See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the compounds and their key intermolecular forces: - (w) Cyclopentanol: MW ~86, has O-H group capable of strong hydrogen bonding (as donor and acceptor). Boiling point ~140°C. - (x) Tetrahydropyran (THP): MW ~86, has oxygen but NO N-H or O-H, so only dipole-dipole and London dispersion forces. Boiling point ~88°C. - (y) 1-Methylpyrrolidine: MW ~85, tertiary amine (N-CH3), NO N-H bond, so only weak dipole and London dispersion. Boiling point ~81°C. - (z) Piperidine: MW ~85, secondary amine with N-H, capable of N-H hydrogen bonding (weaker than O-H but still significant). Boiling point ~106°C. Step 2 - Rank by intermolecular forces: - w (O-H hydrogen bonding, strongest) > z (N-H hydrogen bonding) > x (dipole only, larger ring/more polarizable than y) > y (no H-bonding, tertiary amine, smallest dispersion). Step 3 - Compare x and y more carefully: - Both x and y lack H-bond donors. THP (x) has a more polarizable oxygen contributing to a stronger dipole than the tertiary amine in y. THP also has a 6-membered ring vs. y's 5-membered ring, giving slightly higher London dispersion forces. Thus x > y. Step 4 - Final order: w > z > x > y, which matches option (d). Step 5 - Why other options fail: - (a) w > x > z > y: Places x above z, ignoring that N-H in piperidine enables hydrogen bonding raising its bp above THP. - (b) w > x > y > z: Same error, places x above z and y above z. - (c) w > z > y > x: Places y above x, but THP has stronger intermolecular forces than 1-methylpyrrolidine. Therefore, the correct answer is D.