When the above compound is treated with $\ce{Cl2}$ in the presence of sunlight ($h\nu$), what is the number of monochloro structural isomers formed?
I expected only 1 product becuase I don't believe that monochlorination can take place at the bridgehead or the bridge positions. And rest of the 4 positions, by symmetry, are equivalent. This is because chlorination takes place through free radical mechanism in which a $\ce{sp^2}$ hybridised carbon free radical is formed.
But the answer given in the test's answer key is $3$. I think they have considered the products of bridgehead and bridged positions too but I don't understand how that is possible.