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Color chemistry alcohol -- Eugenol, purple, and the "strength" of the carbon bond.
I've been trying to get an experimental feel for what causes color in organic chemistry, and how I might predict it based on the energy distribution
of the carbon bond.
D-limonene is colorless, even when subjected to a strong acid. However, I found that eugenol (clove oil) turns purple. It doesn't matter if I use
boric acid or sulfuric acid to form an ester with it ... Eugenol basically turns the same color; although it's much more opaque with sulfuric acid.
So, I'm pretty sure that the ester of eugenol is purple; and that the color is mostly independent of the strength of the acid and somehow related to
the -OH radical.
( I realize there can be exceptions, but as a novice I'm talking in general. )
Phenol, on the other hand, also has a ring structure where the active -OH exists.
But, it's color is clear or brown/black in most reactions I have seen.
This got me thinking, how much do bonds that are far away from the -OH radical affect the bond energy and color emitted by the bond?
Is there a general chart for hydrocarbons showing the reactivity of a typical bond in joules ?
By conservation of energy, if I know the energy of local bonds next to carbon can I "add" up the energies in some way to a constant to estimate how
much the -OH radical is affected?
To give an example: Is it more or less likely that the methyl+oxgen bond right next to Eugenol -OH radical affects the color more than the carbon
tail on the other side of the molecule?
Can I do something like, I have a ring of carbon who's default bonding energy is x Joules for each hydrogen. When I add an -OH radical it changes the
bond energy of all hydrogens in the neighborhood of the -OH radical by +- number of joules ?
an O-C-H3, will affect it by +- number of joules.
Or do I have to solve some complicated quantum mechanical equation to even crudely estimate what will happen to the bond energy?
For example:
I know that primary, secondary, and tertiary alcohols show different reactivities and the only difference between them is which carbon the -OH is on.
The difference in energy, though, appears to be VERY repeatable over a wide variety of alcohols regardless of what the rest of the R-, of R-OH,
really looks like.
The victor meyers test, for example, would give a blood red color for a primary alcohol and a blue color for a secondary alcohol ... and no color for
a tertiary. The color is pretty stable regardless of alcohol's exact form ... but very sensitive to primary, secondary, or tertiary.
When I look up the chemistry of monohydric alcohol, the classification depends only on the number of OTHER carbon atoms connected to the carbon atom
which attaches the -OH.
No other carbons -- is primary. ( least reactive/most stable -OH )
One or TWO is -- secondary. ( More reactive).
Three carbons is tertiary. ( unknown to me.)
The literature I see doesn't explain why the change from one to two carbons doesn't affect the energy; but from two to three carbons does.
It also doesn't talk about pi bonds, such as are in carbon ring structures, so that I could guess what phenol or Eugenol might do if subjected to a
victor meyers test.
Any guidance would be appreciated, thank you.
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DraconicAcid
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If you purify the ester of eugenol (whichever ester you make), it's probably going to be colourless or yellow. The purple colour you get in reactions
of phenols are of some oxidized materials which are side products or intermediates.
For colour in organic chemistry, you need delocalized pi electrons. The more, the better. If they can be spread out over nitrogens, even better.
The Victor Meyer test works by turning the alcohol into a nitro compound and then adding funky nitrosyl groups to it. You get completely different
structures from the reactions of primary and secondary alcohols, and that's why they give you completely different colours. A tertiary alcohol will
give you a nitro compound which won't add the funky nitrosyl-like group, so the reaction doesn't continue, and you don't get a coloured product.
The colour has nothing to do with the bond strength.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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semiconductive
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Quote: Originally posted by DraconicAcid ![](images/xpblue/lastpost.gif) | If you purify the ester of eugenol (whichever ester you make), it's probably going to be colourless or yellow. The purple colour you get in reactions
of phenols are of some oxidized materials which are side products or intermediates. |
OK. So, for reference ... purification would mean some kind of distillation?
Or are the oxidized materials generally separable by some kind of solvent?
I don't think simple vacuum filtration is capable of separating the purple color because the ester is way too viscous.
I can imagine trying acetone solution to wash it, and a separatory funnel; afterward I could allow the acetone remaining to evaporate; but i have no
idea what would be a common solvent to try and separate the oxidized products. Is there a simple test I could try?
Water generally ruins esters, and I imagine that driving off the water a second time would make more oxides or intermediate products turning it purple
again ... correct?
But since the purple stuff is possibly oxidized, I suppose I could at least purify the purple stuff by mixing with water and centerfuge it; at which
point it might sink to the bottom because of solubility or mass differences. Eg: mixing oxidized ester with water would likely not cause eugenol to
un-oxidize, correct? (I've got a MSE centaur centerfuge and vials. )
Quote: |
For colour in organic chemistry, you need delocalized pi electrons. The more, the better. If they can be spread out over nitrogens, even better.
The Victor Meyer test works by turning the alcohol into a nitro compound and then adding funky nitrosyl groups to it. |
Right, I understood that much; though I'm going to have to look up the difference between nitro compound and nitrosyl again. For some reason iodine
seems to be needed to catalyze the reaction....?
Quote: |
You get completely different structures from the reactions of primary and secondary alcohols, and that's why they give you completely different
colours.
|
Ok, that's very interesting. Regarding the idea I read about carbon count and alcohol classification:
What would you say is the difference between n-propanol, which is a primary alcohol according to wikipedia; and iso-propyl, which is a secondary
alcohol? Why would one alcohol be much more reactive than the other, given that they have the same number of carbons.
https://en.wikipedia.org/wiki/Category:Secondary_alcohols ( text search iso-propyl)
https://en.wikipedia.org/wiki/Propanol
And regarding "structures" .... where does the color originate from ... because in the meyer's test, the nitro product before the reaction is not
colored, iodine is not in the final product, and only one carbon of the alcohol is affected by the test. Are you saying the reaction between
primary/secondary and nitrosyl somehow creates a delocalized pi-bond in primary/secondary alcohol?
Quote: |
A tertiary alcohol will give you a nitro compound which won't add the funky nitrosyl-like group, so the reaction doesn't continue, and you don't get
a coloured product.
The colour has nothing to do with the bond strength. |
Hmm.... there's something subtle here; I understand (Linus Pauling) that it's bond strength that determines resonance, and resonance determines
photon emission and absorbtion profiles. E=hf. So, I can guess that the bond strength of the carbon-OH radical might not cause the color, but at
very least the bonding energy of the nitrosyl-like group has to have an excitation energy to ground state energy transition that has "purple." In
general I would imagine a quantum well with a ground state energy, and then each excited energy is going to be an integral multiple of the ground
state energy.
So, if "purple" is emitted, then double the wave-length of purple is going to be an allowed energy, and third, quarter, fourth, fifth.... etc. until
we get to the lowest energy state of the pi-bond / or nitrogen bond which causes color.
The only other way I know color can be affected is by interference (physical size) of molecules. If molecules are large enough with regular (crystal
like) spacing, they can create a diffraction grating; etc. The physical size of oil films on water often look rainbow colored because of the
thickness of the oil being near a wavelength of visible light. If I do things to thicken the layer, or thin it too much ... the coloring disappears.
[Edited on 23-7-2020 by semiconductive]
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DraconicAcid
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Quote: | OK. So, for reference ... purification would mean some kind of distillation?
Or are the oxidized materials generally separable by some kind of solvent?
I don't think simple vacuum filtration is capable of separating the purple color because the ester is way too viscous. |
I wouldn't try distilling an ester that large. I'd wash it with cold water to get rid of the excess acid, and sodium bicarbonate solution to make
sure you get all of the acid out. You might be able to crystallize it from a low-polarity solvent, or purify it by chromatography.
Quote: | Water generally ruins esters, and I imagine that driving off the water a second time would make more oxides or intermediate products turning it purple
again ... correct? |
Water will only ruin your ester if it's acidic or basic. Neutral solutions won't generally attack esters at room temperature (I've never had any
problems doing liquid-liquid extractions with water and ethyl acetate).
Quote: | But since the purple stuff is possibly oxidized, I suppose I could at least purify the purple stuff by mixing with water and centerfuge it; at which
point it might sink to the bottom because of solubility or mass differences. Eg: mixing oxidized ester with water would likely not cause eugenol to
un-oxidize, correct? (I've got a MSE centaur centerfuge and vials. ) |
I don't expect that the purple stuff will be isolable.
Quote: | Right, I understood that much; though I'm going to have to look up the difference between nitro compound and nitrosyl again. For some reason iodine
seems to be needed to catalyze the reaction....? |
The alcohol is first turned into an alkyl halide, then a nitroalkane.
A primary alcohol will react to give RCH2NO2, which reacts further to give RC(NO2)=NOH, which gets deprotonated by base to give the dark red coloured
anion.
A secondary alcohol will give R2CHNO2, which will react to give R2C(NO2)-N=O; this will react with base to give a blue anion (although I'm not sure
what it's structure would be).
A tertiary alcohol will give R3CNO2, which doesn't have space on that carbon to add another nitrogen, either as a nitrosyl group (-N=O) or related
group. So we can't make a coloured derivative when we add a base.
Quote: |
What would you say is the difference between n-propanol, which is a primary alcohol according to wikipedia; and iso-propyl, which is a secondary
alcohol? Why would one alcohol be much more reactive than the other, given that they have the same number of carbons. |
The difference is in the number of hydrogens on the carbon with the hydroxyl. If you have fewer hydrogens there to remove, there's more of a limit as
to what you can react the alcohol to form.
It's just like when you oxidize an alcohol- a primary alcohol can be turned into a carboxylic acid or an aldehyde; a secondary one into a ketone. A
tertiary one cannot be simply oxidized (although you can burn it, that's not interesting to the organic chemist). You can't oxidize a secondary
alcohol into a carboxylic acid without breaking carbon-carbon bonds.
Quote: | And regarding "structures" .... where does the color originate from ... because in the meyer's test, the nitro product before the reaction is not
colored, iodine is not in the final product, and only one carbon of the alcohol is affected by the test. Are you saying the reaction between
primary/secondary and nitrosyl somehow creates a delocalized pi-bond in primary/secondary alcohol? |
If you take a hydrogen off of RC(NO2)=NOH, you'll get an anion. You can draw it, and you'll see several conjugated pi bonds, which you can arrange in
several resonance structures. That's what gives the colour.
Quote: | Hmm.... there's something subtle here; I understand (Linus Pauling) that it's bond strength that determines resonance, and resonance determines
photon emission and absorbtion profiles. |
I have no idea what you mean in this statement.
Quote: | So, I can guess that the bond strength of the carbon-OH radical might not cause the color, but at very least the bonding energy of the nitrosyl-like
group has to have an excitation energy to ground state energy transition that has "purple." In general I would imagine a quantum well with a ground
state energy, and then each excited energy is going to be an integral multiple of the ground state energy. |
If you have a double bond ( = ), there is a difference in energy between the pi bonding orbital and the pi* antibonding orbit that usually corresponds
to an ultraviolet photon. Alkenes with isolated double bonds are colourless (unless you can see UV).
If you have two conjugated double bonds ( =-= ), then you get four pi orbitals (two bonding, two antibonding), which are significantly closer together
than the pi orbitals of the isolated double bind. Thus, it will absorb less-energetic photons.
If you get enough conjugation, then the photons getting absorbed appear in the visible spectrum- carotenes have 11 conjugated double bonds, and absorb
violet/blue light, so appear yellow or orange.
If the orbitals get lopsided because you've got nitrogens or oxygens in the structure, then you can get visible absorption with fewer bonds-
nitroaromatic compounds tend to be yellow; nitroaniline is intensely yellow and is used in inks.
Radicals tend to also absorb visible light, even when simple (NO2 is a good example).
Quote: | So, if "purple" is emitted, then double the wave-length of purple is going to be an allowed energy, and third, quarter, fourth, fifth.... etc. until
we get to the lowest energy state of the pi-bond / or nitrogen bond which causes color. |
In organic chemistry (rather than atomic spectroscopy), we're not looking at emitted light, but absorbed light. You shine white light at your sample,
and see which wavelengths come back. A purple compound is absorbing yellow light.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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semiconductive
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Thank you! I appreciate the useful details.
I'm going to have to think about what you've said for a little while.
There are a few comments I'd like to respond to now, though.
Quote: |
Alkenes with isolated double bonds are colourless (unless you can see UV).
|
I'm not sure if you're being sarcastic.... so I will answer assuming you are not. I'm not trying to be offensive....
Yes, I can see some longwave UV. It looks very violet to me, esp. the laser diodes I use in industry. (BTW: do not stare into a laser diode with
your remaining good eye.) I can't see as many shades of colors as the women do. But male Physicists who can barely see longwave U.V. said it was
violet and I agree with that. I can also see some of the shortest wavelength of infra red light; if I turn the lights off, and allow my eyes to
become very dark sensitive ... it's looks red; just much dimmer than the wattage would suggest it should be. At first I wanted to believe the
manufacturer was lying to me about the wattage, until I used a thermal sensor with a black target. Then I took a picture of it with my camera and was
shocked at how BRIGHT it looked to silicon eyes.
I'm Glad I didn't aim that red light directly at either of my perfectly good (but old) eyes.
Regarding colors of hydrocarbons and UV; I would expect color a chemical is to make a definite susceptibility to heating and photo-induced reactions.
I noticed (for example) that HCl (aq) will react (slightly) with alkanes or alkenes when sunlight is present, but not inside the shed under a
fluorescent lamp. ( oh... and HCL after electrolysis can make a BANG when I open the shed door! Useful safety knowledge. )
Since you brought U.V. up ... do you think UV excites and somehow weakens the H-C bond because it's violet colored, or does UV excite the chlorine
and make it attack more vigorously?
I might be able to use iodine to test the difference....
Iodine will react with unsaturated oils but should be a different "color" than chlorine; so I wonder if strong UV will allow Iodine to react with
paraffin.
I have a germicidal mercury lamp....
I'll take your comment seriously and guess that there is shortwave UV coloring of H-C in the *saturated* bond. ( which is colorless to my eyes in
agreement with your claim. )
Quote: |
A purple compound is absorbing yellow light.
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It is absorbing and not-re-emitting yellow light, yes. That's the difference between pigments and television screens. But why restrict the
interpretation to one or the other? In an opaque liquid (as you are talking about) we do "see" reflected light. That's not the same as seeing
transmitted light....
How is light reflected?
Isn't it by the action of atomic processes and orbitals?
As an electrical engineer, I talk about light hitting a mirror (which is a good metallic conductor) and the electrons in the mirror moving in
response to the EM field of light.
I can engineer different amounts of reflectivity (precisely) by controlling altering the conductivity of the metal.
Reflection is caused by the conducting electrons wiggling in response to the electric field made by light (photons); and their free motion creates new
light that goes in the opposite direction as the original light.
To be more precise about what I mean:
The sun, for example, not only emits light due to it's own nuclear energy but also absorbs certain wavelengths of light passing by it (a
specstroscopic example) and some of that light gets re-emitted. The sun does both emission and attenuation not just one or the other.
I'm thinking, in my ester-solution absorbtion of yellow is much greater than re-emission of red and blue; but some atoms are re-emitting red and blue
when free electrons "scatter" incoming light (conductive electrons/and therefore I can't rule out "exicted" electrons).
The difference is a just personal interpretation shift ... or is there a deeper reason for choosing one perspective over the other ?
Again, I have a lot of old information from when I learned chemistry; and I've only had overviews of organic chemistry 25+ years ago. So, I'm
mystified by certain uses of language .... it's more like people are denying the exceptions in chemistry exist, rather than rationally explaining them
or admitting they have a "school" of thought that is just a preference.
In my old chem book, for example, it says monohydric alcohols are classified into primary, secondary, and tertiary alcohols. Those are the only
categories given.
Wikipedia says that nowadays methanol is not called a primary alcohol; though it has been in the past -- 1910 -- encyclopaedia Britanica.
So, let me ask a simple question:
What kind of alcohol is Methanol ? Will it produce "blood red" or "blue" ?
It the definition of primary alcohol based on hydrogen count more accurate than that of carbon count?
Likewise, Phenol or Eugenol -- even I'm wrong in considering them "like" a monohydric alcohol, they still have what I was taught is an -OH radical and
an organic acting "like" an alcohol. R-R or R-OH. (Radical Radical) or (Radical - hydroxyl )
What would you predict?
will the reaction be "blood red" or 'Blue" or "no color change" for methanol and for phenolic-"alcohols" ?
https://byjus.com/chemistry/types-of-alcohols/
![](https://cdn1.byjus.com/chemistry/wp-content/uploads/2016/10/word-image-22.png)
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Opylation
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You’ve never had an issue with using ethyl acetate for liquid-liquid extractions? I tried using it for an extraction of a basic solution and the
sodium hydroxide hydrolyzed it in like 15 seconds. I looked it up and, from what I found, ethyl acetate is only good for neutral extractions. No?
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DraconicAcid
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Quote: Originally posted by Opylation ![](images/xpblue/lastpost.gif) | You’ve never had an issue with using ethyl acetate for liquid-liquid extractions? I tried using it for an extraction of a basic solution and the
sodium hydroxide hydrolyzed it in like 15 seconds. I looked it up and, from what I found, ethyl acetate is only good for neutral extractions. No?
|
Yes- I only use it for neutral extractions, such as washing an esterification mix with sodium bicarbonate solution.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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