The increasing order of the boiling points for the following compounds is :
\(\mathrm{C_2H_5OH}\) (I) \(\mathrm{C_2H_5Cl}\) (II) \(\mathrm{C_2H_5CH_3}\) (III) \(\mathrm{C_2H_5OCH_3}\) (IV)
\(\mathrm{C_2H_5OH}\) (I) \(\mathrm{C_2H_5Cl}\) (II) \(\mathrm{C_2H_5CH_3}\) (III) \(\mathrm{C_2H_5OCH_3}\) (IV)
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- (III) < (IV) < (II) < (I)
- (IV) < (III) < (I) < (II)
- (II) < (III) < (IV) < (I)
- (III) < (II) < (I) < (IV)
The Correct Option is A
Solution and Explanation
(III) \(\mathrm{C_2H_5CH_3}\) = butane: non-polar, only weak London forces, b.p. about 273 K (0 \(^\circ\)C).
(IV) \(\mathrm{C_2H_5OCH_3}\) = methyl ethyl ether: weakly polar, no H-bonding, b.p. about 281 K (7 \(^\circ\)C).
(II) \(\mathrm{C_2H_5Cl}\): polar C-Cl bond with stronger dipole-dipole forces and higher molar mass, b.p. about 285 K (12 \(^\circ\)C).
(I) \(\mathrm{C_2H_5OH}\): extensive intermolecular hydrogen bonding, b.p. 351 K (78 \(^\circ\)C).
Order: (III) < (IV) < (II) < (I).
Answer: (1) (III) < (IV) < (II) < (I)
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