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प्रश्न
Arrange the following.
In increasing order of boiling point C6H5OH, (CH3)2NH, C2H5NH2.
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उत्तर
Since the electronegativity of O is higher than that of N, therefore, alcohols form stronger H-O bonds than amines. In other words, the boiling points of alcohols are higher than those of amines of comparable molecular masses. Therefore the boiling point of C6H5OH (46) is higher than those of (CH3)2NH (45) and C2H5NH2 (45).
Further, the extent of H-bonding depends upon the number of H-atoms on the N-atom. Therefore 1° amine with two H-atoms on the N-atom has higher boiling points than 2° amines having only one H-atom. Therefore the boiling point of C2H5NH2 is higher than that of(CH3)2NH.
The increasing order of boiling point is, (CH3)2NH < C2H5NH2 < C6H5OH.
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संबंधित प्रश्न
In the following

The compound ‘B’ is _______.
(A) Propan–1–amine
(B) Propan–2–amine
(C) Isopropylamine
(D) Dimethylamine
Choose the most correct option.
The hybridisation of nitrogen in primary amine is ____________.
Choose the most correct option.
Identify ‘B’ in the following reactions
\[\ce{CH3 - C ≡ N ->[Na/C2H5OH] A ->[NaNO2/dilHCl]B}\]
Write a short note on the following.
Ammonolysis
Account for the following.
Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Account for the following.
Ethylamine is soluble in water whereas aniline is not.
How will you prepare propan-1-amine from butane nitrile?
How will you convert diethylamine into N, N-diethyl acetamide?
Account for the following:
Aniline does not undergo Friedel-Crafts reaction.
Write a short note on the following.
Ammonolysis
