Question

Why is the bridgehead carbon of bicyclic compounds never $s{p^2}$ or $sp$ hybridized?

Answer 1

Hint: A bicycloalkane with two carbon atoms shared by all three rings in the molecule is known as a bridged bicycloalkane. Bridgehead carbon atoms are the two carbon atoms that are exchanged by all three rings.

Complete step-by-step answer:
We know that the bridgehead carbon has three bonds to other carbon atoms, as well as one bond to a hydrogen atom. It can only be $s{p^3}$ if the carbon has four different bonds. One atom will have to be double bonded to the Carbon in order for $sp^2$ to work. This will result in a trigonal planar structure with 120-degree bond angles. Trying to push certain angles in the smaller ring would place a lot of strain on the molecule. It is possible to have a $sp^2$ hybridised carbon that binds to rings without creating excessive strain in a larger molecule with much longer chains of carbons.

Just two atoms are bound to the carbon in $sp$ hybridization. One triple and one single bond, as in $O = C = O$, or one triple and one single bond, as in $HCN$. In this case, the molecule's component with the $sp$ hybridised carbon must be linear, not a bridgehead, since it would require three atoms to be bonded.

Therefore, because of the extreme rigidity of bicyclic compounds, the bridgehead carbon is never $s{p^2}$ or $sp$ hybridised, and these radicals cannot have a planar form.

Note: Bridgehead carbon atoms are the two carbon atoms that the three rings share whereas a bridge is a bond or a chain of bonds that connects the carbon atoms in the bridgehead.