AMINES

Basic character patterns: - 

Higher Kb value, stronger the basicity

pKb = - log Kb

higher pKb weaker the basicity

lower pKb stronger the basicity

Basicity Order: 

Aniline (4.2x10-10) < N-methylamine (5.0x10-10) < N,N-dimethylamine (11.5x10-10) < Benzylamine (2x10-5) < Ammonia (1.8x10-5) < trimethylamine (0.6x10-4) < methylamine (4.5x10-4) < ethylamine (5.1x10-4) < dimethylamine (5.4x10-4) < triethylamine (5.6x10-4) < diethylamine (10x10-4)

Basicity Order: 

ammonia < trimethylamine < methylamine < dimethylamine

ammonia < ethylamine < triethylamine < diethylamine

basicity depends on the following factors: 

a. +I effect of the alkyl groups

b. Extent of H bonding with water molecules

c. Steric effects of the alkyl groups.

Basicity order of aromatic amines:

o-Chloroaniline (pKb 13.30) <  < o-Nitroaniline (pKb13.22) < Diphenylamine (pKb 13.2) < p-Nitroaniline (pKb13) < m-Nitroaniline (pKb12.64) < m-Chloroaniline (pKb11.52) < p-Chloroaniline (pKb10.0) < m-Anisidine (pKb 9.81) <o-Toluidine (pKb 9.58) < o-Phenylenediamine (pKb 9.52) < o-Anisidine (pKb 9.51) < Aniline (pKb 9.38) < m-Toluidine (pKb 9.34) < N-methylaniline (pKb 9.30) < p-Toluidine (pKb 9.21) < m-Phenylenediamine (pKb 9.00) < N,N-dimethylaniline (pKb 8.92) < p-Anisidine (pKb 8.71) < p-Phenylenediamine (pKb 7.86) < Ammonia (pKb 4.75) < Benzylamine (pKb 4.70) < Methylamine (pKb 3.28)

Intext Questions

Ans. (i) primary (ii) tertiary (iii) primary (iv) secondary

Q2 (i) Write structures of different isomeric amines corresponding to the molecular formula C5H11N

(ii) Write IUPAC names of all the isomers

(iii) What type of isomerism is exhibited by different pairs of amines?

Ans. (i) and (ii) : Primary structures

(a) CH3CH2CH2CH2NH2 (Butanamine)

(b) CH3CH2CH-NH2CH3 (Butan-2-amine)

(c) (CH3)2CHCH2NH2 (2-Methylpropanamine)

(d) (CH3)3CNH2 (2-Methylpropan-2-amine)

Secondary Structures: 

(e) CH3CH2CH2-NH-CH3 (N-Methylpropanamine)

(f) (CH3)2CH-NH-CH3 (N-Methylpropan-2-amine)

(g) CH3CH2-NH-CH2CH3 (N-Ethylethanamine)

Tertiary structure: 

(h) (CH3)2N -CH2CH3 (N,N-Dimethylethanamine)

(iii) Isomerism: 

Position isomers: (a), (b), (e) and (f)

Chain isomers: (a) and (c), (a) and (d), (b) and (c), (b) and (d).

Metamers: (e) and (g), (f) and (g)

Functional isomers: all primary amines are functional isomers of secondary and tertiary amines and vice versa.

Q3: How will you convert: 

(i) Benzene into Aniline

(ii) Benzene into N,N-dimethylaniline

(iii) Cl-(C2H4)2-Cl into hexane-1,6-diamine?

(i) Benzene into Aniline

(ii) Benzene into N,N-dimethylaniline

(iii) Cl-(C2H4)2-Cl into hexane-1,6-diamine?

Q4 Arrange the following in increasing order of their basic strength

Ans.

Q5.

Ans. 

Q6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Q7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Ans.

Q8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Ans. In all four structural isomers are possible. These are: 

Q9 Convert: 

(i) 3-Methylaniline into 3-nitrotoluene 

(ii) Aniline into 1,3,5-tribromoaniline

Ans.

Q1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) (CH3​)2 CHNH2 

(ii) CH3(CH2)2NH2

​(iii) CH3 NHCH(CH3)2

(iv) (CH3)3CNH2

(v) C6H5NHCH3

(iv) (CH3CH2)2NCH3 

(vii) m-BrC6H4NH2

Ans.  

The IUPAC names along with the classification are given below:

(i) 1-Methylethanamine (primary amine)

(ii) Propan-1-amine (primary amine)

(iii) N-Methyl-2-methylethanamine (secondary amine)

(iv) 2-Methylpropan-2-amine (primary amine)

(v) N-methylbenzenamine or N-methylaniline (secondary amine)

(vi) N-Ethyl-N-methylethanamine (tertiary amine)

(vii) 3-Bromobenzenamine or 3-bromoaniline (primary amine)

Q2 Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine        

(ii) Secondary and tertiary amines

(iii) Ethylamine and aniline                           

(iv) Aniline and benzylamine

(v) Aniline and N-methylaniline.

Ans. (i) Methylamine and dimethylamine :             

Carbylamine test.

Methylamine, on heating with alc. KOH solution and chloroform form foul-smelling methyl isocyanide. This test is not given by dimethylamine.

(ii) Secondary and tertiary amines :

Secondary amines give Libermann nitrosoamine test.

On heating with nitrous acid (prepared in situ), they give yellow coloured oily N-nitrosoamine.

Tertiary amines do not give such test.

(iii) Ethylamine and aniline :                          

Azo dye test. 

Aniline on diazotization ( ice cold nitrous acid solution) followed by coupling with 2-naphthol (in alkaline solution) forms brilliant orange or red dye. 

Ethylamine will not form dye. It will give brisk effervescence (due to liberation of nitrogen gas) but solution remains clear.

(iv) Aniline and benzylamine :

Azo dye test.

Aniline on diazotization ( ice cold nitrous acid solution) followed by coupling with 2-naphthol (in alkaline solution) forms brilliant orange or red dye. Benzylamine will not give such test.

(v) Aniline and N-methylaniline :

Carbylamine test.

Aniline, on heating with alc. KOH solution and chloroform form foul-smelling methyl isocyanide. This test is not given by N-methyl aniline.

Q3 Account for the following:

(i) pKb   of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) although amino group is ortho and para directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. 

(v) aniline does not undergo friedel-crafts reaction. 

(vi) diazonium salts of aromatic amines are more stable than those of aliphatic amines. 

(vii) gabriel phthalimide synthesis is preferred for synthesising primary amines

Ans. 

(i) pKb of aniline is more than that of methylamine.

In aniline, the lone pair of electrons on N atom is in resonance with benzene ring. Hence, it cannot be easily donated to an acid. This decreases its basicity. In methyl amine, the +I effect of methyl group increases the electron density on N atom so that the lone pair of electrons on N atom can be easily donated to an acid. Hence, methylamine is more basic than aniline. Higher is the basicity, lower is the pKb and vice versa. 

(ii) Ethylamine is soluble in water whereas aniline is not. With increase in the molecular weight, the solubility decreases. Aniline has higher molecular weight than ethylamine.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Due to the +I effect of −CH3 group, methylamine is more basic than water. Therefore, in water, methylamine produces OH−  ions by accepting H+ ions from water. OH− ions react with ferric chloride to precipitate hydrated ferric oxide. 

(iv) Although the amino group is ortho and para directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

During the Nitration process, the mixture used (Con HNO3+Con H2SO4) protonates anilinium ion and nitroaniline is formed. The anilium ion acts as a deactivating group and directs the NO2 nucleophile to meta position and meta isomer of nitroaniline is formed.

(v) Aniline does not undergo Friedel-crafts reaction.

Aniline, N-substituted and N,N-disubstituted anilines do not undergo Friedel-Craft’s reactions due to salt formation with aluminium chloride which is used as a Lewis acid catalyst. The positive charge on nitrogen is strongly electron-withdrawing and thus deactivates the ring for further acylation or alkylation reactions. During Friedel-Crafts reaction, anhydrous AlCl3 is used as a lewis acid for the generation of electrophile from electrophilic reagent. If aniline is used as base there won’t be any generation of electrophile taking place and hence reaction does not take place.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Diazonium salt of aromatic amine is stable due to the resonance. it carries N atom with a positive charge. such a delocalisation is not possible in aliphatic amines. hence it is less stable

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines

Gabriel pthalmide reaction gives pure primary amines without any contamination in secondary and tertiary amines. hence they are more prefered for the synthesis of primary amines. Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not comprised in this synthesis. Thus, a pure 1° amine can be obtained. Thus, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Q4 Arrange the following

(i) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH & C6H5NH2

(ii) In increasing order of basic strength: 

C6H5NH2, C6H5N(CH3)2, (C2H5)2NH, and CH3NH2

(iii) In increasing order of basic strength: 

     (a) Aniline, p-nitroaniline and p-toluidine 

Ans. 

(i) Stronger base has lower pKb value.

C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH

(ii) Increasing order of basicity is as follows:

C6H5NH2<C6H5NHCH3<CH3NH2<(C2H5)2NH

(iii) Increasing order of basic strength.

(a) p-nitroaniline < aniline < p-toluidine

(b) PhNH2 <PhNHCH3 <PhCH2NH2

​ 

(iv) In decreasing order of basic strength in gas phase:

Et3N>Et2NH>EtNH2 >NH3

​

The stabilization of conjugate acids due to hydrogen bonding is absent.

(v) Increasing order of boiling point

Me2NH < EtNH2 < EtOH

(vi)  Increasing order of solubility in water:

PhNH2 <Et2NH < EtNH2

Q5 How will you convert: 

(i) Ethanoic acid into methanamine 

(ii) Hexanenitrile into 1-aminopentane 

(iii) Methanol to ethanoic acid 

(iv) Ethanamine into methanamine 

(v) Ethanoic acid into propanoic acid 

(vi) Methanamine into ethanamine 

(vii) Nitromethane into dimethylamine 

(viii) Propanoic acid into ethanoic acid

Ans. 

(v)

(vi)

(vii)

(viii)

Q6 Describe a method for the identification of primary, secondary and tertiary amines. Also, write chemical equations of the reactions involved.

Ans. 

Hinsberg's test is used for the identification of primary, secondary, and tertiary amines.

Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl).

It reacts differently with primary, secondary, and tertiary amines.

(i) Hinsberg's reagent reacts with primary amines to form N− alkylbenzenesulphonyl amide which is acidic in nature and soluble in alkali.

Note: N− alkylbenzenesulphonyl amide contains a strong electron-withdrawing sulphonyl group. Due to this, the H− atom attached to nitrogen can be removed easily. Hence, it is acidic.

(ii) Hinsberg's reagent reacts with secondary amines to form a sulphonamide which is insoluble in alkali.

Note: As there is no hydrogen atom attached to the N atom in the sulphonamide, it is not acidic and insoluble in alkali.

(iii) Hinsberg's reagent does not react with tertiary amines.

(ii) Diazotisation

Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.

For example, on treatment with NaNO2 and HCl at 273 - 278 K, aniline produces benzenediazonium chloride, with NaCl and H2O as by-products.

(iii) Hoffmann bromamide reaction

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.

(iv) Coupling reaction

The reaction of joining two aromatic rings through the - N=N - bond is known as coupling reaction. Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form coloured azo compounds.

It can be observed that, the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution.

(v) Ammonolysis

When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino ( - NH2) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines, and also a quaternary ammonium salt as shown.

(vi) Acetylation

Acetylation (or ethanoylation) is the process of introducing an acetyl group into a molecule.

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution when treated with acid chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of - NH2or > NH group by the acetyl group, which in turn leads to the production of amides. To shift the equilibrium to the right hand side, the HCl formed during the reaction is removed as soon as it is formed. This reaction is carried out in the presence of a base (such as pyridine) which is stronger than the amine.

When amines react with benzoyl chloride, the reaction is also known as benzoylation.

For example,

(vii) Gabriel phthalimide synthesis

Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

Q8 Accomplish the following conversions 

 (i) Nitobenzene to benzoic acid  

(ii) Benzene to m-bromophenol 

 (iii) Benzoic acid to aniline  

(iv) Aniline to 2,4,6-tribromofluorobenzene 

 (v) Benzyl chloride to 2-phenylethanamine 

 (vi) Chlorobenzene to p-chloroaniline 

 (vii) Aniline to p-bromoaniline  

(viii) Benzamide to toluene  

(ix) Aniline to benzyl alcohol.

Ans. 

Q9 Give the structures of A, B and C in the following reactions: 

Ans. (i) 

Q10 An aromatic compound A on treatment with aqueous ammonia and heating forms compound B which on heating with Br2 and KOH forms a compound C of molecular  formula C6 H7N. Write the structures and IUPAC names of compounds A,B and C.

Ans. The aromatic compound A is benzoic acid C6H5COOH. On treatment with aq NH3 and heating forms compound B, which is benzamide C6H5CONH2. Benzamide on heating with bromine and KOH forms a compound C of molecular formula C6H7N, which is aniline C6H5NH2. The reaction is called Hoffmann bromamide degradation. 

Q 11 Complete the following reactions: 

(i) C6H5NH2 + CHCl3 + 3 KOH (alc) → 

(ii) C6H5N2Cl + H3PO2 + H2O →

(iii) C6H5NH2 + H2SO4 (Conc) →

(iv) C6H5N2Cl + C2H5OH → 

(v) C6H5NH2 + Br2(aq) → 

(vi) C6H5N2 + (CH3CO)2 →

(vii) C6H5N2Cl (i)HBF4 (ii) NaNO2/ heat 

Q12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Ans. Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

Q13  Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Ans. The reaction of aromatic primary amines with nitrous acid gives aromatic diazonium salts.

The reaction of aliphatic primary amines with nitrous acid gives aromatic diazonium salts which decompose to form a mixture of compounds containing alkyl chlorides, alkenes and alcohols. Usually, alcohols are the major products. The reactions are shown above.

Q 14 Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Ans. 

(i) Amines lose a proton to form amide ion. Alcohols lose a proton to form alkoxide ion.

RNH2 → RNH- + H+ 

R−OH → RO- + H+

O is more electronegative than N, the negative charge is more easily accomodated in RO- than in R−NH- . Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In primary amines, N atoms have two H atoms which results in extensive intermolecular H bonding. In tertiary amines, N atoms do not have H atoms and hydrogen bonding is not possible.

Hence, primary amines have higher boiling point than tertiary amines.

(iii) Aliphatic amines stronger bases than aromatic amines due to following reasons:

(a) Aromatic amines have resonance due to which lone pair of electrons on N atom is delocalized over benzene ring and is less available for protonation.

(b) The stability of aryl amine ions is lower than the stability of alkyl amines. Protonation of aromatic amines is not favoured.