-
Looking for HSC notes and resources? Check out our Notes & Resources page
IR Spectra (1 Viewer)
- Thread starter hs17
- Start date
Good question.
In my opinion, it is reasonable to expect an HSC student to get from the formula C3H6O to the compound being either propanone or propanal. The characteristic carbonyl stretch around 1700 cm-1 makes anything other than those two options unreasonable. The lack of any IR band that might reasonably be a hydroxyl group rules out an alcohol. There is only one O atom so an acid or ester is impossible. The only other option that I can imagine to account for the 1700 cm-1 band is an alkene (though I would expect that signal to have a wavelength closer to 1600 cm-1) that is also an ether - something like methoxyethene (CH2=CH-OCH3) - but I think that unlikely given the position of the signal I would attribute to carbonyl and the possible C-O band being weak and the alkene bend that I would expect around 900 cm-1 being missing.
So, I think concluding propanone or propanal is all that can fairly be expected.
1H or 13C NMR or even MS would give clear evidence to distinguish between the options.
Now, having said all of that... the IR spectrum is unquestionably propanone and not propanal. The reason is that the C-H stretch of an sp3 carbon near 3000 cm-1 (seen in many spectra and usually uninformative) is accompanied by two intense bands between about 2700 and 2900 cm-1 in an aldehyde. As can be seen in the spectra below, propanal shows these prominent aldehydic bans and they are absent in ketones (like propanone, also shown below).
In my opinion, it is reasonable to expect an HSC student to get from the formula C3H6O to the compound being either propanone or propanal. The characteristic carbonyl stretch around 1700 cm-1 makes anything other than those two options unreasonable. The lack of any IR band that might reasonably be a hydroxyl group rules out an alcohol. There is only one O atom so an acid or ester is impossible. The only other option that I can imagine to account for the 1700 cm-1 band is an alkene (though I would expect that signal to have a wavelength closer to 1600 cm-1) that is also an ether - something like methoxyethene (CH2=CH-OCH3) - but I think that unlikely given the position of the signal I would attribute to carbonyl and the possible C-O band being weak and the alkene bend that I would expect around 900 cm-1 being missing.
So, I think concluding propanone or propanal is all that can fairly be expected.
1H or 13C NMR or even MS would give clear evidence to distinguish between the options.
Now, having said all of that... the IR spectrum is unquestionably propanone and not propanal. The reason is that the C-H stretch of an sp3 carbon near 3000 cm-1 (seen in many spectra and usually uninformative) is accompanied by two intense bands between about 2700 and 2900 cm-1 in an aldehyde. As can be seen in the spectra below, propanal shows these prominent aldehydic bans and they are absent in ketones (like propanone, also shown below).