Amine Basicity in Action: Conjugate Acids, Extraction and the Hinsberg Test

Protonating an amine adds one H⁺ to the nitrogen lone pair, so the conjugate acid gains one hydrogen atom and one positive charge relative to the free base, with every other atom in the molecule unchanged. For a base with molecular formula CxHyNOz, the conjugate acid is CxH(y+1)NOz with a plus charge, not simply the same formula with a charge added, and not a formula with an extra oxygen or a removed hydrogen. This applies regardless of how large or complex the rest of the molecule is (an alkaloid like morphine with several oxygen-containing functional groups still only gains the one proton at its basic amine nitrogen), because only the most basic site in the molecule is protonated under normal conditions.

An amide nitrogen lone pair is delocalised into the adjacent carbonyl by resonance, which makes an amide far less basic than a simple aliphatic or aromatic amine; the amide essentially does not protonate under mild aqueous acid. A free amine, by contrast, is readily protonated even by a dilute aqueous acid, converting it into a water-soluble ammonium salt. So when a crude reaction mixture contains a desired neutral amide contaminated with unreacted starting amine, dissolving the mixture in an organic solvent (like ether) and shaking it with a dilute, COLD aqueous acid selectively protonates and extracts only the basic amine into the aqueous layer, leaving the neutral amide behind, dissolved, in the organic layer. Keeping the acid dilute and cold also avoids hydrolysing the amide bond itself, which a hot, strong acid could do. Washing instead with brine (a neutral salt solution) would not remove a basic impurity at all, and washing with aqueous base would do nothing useful here since neither the amine nor the amide is acidic.

The Hinsberg test uses benzenesulfonyl chloride to probe how many N-H bonds a nitrogen has. A primary amine (two N-H) reacts to give a sulfonamide that still has one acidic N-H (made acidic by the adjacent electron-withdrawing sulfonyl group), so that product dissolves in aqueous NaOH. A secondary amine (one N-H) reacts to give a sulfonamide with NO remaining N-H at all (the nitrogen now carries the sulfonyl group and two carbon substituents), so that product is insoluble in NaOH. A tertiary amine (no N-H) does not react with benzenesulfonyl chloride at all under these conditions and remains as the free tertiary amine. When a molecule carries two DIFFERENT amine nitrogens (say, one tertiary and one secondary, in the same chain), only the secondary nitrogen reacts, giving a solid product that is insoluble in NaOH (no acidic N-H, matching the question) but that still dissolves in dilute aqueous acid, because the OTHER, unreacted tertiary amine nitrogen elsewhere in the same molecule is still present and still basic. This is exactly how you identify which candidate structure fits: look for one molecule that has both a tertiary amine site (to stay basic and unreacted) and a secondary amine site (to react and lose its N-H).

Hydrogen-bond donor strength follows how acidic, and how polarised, the X-H bond is. In a protonated catecholamine such as protonated noradrenaline the candidates are the two phenolic OH groups (pKa about 10), the benzylic alcohol OH (pKa about 15 to 16) and the NH₃⁺ group (pKa of the ammonium about 9). The NH₃⁺ hydrogens are the most acidic and also carry a full positive charge, so the ammonium is the strongest hydrogen-bond donor site, ahead of the phenol OH groups, with the alcohol OH the weakest. The order to remember is ammonium N-H > phenol O-H > alcohol O-H. The same acidity of the ammonium ion is why an amine dissolves in dilute acid and is recovered from its salt when base is added.