Assertion (A): Boiling point of propylamine is higher than that of trimethylamine.
Reason (R): In propylamine, intermolecular hydrogen bonding occurs, whereas trimethylamine does not undergo intermolecular hydrogen bonding.
Reason (R): In propylamine, intermolecular hydrogen bonding occurs, whereas trimethylamine does not undergo intermolecular hydrogen bonding.
Show Hint
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
The Correct Option is A
Solution and Explanation
The boiling points of isomeric amines are determined by the strength of their intermolecular forces, specifically the capacity for hydrogen bonding.
Step 2: Meaning
Intermolecular hydrogen bonding in amines requires the presence of at least one hydrogen atom bonded directly to the highly electronegative nitrogen atom (N-H bond).
Step 3: Analysis
Propylamine ($CH_3CH_2CH_2NH_2$) is a primary amine and possesses two N-H bonds, allowing for extensive intermolecular hydrogen bonding. Trimethylamine ($(CH_3)_3N$) is a tertiary amine; all three bonds on nitrogen are to carbon atoms, meaning it has zero N-H bonds and cannot form intermolecular hydrogen bonds with itself.
Step 4: Conclusion
The strong hydrogen bonding network in propylamine leads to a higher boiling point compared to trimethylamine. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.
Final Answer: (A)
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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) pKb of aniline is more than that of methylamine.
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Top CBSE CLASS XII Amines Questions
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
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(v) Aniline does not undergo Friedel-Crafts reaction.
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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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