Concept:For two different molecules to form a hydrogen bond, at least one molecule must possess a strongly electropositive Hydrogen (bonded to N, O, or F), and the other must possess a highly electronegative atom with a lone pair to accept it.
Formula:Not applicable.
Solution:- Aldehydes and ketones (like methanal, ethanal, propanone, acetone) have carbonyl oxygens (acceptors) but lack any Hydrogen attached directly to an oxygen. They cannot hydrogen bond with themselves or each other.
- A tertiary (\( 3^\circ \)) amine has a Nitrogen atom with a lone pair (H-bond acceptor) but no hydrogens attached directly to the Nitrogen, so it cannot H-bond with itself.
- However, when mixed with Water (\( \text{H}_2\text{O} \)), water provides the highly positive Hydrogen (donor), and the \( 3^\circ \) amine provides the lone pair (acceptor). Thus, they can successfully form strong intermolecular hydrogen bonds with each other.
Why other options are incorrect:Options A, B, and D strictly pair aldehydes/ketones together. Since neither molecule in those pairs possesses an \( \text{O-H} \) or \( \text{N-H} \) bond, no hydrogen bonding can possibly initiate between them.
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.