Identify the product in the following reaction. Chlorocyclobutane $\stackrel{\text{(i) KCN}}{\to }$ $\stackrel{\text{(ii) H2O/HCl, }\mathrm{\Delta }}{\to }$ $\stackrel{\text{(iii) Br2/red P}}{\to }$ $\stackrel{\text{(iv) H2O}}{\to }$ Product

Solution & Explanation

Chlorocyclobutane undergoes $\mathrm{S}{}_{\mathrm{N}}$ substitution with $\mathrm{KCN}$ to give cyclobutanecarbonitrile ($\mathrm{R}-\mathrm{CN}$). Acidic hydrolysis converts the nitrile to cyclobutanecarboxylic acid ($\mathrm{R}-\mathrm{COOH}$). $\mathrm{Br}{}_2/$red P is the Hell-Volhard-Zelinsky reagent, which $\alpha$-brominates the carboxylic acid; after the $\mathrm{H}{}_2\mathrm{O}$ work-up the product is 1-bromocyclobutane-1-carboxylic acid, i.e. a cyclobutane ring carrying both $-\mathrm{COOH}$ and $\alpha \text{-}\mathrm{Br}$ on the same carbon.