Consider the above reaction. The number of $$\pi$$ electrons present in the product 'P' is ______

The reaction proceeds through an $$\mathrm{S_N1}$$ mechanism because the solvent $$\mathrm{H_2O}$$ is polar protic.

First, the chloride ion leaves, generating a secondary carbocation.

To attain greater stability, a $$\mathrm{1,2\text{-}Hydride\ Shift}$$ occurs from the adjacent tertiary allylic carbon.

This forms a more stable tertiary allylic carbocation stabilized by resonance and inductive effects.

The nucleophile $$\mathrm{OH^-}$$ then attacks the carbocation to form the alcohol product.

The final product retains one carbon-carbon double bond.

A double bond contains:

$$\mathrm{1\ \sigma\ bond\ and\ 1\ \pi\ bond}$$

Each covalent bond contains two electrons.

Hence, the number of $$\mathrm{\pi}$$ electrons present in product $$\mathrm{P}$$ is:

$$\mathrm{2}$$

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