How can you use uv spectroscopy to help distinguish between 3-methyl-2-cyclohexenone and 4-pentylmethyl ketone?

Answer 1

First of all, you need to clarify your question. 4-pentylmethyl ketone, unfortunately, does not exist. You have made a mistake in the nomenclature.

I am assuming that you are talking about /\/\/*\, which is the carbon skeleton of pentylmethyl ketone, with the * representing C=O. The molecular formula is C7H14O, and it is also known as 2-heptanone. I don't know why you added in the "4-", unless you mean that the methyl is coming off the fourth carbon of the pentyl group. If that is the case, and you have moved methyl off to be a substituent on the pentyl, then the compound should be named an aldehyde and not a ketone.

UV spec gives you a graph with absorbance on the y-axis and wavelength (in nanometers) on the x-axis.

Conjugated systems are where double bonds occur, and are separated by exactly one single bond. In other words C=C-C=C is conjugated. C=C-C-C=C is not.

Conjugation reduces the energy gap between the "excited" state of electrons and the normal state. In other words, it takes less energy (and is easier) to excite electrons in compounds that have conjugated double bonds, such as 3-Methyl-2-cyclohexen-1-one. When these compounds return to their normal state, they emit a photon (light).

This light can have different wavelengths, which is why molecules with conjugated double bonds are often very colorful.

The energy that a photon is:
E = hf
Since f = c/(wavelength)
E = hc/(wavelength)

h is a constant, f stands for frequency, c is the speed of light, and the wavelength is measured in nanometers.

High energy light has high frequency and low wavelength (to make this more intuitive: think of short and fast pulsing ultraviolet radiation from the sun...it is powerful/energetic enough to cause sunburn.)

Therefore, the more conjugated bonds you have, the LESS energy it will take to excite your electrons. This means that if you have more conjugated bonds you have, the compound can give a high signal at higher the wavelength (and lower energies).

Red is very low energy. This is why beta-carotene, which is basically C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C, etc is orange. That's why carrots are orange.

Anyways, this is a lot of detail. Basically, it means that 3-Methyl-2-cyclohexen-1-one will give a peak in your absorbance graph, whereas you other compound probably will not (assuming that I have interpreted its incorrect name).

If you see a compound, the more CONJUGATED double bonds there are, the more you will see peaks at higher wavelengths, because it takes less energy to excite the electrons in those double bonds.

For example: C=C-C=C-C=C-C=C-C=C-C=C will give a signal at a higher wavelength than C=C-C=C=C=C=C=C-C=C. This is because in the second compound, there are fewer CONJUGATED double bonds (even though there are more double bonds total).

Why does this phenomena happen? A double bond is between two pi-orbitals. Basically they look like 8's. One lobe comes up from the carbon atom, and one lobe goes down. The electrons float around in there. If you have a double bond, the pi-orbitals on adjacent carbons share their electrons. Now, if you look at an extended conjugated system, you can see that there is an 8 on every single carbon in line! This means that basically the "electron cloud" is dispersed over the entire molecule. Much more stable, and much less energetic (remember, high energy means not stable).

Hope this helps!

By the way,
http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/UV-Vis/spectrum.htm
might be a good reference for you.

Answer 2

UV spectroscopy can help distinguish between 3-methyl-2-cyclohexenone and 4-pentylmethyl ketone based on their absorption spectra. Each compound will have a unique UV absorption pattern due to differences in their electronic structures. By comparing the peak wavelengths and intensities of their UV absorption bands, you can differentiate between the two compounds.