The increasing order of boiling points of the below mentioned alcohols is
I) 1, 2-dihydroxybenzene
II) 1, 3-dihydroxybenzene
III) 1, 4-dihydroxybenzene
IV) Hydroxybenzene

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IV < I < II < III

I < II < IV < III

I < II < III < IV

IV < II < I < III

answer is D.

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Detailed Solution

Among the given compounds, hydroxybenzene (IV) has least molar mass and therefore possess least boiling point. Among the three isomeric dihydroxybenzenes. 1, 2-dihydroxybenzene (I) forms intramolecular H-bonding with the result that it will not form intermolecular H-bonding leading to lowest boiling point. On the other hand 1, 3-dihydroxybenzene (II) and 1, 4-dihydroxybenzene (III) does not undergo chelation, hence they will involve extensive intermolecular H-bonding leading to higher boiling point. Further intermolecular hydrogen bonding is stronger in the p-isomer than the m-isomer hence former has highest b.p. Thus the decreasing order of boiling points is III > II > I > IV.

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