# Aromatic Stretches C=C in Salicylamide

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 $$1580 - 1630 cm^{-1}$$?

• $\pu{2736 cm-1}$ is more like noise to me. Where do you expect peak for $\ce{C-H}$ (aromatic) stretching would appear? – Mathew Mahindaratne Mar 6 '20 at 0:02