# O-H infrared band dependence of alcohol concentration in water

With regard to IR spectroscopy, how would the O-H band change as an alcohol solution was diluted with water? I am referring to any alcohol, but for the sake of example we can use ethanol. I think first that the $\ce{O-H}$ band shown would increase in intensity. I believe this is the case because the water is becoming a greater proportion of the mixture and water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule. Is this correct? Is there also a wave number shift that would occur too that I am missing?

The easiest thing to do is look at some example spectra. All of them are taken from the Spectral Database for Organic Compounds.

This is 1-propanol, first as a liquid film and then in $\ce{CCl4}$.

In the liquid, the broad peak at $\pu{3333 cm^{-1}}$ is due to intermolecular hydrogen bonding, just as you would see in an IR spectrum of bulk pure water. The $\ce{CCl4}$ spectrum shows that as you reduce the probability of 1-propanol (or any alcohol) molecules contacting each other, the free stretch appears at $\pu{3637 cm^{-1}}$. This is similar to water in a gas-phase spectrum.

water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule

Close. It isn't about the fraction of total bonds, in this case it's simply that water has two hydroxyl stretches and ethanol only has one. As water is added, the intensity of the hydrogen bonding $\ce{O-H}$ band should increase over the pure alcohol hydrogen bonding band, but we cannot know quantitatively what the increase will be without performing an experiment.

Is there also a wave number shift that would occur too that I am missing?

Yes. In the gas phase, the water symmetric and asymmetric $\ce{O-H}$ stretches are at $\pu{3657 cm^{-1}}$ and $\pu{3756 cm^{-1}}$, which then red-shifts to $\pu{3404 cm^{-1}}$ in the bulk due to hydrogen bonding. Compare this to 1-propanol, where the "free" and hydrogen bonding values are $\pu{3637 cm^{-1}}$ and $\pu{3340 cm^{-1}}$, respectively. The water stretches are blue-shifted (higher in energy) compared to the alcohol.

Water vibrational frequency values taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html.

• So just so I understand. It "blue shifts" or shifts towards a higher frequency because the O-H bond in water is more polar than the O-H bond in alcohol? is that correct? thanks so much for your help also. – bruh Oct 15 '17 at 3:45
• I'm not sure about the general case, but I know that charge transfer (additional electron density) into the $\ce{O-H}$ bond causes a red-shift due to populating previously empty antibonding orbitals. If you assume that a decrease in frequency is purely due to an increase in the bond's electron density, it's consistent with a little more electron donation from a $\ce{CH3}$ group than a hydrogen atom, like en.wikipedia.org/wiki/Electrophilic_aromatic_directing_groups. – pentavalentcarbon Oct 15 '17 at 16:19
• It doesn't look like there are large changes going from 1-propanol to 1-butanol, but the broad band in methanol (liquid) is at 3347, compared to 3333 in 1-propanol, presumably because the electronic structure of $\ce{CH3CH2CH2\bond{-}}$ isn't very different from $\ce{CH3\bond{-}}$. – pentavalentcarbon Oct 15 '17 at 16:23

The water and ethanol -OH stretch bands are too close to be distinguished in a system with hydrogen bonding (liquid systems). You could expect a broadening in the signal as you add water since you will have the two compounds in your system, then the ditribution of frequencies will be broadened by both water and ethanol absorptions. You are right on the assumption that the intensity will be larger since water has 2 OH groups while an alcohol only has 1. As your system becomes very diluted, it will eventually show the IR spectrum of pure water, with a small wavenumber shift towards the water -OH stretch (3700-3100 cm-1).

The situation stated above is for alcohols such as ethanol that are capable of doing hydrogen bonding. Some alcohols don't have such capability and the -OH stretch band in a sample of this pure alcohol would show a sharp and intense -OH band, such as the remarkable -C=O band. If you would add water to such substance, then you would observe the disappearance of this band and the appearance of the "upside down bell" band characteristic of water. (Both assumptions are for liquid systems, not for gases).

• In pure alcohol the band is narrower than in alcohol/water however indeed is not sharp. Broadening occurs because not only there are two similar vibrations within alcohol and water, but because an ensemble of new vibrations occurs, those due to alcohol - water hydrogen bonds. – Alchimista Oct 14 '17 at 9:53