How to decide which hydrogen would be the leaving group in compounds such as bicyclo[3.3.1]nonane?More specifically why does the bromine substitution occur at the hydrogen attached to the bridgehead and not the other hydrogens attached to the carbons .

Also calculate the charges of each atom

Aug 4, 2017

At least in the first mechanistic step, everything is neutral... it's a radical reaction, i.e. homolytic cleavage occurs (or should occur) all the way through.

Okay, so you're talking about the product:

and why the hydrogen radical, $\text{H} \cdot$, that left from the compound bicyclo alkane is one of the ones at the bridgehead carbon. Well, it's not quite as complicated as it seems.

If you notice the number of bonding groups around the bridgehead carbons, there are four:

• one from each side
• one from the bridge
• one from the $\text{H}$ (or in the product, a $\text{Br}$)

That generates a tertiary carboradical if the $\text{H} \cdot$ leaves, compared to secondary carboradicals off of any other carbon center:

Carboradicals follow the same thermodynamic stabilization trend as carbocations. They are roughly planar.

(Granted, the differences in stability are subtle. Being within $\text{1 kcal/mol}$ is considered chemical accuracy.)

This is because hyperconjugation from an adjacent $\text{C"-"H}$ bond delocalizes the electron density into the central nonbonding $p$ orbital, whether it is empty (in carbocations) or half-filled (in carboradicals).

And since the compound was originally symmetrical (${C}_{2 v}$ point group, bisected in the plane of the screen), either of the two bridgehead carbons could have formed the carboradical.