid
TemperatureKept below 10^\circ\mathrm{C}Stable below 10^\circ\mathrm{C}, decomposes easily above 10^\circ\mathrm{C}
Worked example: aniline and nitrous acid
Aniline is treated with nitrous acid and hydrochloric acid. Write the equation, name the product, and explain why the mixture must be kept below 10^\circ\mathrm{C}.
  1. A primary amine reacts with \ce{HNO2} in the presence of \ce{HCl} to form a diazonium salt and water. For aniline:
    \ce{C6H5NH2 + HNO2 + HCl -> [C6H5-N#N]+Cl- + 2H2O}
  2. Check the balance: \ce{H} 5 + 2 + 1 + 1 = 9 = 5 + 4; \ce{N} 1 + 1 = 2; \ce{O} 2 = 2; \ce{Cl} 1 = 1; \ce{C} 6 = 6.
  3. The product is benzene diazonium chloride. The delocalised electrons of the ring stabilise the ion, so it can be used below 10^\circ\mathrm{C}.
  4. Above 10^\circ\mathrm{C} it decomposes easily, so the mixture is kept cold (in ice) and never warmed. An alkyl diazonium salt is too unstable to be used even in this way.
  5. Combined with phenol, the diazonium salt gives an orange-red dye.
Benzene diazonium ion
\text{Benzene diazonium chloride, kept below }10^\circ\mathrm{C}

Phenyl azo compounds: dyes and pigments

Benzene diazonium chloride is a colourless crystalline solid. When it is combined with phenol it forms an orange-red dye. Phenyl azo compounds are used as dyes and pigments in the textile, cosmetics, paint and leather industries, and phenyl azo dyes have antibacterial and antifungal properties.

Industries named in the book that use phenyl azo dyes and pigments.
Use of phenyl azo compoundsIndustry
Dyes and pigmentsTextile
Dyes and pigmentsCosmetics
Dyes and pigmentsPaint
Dyes and pigmentsLeather
Summary of the reactions of amines in section 20.5.
ReagentAmine reacts asProduct
Acid halide \ce{R-CO-Cl}NucleophileSecondary amide \ce{R-CO-NHR} and HCl
Alkyl halide \ce{R-X}NucleophileSecondary or tertiary amine, then the quaternary ammonium salt
\ce{HNO2} + HCl (below 10^\circ\mathrm{C})Primary amineDiazonium salt and water

Formula sheet

General formula of a primary amine
\ce{R-NH2}
\ce{R} is an alkyl or aryl group; one group is linked to nitrogen.
no unit
General formula of a secondary amine
\ce{R-NH-R}
Two alkyl groups are linked to nitrogen.
no unit
General formula of a tertiary amine
\ce{(R)3N}
Three alkyl groups are linked to nitrogen.
no unit
General formula of a nitrile
\ce{R-C#N}
\ce{R} is an alkyl or aryl group; reduction gives \ce{R-CH2-NH2}.
no unit

Key terms

Amine
An organic compound containing nitrogen linked with alkyl or aryl groups, with a lone pair on the nitrogen atom; a derivative of ammonia.
Primary amine
An amine in which only one alkyl group is linked to the nitrogen atom, general formula \ce{R-NH2}.
Secondary amine
An amine in which two alkyl groups are linked to the nitrogen atom, general formula \ce{R-NH-R}.
Tertiary amine
An amine in which three alkyl groups are linked to the nitrogen atom, general formula \ce{(R)3N}.
Nucleophile
A species that donates an electron pair to an electron-deficient centre; amines act as nucleophiles through the lone pair on nitrogen.
Nitrile
A compound with the general formula \ce{R-C#N}; reduction gives a primary amine.
Quaternary ammonium salt
The salt \ce{[R4N]+ X-} formed when a tertiary amine reacts with an alkyl halide, for example \ce{[(C2H5)4N]+ I-}.
Alkylation
Replacement of a hydrogen on nitrogen by an alkyl group when an amine reacts with an alkyl halide.
Hofmann degradation
Treatment of an amide with bromine and KOH to give a primary amine.
\mathrm{p}K_b
The measure of the strength of a base; the smaller the \mathrm{p}K_b, the stronger the base.
Diazonium salt
The compound \ce{[R-N#N]+Cl-} formed when a primary amine reacts with nitrous acid and hydrochloric acid below 10^\circ\mathrm{C}.
Phenyl azo compounds
Compounds obtained from benzene diazonium chloride, used as dyes and pigments and having antibacterial and antifungal properties.

Multiple choice questionsp. 378

i
General formula for secondary amine is
  1. A\ce{RNH2}
  2. B\ce{RNHR}Answer
  3. C\ce{NH3}
  4. D\ce{RCONH2}
Why

A secondary amine has two alkyl groups linked to the nitrogen atom and one hydrogen left on it, so its general formula is \ce{R-NH-R}, written here as \ce{RNHR}.

\ce{RNH2} is the primary amine (only one alkyl group), \ce{NH3} is ammonia with no alkyl group, and \ce{RCONH2} is an amide, which is not an amine.

ii
General formula for nitriles is
  1. A\ce{RNH2}
  2. B\ce{RNO3}
  3. C\ce{RCN}Answer
  4. D\ce{ROH}
Why

A nitrile contains the carbon-nitrogen triple bond, \ce{C#N}, attached to an alkyl or aryl group, so its general formula is \ce{R-C#N}, written \ce{RCN}.

\ce{RNH2} is the amine that a nitrile gives on reduction (\ce{RCN -> RCH2NH2}), not the nitrile itself. \ce{RNO3} is not a nitrile, and \ce{ROH} is an alcohol.

iii
Which of the following is a reducing agent?
  1. A\ce{KMnO4}
  2. B\ce{K2Cr2O7}
  3. C\ce{LiAlH4}Answer
  4. D\ce{NaOH}
Why

\ce{LiAlH4} is a reducing agent: it adds hydrogen (hydride ions) to the nitrogen-containing groups, which is how the book reduces nitriles, nitro compounds and amides to amines (Section 20.3).

\ce{KMnO4} and \ce{K2Cr2O7} are oxidizing agents, which remove electrons, so they are the opposite of reducing agents. \ce{NaOH} is a base and does not reduce these compounds.

iv
Which can donate an electron pair?
  1. A\ce{RX}
  2. B\ce{RCOR}
  3. C\ce{RCHO}
  4. D\ce{RNH2}Answer
Why

An amine has a lone pair on the nitrogen atom that it can donate to an electron-deficient centre. This is why amines are nucleophiles (Key points, and Section 20.2).

In \ce{RX} the carbon is the electron-deficient site that is attacked, so it does not donate a pair. \ce{RCOR} and \ce{RCHO} are also attacked at their electron-deficient carbonyl carbon.

Note. The book prints no key for this question. The oxygen atoms of \ce{RCOR} and \ce{RCHO} also carry lone pairs; the intended answer is the amine, because the unit's point is that amines donate their electron pair. Option D is the answer in this unit's sense.

Short questionsp. 378–379

i3 marks
What are primary, secondary and tertiary alkyl amines?
Solution

Amines are classified by the number of carbon atoms of alkyl groups that are attached directly to the nitrogen atom. A primary amine has one alkyl group, \ce{R-NH2} (for example methylamine, \ce{CH3NH2}). A secondary amine has two alkyl groups, \ce{R-NH-R} (for example dimethylamine, \ce{(CH3)2NH}).

A tertiary amine has three alkyl groups linked to the nitrogen, \ce{R3N} (for example triethylamine, \ce{(C2H5)3N}). Primary amines keep two hydrogen atoms on the nitrogen, secondary amines keep one, and tertiary amines have none.

All three types have the nitrogen atom with a lone pair, so the classes differ only in how many hydrogen atoms are replaced by alkyl groups.

ii3 marks
What is a diazonium salt?
Solution

A diazonium salt has the general formula \ce{R-N2+ X-}: an \ce{N2+} group, in which two nitrogen atoms are joined by a triple bond, is attached to an alkyl or aryl group and balanced by a halide ion such as \ce{Cl-}. The book's example is benzene diazonium chloride, \ce{C6H5N2+ Cl-}.

Benzene diazonium ion

Alkyl diazonium salts are generally too unstable for practical use. In the benzene diazonium ion the delocalized electrons of the ring stabilize the positive charge to some extent, so it is far more stable than the aliphatic salts. Benzene diazonium chloride is a colourless crystalline solid that is stable below 10 °C and decomposes easily above 10 °C.

Diazonium salts are important because benzene diazonium chloride combines with phenol to form an orange-red azo dye (Section 20.5).

iii3 marks
Amines are more basic than ammonia.
Solution

Basicity is the ability to accept a proton. Ammonia and methylamine both react with hydrochloric acid, \ce{NH3 + HCl -> NH4+ + Cl-} and \ce{CH3-NH2 + HCl -> CH3-NH3+ + Cl-}. The strength of a base is given by pK_b, and the smaller the pK_b, the stronger the base. Ammonia has pK_b = 4.76 and methylamine has pK_b = 3.38, so methylamine is the stronger base.

Alkyl groups are electron-donating: they push electron density towards the nitrogen atom. The lone pair on the nitrogen of an amine is therefore richer in electron density than the lone pair of ammonia and accepts a proton more readily. Each alkyl group adds to this effect, which is why secondary amines (two alkyl groups) are stronger bases than primary amines (Section 20.4).

Note. Printed as a statement with no question word; answered as 'explain why amines are more basic than ammonia'. The book's explanation in Section 20.4 (electrons attracted by the s-orbitals of hydrogen) is loosely worded; the answer uses the electron-donating effect of alkyl groups that the book gives for basicity.

iv3 marks
What are amines?
Solution

Amines are organic compounds derived from ammonia, in which one or more hydrogen atoms of \ce{NH3} are replaced by alkyl or aryl groups. They contain nitrogen linked with alkyl or aryl groups and a lone pair on the nitrogen atom, so they are derivatives of ammonia.

The nitrogen atom in an amine is sp^3-hybridized and has a nearly tetrahedral shape. Three of its sp^3 orbitals form sigma bonds to the attached groups and the fourth holds the lone pair. Because the lone pair can be donated, amines act as bases and nucleophiles.

Methylamine
Dimethylamine
v3 marks
Amines have high melting and boiling points as compared to other analogous alkanes. Why?
Solution

The N–H bond is polar, because nitrogen is more electronegative than hydrogen. The hydrogen atom of one amine molecule is therefore attracted to the nitrogen atom of a neighbouring molecule, and these hydrogen bonds hold amine molecules together.

An alkane of similar molar mass has no polar bond and no hydrogen bonding, so its molecules are held only by weak van der Waals forces. More energy is needed to separate amine molecules, so their melting and boiling points are higher.

The same hydrogen bonding with water makes amines highly soluble in water (Section 20.2).

vi3 marks
Methylamine is a stronger base than ammonia. Explain.
Solution

Methylamine has pK_b = 3.38 and ammonia has pK_b = 4.76. A smaller pK_b means a stronger base, so methylamine is the stronger base. Both accept a proton from HCl:

\ce{NH3 + HCl -> NH4+ + Cl-}\qquad\ce{CH3-NH2 + HCl -> CH3-NH3+ + Cl-}

The methyl group is electron-donating, so it pushes electron density towards the nitrogen atom. The lone pair on the nitrogen of methylamine is therefore richer in electron density than that of ammonia and binds the proton more strongly, which makes methylamine the stronger base.

Note. Repeats the idea of question (iii) with the specific pair of bases; answered in full here as the book asks it.

vii3 marks
How would you prepare a diazonium salt?
Solution

A diazonium salt is prepared by treating a primary amine with nitrous acid in the presence of hydrochloric acid. The mixture must be kept below 10 °C.

\ce{RNH2 + HNO2 + HCl -> R-N2+ Cl- + 2H2O}\qquad(\text{below }10\,^\circ\mathrm{C})

For aniline (phenylamine) the reaction is \ce{C6H5NH2 + HNO2 + HCl -> C6H5N2+ Cl- + 2H2O}, giving benzene diazonium chloride. The cold is essential, because the product decomposes easily above 10 °C.

Aniline (phenylamine)

Long questionsp. 379

38 marks
Discuss the methods of preparation of amines.
Solution

Overview

The book describes three groups of methods: the reaction of alkyl halides with ammonia, the reduction of nitrogen-containing functional groups (nitriles, nitro compounds and amides), and the Hoffmann degradation of amides. The figure below summarizes the routes.

Routes to amines: starting groups on the left, the amine on the right. The alkyl halide route gives a mixture of primary, secondary and tertiary amines and a quaternary salt.

1. From alkyl halides and ammonia

When an alcoholic solution of ammonia is heated with an alkyl halide in a sealed tube at 100^\circ\mathrm{C}, the ammonia lone pair attacks the alkyl halide. Each amine formed still has a lone pair, so it can react again. The sequence with ethyl iodide is:

  1. Ammonia gives the primary amine, ethylamine.
    \ce{C2H5I + NH3 -> C2H5NH2 + HI}
  2. Ethylamine gives the secondary amine, diethylamine.
    \ce{C2H5I + C2H5NH2 -> (C2H5)2NH + HI}
  3. Diethylamine gives the tertiary amine, triethylamine.
    \ce{C2H5I + (C2H5)2NH -> (C2H5)3N + HI}
  4. Triethylamine gives the quaternary ammonium salt.
    \ce{(C2H5)3N + C2H5I -> [(C2H5)4N]+ I-}

The product is a mixture of primary, secondary and tertiary amines and a quaternary salt. Adding a strong alkali such as KOH liberates the free amines from their salts, but the quaternary salt is unaffected. The three amines are then separated by fractional distillation.

2. By reduction of nitrogen-containing functional groups

Nitriles, nitro compounds and amides are reduced to the amino group by suitable reducing agents. In each case the nitrogen is already in the molecule; reduction adds hydrogen, shown as [\mathrm{H}].

a. Reduction of nitriles

Alkyl or aryl nitriles give primary amines on reduction, with \ce{LiAlH4} or with sodium in ethanol. Hydrogen gas over the catalysts Rh–Al₂O₃, Pt or Raney nickel also works.

\ce{R-C#N ->[LiAlH4] R-CH2-NH2}
Ethanenitrile

Examples from the book: ethanenitrile is reduced by catalytic hydrogenation to aminoethane, and benzyl cyanide gives phenylethylamine.

\begin{gathered}\ce{CH3CN + 2H2 ->[Rh-Al2O3] CH3CH2NH2}\\ \ce{C6H5CH2CN + 2H2 ->[Ni][300^\circ C] C6H5CH2CH2NH2}\end{gathered}
b. Reduction of nitro compounds

Nitro compounds give primary amines. Nitrobenzene is reduced to aniline (phenylamine) by tin and concentrated HCl, or by hydrogen with a palladium catalyst (H₂/Pd). The book also reduces 4-nitroaniline to 1,4-diaminobenzene with iron and \ce{H2SO4}.

\ce{C6H5NO2 + 6[H] ->[Sn + HCl] C6H5NH2 + 2H2O}
Nitrobenzene
Aniline
c. Reduction of amides

Amides are reduced by \ce{LiAlH4}. The \ce{-CO-} group of the amide is reduced to \ce{-CH2-}, the carbon chain stays intact and a primary amine is formed.

\ce{CH3CONH2 + 4[H] ->[LiAlH4] CH3CH2NH2 + H2O}
Acetamide
d. Hoffmann degradation

Treating an amide with bromine in KOH removes one carbon atom from the chain, and the product is a primary amine with one carbon fewer (Hoffmann degradation).

\ce{CH3CONH2 + Br2 + 4KOH -> CH3NH2 + K2CO3 + 2KBr + 2H2O}

Summary

MethodStarting materialReagent or conditionProduct
Alkyl halide and ammonia\ce{R-X}alcoholic \ce{NH3}, sealed tube, 100 °Cmixture of 1°, 2°, 3° amines and quaternary salt
Reduction of nitrile\ce{R-C#N}\ce{LiAlH4}, Na in ethanol, or \ce{H2} with Ni, Pt, Rhprimary amine \ce{R-CH2-NH2}
Reduction of nitro compound\ce{Ar-NO2}Sn + HCl, Fe + \ce{H2SO4}, or \ce{H2}/Pdprimary amine \ce{Ar-NH2}
Reduction of amide\ce{R-CONH2}\ce{LiAlH4}primary amine, same carbon chain
Hoffmann degradation\ce{R-CONH2}\ce{Br2} + KOHprimary amine with one carbon less

Note. The printed heading reads 'methods of preparation on amines'; read as 'of'. The book's Hoffmann equation is unbalanced (see the callout). The figure is a summary drawn from the book's routes, not a book figure.

46 marks
What are nitriles? How can they be reduced?
Solution

Nitriles

Nitriles are organic compounds that contain the cyano group, \ce{-C#N}, attached to an alkyl or aryl group. Their general formula is \ce{R-C#N}, written \ce{RCN}. Ethanenitrile (\ce{CH3CN}) and benzyl cyanide (\ce{C6H5CH2CN}) are examples.

Ethanenitrile
Benzyl cyanide

Reduction of nitriles

Reduction converts the \ce{C#N} group into a \ce{CH2-NH2} group, so alkyl or aryl nitriles give primary amines. The carbon of the nitrile group becomes the \ce{CH2} carbon of the amine, so the number of carbon atoms is unchanged. Three methods are given in the book:

  1. Lithium aluminium hydride, \ce{LiAlH4}, as the reducing agent: \ce{R-C#N ->[LiAlH4] R-CH2-NH2}.
  2. Sodium in ethanol as the reducing agent.
  3. Catalytic hydrogenation: hydrogen gas with the catalysts Rh–Al₂O₃, Pt or Raney nickel.

Worked examples

  1. Catalytic reduction of ethanenitrile with two molecules of hydrogen. Four hydrogen atoms are added: two to the carbon, giving \ce{CH2}, and two to the nitrogen, giving \ce{NH2}.
    \ce{CH3CN + 2H2 ->[Rh-Al2O3] CH3CH2NH2}
  2. Reduction of benzyl cyanide with hydrogen over nickel at 300 °C gives phenylethylamine. Again two molecules of hydrogen are added across the \ce{C#N} group.
    \ce{C6H5CH2CN + 2H2 ->[Ni][300^\circ C] C6H5CH2CH2NH2}
  3. Both products are primary amines, because the \ce{NH2} group is attached to a single carbon.
Phenylethylamine
\ce{R-C#N ->[LiAlH4] R-CH2-NH2}
58 marks
Identify the differences between different types of amines in terms of their structures and chemical properties.
Solution

Structure

The types differ in how many alkyl groups are attached to the nitrogen atom, and so in how many hydrogen atoms remain on it. In all three the nitrogen is sp^3-hybridized, has a nearly tetrahedral shape and carries one lone pair.

Structural differences between the three classes of amines
FeaturePrimary amineSecondary amineTertiary amine
Alkyl groups on Nonetwothree
General formula\ce{R-NH2}\ce{R-NH-R}\ce{R3N}
Hydrogen atoms on Ntwoonenone
Example\ce{C2H5NH2}, ethylamine\ce{(C2H5)2NH}, diethylamine\ce{(C2H5)3N}, triethylamine
Lone pair on Noneoneone
Ethylamine
Diethylamine
Triethylamine

Chemical properties

Chemical behaviour of primary, secondary and tertiary amines
PropertyPrimary amineSecondary amineTertiary amine
Basicityweaker base than a secondary aminestronger base, because two alkyl groups donate electronsnot compared in the book
Alkylation by an alkyl halide\ce{RNH2 + R'I -> RNHR' + HI}, giving a secondary amine, then a tertiary amine and a quaternary salt\ce{RNHR + R'I -> R3N + HI}, giving a tertiary amine, then a quaternary salt\ce{R3N + R'I -> [R4N]+ I-}, a quaternary ammonium salt
Acid halide\ce{RCOCl + R'NH2 -> RCONHR' + HCl}, a secondary amidenot described in the booknot described in the book
Nitrous acid with HCl, below 10 °C\ce{RNH2 + HNO2 + HCl -> R-N2+ Cl- + 2H2O}, a diazonium saltnot described in the booknot described in the book

A primary amine has two N–H bonds and so can be alkylated twice before it becomes quaternary. A secondary amine has one N–H bond, so it gives a tertiary amine on alkylation. A tertiary amine has no N–H bond and can only form the quaternary salt.

Note. The table cells marked 'not described in the book' are deliberate: the book does not give those reactions for secondary or tertiary amines.

66 marks
Evaluate the importance of diazonium salts in everyday life.
Solution

Formation and stability

Diazonium salts form when amines react with nitrous acid in the presence of hydrochloric acid, below 10 °C. Alkyl diazonium salts are too unstable for practical use, so the useful salts are aromatic. The benzene diazonium ion is stabilized by the delocalized electrons of the ring, and benzene diazonium chloride is a colourless crystalline solid that is stable below 10 °C.

Uses

  • Dyes and pigments. Benzene diazonium chloride combines with phenol to give an orange-red dye. Phenyl azo compounds are used as dyes and pigments in the textile, cosmetics, paint and leather industries.
  • Antibacterial and antifungal properties. Phenyl azo dyes have antibacterial and antifungal properties, which adds to their value beyond colour.

Limitations

  • The reaction mixture must be kept below 10 °C, because benzene diazonium chloride decomposes easily above 10 °C.
  • Alkyl diazonium salts are generally too unstable to be used in practice, so their use is limited to the aromatic salts.

Evaluation

The importance of diazonium salts rests on the azo dyes and pigments they make, which colour many everyday products such as textiles, cosmetics, paints and leather goods, and on the antibacterial and antifungal properties of those dyes. The main practical limit is their low-temperature stability: they must be made cold, but this does not prevent their use.

76 marks
Make a comparison between primary amines and secondary amines.
Solution

Comparison

FeaturePrimary amineSecondary amine
Alkyl groups on Nonetwo
General formula\ce{R-NH2}\ce{R-NH-R}
N–H bondstwoone
Example\ce{C2H5NH2}, ethylamine\ce{(C2H5)2NH}, diethylamine
Physical propertieshigh melting and boiling points, soluble in water (hydrogen bonding)high melting and boiling points, soluble in water (hydrogen bonding)
Basicityweaker basestronger base, two electron-donating alkyl groups
Alkylationgives a secondary amine, then a tertiary amine and a quaternary saltgives a tertiary amine, then a quaternary salt
Acid halidegives a secondary amidenot described in the book
Nitrous acidgives a diazonium salt below 10 °Cnot described in the book

Summary

Both classes have the same nitrogen centre with a lone pair, so both are bases and nucleophiles. The difference lies in the number of N–H bonds. The primary amine has two, so it can react with more alkyl groups and gives the diazonium salt with nitrous acid. The secondary amine has one N–H bond and is the stronger base, because its two alkyl groups push more electron density onto the nitrogen.

Note. Cells marked 'not described in the book' are left deliberately; the book does not give those reactions for secondary amines.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 131 mark
Amines are ______ in nature.
  1. AAcidic
  2. BNeutral
  3. CMetallic
  4. DBasicAnswer
Why

Amines are derivatives of ammonia with a lone pair of electrons on the nitrogen atom. They can donate this pair to \ce{H+} or to another electron-deficient centre, so they are basic (and nucleophilic) in nature. For example \ce{CH3NH2 + HCl -> CH3NH3+ + Cl-}.

Acidic is the tempting wrong choice only because of the \ce{-NH2} group's hydrogen atoms; but an amine does not release \ce{H+} in water, it accepts it. Amines are organic compounds, so 'metallic' makes no sense, and the lone pair makes them far from neutral.

2025 Annual I, Set B, Q.2 (viii) OR3 marks
How can ethyl amine be prepared from: (a) an alkyl halide (b) a nitrate (c) an amide
Solution

Ethylamine, \ce{CH3CH2NH2}, is a primary amine.

(a) From an alkyl halide (1 mark)

Ethyl halide is heated with excess alcoholic ammonia in a sealed tube at 100^\circ\mathrm{C} (nucleophilic substitution). Excess ammonia keeps over-alkylation low.

\ce{C2H5Cl + NH3 -> C2H5NH2 + HCl}
(b) From a nitrile (the printed 'nitrate'; 1 mark)

The reduction of a nitrile gives a primary amine with one more carbon than the alkyl group of the halide. Ethylamine needs the nitrile \ce{CH3CN} (ethanenitrile), reduced by \ce{LiAlH4}, by sodium in ethanol, or catalytically by \ce{H2} with Ni, Pt or Rh-\ce{Al2O3}.

\ce{CH3-C#N + 4[H] ->[LiAlH4] CH3CH2NH2}
\ce{CH3CN + 2H2 ->[Ni] CH3CH2NH2}
(c) From an amide (1 mark)

Acetamide is reduced by \ce{LiAlH4}: the \ce{C=O} group becomes \ce{CH2} and the carbon chain stays intact.

\ce{CH3-CO-NH2 + 4[H] ->[LiAlH4] CH3CH2NH2 + H2O}
Ethanenitrile
Acetamide
Ethylamine

Note. The paper prints 'nitrate'; the nitrile is meant (see solution).

2026 Annual I, Set B, Q.2 (ix) OR1.5+1.5 = 3 marks
Complete the following reactions: (a) \ce{C6H5NO2 ->[Sn+HCl][H]} (b) \ce{NH2-C6H4-NO2 ->[Fe+H2SO4]}
Solution
(a) Nitrobenzene to aniline (1.5 marks)

Nitrobenzene is reduced by tin and concentrated hydrochloric acid (nascent hydrogen [\mathrm{H}]). The \ce{-NO2} group becomes \ce{-NH2} and aniline (phenylamine) is formed.

\ce{C6H5NO2 ->[Sn + HCl][H] C6H5NH2}
\ce{C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O}
Nitrobenzene
Aniline (phenylamine)
(b) 4-Nitroaniline to 1,4-diaminobenzene (1.5 marks)

Iron with sulphuric acid also reduces a nitro group to an amino group. Only the \ce{-NO2} group is reduced; the \ce{-NH2} group already on the ring is unchanged, so the product is 1,4-diaminobenzene (p-phenylenediamine).

\ce{H2N-C6H4-NO2 ->[Fe + H2SO4][H] H2N-C6H4-NH2}
4-Nitroaniline
1,4-Diaminobenzene
2026 Annual I, Set D, MCQ 131 mark
Methylamine is a stronger base than:
  1. A\ce{NH3}Answer
  2. BDimethyl amine
  3. CMethyl ethyl amine
  4. DDiethyl amine
Why

Basic strength depends on how easily the nitrogen lone pair is donated. Ammonia has three H atoms and no electron-donating group, so it is the weakest of these: \mathrm{p}K_b of \ce{NH3} is 4.76 against 3.38 for \ce{CH3NH2}. A smaller \mathrm{p}K_b means a stronger base, so methylamine is stronger than ammonia.

The other three options are secondary amines (dimethylamine, methylethylamine, diethylamine) with two electron-donating alkyl groups on nitrogen. A secondary amine is a stronger base than a primary amine such as methylamine, so methylamine is weaker than each of them.

2025 Annual I, Set D, MCQ 171 mark
The reduction of \ce{C6H5-NO2} by Sn/HCl results in the formation of:
  1. A\ce{C6H6}
  2. B\ce{C6H5-NH2}Answer
  3. C\ce{C6H5-OH}
  4. D\ce{C6H5-Cl}
Why

Tin and concentrated hydrochloric acid supply nascent hydrogen, which converts the nitro group into an amino group: \ce{C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O}. The product is aniline (phenylamine), \ce{C6H5-NH2}.

Benzene (\ce{C6H6}) would need the whole \ce{-NO2} group removed, phenol (\ce{C6H5OH}) would have the nitrogen replaced by oxygen, and chlorobenzene does not form because HCl here only supplies acid and, with Sn, hydrogen. The ring and the C-N bond stay intact.

2026 Annual I, Set D, Q.2 (ix) OR1.5+1.5 = 3 marks
Describe the preparation methods of primary and secondary amines by nucleophilic substitution reaction.
Solution
Principle (1 mark)

An alkyl halide is heated with an alcoholic solution of ammonia in a sealed tube at 100^\circ\mathrm{C}. The nitrogen lone pair of ammonia attacks the carbon bonded to the halogen, so ammonia is the nucleophile and the halide ion is replaced. A mixture of primary, secondary and tertiary amines and a quaternary ammonium salt results.

Primary amine (1 mark)

Ammonia with ethyl iodide gives the primary amine, ethylamine.

\ce{C2H5I + NH3 -> C2H5NH2 + HI}
Secondary amine (1 mark)

The primary amine still has a lone pair, so it acts as a nucleophile toward more alkyl halide and gives the secondary amine, diethylamine.

\ce{C2H5I + C2H5NH2 -> (C2H5)2NH + HI}

The reaction goes on to the tertiary amine \ce{(C2H5)3N} and the quaternary salt \ce{[(C2H5)4N]+ I-}. Adding a strong alkali such as KOH frees the amines from their salts and the amines are separated by fractional distillation; using excess ammonia favours the primary amine.

Ethylamine (primary)
Diethylamine (secondary)
2025 Annual I, Set D, Q.2 (x) OR1.5+1.5 = 3 marks
Write down the reactions of ethyl amine (\ce{C2H5-NH2}) with: (a) \ce{CH3CH2-Cl} (b) \ce{HNO2/HCl}
Solution
(a) With ethyl chloride (1.5 marks)

Ethylamine is a primary amine with a lone pair on N. It attacks ethyl chloride (alkylation of an amine) and gives the secondary amine diethylamine. With more ethyl chloride the reaction continues to triethylamine and then to the quaternary salt.

\ce{CH3CH2Cl + C2H5NH2 -> (C2H5)2NH + HCl}\quad(\text{diethylamine})
\ce{(C2H5)2NH + C2H5Cl -> (C2H5)3N + HCl}\quad(\text{triethylamine})
Diethylamine
(b) With nitrous acid and hydrochloric acid (1.5 marks)

A primary amine reacts with nitrous acid in the presence of HCl, keeping the mixture below 10^\circ\mathrm{C}, to form a diazonium salt. The alkyl diazonium salt is unstable, unlike the benzene diazonium ion.

\ce{C2H5NH2 + HNO2 + HCl -> [C2H5-N#N]+Cl- + 2H2O}\quad(\text{ethyl diazonium chloride})

Exam focus

  1. Preparation of primary and secondary amines: alkyl halide with ammonia (nucleophilic substitution), stepwise alkylation, role of excess ammonia. (short)
    Asked in 2026 (Set D Q.2 (ix) OR; Set D Q.2 (iii) (a) is solved in unit 17) and 2025 (Set B Q.2 (viii) OR (a)); also the book's long question 3.
    Go to answer
  2. Preparation of ethylamine by reduction of a nitrile or an amide, with reagents (LiAlH4, H2/Ni) and equations. (short)
    Asked in 2025 (Set B Q.2 (viii) OR (b) and (c)); book long questions 3 and 4.
    Go to answer
  3. Reduction of nitro compounds to amines: nitrobenzene to aniline with Sn/HCl, 4-nitroaniline with Fe/H2SO4. (short)
    Asked in 2026 (Set B Q.2 (ix) OR) and 2025 (Set D MCQ 17).
    Go to answer
  4. Why amines are basic, and why methylamine is a stronger base than ammonia (electron-donating alkyl group, pKb 3.38 against 4.76); secondary stronger than primary. (mcq)
    Asked in 2026 (Set B MCQ 13 and Set D MCQ 13); book short questions iii and vi.
    Go to answer
  5. Reactions of ethylamine: alkylation with alkyl halide to a secondary amine, and the reaction with HNO2/HCl to a diazonium salt below 10 degrees C. (short)
    Asked in 2025 (Set D Q.2 (x) OR); book short question vii.
    Go to answer
  6. Preparation, stability and uses of diazonium salts and azo dyes. (long)
    Not in the 2025 or 2026 paper beyond Set D 2025 Q.2 (x) OR (b); book short questions ii and vii and long question 6.
    Go to answer
  7. Compare primary and secondary amines, and different amines in structure and properties. (long)
    Not in the 2025 or 2026 paper; book long questions 5 and 7 and an SLO of the unit.
    Go to answer
  8. High boiling points and water solubility of amines (hydrogen bonding), and the sp3 structure with a lone pair. (short)
    Not in the 2025 or 2026 paper; book short question v.
    Go to answer

Mark this unit as done to fill the Class 11 Chemistry ring.

Next, unit 21
Organic synthesis

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