$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_{12}\left(\mathrm{A}\right)\frac{{\mathrm{Br}}_2/\mathrm{hv}}{\text{ Monobromination }}\text{ One isomer (B) } \\ {\mathrm{C}}_6{\mathrm{H}}_{12}\left(\mathrm{C}\right)\frac{{\displaystyle {\mathrm{Br}}_2/\mathrm{hv}}}{{\displaystyle \text{ Monobromination }}}\mathrm{Number} \mathrm{of} \mathrm{isomers} possible with same ring size(structural and geometrical) \end{gathered}$$

Both (A) and (C) do not decolourise Baeyer's reagent or Br₂ solution.

Which of the statements is/are correct?

The correct statements are:

1. (A): Compounds (A) and (B) are, respectively, cyclohexane and bromocyclohexane.
2. (B): Compound (C) is methylcyclopentane.
3. (C): The total number of isomers obtained by monobromination of (C) is six, including geometrical isomers.
4. (D): The major product of monobromination of (C) is 1-bromo-1-methylcyclopentane.

Concept Behind the Question:

This question is based on the reactions of hydrocarbons under free radical substitution conditions (Br₂/hv). Below is a detailed breakdown:

• (A) and (B):

  Compound (A) is cyclohexane, and upon monobromination using Br₂/hv, it forms bromocyclohexane (B). Neither of these compounds decolourises Baeyer's reagent or Br₂ solution, indicating the absence of double bonds.

• (B) and (C):

  Compound (C) is methylcyclopentane. Its structure is different from cyclohexane due to the pr

  esence of a methyl group on the cyclopentane ring.

• Number of Isomers:

  During the monobromination of (C), six isomers are possible, including structural and geometrical isomers, due to the multiple positions available for substitution and the methyl group’s presence.

• Major Product:

  The major product of monobromination of (C) is 1-bromo-1-methylcyclopentane, where bromine substitutes the most stable carbon atom (tertiary carbon) via the free radical mechanism.