Angular momentum in molecules

When using the Born-Oppenheimer approximation to find the electronic energy levels of a diatomic molecule, $L^2$ doesn’t commute with the Hamiltonian i.e. $l$ is not a good quantum number, but its projection on the internuclear axis is. If we are in a singlet state, such that the total intrinsic spin, S is 0, then we have that the total angular momentum of the molecule (we are looking only at the electronic energy for now, so no rotation or vibration) is $J=L+S=L$. However this means that $J^2$ doesn’t commute with the Hamiltonian. But, I thought that in a closed system, the Hamiltonian is always commuting with the total angular momentum. Why is it not the case here?