JEE Main 2026 · 6 Apr · Shift 2 · Chemistry Q69 · Dronacharya Exams

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ChemistrySingle correct+4 / -1Question 69 of 75

Consider the following reactions. Total number of electrons in the $\pi$ bonds and lone pair of electrons in the product (X) is :

Open-chain glucose, CHO-(CHOH)₄-CH₂OH, treated with (i) HI, Δ, (ii) V₂O₅/10-20 atm/773 K, (iii) benzoyl chloride/anhydrous AlCl₃ gives the major product (X).

A

12

B

16

C

14

D

18

Answer and worked solution

Official answer: D

Sequence.

HI, heat reduces glucose to n-hexane.
$\ce{V2O5}$ at 773 K and 10-20 atm aromatises it to benzene.
Friedel-Crafts benzoylation gives benzophenone, $\ce{(C6H5)2CO}$ .

Electrons in X.

π bonds: $2\times6$ in the rings, plus 2 in C=O: 14.
Lone pairs on O: 2 pairs, 4 electrons.

Total: 18.

(D)

Topics:Carbohydrates Aromatic hydrocarbons and electrophilic substitution

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