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What I know about $\ce{^{.}CR3}$ radical is that it has both $\ce{sp^2}$ and $\ce{sp^3}$ character but the $\ce{sp^2}$ character dominates and so the radical is $\ce{sp^2}$ in nature.

So, suppose we have the above situation. In this, a radical is formed at the carbon of the 6 membered ring at which one $\ce{-CH3}$ is attached since the bromination is selective, but will it form enantiomers? That is, can the $\ce{^{.}Br}$ radical attack from both sides?

In this question, the answer given is (c), but I think that enantiomers can not form, as the electron of the radical would be present only on one side of the radical carbon, and so cannot migrate to other side.

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Reaction mechanism

In the first step, due to the selectivity of bromine in a free radical reaction, the tertiary carbon which has a $\ce{-CH3}$ group attached to it forms the radical.

Now, since the radical is a $\ce{^{.}CR3}$ radical, the carbon radical in the intermediate becomes $\ce{sp^2}$. This means that $\ce{^{.}Br}$ can attack from both the top and the bottom since this lone electron in the $\ce{^{.}CR3}$ radical is in a p-orbital. This is because the electron is equally likely to be found in both lobes of the orbital.

Now after the attack of the $\ce{^{.}Br}$ radical, you get two products. The first one shows the product formed when $\ce{^{.}Br}$ attacks from the bottom and the second one shows the product formed after the $\ce{^{.}Br}$ attacks the intermediate from the top.

You may have noticed that both products are enantiomers of each other and hence the answer would be (c).

Reference Chem Libretexts

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