JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

Step 1: The starting material is an aldohexose (like D-glucose). On prolonged heating with HI, it undergoes reduction to form n-hexane. This indicates that all six carbon atoms are linked in a straight chain. CHO -( CHOH )_4- CH _2 OH HI, CH _3- CH _2- CH _2- CH _2- CH _2- CH _3 (n-hexane) Step 2: n-Hexane, when passed over heated V_2O_5 at 773 K and 10-20 atm pressure, undergoes aromatization (dehydrogenation and cyclization) to form benzene. n-hexane V_2O_5, 773 K, 10-20 atm C _6 H _6 (Benzene) Step 3: Benzene reacts with benzoyl chloride ( C _6 H _5 COCl ) in the presence of anhydrous AlCl _3 to form benzophenone. This is a Friedel-Crafts acylation reaction. C _6 H _6 + C _6 H _5 COCl Anhyd. AlCl _3 C _6 H _5- CO - C _6 H _5 (Benzophenone) Step 4: Analyze the product (X), which is benzophenone. The structure of benzophenone consists of two phenyl rings attached to a carbonyl group. - Each phenyl ring contains 3 bonds, so two rings have 6 bonds. - The carbonyl group ( C=O ) contains 1 bond. - Total number of bonds = 6 + 1 = 7. - Total number of electrons = 7 2 = 14. - The oxygen atom in the carbonyl group has 2 lone pairs of electrons. - Total number of lone pair electrons = 2 2 = 4. Total number of electrons in the bonds and lone pairs = 14 + 4 = 18. Answer: 18