See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Fischer esterification mechanism under acid catalysis (H2SO4). The reaction is between trifluoroacetic acid (CF3COOH) and isopropanol (CH3CH(OH)CH3) to give the ester CF3COOCH(CH3)2 and water. Step-by-step mechanism: 1. Protonation of the carbonyl oxygen of CF3COOH by H2SO4 gives the activated acylium-like species: CF3C(=OH+)-OH. This is option (a) — a valid intermediate (the protonated carbonyl). 2. Nucleophilic attack by the oxygen of isopropanol on the electrophilic carbonyl carbon produces a tetrahedral intermediate with four groups on the central carbon: CF3, OH (from original acid), OH (protonated from attack, then deprotonated or still +OH2 at oxygen), and OCH(CH3)2. The neutral tetrahedral intermediate with two OH groups and one OCH(CH3)2 is option (d) — a valid intermediate. 3. Protonation of one OH group gives the +OH2 leaving group, yielding the intermediate in option (c) — also a valid intermediate, which then loses water to regenerate a protonated ester, completing the mechanism. 4. Option (b) shows a tetrahedral intermediate bearing a negatively charged oxygen (O-), an OCH(CH3)2 group, and an OH group. Under acidic conditions (H2SO4), formation of a negatively charged oxyanion intermediate is highly unfavorable — this would be characteristic of base-catalyzed (or uncatalyzed nucleophilic acyl substitution) conditions, not acid-catalyzed esterification. In the acid-catalyzed mechanism, all intermediates are neutral or positively charged; no anionic (O-) intermediates are formed. Why option (b) is NOT an intermediate: In acid-catalyzed esterification, the mechanism never involves a negatively charged tetrahedral intermediate. The oxyanion shown in (b) would only arise in base-catalyzed conditions. Under H2SO4 catalysis, the oxygen atoms remain protonated or neutral throughout the mechanism. Therefore, the correct answer is B.