Rule of mutual exclusion: is the converse true? There is a "famous" rule in spectroscopy,1 that goes this way:


  If a compound is centrosymmetric, then its normal vibrational modes cannot be simultaneously IR and Raman active.


and this is simple enough to prove: for a vibrational mode to be IR active, it has to transform under the same irrep as (one of) the cartesian axes $x$, $y$ or $z$, which are all ungerade. For a vibrational mode to be Raman active, it has to transform under the same irrep as one of $(x^2, y^2, z^2, xy, yz, xz)$, and since $u \otimes u = g$, these irreps cannot be the same as the irreps of $(x,y,z)$. Hence, a vibrational mode, which transforms under a single irrep, cannot simultaneously satisfy both conditions at once.

However, I am curious about the converse statement. 


  If a compound does not have any vibrational modes which are simultaneously IR and Raman active, is it necessarily centrosymmetric?


If the answer is yes, I would like a proof; if it is no, a counterexample, please.

(Just in case anybody is confused, because I am: A counterexample would be a molecule without a centre of inversion, which has no vibrational mode which is simultaneously IR and Raman active.)



1 Famous enough for me to know, at least.
If there is a molecule of $O$ (as opposed to $O_h$) or $I$ (as opposed to $I_h$) symmetry, it would be such a counter example (no mode that is both IR and raman active, and not centrosymmetric). 

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=903&amp;option=4

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=905&amp;option=4

Apparently a snub cube is an example of the $O$ point group and in Chemistry for the 21st Century it is stated:


  9.5.2.5 Snub Cube (6) We have recently demonstrated the ability of six resorcin[4]arenes and eight water molecules to assemble in apolar media to form a spherical molecular assembly which conforms to a snub cube (Fig. 9.3)