The major product 'A' of the following reaction has ______ sp2 hybridized carbon atoms.
2,7-dimethylocta-2,6-diene H+ A (major product)
Answer: 2
Solution
The substrate 2,7-dimethylocta-2,6-diene, (CH3)2C=CH−CH2−CH2−CH=C(CH3)2, carries two trisubstituted double bonds separated by just two CH2 groups, which is exactly the spacing needed for an acid-promoted ring closure. Protonation of one double bond follows Markovnikov's rule and places the charge on the more substituted carbon, giving a stable tertiary carbocation. The second, still-intact double bond now behaves as the nucleophile and attacks this cation from within the same molecule; because that attack closes a five-membered carbocycle it is strongly favoured, and the fresh positive charge again settles on a tertiary carbon lying just outside the new ring. Loss of a proton from the ring carbon next to that cation sends the electrons back to form a single carbon-carbon double bond exocyclic to the cyclopentane ring. The major product is therefore an alkylidenecyclopentane in which only this one π bond survives, so the two carbons joined by that double bond are the only sp2 centres in the molecule and the number of sp2 hybridized carbons is 2.
JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer
