  We perform shell model calculations in odd-odd nuclei using a
quadrupole-quadrupole interaction and single-particle splittings chosen so as
to obtain the SU(3) results. Elliott had shown that such an interaction gives
rotational bands for which the energies go as I(I+1). This is certainly true
for eve-even and for odd-even or even-odd nuclei with $K\neq 1/2$. We have
looked at odd-odd nuclei e.g $^{22}Na$ and found somewhat different behaviour.
In $^{22}Na$ the $I=1^+_1 T=0$ and $I=0^+_1 T=1$ states are degenerate, and a
rotational band built on the $I=0^+_1 T=1$ state behaves in a normal fashion.
For the $I+1^+_1 T=0$ band however, we find that the energy is given by
$E(I)-E(1^+_1)=AI(I+1)$. This differs from the `normal' behaviour which would
be $E(I)-E(1^+_1)=AI(I+1)-2A$.
