I want to ask a question about the C=C streching modes of Salicylamide.
I was presented with the following spectra and asked to comment on the bonds causing the peaks.
I identified the molecule from this and its NMR spectra to be salicylamide.
Here is my understanding of the IR spectra (solid state, ATR).
- $3391.66 cm ^{-1}$ is the asymmetric stretch of the $\ce{N-H}$ in the amide.
- $3184.80 cm^{-1}$ is the symmetric stretch of the $\ce{N-H}$ in the amide.
From previous understanding, I was aware that the asymmetric stretch requires more energy, and hence a higher wavenumber than the symmetric stretch.
- $2736.27 cm^{-1}$ is most possibly the $\ce{O-H}$ peak of the phenol. I attributed this to a few reasons, possibly due to hydrogen bonding or due to the dissociation of the $\ce{O-H}$ bond, which is driven to the right hand side of the equilibrium by the presence of the EWG amide group.
- $1672.98 cm^{-1}$ is due to the $\ce{C=O}$ stretch of the amide.
- $1628.09$ and $1589.49 cm^{-1}$ are due to $\ce{C=C}$ aromatic stretches.
I tried to find information to help me describe the $\ce{C=C}$ stretches (i.e. what orientation, any dipole arrows) but I failed to find any information to support my argument.
I recall in ferrocene and acetalferrocene, the ring breathing effect is observed, whereby ferrocene has fewer IR peaks than acetalferrocene due to a lack of a net dipole moment being non-zero, and I wondered whether there was a similar vibrational mode to apply for these two $\ce{C=C}$ peaks, but I was not sure.
What are the directions of the C=C stretches in the molecule of salicylamide in the range $1580 - 1630 cm^{-1}$?
