# Radical monobromination of cycloalkane

I'm trying to figure out the isomeric products of the radical monobromination of bicyclo[2.2.1] heptane. I'm not sure at which site the bromination would happen and what makes one site better than the other (how can one determine this without given the reaction enthalpy ?). I was also given the hint that there are stereoisomers but I can't think of one.

• Would help the discussion if you supplied a drawing with atom numbering – Waylander Jul 17 '19 at 13:27

The radical monobromination of bicyclo[2.2.1] heptane could give $$\ce{1}$$ ,$$\ce{2}$$ and $$\ce{3}$$.
However , $$\ce{2}$$ and $$\ce{3}$$ , are unstable .On observing $$\ce{2}$$ , planar radical at bridgehead is unstable due to angular strain introduced by planarity of radical at bridgehead .
Similarly in $$\ce{3}$$ , due to angular strain introduced in to the neighboring bonds,this radical also is unstable.
The situation leads to product from $${1}$$ , that gives stereoisomers as shown below.