QInteger TypeOrganic Chemistry
'x' is obtained from benzene by reacting with CO and HCl in presence of CuCl (Gattermann-Koch reaction). 'y' is the major product from benzene with ethanoyl chloride (CH₃COCl) in presence of anhydrous AlCl₃ (Friedel-Crafts acylation). The major product (z) obtained by heating x and y in presence of alkali is z. Total number of π (pi) electrons in z is __________.
✅ Correct Answer
16
Solution
1
Identify x and y

x — Gattermann-Koch reaction: Benzene + CO + HCl → C₆H₅CHO (benzaldehyde). This reaction introduces an aldehyde group directly onto the benzene ring using CO/HCl with CuCl/AlCl₃ as catalyst.

y — Friedel-Crafts acylation: Benzene + CH₃COCl → C₆H₅COCH₃ (acetophenone). The acetyl group (COCH₃) is introduced by electrophilic aromatic substitution.

2
Identify product z (Claisen-Schmidt condensation)

Benzaldehyde (x) has no α-hydrogen. Acetophenone (y) has α-hydrogens on the CH₃ group. With NaOH (alkali), this is a Claisen-Schmidt condensation (cross-aldol between an aldehyde and a ketone with no enolisable protons on the aldehyde side).

C₆H₅CHO + C₆H₅COCH₃ → C₆H₅−CH=CH−CO−C₆H₅ + H₂O

Product z = Benzalacetophenone (Chalcone)
C₆H₅−CH=CH−CO−C₆H₅
3
Count all pi electrons in chalcone
Phenyl ring 1 (C₆H₅−): 3 double bonds × 2π electrons = 6 π electrons
C=C double bond (−CH=CH−): 1 × 2 = 2 π electrons
C=O carbonyl (−C=O−): 1 × 2 = 2 π electrons
Phenyl ring 2 (−C₆H₅): 3 double bonds × 2π electrons = 6 π electrons

Total = 6 + 2 + 2 + 6 = $\boxed{16}$ π electrons
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Theory & Key Concepts
Claisen-Schmidt Condensation and Pi Electron Count

Gattermann-Koch reaction: Benzene + CO + HCl → Benzaldehyde, using CuCl/AlCl₃ as Lewis acid catalyst. This is a specialized electrophilic aromatic substitution that introduces −CHO directly. The formyl cation (CHO⁺) is the electrophile.

Friedel-Crafts acylation: Benzene + RCOCl → ArCOR. The acylium ion (RCO⁺) is the electrophile. With CH₃COCl, the product is acetophenone (C₆H₅COCH₃).

Claisen-Schmidt condensation: A special cross-aldol between an aromatic aldehyde (no α-H) and a carbonyl compound with α-H, under basic conditions. The aldol product immediately dehydrates to give an α,β-unsaturated carbonyl compound (chalcone here).

Chalcone (benzalacetophenone): Structure is C₆H₅−CH=CH−CO−C₆H₅. It is an enone (α,β-unsaturated ketone) with extended conjugation. Used as a precursor in synthesis of flavonoids.

Pi electrons counting: Each C=C counts 2, each C=O counts 2, each benzene ring counts 6 (3 double bonds). Do NOT count lone pairs on oxygen — only π bonds from double bonds.

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Important Concepts
Gattermann-Koch Reaction

Benzene + CO + HCl → benzaldehyde, using CuCl + AlCl₃. The active electrophile is the formyl cation CHO⁺ (or its equivalent). This is one of the few ways to directly introduce −CHO onto a benzene ring without side chain.

Claisen-Schmidt vs Normal Aldol

Normal aldol: two molecules of the same carbonyl compound. Claisen-Schmidt: aromatic aldehyde (no α-H, cannot form enolate) + aliphatic/aryl ketone (has α-H). The aromatic aldehyde acts as electrophile; enolate of ketone attacks it. Product always dehydrates to give conjugated enone.

Why Chalcone Has Extended Conjugation

In chalcone C₆H₅−CH=CH−CO−C₆H₅, all π systems (both rings, C=C, C=O) are conjugated through the carbon chain. This extended conjugation is responsible for the yellow color of chalcones and their UV absorption.

Counting Pi Electrons Systematically

For any organic molecule: count every π bond. Benzene ring = 3 C=C = 6π. Alkene C=C = 2π. Carbonyl C=O = 2π. Sum all. In chalcone: 6+2+2+6=16π. Note: lone pairs on O are NOT π electrons unless explicitly forming a π system.

FAQs
1
What is x in this problem?
x = benzaldehyde (C₆H₅CHO). Made by Gattermann-Koch reaction: benzene + CO + HCl with CuCl/AlCl₃.
2
What is y?
y = acetophenone (C₆H₅COCH₃). Made by Friedel-Crafts acylation: benzene + CH₃COCl with anhydrous AlCl₃.
3
What is z?
z = chalcone (benzalacetophenone): C₆H₅−CH=CH−CO−C₆H₅. Formed by Claisen-Schmidt condensation between benzaldehyde and acetophenone in NaOH solution.
4
How many pi electrons does each group contribute?
Benzene ring: 6. Alkene C=C: 2. Ketone C=O: 2. Chalcone has two phenyl rings + one C=C + one C=O: 6+6+2+2=16.
5
Is this from JEE Main 2026?
Yes, this appeared in JEE Main 2026 Chemistry Section B.
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