Question

In the following sequence of reactions the maximum number of atoms present in molecule 'C' in one plane is _________.

(A is a lowest molecular weight alkyne)

Hint:

The lowest alkyne is ethyne; a red-hot Cu tube trimerises it to benzene, and Friedel-Crafts methylation gives toluene.

Answer 1

Correct Answer: 13

A = lowest molecular weight alkyne = ethyne, HC≡CH.
Red hot Cu tube: 3 C₂H₂ → C₆H₆, so B = benzene.
Benzene + CH₃Cl (1 eq.)/anhydrous AlCl₃ (Friedel-Crafts alkylation): C = toluene, C₆H₅CH₃.
In toluene all 6 ring carbons, 5 ring hydrogens and the methyl carbon lie in the plane of the ring: 6 + 5 + 1 = 12 atoms. The methyl carbon is sp³, so of its 3 hydrogens at most 1 can lie in the ring plane: +1 = 13.
Answer: 13