See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Boiling points of amines depend on their ability to form hydrogen bonds. Primary amines (RNH2) have two N-H bonds and can form strong intermolecular hydrogen bonds. Secondary amines (R2NH) have one N-H bond and form weaker hydrogen bonding than primary amines. Tertiary amines (R3N) have no N-H bonds and cannot act as hydrogen bond donors, so they have the lowest boiling points among isomers of similar molecular weight. Step 1: Identify the class of each amine. (a) 4-methylpiperidine: The nitrogen has one H attached (N-H present in the ring), making it a secondary amine — can form hydrogen bonds. (b) N-methylcyclopentylamine: The nitrogen has one H (NHCH3), making it a secondary amine — can form hydrogen bonds. (c) 1-methylpiperidine: The nitrogen has a methyl group and is part of the piperidine ring with no N-H bond, making it a tertiary amine — cannot donate hydrogen bonds. (d) Cyclohexylamine: The nitrogen has two H atoms (NH2), making it a primary amine — strongest hydrogen bonding among these options. Step 2: Rank boiling points. Primary amine (d) > Secondary amines (a, b) > Tertiary amine (c). Tertiary amine (c), 1-methylpiperidine, has no N-H bond, so it cannot form hydrogen bonds as a donor, resulting in the lowest intermolecular attractive forces and thus the lowest boiling point among these isomers. Step 3: Why other options fail. (a) and (b) are secondary amines with one N-H bond, so they have higher boiling points than the tertiary amine. (d) is a primary amine with two N-H bonds, giving it the highest boiling point. Therefore, the correct answer is C.