Chemistry Q69 – JEE-MAIN 2026

Home Chemistry Basic Organic ChemistryQuestion #129

Consider the following reactions. Total number of electrons in the

π

bonds and lone pair of electrons in the product (X) is :

{CHO(CHOH)}_{4} {CH}_{2} OH \overset{\begin{matrix} (i) HI, Δ \\ {(ii) V}_{2} O_{5} /10-20 atm/773 K \\ {(iii) Benzoyl chloride/Anhyd. AlCl}_{3} \end{matrix}}{\to} (X) Major product

Clue 1 of 3: Reaction Sequence Analysis

Identify the type of reaction occurring at each step and the intermediate products formed.

🧠 Full Solution Path

Step 1: Determine the product of each reaction step✦ Active

The starting material is an aldohexose, ${CHO(CHOH)}_{4} {CH}_{2} OH$ (e.g., D-glucose).

Step (i): $HI, Δ$ . This is a strong reducing agent that reduces all functional groups of the sugar to an alkane. For a 6-carbon aldohexose, the product is n-hexane.

{CHO(CHOH)}_{4} {CH}_{2} OH \overset{HI, Δ}{\to} {CH}_{3} {CH}_{2} {CH}_{2} {CH}_{2} {CH}_{2} {CH}_{3} (n-hexane)

Step (ii): $V_{2} O_{5} /10-20 atm/773 K$ . Under these conditions, n-hexane undergoes oxidative dehydrocyclization to form benzene.

{CH}_{3} {CH}_{2} {CH}_{2} {CH}_{2} {CH}_{2} {CH}_{3} \overset{V_{2} O_{5} /10-20 atm/773 K}{\to} Benzene

Step (iii): Benzoyl chloride ( $C_{6} H_{5} COCl$ )/Anhyd. ${AlCl}_{3}$ . This is a Friedel-Crafts acylation reaction. Benzene reacts with benzoyl chloride to form benzophenone.

Benzene + C_{6} H_{5} COCl \overset{{Anhyd. AlCl}_{3}}{\to} C_{6} H_{5} {COC}_{6} H_{5} (Benzophenone)

Step 2: Identify the final product (X)○ Expand

The major product (X) of the entire reaction sequence is Benzophenone ( $C_{6} H_{5} {COC}_{6} H_{5}$ ). Its structure consists of two phenyl rings attached to a carbonyl group.

Step 3: Count

π

electrons and lone pair electrons in Benzophenone○ Expand

In Benzophenone ( $C_{6} H_{5} {COC}_{6} H_{5}$ ):

1. ** $π$ electrons from phenyl rings:** Each phenyl ring has 3 double bonds, contributing $3 \times 2 = 6$ $π$ electrons. For two phenyl rings, the total is $6 + 6 = 12$ $π$ electrons.

2. ** $π$ electrons from carbonyl group (C=O):** The carbon-oxygen double bond has 1 $π$ bond, contributing $1 \times 2 = 2$ $π$ electrons.

3. **Lone pair electrons on oxygen:** The oxygen atom in the carbonyl group has two lone pairs. Each lone pair consists of 2 electrons, so $2 \times 2 = 4$ lone pair electrons.

Total number of electrons = ( $π$ electrons from rings) + ( $π$ electrons from C=O) + (lone pair electrons on O)

Total electrons = 12 + 2 + 4 = 18

💡 Teacher's Secret Hint

Ensure all $π$ bonds (including those in aromatic rings) and all lone pairs on heteroatoms are counted.

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Chemistry Question 69 – JEE-MAIN 2026

Clue 1 of 3: Reaction Sequence Analysis

🧠 Full Solution Path