Bihar Board Class 12 Chemistry Solutions Chapter 13 Amines
BSEB Bihar Board Class 12 Chemistry Solutions Chapter 13 Amines
Bihar Board Class 12 Chemistry Amines Intext Questions and Answers
Question 1.
Classify the following amines as primary, secondary or tertiary?
(iii) (C2H5)2 CHNH2,
(iv) (C2H5)2 NH.
Answer:
(i) Primary amine,
(ii) tertiary amine,
(iii) primary amine,
(iv) secondary amine.
Question 2.
(i) Write structures of different isomeric amines corresponding to the molecular formula C4H11N.
(ii) Write IUPAC names of all isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:
(i) C4H11N stands for the following isomeric amines
(a) Primary amines
(ii) IUPAC names of all isomeric amines have been given in brackets along with their structures.
(iii) (a) Chain isomerism -CH3-CH2-CH2-CH2NH2 and
Question 3.
How will you convert (i) Benzene into aniline.
(ii) Benzene into N, N-dimethylaniline.
(iii) Cl-(CH2)4-Cl into hexane-1,6-diamine?
Answer:
(i) Benzene into aniline-
Question 4.
Arrange the following in increasing order of their basic strength.
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, and (C2H5)2NH.
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2.
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2.
Answer:
(i) C6H5NH2 < NH3 < C6H5CH2 NH2 < C2H5NH2 < (C2H5)2 NH.
(ii) C6H5NH2 < C2H5 NH2 < (C2H5)3 N < (C2H5)2 NH.
(iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3 N < CH3NH2 < (CH3)2 NH.
Question 5.
Complete the following acid-base reactions and name the products:
(i) CH3 CH2 CH2 NH2 + HCl →
(ii) (C2H5)3N + HCl →
Answer:
(i) CH3 CH2 CH2 NH2 + HCl →CH3 CH2 CH2 NH3Cl– n-Propylinium chloride
(ii) (C2H5)3N + HCl → [(C2H5)3 Triethyl ammonium chloride
Question 6.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Answer:
Question 7.
Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:
Question 8.
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.
Answer:
Question 9.
Convert (i) 3-Methylaniline into 3-nitrotoluene, (ii) Aniline into 1,3,5-tribromobenzene.
Answer:
(i) 3-Methylaniline into 3-nitrotoluene-
Bihar Board Class 12 Chemistry Amines Text Book Questions and Answers
Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiery 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) m- BrC6H4NH2.
Answer:
Question 2.
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-methlyaniline.
Answer:
(i) Methylamine and dimethylamine-Methylamine-a primary amine respond to Carbylamine reaction whereas dimethylamine does not.
(ii) Secondary and tertiery amines-A secondary amine reacts with benzene sulphonyl chloride to form N, N-dialkyl benzene sulphonamide which does not dissolve in KOH. A tertiary amine does not react with benzene sulphonyl chloride.
(iii) Ethylamine and aniline-Ehtylamine on reaction with nitrous acid (HNO2) gives out N2 gas, whereas aniline does not.
(v) Aniline and N-methyl aniline-Aniline is a primary aromatic amine, N-methyl aniline is a secondary aromatic amine. They may be distinguished by Carbylamine test.
Aniline gives phenyl isocyanide on heating with chloroform (CHCl3) and alcoholic KOH which is extremely foul-smelling while N-methyl aniline does not give this test.
Question 3.
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 o- and p- 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.
Answer:
(i) Aniline is a much weaker amine than methylamine. In methylamine, the lone pair of electrons on N of 
is made many phones readily available for sharing due to the +1 effect exerted by – CH3 group thus making its KJ, large or pkb small. On the other hand, the benzene ring present in aniline withdraws electrons towards itself making the lone pair of electrons less available on N of NH2 group. Here its pkb is large.
(ii) Ethylamine is soluble in water because it can form H-bonds with water molecules.
On the other hand aniline is insoluble in water due to the larger hydrocarbon part [C6H5 as compared to C2H5 of ethylamine] which tends to retard the formation of H-bonds.
(iii) Methylamine like ammonia reacts with transition metal ions in aqueous solutions like Fe3+ [of ferric chloride] to form complex like [Fe (NH3)6]3+ as methylamine in aqueous solution is a good nucleophile and hence attacks metal ions like Fe3+ which are electrophiles. The iron complex in aqueous solution precipitates as hydrated ferric oxide.
(iv) Aniline, in the strongly acidic medium (like in a mixture of cone. HNO3 and cone. H2SO4) is protonated to form anilinium ion C6H5 NH3 which is meta-directing. That is why NH2 group in spite of being 0- and p-directing also forms significant amount of meta-derivative during nitration in strongly acidic medium in electrophilic substitution reactions.
(v) Aniline does not undergo Friedel Crafts reaction like alkylation and acetylation because it forms salt with aluminium chloride which is used at a catalyst and is a Lewis acid. Due to this, nitrogen of – NH2 of aniline acquires a positive charge and hence acts as a strong deactivating group for further reaction.
(vi) Diazonium salts of aromatic amines are more stable than that of aliphatic amines. It is because even at low-temperature diazonium salts of aliphatic amines decompose to give out N2 gas.
(vii) Gabriel’s phthalimide synthesis is preferred for synthesis of primary aliphatic amines because primary amines are obtained in the pure form by this method. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Question 4.
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)2NH 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)2NH, (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, C2H5NH2
Answer:
(i) In decreasing order of pkb values
C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2 NH.
(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)2 NH < C2H5NH2.
Question 5.
How will you convert
(i) Ethanoic acid into methenamine,
(ii) Hexanenitrtile into 1-amino pentane,
(iii) Methanol to ethanoic acid,
(iv) Ethanamine into methenamine,
(v) Ethanoic acid into propanoic acid,
(vi) Methanamine into ethanamine,
(vii) Nitromethane into dimethylamine,
(viii) Propanoic acid into ethanoic acid?
Answer:
(i) Ethanoic acid into methanamine:
(iii) Methanol to ethanoic acid:
Question 6.
Describe the method for the identification of primary, secondary and tertiery amines. Also write chemical equations of the reactions involved.
Answer:
Hinsberg’s Method-This is an excellent test for distinguishing primary, secondary and tertiery amines. The amine is shaken with benzene sulphonyl chloride in the presence of aqueous KOH solution.
(i) A primary amine gives a clear solution which an acidification gives an insoluble N-alkyl benzene sulphonamide.
Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis
Answer:
(i) 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.
In contrast, secondary and tertiery amines (both aliphatic and aromatic) do not give this test.
(ii) Diazotisation-The reaction of converting aromatic primary amines into diazonium salts with a cold (273-278 K) solution of nitrous acid is called diazotisation.
(iii) Hoffmann bromamide reaction-The reaction of a higher add amide with Br2 and KOH to give a lower amine is called Hoffman bromamide reaction. It gives an amine-containing one carbon atom less than the original amide.
(iv) 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.
(v) 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.
(vi) 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.
(vii) 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 8.
Accomplish the following conversions:
(i) Nitrobenzene 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-phenylethanolamine.
(vi) Chlorobenzene to p-chloroaniline.
(vii) Aniline to p-bromoaniline.
(viii) Benzamide to toluene (ix) Aniline to benzyl alcohol.
Answer:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol-
(v) Benzyl chloride to 2-phenyl ethanamine-
(vii) Aniline top-bmmoaniline-
Question 9.
Give the structures of A, B and C in the following reactions:
Question 10.
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 C6H7N. Write the structures and IUPAC names of compounds A, B and C.
Answer:
Since it is an aromatic compound, it has a benzene ring in it. B on heating with Br2 and KOH [Hoffmann bromamide reaction] forms a compound ‘C’ of molecular formula C6H7N. Now only higher amides on Hoffman bromamide reaction [Br2 + KOH treatment] give rise to lower amine. Therefore B is C6H5CONH2 and ‘C’ with the molecular formula is C6H5NH2. Since the compound C6H5CONH2 is obtained from A on treatment with aqueous ammonia and heating; A has to be a carboxylic acid C6H5COOH. The sequence of reactions along with names (IUPAC) of A, B and C are as follows :
Question 11.
Complete the following reactions:
(i) C6H4NH2 + CHCl3 + ale. KOH
(ii) C6H5N2Cl + H3PO2+ H2O →
(iii) C6H5NH2+ H2SO4 (cone.) →
(iv) C6H5N2Cl + C2H5OH →
(v) C6H5NH2 + Br2 (aq) →
(vi) C6H5NH2+(CH3CO)2O →
Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer:
Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide, that is,
Question 13.
Wite the reaction of
(i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Answer:
Aromatic primary amines react with nitrous acid in the presence of dil. HCl to form aromatic diazonium chloride.
Question 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 than tertiery amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(i) The alkoxide ion 
left after the removal of a proton from an alcohol ROH is more stable than RNH left after the removal of a proton from an amine RNHr This is due to the fact that oxygen atom in alcohols is more electronegative than N atom in amines and thus can accommodate the negative charge better.
(ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiery amines due to the absence of a H-atom on the N-atom, do not undergo hydrogen bonding. As a result, primary amines have higher boiling points than tertiery amines of comparable molecular mass, e.g., B.Pt of n- Butylaniline (351K) is much higher than that of tert-butylamine (319 K).
(iii) Aliphatic amines are stronger bases than aromatic amines. In aliphatic amines, alkyl groups present exert +1 effect
(electron releasing) thus make the lone pair of electrons more readily available for sharing by an acid, whereas in aromatic amines, the presence of benzene ring exerts -1 effect (electron-withdrawing) due to which the availability of lone pair of electrons on N in NH2 decreases thus making it less basic.