Class 12 Chemistry - Chapter Amines NCERT Solutions | Account for the following: (i) pKb of
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, 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.
(i) pKb of aniline is more than that of methylamine:
Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.
On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not:
Ethylamine when added to water forms intermolecular H - bonds with water. Hence, it is soluble in water.
But aniline does not undergo H - bonding with water to a very large extent due to the presence of a large hydrophobic - C6H5 group. Hence, aniline is insoluble in water.
(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.
Ferric chloride (FeCl3) dissociates in water to form Fe3+ and Cl - ions.
Then, OH - ion reacts with Fe3+ ion to form a precipitate of hydrated ferric oxide.
(iv) Although amino group is o,p - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline:
Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).
For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction:
A Friedel-Crafts reaction is carried out in the presence of AlCl3. But AlCl3 is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt (as shown in the following equation).
Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo the Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
The diazonium ion undergoes resonance as shown below:
This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines:
Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.
More Questions From Class 12 Chemistry - Chapter Amines
- Q:-
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.
- Q:-
How will you convert?
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) Cl-(CH2)4-Cl into hexan-1, 6-diamine?
- Q:-
(i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N
(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
- Q:-
Arrange the following:
(i) In decreasing order of the pKbvalues:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and 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
(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) In increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2.
- Q:-
Convert
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5-tribromobenzene.
- Q:-
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
- Q:-
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.
- Q:-
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
- Q:-
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.
- Q:-
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-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol.
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2 Comment(s) on this Question
i like your all answer. thank u very much
Why ortho product in little amount is also formed in nitration of aniline?
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