Question

The IR spectra of the bis-dimedone derivative and the octahydroxanthenedione derivative should have significant differences in the frequency of the carbonyl stretch. Explain the reason for this difference.

Answer 1

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To explain this question you have to observe the structures of each of the compounds. Bis-dimedones are prepared starting with dimedone:

We don’t need to focus on the group Ar because we want of figure out about the IR carbonyl bands.

Structure of the octahydroxanthenedione derivative is different:

It is a tricyclic structure much more rigid that the other structure. In fact the IR stretch for this compound should be as a cyclic a,b-unsaturated ketone. This stretch frequently presents as two bands over the 1700-1580 cm^-1 region.

Bis-dimedones are more flexible compounds and they have a three dimensional structure that changes the way the carbonyl band appears. I build a model for you to see what is going on:

This structure is stabilized by two hydrogen bonds. In this structure there is a presence of the keto-enol form. This enolic form does not show on the normal absorption of conjugated ketones. Instead a broad band appears in the 1640-1580 cm^-1 region, several times more intense that the normal carbonyl absorption. This is a result of the intra-molecular hydrogen bonding.

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