5
$\begingroup$

In an IR spectrum, all sources describe the position of a peak as a range. The ranges seem to be defined arbitrarily. Take, for example, the IR spectrum of ethanol (which I can't upload for some reason); why is the $\ce{O-H}$ stretch defined as having a range of $3200$ to $3750~\mathrm{cm}^{-1}$. If it was up to me, I would say that the trough goes from $3100$ to $3650~\mathrm{cm}^{-1}$. And if the $\ce{C-H}$ stretch wasn't present, I would say that the $\ce{O-H}$ trough extends to below $3100~\mathrm{cm}^{-1}$.

Moreover, sometimes there are very specific values given. These also seem to be arbitrarily defined. How do I determine the exact specific value that corresponds to a peak?

And finally, what is the correct way to identify the position of a trough on an IR spectrum? Is it a range? Or do I have to give a specific value?

P.S. I know I asked three questions, but they all serve the overall purpose to help me define how a trough is defined within an IR spectrum.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ If your instrument doesn't have a peak/trough picking routine then numerically differentiate the spectrum and look for zero crossings (where the gradient is zero), some smoothing may be necessary, then plot this together with the normal spectrum and so identify peaks and troughs. $\endgroup$
    – porphyrin
    Mar 7, 2017 at 13:09
  • $\begingroup$ Here is a link to a IR spectra of ethanol. webbook.nist.gov/cgi/… $\endgroup$
    – MaxW
    Mar 7, 2017 at 16:14
  • $\begingroup$ Phase transitions also have a huge effect on the spectra: crystalline phases are well ordered, thus all molecules have very similar surroundings and the bands are narrow (but you may be able to observe differences in the spectra in the volume of a bulk phase vs. surface). Liquid but pure substances have much broader bands (as different molecules have different environments), and also band position may change respectively. Gas phase typically has approximately no intermolecular forces, and the spectra again have very narrow bands (narrow enough to observe rotation in addition to vibration). $\endgroup$
    – cbeleites unhappy with SX
    Mar 7, 2017 at 19:45
  • $\begingroup$ A liquid mixtur like ethanol + water has yet different band positions and band width, as again the molecular environment is different. Note: the IUPAC defines chemical species via "having different spectra" - so you can say that all those different spectra of ethanol correspond to different ethanol species - and you have to expect different spectra if you have different species (but not all bands / spectral ranges need to be affected). $\endgroup$
    – cbeleites unhappy with SX
    Mar 7, 2017 at 19:45

2 Answers 2

Reset to default
5
$\begingroup$

Generally, the width of IR absorption peaks depends on the environment of the target molecule. Intermolecular forces alter vibrational modes.

If intermolecular attractions are fairly strong (like the hydrogen bonding present in liquid ethanol), the number and strength of interactions can vary greatly from molecule to molecule, causing the vibrational modes observed in an IR absorption spectrum to widen.

As for the reported width of the peak, the convention is to report the width at 1/2 the height of the peak.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Just to add to your last sentence- this convention is called "Full width at half max" (FWHM) and is a pretty common way to define peak widths for all kinds of spectra across different fields (this is also used in physics and astronomy, for example). See: en.wikipedia.org/wiki/Full_width_at_half_maximum $\endgroup$
    – spacetyper
    Mar 7, 2017 at 15:21
  • 1
    $\begingroup$ FWHM is only meaningful if you have one peak. If you have multiple overlapping peaks it is meaningless. $\endgroup$
    – MaxW
    Mar 7, 2017 at 16:04
  • $\begingroup$ Point is that in IR spectra nice clear peaks are rare. $\endgroup$
    – MaxW
    Mar 7, 2017 at 16:15
  • $\begingroup$ Also changes in intramolecular forces (conformation) change vibrational spectra. $\endgroup$
    – cbeleites unhappy with SX
    Mar 7, 2017 at 19:36
5
$\begingroup$

In spectroscopy (and spectrophotometry), the position of a peak/trough is defined by the location of the tip of the peak or trough. This is why you see such precise values given for their locations.

The ranges you refer to (e.g., $3200$ to $3750\pu{ cm^{-1}}$ for the $\ce{-OH}$ stretch) are the ranges of wavenumber values where the tip of the relevant peak/trough is usually found, and not the range of values over which the peak/trough spreads.

Taking IR spectroscopy as representative, since that's the technique you're asking about, the following is from LibreTexts:

[This equation] gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light. $$\nu={1\over2\pi}\left({k\over\mu}\right)^{1\over 2} \\ \nu = \text{ frequency in cm}^{-1} \\ k = \text{ force constant in N/cm} \\ \mu = \text{ reduced mass in kg}$$

Thus, in theory a given vibration corresponds to an exact, specific wavelength of light that would be absorbed when recording an IR spectrum. However, due to effects like thermal excitation and solvation, we don't observe "infinitely narrow" absorption lines, but instead peaks of finite width. From that same LibreTexts reference:

In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1 shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.

In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol and hexanoic acid. When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.

Thus, if not for environmental effects, every absorption peak would be an absorption line. But, environmental effects cause the absorption to "spread" around that theoretical value. The peak is still considered to be located at that central point, though.

This helps to explain why, for example, the $\ce{-OH}$ stretch peak/band tends to be extraordinarily broad (lots of local variations in the hydrogen bonding environment), whereas the carbonyl $\ce{>C=O}$ stretch is often quite narrow (fewer environmental effects are able to perturb its frequency).

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thank you very much for that, but may you please provide a source to prove that? I have been trying to look for a source, but couldn't find one. You and Dan Burden seem to disagree on the fact that the range values are where the tip of the trough is to be found. Also, see the description of the IR spectrum for lactic acid in this link: chemguide.co.uk/analysis/ir/interpret.html $\endgroup$
    – Mathematician
    Mar 7, 2017 at 14:26
  • $\begingroup$ @Mathematician You're right, it is hard to find a source that says it explicitly. See my edit; is that argument enough? $\endgroup$
    – hBy2Py
    Mar 7, 2017 at 14:51
  • $\begingroup$ @Mathematician I read the description of the lactic acid spectrum at that link as being fully consistent with my answer. I still could be wrong, though! $\endgroup$
    – hBy2Py
    Mar 7, 2017 at 14:53
  • $\begingroup$ @Mathematician In Dan's answer, when he refers to "width" and "1/2 height of the peak," he's talking about the standard method for estimating the area of a peak (see here), and not about the range of wavenumbers where a peak is considered to be located. $\endgroup$
    – hBy2Py
    Mar 7, 2017 at 14:57
  • $\begingroup$ I am a high school student, and your knowledge of chemistry seems much better than mine. Your source and deduction seem very logical. I trust that you are indeed correct, even more so supported by other questions that I've seen on this site before i asked this question. Lastly, just to make sure, take a look at question 26 on the paper in the next comment. Do you agree that the O-H stretch should be at around 3330 rather than at 3400? $\endgroup$
    – Mathematician
    Mar 7, 2017 at 15:54

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.