Consider the following chemical reaction.
CH≡CH →(1) Red hot Fe tube, 873 K) Product
The number of sp² hybridized carbon atom(s) present in the product is ________.

Solution

Chemistry ma’am · 4 steps

Three acetylenes make benzene, and Gattermann-Koch adds a CHO. Now count every carbon with a double bond.

1. Red hot iron tube. At 873 K, three molecules of ethyne join into a ring. This cyclic trimerisation gives benzene.

   3 HC≡CH →(red hot Fe, 873 K) C₆H₆

2. Gattermann-Koch reaction. CO and HCl with anhydrous AlCl₃ put a -CHO group on the ring. Benzene becomes benzaldehyde, C₆H₅CHO.
3. Count sp2 carbons. All 6 ring carbons are sp², because each forms three σ bonds and takes part in the ring's π system. The aldehyde carbon has a C=O double bond and three σ bonds, so it is sp² too.
4. So. Benzaldehyde has 6 + 1 = 7 sp² hybridised carbon atoms.

The common slip: 6. Counting only the benzene ring and forgetting the aldehyde carbon gives 6, not 7. The C=O carbon is sp² as well.

The idea to keep. A carbon with three σ bonds and one π bond is sp². That covers every aromatic ring carbon and every carbonyl carbon, so count both.

Answer from the NTA final answer key. Solution by PrepWiser.