I'm not sure about the attributions of 3 bands of my ATR FT-IR spectrum of an extract of vanilla sugar, that probably contains mainly vanillin and some impurities. The extraction solvent was ethyl acetate and was evaporated with a rotary evaporator. The final extract was a dark yellow liquid and vanillin is normally a clear solid. These impurities were probably due to the fact that the vanilla sugar uses "natural" vanilla extract.

Vanillin has an aromatic ring, an aldehyde, an ether and aromatic alcohol.

So back to the IR spectrum : IR Sec

Vanillin is an aldehyde and elongation of C-H bond of aldehyde usually appear as a two bands (asymmetric and symmetric elongation) between $\pu{2650}$ and $\pu{2800 cm-1}$ (from my IR table). Before I determined the positions of the band with my ruler and my square (so it's less precise then what a computer would do), I thought the "aldehyde bands" were the $\pu{2911 cm-1}$ and $\pu{2727 cm-1}$ bands. But $\pu{2911 cm-1}$ I think is too high to be an aldehyde band, whereas the band of weak intensity at $\pu{2727 cm-1}$ is in the right position. But that $\pu{2727 cm-1}$ band seems of too weak intensity. Then the $\pu{2911 cm-1}$ band would be the $\ce{C-H}$ elongation from the methoxy group ($\ce{O-CH3}$) of vanillin.

So which is the right one? If neither, can aldehyde shows just one band at $\pu{2839 cm-1}$ instead of two (for asymmetric and symmetric elongation)?

  • 2
    $\begingroup$ For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations and with pu{} one for a non-breakable space between figure and unit. $\endgroup$
    – Buttonwood
    Jan 14 at 14:17

1 Answer 1


The freely accessible SDBS holds three records of transmission IR spectra for vaniline (CAS 121-33-5) as record SDBS 726: a $\ce{KBr}$ disc, in solution of $\ce{CCl4}$, and in nujol. Only the recording in tetra displays absorption bands one can attribute to $\ce{C-H}$ vibration and these are (in comparison to the other ones) of weak to moderate intensity only:

enter image description here

Compare this with the one in nujol:

enter image description here

Especially if there are complementary data by NMR and MS available, I would not invest too much time to assign each band in an IR spectrum. With keyword "aldehyde", I associate the motif of $\ce{C=O}$ and hence seek in the spectrum for the presence of its absorption band around $\pu{1725 cm-1}$ (value from Wikipedia's compilation) of at least moderate if not strong intensity.

The position of IR absorption bands depends a bit on the chemical environment, so an alkyl $\ce{R-CHO}$ may show at higher numbers, than an aldehyde bond to an aryl, or a $\alpha$,$\beta$ insaturated chain. There equally is some influence in position by recording sample in nujol or tetra solution (an apolar matrix) or in a pellet of $\ce{KBr}$ (polar matrix) in transmission IR spectroscopy, and the neat sample (typical for ATR). Second (and sometimes operators forget this), the absorption intensities recorded on an ATR spectrometer have to be corrected for the difference in refractive indices of the sample and the window of the spectrometer (e.g., Ge, Si, Diamond) because the penetration depth of the evanescent wave varies over wavelength. This limits the comparison absorption bands' intensity between data recorded by transmission, and one by ATR.

Discounting information in the fingerprint region, the one on $\pu{1697 cm-1}$ (your reading) possibly is the signal indicative enough for the presence of the $\ce{C=O}$ motif. There may be an individual assignment of the bands in the literature but if the extract still is a yellowish liquid instead of a colorless solid (mp $\pu{81 ^\circ{}C}$, property box of Wikipedia) your sample might not yet be pure enough to merit this investment.

If you can influence the recording and subsequent plot of the IR spectrum, seek to split the scale of the abscissa e.g., around $\pu{2000 cm-1}$ for a subsequent expansion (cf. the SDBS spectra).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.