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

Question

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.

Answer 1

OH str for beta diketones is tabulated from 3200 to 2400 1/cm. The shift is explained by resonance of the hydrogen bonded structure.

Its absorption features is small or unseen as for

• commonly the ketonic form is highly dominant; i.e. we won't expect to see OH str in the acetaldehyde spectrum, even

• owing to the above mentioned resonance, the dipole change associated to symmetric OH str is null, and it is minimal for the AS one.

The symmetry of 2,4-pentanedione makes the last point even more important.

Concerning the second part. The fact that the O of the carbonyl of the enoform is H bonded does further shift down the related CO str. Moreover, in addition to resonance with the C=C, note that again there will be resonance with the CO in beta as above, and this further shift the frequency to lower wn.