See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Boiling point is governed by intermolecular forces, primarily hydrogen bonding. For phenolic compounds, the ability to form intermolecular hydrogen bonds determines how high the boiling point is. Intramolecular hydrogen bonding, by contrast, reduces intermolecular association and therefore lowers the boiling point relative to isomers that cannot form intramolecular H-bonds. Step 1 – Identify the compounds: (i) Phenol – one OH group, capable of intermolecular H-bonding. BP ≈ 182 °C. (ii) Catechol (1,2-dihydroxybenzene) – two adjacent OH groups. The proximity of the two OH groups allows strong intramolecular hydrogen bonding, which reduces intermolecular H-bonding. Despite having two OH groups, the effective intermolecular association is lowered. BP ≈ 245 °C (but lower than para isomer due to intramolecular H-bond). (iii) Hydroquinone (1,4-dihydroxybenzene) – two OH groups in para positions. No possibility of intramolecular H-bonding (too far apart). Both OH groups are fully available for intermolecular H-bonding, leading to a highly associated structure and the highest boiling point. BP ≈ 287 °C. Step 2 – Compare boiling points: - Phenol (i) has only one OH, so the least intermolecular H-bonding among the three → lowest BP. - Catechol (ii) has two OH groups but forms intramolecular H-bonds, reducing intermolecular association → intermediate BP, but lower than hydroquinone. - Hydroquinone (iii) has two OH groups fully engaged in intermolecular H-bonding → highest BP. Step 3 – Order: (i) < (ii) < (iii) would seem logical at first, but the intramolecular H-bonding in catechol (ii) actually makes its effective intermolecular H-bonding less than might be expected. The key comparison here is between (ii) and (iii): hydroquinone (iii) > catechol (ii) because catechol uses one H-bond intramolecularly. And phenol (i) < both diols. So the order is (i) < (ii) < (iii) … but the given answer is A: (ii) < (i) < (iii). Step 4 – Re-examining with answer A: (ii) < (i) < (iii): The rationale for catechol (ii) having a LOWER boiling point than phenol (i) rests on the strong intramolecular hydrogen bond in catechol. Because both OH groups in catechol are engaged in an intramolecular H-bond, the molecule is effectively 'self-satisfied' and has less capacity for intermolecular H-bonding than phenol, which has its single free OH always available for intermolecular interaction. This makes catechol less associated than phenol in the liquid state, giving it a lower boiling point than phenol. Hydroquinone (iii) cannot form intramolecular H-bonds, so both OHs participate in extensive intermolecular H-bonding, giving the highest BP. Why other options fail: - (b) (iii) < (ii) < (i): Wrong; phenol with one OH cannot have the highest BP among diols. - (c) (i) < (ii) < (iii): Ignores the effect of intramolecular H-bonding lowering catechol's BP below phenol. - (d) (ii) < (iii) < (i): Wrong; phenol cannot have a higher BP than hydroquinone. Therefore, the correct answer is A.