The correct order of decreasing basic strength of $CH_3NH_2$, $(CH_3)_2NH$ and $(CH_3)_3N$ in aqueous solution is:
Show Hint
- $(CH_3)_3N \gt (CH_3)_2NH \gt CH_3NH_2$
- $CH_3NH_2 \gt (CH_3)_2NH \gt (CH_3)_3N$
- $(CH_3)_2NH \gt CH_3NH_2 \gt (CH_3)_3N$
- $(CH_3)_2NH \gt (CH_3)_3N \gt CH_3NH_2$
The Correct Option is C
Solution and Explanation
Basicity of amines in aqueous solution is governed by three competing factors: the inductive $+I$ effect of methyl groups, steric hindrance preventing solvation, and the extent of hydrogen bonding of the conjugate acid with water.
Step 2: Analysis
Dimethylamine has two methyl groups (strong $+I$ effect) and its conjugate acid ($Me_2NH_2^+$) has two N-H bonds for effective hydration. Methylamine has one methyl group and three N-H bonds for hydration, but less $+I$ effect than dimethylamine. Trimethylamine has the maximum $+I$ effect but its bulky structure provides severe steric hindrance, and its conjugate acid has only one N-H bond for very poor solvation.
Step 3: Conclusion
The combined result of $+I$ effect and solvation places dimethylamine as the strongest base, methylamine next, and trimethylamine as the weakest in aqueous solution.
Final Answer: (C)
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
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(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
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(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.
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(v) Aniline does not undergo Friedel-Crafts reaction.
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
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(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
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(iv) Aniline and benzylamine
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(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.
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) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine Exercises Rationalised 2023-24 279 Amines (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 .
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?
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