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.