[simonsci]

Hi everyone! Please have a look at the question below from Organic Chemistry (7^th edition) by Bruice.

'Which would show an absorption band at a larger wavenumber: a carbonyl group bonded to an sp³ carbon or a carbonyl group bonded to an sp² carbon of an alkene?'

The solution is the one bonded to an sp³ carbon as it won't undergo electron delocalization. Let's consider the molecule on the right. I think no electrons will delocalize so I think the two carbonyl groups will show absorption bands at equal wavenumbers in some situations. Am I right?

[Dan]

Ketene C=O stretches are vastly different from ketone (unsaturated or not) stretches.

Consider the resonance structures of the ketene you drew - do you think the frequency of the C=O stretch will be higher or lower than that of a "standard" ketone?

A good exercise would be to put acetone, butenone and dimethylketene in order of C=O stretching frequency.

[simonsci]

Thank you for your help. I think the carbonyl group stretch of acetone occurs at a higher wavenumber than that of butenone, as it doesn't engage in electron delocalization. But I believe the carbonyl group stretch of dimethylketene occurs at similar wavenumber as acetone, since it doesn't undergo electron delocalization too. So my answer is:

Acetone ≈ Dimethylketene > Butenone

Is it correct?

[Dan]

I believe the carbonyl group stretch of dimethylketene occurs at similar wavenumber as acetone, since it doesn't undergo electron delocalization too. So my answer is:

Acetone ≈ Dimethylketene > Butenone

Is it correct?

No - as I already mentioned, the C=O stretch of ketenes is very different from acetone and butenone.

It is possible to draw ketene resonance structures to account for this - start at the O lone pair.