JEE Main 2026 — Alcohols Phenols and Ethers Question with Solution

Step 1: The first reagent is Cu at $573\text{ K}$. When a tertiary alcohol like tert-butyl alcohol is passed over heated copper at $573\text{ K}$, it undergoes dehydration rather than dehydrogenation to form an alkene.

$(C{H}_3{)}_{3}C-OH\stackrel{Cu/573\text{ K}}{\to }C{H}_2=C(C{H}_3{)}_2+H_{2}O$
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:
$C{H}_2=C(C{H}_3{)}_2+H^{+}\rightleftharpoons (C{H}_3{)}_3{C}^{+}$

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:
$PhCOOH+(C{H}_3{)}_3{C}^{+}\rightleftharpoons Ph-CO-O^{+}(H)-C(C{H}_3{)}_3\rightleftharpoons Ph-CO-O-C(C{H}_3{)}_3+H^{+}$

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: