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To be IR active, the vibration of a bond must result in a substantial change in dipole moment. Since trans-2-butene is symmetrical, will the C=C stretch show up on IR?

Something tells me no because if the C=C bond vibrates, there will be no net change in the bond dipole moment, unlike in cis-2-butene. Any polarization one way in trans-2-butene is cancelled out by a polarization in the opposite direction in trans-2-butene. Cis-2-butene, however, has a permanent dipole and any vibration only exacerbates this dipole.

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    $\begingroup$ What about the C-H bonds? They will show in the IR. $\endgroup$ – LDC3 Sep 8 '14 at 2:52
  • $\begingroup$ I was thinking specifically of the C=C bond. $\endgroup$ – Dissenter Sep 8 '14 at 2:53
  • $\begingroup$ How come something such as 2-butyne is IR inactive according to my book? Couldn't the methyls be moving in different directions here too? $\endgroup$ – Dissenter Sep 8 '14 at 3:11
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – LDC3 Sep 8 '14 at 3:27
  • $\begingroup$ Dissenter I think Martin has answered your first comment below. If you would, add a separate question about what you and @LDC3 came up with in chat. I'm going to leave the second comment for now. $\endgroup$ – jonsca Sep 8 '14 at 20:12
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The $\ce{C=C}$ stretch will not show in IR. The frequency is $\nu_{\ce{C=C}}=1700~\mathrm{cm^{-1}}$ and has A gerade symmetry (DF-BP86/def2-SVP). There is no change in the dipole moment.
c=c stretch

However, some bending frequencies involving the $\ce{C=C}$ double bond will show up. For example $\nu_{\ce{HC=CH}}=963~\mathrm{cm^{-1}}$ with a medium intensity.
c=c bend

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