Class 12 Chemistry - Chapter Amines NCERT Solutions | Arrange the following in increasing orde
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.
(i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as:
NH3 2H5NH2 <(C2H5)2NH
Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have:
C6H5NH2 3 2H5NH2 <(C2H5)2NH
Due to the - I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as:
C6H5NH2 3 2H5NH2 C6H5CH2NH22H5NH2<(C2H5)2NH
(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2 H5)2NH2, and their basic strengths as follows:
C2H5NH2 <(C2 H5)3N <(C2 H5)2NH
Again, due to the - R effect of C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:
C2H5NH2 2 H5NH2 <(C2 H5)3N <(C2 H5)2NH
(iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as:
(CH3)3N 3NH2 (CH3)2NH
In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N - atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CH2 NH2 is more basic than C6H5NH2.
Again, due to the - I effect of C6H5 group, the electron density on the N - atom in C6H5CH2NH2 is lower than that on the N - atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.
C6H5NH2 6H5CH2NH2 <(CH3)3N 3NH2 <(CH3)2NH
More Questions From Class 12 Chemistry - Chapter Amines
- 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.
- Q:-
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.
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2 Comment(s) on this Question
The answers of (I)st and (ii)nd are wrong apart from the (iii)rd one. Kindly update the answers with the correct order
Recheck your answers plz...
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