JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Haloalkanes and Haloarenes

The substrate is (1-bromoethyl)benzene, ${\text{C}}_6{\text{H}}_5{\text{CHBrCH}}_3$, and the target is phenylacetic acid, ${\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{COOH}$, so a carboxyl group must appear one carbon from the ring. Sodium ethoxide brings about dehydrohalogenation to give styrene, ${\text{C}}_6{\text{H}}_5\text{CH}={\text{CH}}_2$. Hydroboration with diborane followed by alkaline hydrogen peroxide adds water in anti-Markovnikov fashion, placing the hydroxyl on the terminal carbon to give 2-phenylethanol, ${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2\text{OH}$. Oxidation of this primary alcohol with Jones reagent converts it to the carboxylic acid. The correct sequence of reagents is (i) NaOEt, (ii) ${\text{B}}_2{\text{H}}_6/{\text{H}}_2{\text{O}}_2$, (iii) Jones reagent.