Animated Solution for Chemistry - Organic Compounds Containing Oxygen: Consider the following molecules and statements related to them (I) (B) is more likely to be crystalline than (A). (II) (B) has higher boiling point than (A). (III) (B) dissolves more readily than (A) in water. Identify the correct option from below :

Ortho vs. Para

When we look at organic molecules, it is fascinating how a simple shift in the position of a functional group can drastically alter the physical properties of the entire substance. In this problem, we are comparing two isomers of hydroxybenzoic acid: $o$-hydroxybenzoic acid (commonly known as salicylic acid, labeled as Molecule A) and $p$-hydroxybenzoic acid (labeled as Molecule B).

Both molecules possess the exact same functional groups: a hydroxyl group ($-OH$) and a carboxyl group ($-COOH$). Because these groups contain highly electronegative oxygen atoms bonded to hydrogen, they are prime candidates for hydrogen bonding. However, how they hydrogen bond makes all the difference.

The Extrovert

Let's examine Molecule A ($o$-hydroxybenzoic acid). Because the $-OH$ and $-COOH$ groups are situated at the ortho position, they are right next door to each other. This proximity allows the hydrogen atom of the hydroxyl group to form a strong hydrogen bond with the carbonyl oxygen of the carboxyl group within the same molecule. This is known as intramolecular hydrogen bonding. You can think of Molecule A as an introvert, wrapping its arms around itself and forming a stable, discrete, chelated ring structure.

Now, let's look at Molecule B ($p$-hydroxybenzoic acid). Here, the functional groups are at the para position, situated at opposite ends of the benzene ring. They are simply too far apart to interact with each other. Instead, they reach out and form hydrogen bonds with the functional groups of neighboring molecules. This is called intermolecular hydrogen bonding. Molecule B acts like an extrovert, holding hands with its neighbors to form a massive, interconnected polymeric network.

With this structural understanding, we can easily evaluate the given statements.

Statement II: Boiling Point Boiling a liquid requires supplying enough thermal energy to overcome the intermolecular forces holding the molecules together. Because Molecule B forms an extensive network of intermolecular hydrogen bonds, it takes a significant amount of energy to break these molecules apart. Molecule A, on the other hand, is busy hydrogen-bonding with itself, leaving fewer opportunities to bond strongly with its neighbors. Consequently, the intermolecular forces in A are weaker. Therefore, Molecule B has a higher boiling point than Molecule A, making Statement II True.

Statement III: Solubility in Water For an organic molecule to dissolve in water, it must be able to form hydrogen bonds with the $H_{2}O$ molecules. In Molecule A, the polar groups are already "locked up" in their internal hydrogen bond, making them less available to interact with water. In Molecule B, the polar groups are exposed and readily available to form hydrogen bonds with the solvent. Thus, Molecule B dissolves more readily in water than Molecule A, making Statement III True.

Statement I: Crystallinity This is where many students fall into a trap. You might assume that the extensive network of Molecule B makes it more crystalline. However, crystallinity is determined by how efficiently and uniformly molecules can pack together into a solid lattice. Because Molecule A forms an intramolecular hydrogen bond, it adopts a very neat, compact, and discrete shape. These uniform, compact units can stack together incredibly efficiently, leading to a highly ordered crystal lattice. The extended networks of Molecule B can sometimes be more difficult to pack perfectly compared to the discrete units of A. Therefore, Molecule A is actually more likely to be highly crystalline than Molecule B, making Statement I False.

Since Statement II and Statement III are true, while Statement I is false, the correct choice is Option (c). This problem beautifully illustrates how the spatial arrangement of atoms dictates the macroscopic physical properties we observe in the lab.