Anhydrous glucose has an absorption spectra (glucose IR spectra) that is quite different from the spectra of water (Liq. water absorption spectra). However, when a solution of glucose is prepared in water, the absorption spectra is almost indistinguishable from the absorption spectra of water alone. Is this because glucose forms additional bonds when in water? Pointers to links that explain this in detail is also appreciated. Most links I searched on Google just stated this as an observation without explaining the underlying reason.
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4$\begingroup$ In order to answer accurately more details would be needed on exactly how the spectra were taken, the concentration of the glucose solution and so on. // The gist of the problem is signal to noise. There is a lot of water and little glucose. So you have to detect a relatively small glucose signal in a large water signal. $\endgroup$– MaxWJan 7, 2016 at 18:10
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$\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. Visit the help center to learn more about how it works. As Max said, this question would greatly benefit from an IR spectrum of the actual solution. $\endgroup$– JanJan 7, 2016 at 18:13
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$\begingroup$ @MaxW I am referring to the plots in this link. So if I am getting this right, we would need large amount of glucose in water to be able to see peaks from glucose? At 4.4mol/lt, the difference in absorption spectra is very little. $\endgroup$– user1155386Jan 7, 2016 at 18:16
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$\begingroup$ The article is doing in-vivo monitoring, so you can't just dissolve more glucose. So yes, there is a poor S/N ratio. $\endgroup$– MaxWJan 7, 2016 at 18:27
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$\begingroup$ Thanks! Is there a website that holds data for uv-visible-IR absorption spectrum of common chemicals? The NIST database lists only the IR spectrum. $\endgroup$– user1155386Jan 7, 2016 at 20:00
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3 Answers
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I see the spectra in the following way:
- The sample and measurement: glucose in water and IR absorption measurement
Considering free glucose molecule, we should expect the vibrations from $C-H$, $O-H$, $C-O$ and $C-C$ bonds and other associated vibrations. The approximate vibrational frequencies are :
a) $C-H$ = ~2800 - 3300 $cm^{-1}$ (stretching)
b) $O-H$ = ~3800 - 3200 $cm^{-1}$ (stretching)
c) $C-O$ = ~1200 - 1030 $cm^{-1}$ (stretching)
d) $C-C$ = usually not good for interpretation
e) $C-H$, $CH_{2}$ and $O-H$ bend(in plane) = 1500-1200 $cm^{-1}$
and now considering water, the $O-H$ stretching band spectrally overlaps with the bands of glucose (~3750 - 3200). The combination band (~2300) and scissor mode (~1600) are also relatively broad and strong. Due to this spectral overlap and also larger infrared absorption cross-section of water, glucose is not clearly identifiable in solution.
Instrument and spectral quality : For obtaining a high resolution spectra with good s/n ratio the instrument should be good (FT-IR) and long exposure is needed.
The spectra shown by @user1155386 has the $x-axis$ as nanometers (not wavenumbers ! ), indicating that the spectral resolution is low. It is noisy showing that exposure time can be improved (the exposure time shown there is only 10 milliseconds).
However, still it would be difficult (but not impossible) to differentiate glucose solution and water. The main hurdle would be get a good spectra and the reason 1. mentioned above.
Additionally, I should mention Raman spectroscopy is better for aqueous solution measurements.
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Like @MaxW said, in this case it is probably a problem with the SNR ratio of the glucose to water. Of course there could be some differences in a solid state spectra to a spectra of a liquid. This could be caused by interactions with the solvent.
An additional database for IR-spectra (and more) is SDBS. There are more databases like reaxys, but it is not for free. Maybe you university or library owns an account.
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Take a good spectrum (not too high resolution, more than a few scans if your spectrometer is not in top shape) of your solution, and an evenly good spectrum of pure water, and subtract the two from each other. The glucose spectrum (i guess looking quite different from the spectrum of the solid) should appear as plain as day.
Try to scale the relative intensity of the two spectra so as to minimise the residual difference. The absolute intensity signal intensity in IR spectrometers tends to drift quite heavily (compared to the very small signal difference you want to observe) on a timescale of minutes.
