Question 1 of 13: Bronsted–Lowry Acidity/Basicity of Drug Molecules

Question 1: Bronsted–Lowry Acidity/Basicity of Drug Molecules (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) Ibuprofen's most acidic proton. Ibuprofen carries a single carboxylic acid group (–COOH) hanging off the benzylic-type stereocentre; every other C–H in the molecule (aromatic ring, isobutyl chain, the benzylic methine) is far less acidic (pK_a > 40) because none of those carbanions are stabilised by an adjacent electronegative, resonance-withdrawing group.

Deprotonating the O–H gives a carboxylate anion whose negative charge is delocalised by resonance over both oxygens of the –CO₂⁻ group (two equivalent resonance structures, C–O bond orders both ½ extra) — this resonance stabilisation is exactly why the carboxylic acid proton (pK_a≈4.4) is enormously more acidic than any C–H in the rest of the molecule.

b) Cocaine's most basic electron pair. Cocaine has two types of basic sites: the tertiary amine nitrogen (bridgehead N–CH₃ of the tropane ring) and the lone pairs on the ester/ether oxygens. The amine nitrogen's lone pair is by far the most basic — oxygen lone pairs in esters are strongly delocalised into the adjacent carbonyl (poor bases), while the amine nitrogen's lone pair sits in an sp³ orbital with no competing resonance delocalisation and is free to accept a proton.

Protonating the tertiary amine nitrogen gives the corresponding ammonium cation (this is, in fact, exactly how cocaine is formulated pharmaceutically/illicitly as a water-soluble hydrochloride salt — "crack" free-base vs. powder salt is precisely this acid–base pair). The ester oxygens are not protonated preferentially because protonating them would generate a destabilised acylium-adjacent oxocarbenium without the amine's clean, localised lone pair.