JEE Advanced · Chemistry · 25. Alcohols, Phenols & Ethers
(I) 1, 2-dihydroxy benzene
(II) 1, 3-dihydroxy benzene
(III) 1, 4-dihydroxy benzene
(IV) Hydroxy benzene
The increasing order of boiling points of above mentioned alcohols is
- A I \( < \) II \( < \) III \( < \) IV
- B I \( < \) II \( < \) IV \( < \) III
- C IV \( < \) I \( < \) II \( < \) III
- D IV \( < \) II \( < \) I \( < \) III
Answer & Solution
Correct Answer
(C) IV \( < \) I \( < \) II \( < \) III
Step-by-step Solution
Detailed explanation
1, 4-dihydroxy benzene shows the highest boiling point among given compounds because it form strong intermolecular hydrogen bond (It does not form intramolecular \(\mathrm{H}\)-bonding).

Order of H-bonding in \(o, m\) and \(p\)-isomers of a compound is given below.
Intermolecular \(\mathrm{H}\)-bonding \(0 < m < p\)-isomers
Intramolecular \(\mathrm{H}\)-bonding \(o>m>p\)-isomers
Hydroxy benzene do not form a chain of \(\mathrm{H}\)-bonding. Hence, intermolecular \(\mathrm{H}\)-bond is stronger then intermolecular \(\mathrm{H}\)-bond, so the stability of 1, 4-dihydroxy benzene is highest. Hence, its boiling point is highest.
The increasing order of the boiling points of the given compounds IV \( < \) I \( < \) II \( < \) III.

Order of H-bonding in \(o, m\) and \(p\)-isomers of a compound is given below.
Intermolecular \(\mathrm{H}\)-bonding \(0 < m < p\)-isomers
Intramolecular \(\mathrm{H}\)-bonding \(o>m>p\)-isomers
Hydroxy benzene do not form a chain of \(\mathrm{H}\)-bonding. Hence, intermolecular \(\mathrm{H}\)-bond is stronger then intermolecular \(\mathrm{H}\)-bond, so the stability of 1, 4-dihydroxy benzene is highest. Hence, its boiling point is highest.
The increasing order of the boiling points of the given compounds IV \( < \) I \( < \) II \( < \) III.
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