See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: When analyzing conformational stability, we must consider both steric interactions and intramolecular interactions such as hydrogen bonding. The compound is 3-hydroxypropanal: HO-CH2-CH2-CHO. We are asked to consider rotation about the C2-C3 bond. Step 1: Identify the groups on C2 and C3. - C3 bears: OH, H, H - C2 bears: CHO (via C1), H, H Step 2: Consider possible conformers about C2-C3. The Newman projection looking down C2-C3 would have: - Front carbon (C2): CHO group and two H's - Back carbon (C3): OH group and two H's Step 3: Evaluate stability considerations. In the anti (fully staggered) conformer, the OH and CHO groups are 180° apart — maximum separation but no opportunity for intramolecular hydrogen bonding. In the gauche conformer, the OH and CHO groups are approximately 60° apart. This proximity allows the hydroxyl group (OH) to form an intramolecular hydrogen bond with the carbonyl oxygen (C=O) of the aldehyde group. This intramolecular H-bond stabilizes the gauche conformer significantly. Step 4: Why other options fail. - (a) Staggered (anti): Although it minimizes steric strain, it does not permit intramolecular hydrogen bonding between OH and C=O, so it is less stable than gauche for this specific molecule. - (c) Fully eclipsed: Maximum torsional and steric strain; least stable. - (d) Partially eclipsed: Still has torsional strain; not the most stable. Step 5: Conclusion. The gauche conformer is most stable because it allows intramolecular hydrogen bonding between the -OH group on C3 and the carbonyl (C=O) on C1, which provides extra stabilization that outweighs the slightly increased steric interaction compared to the anti conformer. Therefore, the correct answer is B.