Consider the following molecules and statements: (A) 2-hydroxybenzoic acid (salicylic acid) (B) 4-hydroxybenzoic acid…
Chemical Bonding · Class 11 · JEE Main Previous Year Question
Consider the following molecules and statements:
(A) 2-hydroxybenzoic acid (salicylic acid) (B) 4-hydroxybenzoic acid (para-hydroxybenzoic acid)
(a) (B) is more likely to be crystalline than (A) (b) (B) has higher boiling point than (A) (c) (B) dissolves more readily than (A) in water
Identify the correct option:
- a
(a) and (b) are true
- b✓
(a) and (c) are true
- c
Only (a) is true
- d
(b) and (c) are true
(a) and (c) are true
🧠 Compare 2-hydroxybenzoic acid (salicylic acid, A, ortho) and 4-hydroxybenzoic acid (para-HBA, B) using intramolecular vs intermolecular H-bonding.
🗺️ Key structural difference:
A (salicylic acid, ortho): The –OH and –COOH groups are adjacent → form a 6-membered intramolecular H-bond ring → the –OH is locked internally → less ability to form intermolecular H-bonds with other molecules or with water.
B (para-hydroxybenzoic acid): –OH and –COOH are on opposite ends → no intramolecular H-bonding → both groups free for intermolecular H-bonds.
Evaluate statement (a) — B is more likely crystalline than A: B has more intermolecular H-bonding → forms a more ordered solid lattice → more crystalline, higher mp → (a) is TRUE ✓
Evaluate statement (b) — B has higher boiling point than A: B does have more intermolecular interactions... but salicylic acid's intramolecular ring creates a rigid compact structure. The actual boiling point difference depends on multiple factors. (b) is considered FALSE in this context.
Evaluate statement (c) — B dissolves more readily than A in water: A's intramolecular H-bond reduces availability of –OH to H-bond with water → A is less water-soluble. B has free –OH and –COOH → more H-bonding with water → (c) is TRUE ✓
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