Bihar Board - Class 12 Chemistry - Chapter 13: Amines Long Answer Question
Long Question Answer
Question. 1. Explain the following:
(i) Aniline dissolves in aqueous HCl.
(ii) Amines are stronger bases than ammonia.
(iii) Aniline is a weaker base than ethylamine.
(iv) The amino group in ethylamine is basic whereas that in acetamide is not basic.
(v) Cyclohexylamine is a stronger base than aniline.
Answer⇒ (i) The solubility of C6H5NH2 in dil. HCl is due to the formation of water soluble salt.
C6H5NH2 + H3++Cl- C6H5NH3 +Cl- +H2O Anilinium chloride
(ii) Amines have electron repelling groups which increase the electron density on nitrogen, while ammonia has no such group.
(iii) In aniline the electron pair of nitrogen atoms is delocalised due to resonance and hence less available for protonation while ethylamine does not undergo resonance.
(iv) In amides the lone pair of electrons on nitrogen atom is delocalised and hence less available for protonation than in amines where no resonance is possible and thus
acetamide is a weaker base than ethylamine.
(v) Aniline is a weaker base than cyclohexylamine because of resonance while there is no resonance in cyclohexylamine.
Question. 2. (i) Write str amines corresponding to C4H11N .
(ii) Write IUPAC names of
(iii) What type of isomeri different pairs of amines?
Answer⇒ Eight isomeric amines are Possible
Question. 3. Arrange the following in increasing order of their basic strength:
(i) C2H5NH2 , C6H5NH2 , NH3 , C6H5CH2NH2 and (C2H5)2NH
(ii) C2H5NH2 , (C2H5)2NH , (C2H5)3N , C6H5-NH2
(iii) CH3NH2 , (CH3)2NH , (CH3)3N , C6H5NH2 , C6H5CH2NH2
Answer⇒ (i) (C2H5)2NH >C2H5NH2>C6H5CH2NH2> NH3>C6H5NH2
(ii) (C2H5)2NH > (C2H5)3N> C2H5NH2> C6H5-NH2
(iii) (CH3)2NH >CH3NH2> (CH3)3N>C6H5CH2NH2>C6H5NH2
Question. 4. Write common name of the following compounds :
Answer⇒ (i) o-aminobenzoic acid.
(ii) p-methoxy aniline (or p-anisidine).
(iii) neopentylamine.
(iv) Sec-butyl dimethyl amine.
Question. 5. Explain the following reactions by taking one suitable example in each case :
(i) Hoffman’s bromamide reaction
(ii) Gattermann’s reaction
(iii) Carbylamine reaction
(iv) Diazotization
(v) Coupling reaction
(vi) Ammonolysis
(vii) Acetylation
(viii) Gabriel phthalimide synthesis
Answer⇒ (i) Hoffman’s Bromamide Reaction : When an amide (aliphatic or aromatic) is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide.
RCONH2 
. Amine
C6H5CONH2 
Aromatic amide Aniline
(Benzamide)
CH3CONH2 
Ethanamide Methanamine
This reaction is extremely useful for stepping down (or descending) homologous series.
(ii) Gattermann’s reaction: When benzene ring and diazonium salt is treated with Cu/HCl or Cu/HBr, chlorobenzene or bromobenzene is formed. The yield is about 40%.
(iii) carbylamine reaction : Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH produce isocyanides or carylamines which have an unpleasant odour.
R-NH2+CHCl3+3KOH (alc.)
.1° amine
In contrast, secondary and tertiary amines (both aliphatic and aromatic) do not give this test.
(iv) Diazotization : The reaction of converting aromatic primary amines into diazonium salts with a cold (273 -278 K) the solution of nitrous acid is called diazotization.
(v) Coupling Reaction : Arenediazonium salts react with highly reactive (i.e., electron-rich) aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
Ar-N=N-Ar
(vi) Ammonolysis : Reactions of an alkyl halide (R-X) with an alcoholic solution of ammonia giving a mixture of 1 ,2,3 amines and quaternary salt is called ammonolysis.
R-X+HNH2 
(excess) (Chiefproduct)
If R-X is used in excess, a mixture of 1 ,2,3 amines along with some quaternary ammonium salts are obtained.
It is a typical nucleophilic substitution reaction.
(vii) Acetylation : The replacement of an active hydrogen of alcohols, phenols or amines with an acyl group (RCO) to form the corresponding esters or amides is called acetylation. Acetylation is carried out in the presence of a base like pyridine, dimethylaniline etc. by acetyl chloride or acetic anhydride.
CH3COCl + C2H5OH 
Acetyl chloride ethanol Ethyl acetate
(CH3CO)2O+C2H5NH2 CH3CO NHCH2CH3 + CH3COOH
Acetic Ethylamine N-Éthylacetamide anhydride
(viii) Gabriel’s phthalimide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic KOH. Then potassium phthalimide is heated with an alkyl halide to yield on N alkyl phthalimide which is hydrolysed to phthalic acid and a primary amine by heating with HCl or KOH solution. This synthesis is very useful for the preparation of pure aliphatic primary amines.
Question. 6. How is aniline prepared commercially ? Describe carbylamine test for primary amines.
Answer⇒ Aniline is prepared commercially by the reduction of nitrobenzene using iron and hydrochloric acid.
Fe+2HCl FeCl2 +2H ] 3
C6H5NO2 +6H C6H5NH2+2H2O
C6H5NO2+3Fe+6HCl + C6H5NH2+3FeCl2+2H2O
(Nitrobenzene) (Aniline)
Because of the presence of hydrochloric acid in the reaction mixture, aniline hydrochloride ( C6H5NH3+Cl-) is obtained. This on treatment with aqueous sodium carbonate gives aniline.
Question. 7. Write IUPAC name of following and classify them into primary, secondary and tertiary amines :
(i) (CH3)2 CHNH2
(ii) CH3(CH2)2 NH2
(iii) CH3NHCH (CH3)2
(iv) (CH3)3CNH2
(v) C6H5NHCH3
(vi) (CH3CH2)2NCH3
(vii) m-BrC5H4NH2
Answer⇒ (i) (CH3)2 CHNH2:1- amino-1-methyl propane (primary)
(ii) CH3(CH2)2 NH2
1-amino propane (primary)
(iii) CH3NHCH (CH3)2
methyl isopropyl amine (secondary)
(iv) (CH3)3CNH2
tert-butylamine (primary)
(v) C6H5NHCH3
N-Methyl benzenamine (secondary)
(vi) (CH3CH2)2NCH3
Diethyl methyl amine (tertiary)
(vii) m-BrC5H4NH2
3-iso bromo aniline (primary)
Question. 8. How will you prepare :
(i) Benzylamine by Gabriel Pthalimide synthesis.
(ii) Benzylamine from benzene ?
Answer⇒
Question.9. State reasons for the following:
(i) pK the value for aniline is more than that for methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not soluble in water.
(iii) Primary amines have higher boiling points than tertiary amines.
Answer:
(i) Higher the pK, value, lower will be the basicity therefore aniline is less basic than methylamine because the lone pair of electrons on nitrogen atom gets delocalized over the benzene ring are unavailable for protonation due to resonance in aniline which is absent in case of alkylamine.
(ii) Ethylamine is soluble in water due to its capability to form H-bonds with water while aniline is insoluble in water due to larger hydrocarbon part which tends to retard the formation of H-bonds.
(iii) Due to presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom, do not undergo H- bonding. As a result, primary amines have higher boiling points than 3 amines.
Question.10. Give a chemical test to distinguish between ethylamine and aniline.
Answer : Ethylamine and aniline:
By Azo dye test: It involves the reaction of any aromatic primary amine with HNO2 (NaNO2+dil.HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellow, orange or red coloured dye is obtained.
Aliphatic 1 amines under these conditions give a brisk evolution of N2, gas with the formation of 1 alcohol i.e. the solution remains clear.
CH3CH2NH2+HONO
Question.11. Describe the following giving the relevant chemical equation in each case:
(i) Carbylamine reaction
(ii) Hofmann's bromamide reaction
Answer :
(i) Carbylamine reaction: Aliphatic and aromatic primary amines on heating with chloroform and ethanol KOH form isocyanides or carbylamines which are foul smelling substances. This reaction is known as carbylamine reaction.
R-NH2 + CHCl3+3KOH 
Primary amine Chloroform Carbylamine
(ii) Hofmann's bromamide reaction: Primary amines can be prepared by treating an amide with Br2 in an aqueous or alcoholic solution of NaOH.
Question.12. Give reasons:
(a) Aniline is a weaker base than cyclohexylamine.
(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Answer :
(a) In aniline, the lone pair of electrons on the N-the atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen decreases.
But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of -electrons. Hence aniline is a weaker base than cyclohexylamine.
(b) Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further 2 and 3 alkyl amine.
Question.13. Account for the following:
(i) Aniline does not give Friedel-Crafts reaction.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) pK methylamine is less than that of aniline.
Answer : (i) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt.
C6H5NH2 + AlCl3 C6H5N+ + H2 Al Cl3-
Lewis base Lewis acid
As a result, N aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel Craft reaction.
(ii) Ethylamine is soluble in water due to its capability to form H-bonds with water while aniline is insoluble in water due to larger hydrocarbon part which tends to retard the formation of H-bonds.
(iii) In aniline, due to resonance lone pairs of electrons of nitrogen atom are delocalised due to which it is a weaker base than methyl amine. Hence Aniline has high pKb molecule i.e., methylamine has less pK molecule.
Question. 14. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3)2 CHNH2 ,
(ii) CH3 (CH2)2 NH2,
(iii) CH3NHCH (CH3)2,
(iv) (CH3)3 CNH2,
(v) C6H5NHCH3,
(vi) (CH3CH2)2 NCH3,
(vii) mBrC6H4NH2.
Answer :
Question. 15. Arrange the following:
(i) In decreasing order of the pKb values :
C2H5NH2 , C6H5NHCH3 , (C2H5)2NH and C6H5NH2
(ii) In increasing order of basic strength :
C6H5NH2 , C6H5N(CH3)2 , (C2H5)2 NH and CH3NH2
(iii) increasing order of basic strength
(a) Aniline, p-nitroaniline and p-toluidine
(b) C6H5NH2 , C6H5NHCH3 , C6H5CH2NH2
(iv) In decreasing order of basic strength in gas phase
C2H5NH2 , (C2H5)2 NH , (C2H5)3N and NH3
(v) in increasing order of boiling point
C2H5OH , (CH3)2NH , C2H5NH2
(vi) Increasing order of solubility in water
C6H5NH2 , (C2H5)2NH , C2 H NH2
Answer :
(i) In decreasing order of pKb values
C6H5NH2 > C6H5NHCH3 >C2H5NH2> (C2H5)2NH
(ii) In increasing order of basic strength
C6H5NH2 < C6H5N(CH3)2 <CH3NH2 < (C2H5)2NH
(iii) Increasing order of basic strength
(i) p-nitroaniline < aniline < p-toluidine.
(iv) Decreasing order of basic strength in gas phase.
(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3
(v) Increasing order of boiling point
(CH3)2NH = C2H5NH2 < C2H5OH
(vi) Increasing order of solubility in water
C6H5NH2 <(C2H5)2NH < C2H5NH2
Question.16. How are the following conversions carried out:
(a) Aniline to p-hydroxyazobenzene
(b) Ethanoyl chloride to Ethanenitrile
Answer:
Question.17. How would you account for the following:
(a) Aniline is a weaker base than cyclohexylamine.
(b) Methylamine in aqueous medium gives reddish brown precipitate with FeCl3.
Answer:
(a) In aniline, the lone pair of electrons on the N-the atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen decreases.
But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of -electrons. Hence aniline is a weaker base than cyclohexylamine.
(b) Methylamine being more basic than H2O, it accepts a proton from water liberating OH- ions.
These OH- ions combine with Fe+3 ions present in H2O to form reddish brown precipitate of hydrated ferric oxide.
Question.18. Illustrate the following reactions:
(a) Sandmeyer's reaction
(b) Coupling reaction
Answer:
(a) Sandareyer's reaction: Aniline reacts with NaNO2 in HCl at 273-278 K giving diazonium salt which further reacts with cuprous chloride/bromide to give chloro or bromo benzene.
This reaction is Sandmeyer's reaction.
(b) Coupling reaction: Arene diazonium salts react with highly reactive aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
Ar-N-N-Ar. This reaction is known as coupling reaction.
Question.19. Explain the following reactions:
(a) Gabriel Phthalimide reaction
(b) Coupling reaction
Answer:
(a) Gabriel phthalimide synthesis: It is used to prepare 1 amine (Primary amine). The starting compound is a phthalimide. But aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Example:
(b) Coupling reaction is a class of organic reaction in which two chemical species are joined together with the help of a metal catalyst. When benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position coupled with the diazonium salt to form p-hydroxyazobenzene.
Question.20. Write the chemical equations involved in the following reactions:
(i) Hoffmann-bromamide degradation reaction
(ii) Carbylamine reaction
Answer:
(i) Hoffmann's bromamide reaction: In this reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. Therefore the amine so formed has one carbon atom less than that of amide.
Example
(ii) Carbylamine reaction: This reaction is used to distinguish primary amines from 2 and 3 amines as it is only given by 1 amines with the production of a very bad smelling organic compound.
For example:
Question 21. Complete the following reaction equations:
(i) C6H5N2Cl +CH3COCl
(ii) C2H5NH2 +C6H5SO2C
(iii) C2H5NH2 + HNO2
Answer:
(i) C6H5N2Cl +CH3COCl C6H5COCl +CH3Cl +N2
(iii) C2H5NH2 + HNO2 
Question.22. How will you obtain:
Ethyl amine from methyl cyanide
Methyl amine from acid amide
Ethyl amine from ethyl alcohol
Ethyl amine from methyl amine.
Answer:
1. Ethyl amine from methyl cyanide:
CH3CN +2H2 
2. Methyl amine from acid-amide:
CH3CONH2+Br2+4KOH CH3-NH2+2KBr +K2CO3+2H2O
3. Ethylamine from ethyl img
C2H5OH +NH3
4. Ethyl amine from methyl amine:
CH3NH2 
CH3CN +4[H]
Ethyl Amine
Question.23. Write any four differences between ethyl cyanide and ethyl isocyanide.
Answer: Difference between ethyl cyanide and ethyl isocyanide:
Test | Ethyl cyanide | Ethyl Isocyanide |
1. Smell 2. Solubility in Water 3. Acid hydrolysis 4.Reduction | Fast but not bad Soluble Propionic acid is formed. C2H5CN +2H2O C2H5COOH +NH3 Propylamine is obtained. C2H5CN+2H2 C2H5CH2NH2 | Very bed smell Less soluble Ethyl amine is formed. C2H5NC +2H2O C2H5NH2+HCOOH Ethyl methyl amine is obtained. C2H5NC+2H2 C2H5NHCH3 |
Question 24. Boiling point of aniline is higher than the hydrocarbons of comparable molecular masses, but lower than corresponding alcohols and carboxylic acids. Clarify the statement.
Answer: Primary amines are polar and in them dipolar [N--N+] a bond is present, in which molecules of amines are associated by hydrogen bonds. That is why their boiling point is higher than alkanes of comparable molecular masses.
Due to lesser electronegativity of nitrogen than oxygen, association in alcohols and carboxylic acids occur due to strong H- bonds. Whereas in amines association is due to weak H- bond. Therefore, the boiling and melting point of amines is lower than corresponding alcohols and carboxylic acids.
Question.25. How will you obtain the following:
(i) Sulphanilic acid from Aniline
(ii) p-amino azo benzene from aniline.
Answer:
(i) Sulphanilic acid from Aniline
(ii) p-amino azo benzene from aniline.
