Among the following, which is the strongest base?
Show Hint
- \(p\)-Nitroaniline
- Benzylamine
- \(p\)-Toluidine
- Aniline
The Correct Option is B
Solution and Explanation
Basic strength of amines depends on the availability of the lone pair of electrons on nitrogen. If the lone pair is easily available for donation, the compound is more basic. If the lone pair is involved in resonance, basicity decreases.
Step 1: Compare aniline-type compounds.
In aniline, the lone pair on nitrogen is delocalised into the benzene ring by resonance. \[ C_6H_5NH_2 \] Because the lone pair is less available for protonation, aniline is less basic than aliphatic amines.
Step 2: Effect of substituents on aniline.
In \(p\)-nitroaniline, the \(-NO_2\) group is strongly electron withdrawing. It decreases electron density on nitrogen. So it is very weakly basic. \[ -NO_2 \Rightarrow -I,\;-R \] In \(p\)-toluidine, the \(-CH_3\) group is electron releasing. It increases basicity compared with aniline, but the nitrogen lone pair is still involved in resonance.
Step 3: Compare benzylamine.
Benzylamine is: \[ C_6H_5CH_2NH_2 \] Here, the \(-NH_2\) group is not directly attached to the benzene ring. The lone pair on nitrogen is not delocalised into the ring. Therefore, the lone pair is more available for protonation. So benzylamine is the strongest base among the given options. Hence: \[ \boxed{\text{(B) Benzylamine}} \]
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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
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-methylanilineAccount 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.
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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
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
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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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