Consider the vibrational mode that corresponds to the boat-like bending of a benzene ring. Is it (i) Infrared active? (ii) Raman active?

This question was asked in an assignment in my college. However, all we've been taught is how to distinguish functional groups in organic molecules using characteristic IR frequencies, nothing else. From this I can tell that the C=O stretch in a carbonyl compound, for example, is IR-active. But how do we determine this for any molecule and any normal mode in general?

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    $\begingroup$ This appears to be a honework question. You must show some effort towards solving it, and then ask for help. In its current form it doesn't comply with our homework policy, and will be closed. Welcome to ChemistrySE $\endgroup$ – getafix Sep 21 '16 at 3:14
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    $\begingroup$ If what you say is true, the question appears to be asked by mistake. Knowing which modes are active in IR and which are in Raman involves a substantial body of knowledge (symmetry groups, irreducible representations, characters, etc). There is no way it can be squeezed into a single answer. $\endgroup$ – Ivan Neretin Sep 21 '16 at 8:23
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    $\begingroup$ There is a simpler way to find this out. Whether the vibrational mode is IR active depends on whether there is a change in the molecular dipole moment upon vibration. This is equivalent to asking whether there is a dipole moment in the boat-like conformation, since the ground state planar conformation has no dipole moment. Next, because the benzene molecule is centrosymmetric (i.e. possesses a point of inversion), IR-active modes cannot be Raman-active, and vice versa. $\endgroup$ – orthocresol Sep 21 '16 at 10:52
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    $\begingroup$ [...] of course, the "rule of mutual exclusion" that I described has its origins in symmetry (Cartesian axes being ungerade and their products being gerade means that the IR-active irreps cannot be the same as the Raman-active irreps), and a full explanation of why a vibrational mode is IR/Raman active is of course dependent on symmetry and some quantum mechanics. So in general, with no foundational knowledge of why a mode is IR/Raman active (I presume OP has not been taught these yet), one cannot tell. $\endgroup$ – orthocresol Sep 21 '16 at 10:55
  • $\begingroup$ @orthocresol I think that if you write a short answer focusing on the change in dipole moment and ignore symmetry, it's a good start. $\endgroup$ – pentavalentcarbon Sep 29 '16 at 2:51

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