JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

Adding HCN across the ketone carbonyl of $(\pm )\mathrm{Ph}\text{-}\mathrm{CO}\text{-}\mathrm{C}(\mathrm{OH})(\mathrm{CN})\text{-}\mathrm{Ph}$ turns that carbon into a second $\mathrm{C}(\mathrm{OH})(\mathrm{CN})\mathrm{Ph}$ centre, so the product has two constitutionally identical stereocentres, each bearing $\mathrm{Ph}$, $\mathrm{OH}$, $\mathrm{CN}$ and the other equivalent carbon. A molecule with two such equivalent stereocentres behaves like tartaric acid: the $(R,R)$ and $(S,S)$ forms make up one enantiomeric pair, while $(R,S)$ is a single meso compound, so only three stereoisomers exist rather than four.