Consider the following molecules and statements related to them :
(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 from below :

Solution

Chemistry ma’am · 3 steps

In (A) the OH bonds to its own COOH; in (B) it can only reach the next molecule. That decides all three.

1. Spot the H-bonds. (A) is ortho-hydroxybenzoic acid (salicylic acid). Its OH sits right next to the COOH, so the two form an intramolecular hydrogen bond inside the same molecule. (B) is para-hydroxybenzoic acid. Its OH and COOH are on opposite ends of the ring and cannot reach each other, so both groups form intermolecular hydrogen bonds with other molecules.
2. What that changes. In (A), part of the hydrogen-bonding ability is used up inside each molecule, so fewer bonds link one molecule to the next. In (B), every OH and COOH is free to bond outward, to neighbouring (B) molecules or to water.

   Statement	Reason	Verdict
   (a) (B) more crystalline	Intermolecular H-bonds in (B) build an extended, ordered network that holds the lattice together	True
   (b) (B) higher boiling point	Separating (B) molecules means breaking many intermolecular H-bonds, so more energy is needed	True
   (c) (B) dissolves more readily in water	The free OH and COOH groups of (B) H-bond with water; in (A) they are partly tied up with each other	True

3. So. All three statements are true, which is option (b).

The idea to keep. Ortho groups that can bond to each other turn inward: lower boiling point, lower water solubility, more volatile. Para isomers bond outward, to each other and to water.

Answer from the NTA final answer key. Solution by PrepWiser.