# Assignment of the IR spectrum of 2,4-pentanedione (acetylacetone)

The IR spectrum is shown here:

My assignments so far:

• The 1728 and 1708 stretches correspond to $\ce{C=O}$ stretches in the keto form - the symmetric/asymmetric stretches respectively.

• The 3003 stretches are typical $\ce{C-H}$ stretches.

• The 1606 stretch corresponds to the $\ce{C=C}$ in the enol form. However, hidden under this peak there is also the $\ce{C=O}$ stretch in the enol form, which is lowered by conjugation to the $\ce{C=C}$ and the $\ce{O}$ of the $\ce{-OH}$ respectively.

I have two questions about the spectrum, however:

1. Why is there no $\ce{-OH}$ peak? Even if the enol is hydrogen bonded to itself, there should still be an $\ce{-OH}$ peak in the $\pu{3400 cm^{-1}}$ range?

2. Does the fact that the carbonyl peak in the enol form is hydrogen bonded change its shift? Assuming the only contribution to the lowering of the $\ce{C=O}$ stretching frequency is conjugation to the one $\ce{C=C}$ bond, this would lower the peak to ~$\pu{1655 cm^{-1}}$: Surely this would be seen as a peak in its own right? It seems to be shifted even lower to be fully hidden by the 1606 peak.

• You're looking for the OH stretch, but what does that look like in this molecule? Because of internal hydrogen bonding, you might expect the bond strength to be very different, so the stretch frequency is not going to be where you would expect... – Zhe Jan 28 '18 at 17:36