The major product 'A' of the following given reaction has $$sp^2$$ hybridized carbon atoms.
$$2,7\text{-Dimethyl-2,6-octadiene} \xrightarrow{H^+} A$$ (Major Product)
The number of $$sp^2$$ hybridized carbon atoms in 'A' is ______

The reactant is:

$$\mathrm{2,7\text{-}Dimethyl\text{-}2,6\text{-}octadiene}$$

In presence of acid $$\mathrm{(H^+)}$$, protonation of one double bond occurs according to Markovnikov’s rule, generating a tertiary carbocation.

$$\mathrm{(CH_3)_2C^+ - CH_2 - CH_2 - CH_2 - CH=C(CH_3)_2}$$

The remaining double bond attacks this carbocation intramolecularly, causing cyclization.

Formation of the $$\mathrm{5}$$-membered ring is favored because it generates a more stable tertiary carbocation intermediate.

Finally, deprotonation occurs according to Zaitsev’s rule, forming the most stable alkene.

The final major product contains only one double bond.

Each carbon atom involved in a double bond is $$\mathrm{sp^2}$$ hybridised.

Since one double bond contains exactly two carbon atoms, the number of $$\mathrm{sp^2}$$ hybridised carbon atoms is: $$\mathrm{2}$$

Correct Answer: $$\mathrm{2}$$