JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

Step 1: The first reagent is Cu at 573 K . When a tertiary alcohol like tert-butyl alcohol is passed over heated copper at 573 K , it undergoes dehydration rather than dehydrogenation to form an alkene. (CH_3)_3C-OH Cu / 573 K CH_2=C(CH_3)_2 + H_2O The intermediate product is isobutylene (2-methylpropene). Step 2: The second set of reagents is H^+ and PhCOOH (benzoic acid). This is an acid-catalyzed addition of a carboxylic acid to an alkene, which is a standard method for synthesizing tert-butyl esters. First, the alkene is protonated by the acid to form the most stable carbocation, which is the tertiary butyl carbocation: CH_2=C(CH_3)_2 + H^+ (CH_3)_3C^+ Next, the carboxylic acid acts as a nucleophile and attacks the carbocation. The oxygen atom of the benzoic acid attacks the tert-butyl carbocation, followed by deprotonation: PhCOOH + (CH_3)_3C^+ Ph-CO-O^+(H)-C(CH_3)_3 Ph-CO-O-C(CH_3)_3 + H^+ The final major product is tert-butyl benzoate. Electrophilic aromatic substitution (Friedel-Crafts alkylation) on the benzene ring does not occur here because the -COOH group is strongly deactivating, making O-alkylation (ester formation) the highly preferred pathway. Comparing this with the given options, Option (C) represents the structure of tert-butyl benzoate. Answer: