Sciencemadness Discussion Board

Glycerol to Propionic Acid: A Chemical Safari Tour.

PrimoPyro - 26-9-2002 at 06:28

Hi. :)

As a prelude to my next thread series (coming soon), here is an example of a simple reaction, that when looked at in Inorganic mindset, might prove to seem very abstract or puzzling, but when explained with very basic Organic principles, step by step, becomes obvious and simple. And highly interesting if you are interested in useful chemistry, which I am sure you all are, or else why are you here? Right? Right.

This is only the second (I believe so) thread I've created at this forum, because I don't really know what I am supposed to write about. As instructed by Polverone, I will be making more posts more frequently. ;)

I have decided that I would like to make myself useful here by relating the intricacies of common Organic Chemistry in a series of tutorial threads, with a reaction example safari twist to it, so you can see real examples of the principles I will be explaining.

This thread is actually out of place, as it doesn't start with the very basics, but I want to post it anyway, as a prelude and interest-getter. The "Introduction to Organic Chemistry" series will soon follow in the next few days, perhaps with the first installment later today. So stay tuned, information is afoot, and you won't want to miss it. ;)

Glycerol (1,2,3-trihydroxypropane) ----> Acrolein, a Chemical Safari Tour:

The reaction is a simple dehydration followed by Carbocationic Rearrangement. First, recognize a couple very basic principles that are very important:

1.Oxygen is much more electronegative than carbon. C=O groups are not neutral, but are positively polarized at the carbon and negatively polarized at the oxygen. This does not mean that C=O exists as the ionic +C-O- but just means that the electrons reside closer to the oxygen.

2.C+ species are called carbocations. Carbocations are self explanatory: Carbo = carbon, and cation = positive ion, +. Carbocation = C+ species. A C- species is a carbanion. It is not pronounced like carbo-cashin or carbo-cashun or carbo-kay-shun, etc. It is carbo-cat-ion, with emphasis on the cat. Not catION, but CATion. Same thing with ANion, pronounced like "an ion" but together: ANion. carbANion/carboCATion. Got it? Cool.

3.Carbocations are prone to rearrangements, whereas in msot cases carbanion are not. This means that in carbon chains and rings, a C+ species will shift when possible, until it reaches the most stable carbocation. This is incredibly important. It is both the reason many reactions occur, and the reason why others cannot. For example, In a phenyl-propane, which is Ph-CH2-CH2-CH3 where Ph = a benzene ring, if one makes a carbocation at the secondary carbon, one gets Ph-CH2-C+H-CH3. However, this almost always instantly rearranges to Ph-C+H-CH2-CH3 because this is much more stable. This carbon that is next to the phenyl group (the benzene ring) is called the benzyllic carbon, and it's chemistry is very important.

This brings us to our next principle.

4.Carbocations have fairly predictable stability. For alkanes and alkenes, the more substituted a carbon atom is, the more stable that carbocation will be. Substituted means having groups other than H attached to the carbon. So primary carbocations are the least stable, then secondary, tertiary are very stable, and quaternay carbons cannot form carbocations as long as they remain quaternary.

The presence of EWGs (electron withdrawing groups, which are very electronegative species like F or NO2 or Br or Ph or C=O or CN, etc) stabilizes carbocations. So if you have a secondary carbocation, R2C+H, it is semistable. But if one R is a NO2, that carbocation will be more stable than if that R were a methyl, CH3.

Delocalized systems stabilize carbocations, like benzene rings, indole rings, etc. In carbon chemistry, "localized bonding" translates to the practical use of "Point of Atack" for your methods. Delocalized bonds are strong, localized bonds are easy exploitation sites for you to alter the molecule. Benzene is a delocalized system. That is why our Ph-CH2C+H-CH3 rearranged to Ph-C+H-CH2-CH3. The presence of the electronegative, delocalized benzene ring system makes that benzyllic carbon very stable, moreso than the carbon next to it.

5.Water exists as H2O, but also self ionizes to H+ and OH-. Remember this. In every water solution there are both free protons and free hydroxide ions. Remember that this is an equilibrium reaction as well. This equilibrium can be shifted in either direction by addition of proton donors or accepters, as well as by hydroxide accepters and donors.

Important Fact: Equilibrium is the God of Organic Chemistry. I *think* it is the same for Inorganic Chemistry as well.

All reactions, ALL REACTIONS, did I mentions all reactions, are rooted around an equilibrium factor of some sort. This is amazingly important to remember in Organic Chemistry. The same reagents can be used to do opposite, or just completely different reacitons on the same molecule, or skeleton, or new substrate entirely, by simply shifting the equilibrium of the system in one direction or another.

For competing mechanisms, the produced product is often dependent on the equilibrium of the system, making one reaction more common than the other. In Organic Chemistry, it usually goes another step further: Since in many cases, a reaction both consists of multiple steps, and each step is often reversible, it is common to exploit a difference in equilibrium to eliminate an unwanted side reaction, by making it so that the desired steps happen more readily, and the unwanted steps have lesser chance of occuring, and also a better chance of reversing.

If this is done properly, the substrate will react the desired way, and the equilibirum forces the intermediates to react further until completion, and inhibits the side reaction from progressing by both slowing formation of its intermediates, and also reversing those intermediates back again, freeing up more substrate for the desired reaction pathway. Eventually, the desired reaction is the only one taking place. :)

Alright, I think that should do it for this round, at least.

Now to the reaction: CH2(OH)-CH(OH)-CH2(OH) is glycerol, or trihydroxypropane. Our compound, trihydroxypropane, hereafter referred to as glycerol, is a triol, which is an alcohol molecule with three -OH groups. Alcoholic OH groups are not acidic like carboxyl OH groups are. Alcoholic OH groups are not basic like metallic hydroxides are either. They are rather neutral. There is some minor polarization between C and O, but this is generally very weak in common systems, unless of course the equilibrium is changed, like we will do in a moment.

It is a very commonly known reaction to dehydrate glycerol with KHSO4 to acrolein, which is rightly called propenal: CH2=CH-CHO. Propenal is an aldehyde, containing the CHO group, which is like a carboxyl group, but where the C=O is attached to an -H instead of an -OH. Aldehydes are not acidic like carboxyls are. This reaction may seem mundane, but it is very interesting when you decode the intricacies of what is going on, and serves as a wonderful template for realizing the details of similar reactions on different compounds.

To simplify things, I am going to name the carbons. There are three in glycerol. We have the middle carbon, that the two others are both attached to, I will call this the iso-carbon, since substitutions at this carbon are called isopropyl substitutions. The carbon that will turn into our aldehyde will be called the "head carbon" since this will turn into the molecules defining functional group, or head. The last carbon of the chain (opposite of the head) will be called the "temrinal carbon" because when the head is formed, this carbon will become the "end of the chain" and hence terminates the chain.

For your visualization, I am speaking as though the terminal carbon is on the left, iso in the center, and head on the right.

This reaction can be carried out with some H2SO4 as well, but side reactions do occur. The reaction with KHSO4 yields nearly pure acrolein, or if desired and changes made, allyl alcohol. We will be making acrolein, CH2=CH-CHO. Then we will use the opposite mechanism to change this to propionic acid, CH3-CH2-COOH.

The reaction proceeds by addition of dry (need not be anhydrous) glycerol to dry potassium hydrogensulfate (potassium sulfate, K2SO4, is not an acceptable substitute) and minor warming. Additional H2SO4 is optional and raises yields somewhat.

What happens first is that the alcohols in glycerol get protonated by the KHSO4 and/or H2SO4, forming a H2O+ group, attached to their respective carbon atom. Not all three get protonated at the same time because of electrostatic repulsion between the H2O+ group on the carbon, and any H+ atoms in the vicitnity, preventing further protonations as long as the H2O+ group is present.

The alcohols get protonated, and some then release the protons (only stays protonated for a couple microseconds) and this equilibrium happens repeatedly. But because the equilibrium is shifted so far onto the dehydrating side because of the KHSO4 and H2SO4 having dehydrating powers, the reaction tends to proceed rather than be happy where it is protonating and deprotonating the alcohols repeatedly.

What happens is that the protonated alcohol, C-OH2+, then loses H2O due to the equilibrium, to form C+ + H2O. This is our first carbocation. Which alcohol gets dehydrated? Well, which one is the most stable? Both the head and terminal alcohols are primary, and the iso carbon alcohol is secondary. The iso-alcohol is more substituted, therefore it's formed carbocation is more stable. The others will get dehydrated to some extent, but with the conditions we have created, the most common reaction that occurs is the one that can lose water the fastest, which is the one that can get protonated and then extrude H2O the fastest, which is the one that will form the most stable carbocation, which is the iso-alcohol.

So our first reaction step proceeds as:

CH2(OH)-CH(OH)-CH2(OH) + H+ ----> CH2(OH)-CH(OH2+)-CH2(OH)

Then the next step is loss of H2O, forming the carbocation:

CH2(OH)-CH(OH2+)-CH2(OH) ----> CH2(OH)-C(+)H-CH2(OH) + H2O

Now what happens next is a carbocation rearrangement to a more stable carbocation. But, "Wait a second." you say? "You said the iso-carbon formed the carbocation because it was the most stable. So why would it rearrange if the other two options are less stable?" Ah, good question, and a VERY important answer. ;)

Because, it just does. LOL, just kidding! :P

Seriously, because we have changed the system now. Before, the conditions were different than they are now. Let's look at the details for carbocation formation "before" we had our dehydration. You have a terminal and head/carbon alcohol, which right now are mirror images, and can be treated as the same, for now. If it got protonated, it would not be the most stable carbocation, because the formed carbocation would be primary. Not only this, but the other option looks much more favorable: The iso-carbon is both more substituted, and has a slightly electron withdrawing group on it:the oxygen atom. Our current primary carbocation has nothing. This is why it is not as stable in the formation of the carbocation.

But now we have changed the system. Now, our secondary carbocation is next to two carbons that both have an electronegative atom on them. Well, like in life, the grass is always greener on the other side, and remember that we lost our alcohol group on the iso-carbon to form that H2O. The carbocation rearranges to the more stable one, the one where it can have the -OH group on it.

So while our secondary iso-carbon was more stable before, now the primary terminal/head carbons are more stable because of the OH groups on them. They did not form stable carbocations because to do so they had to give up their OH groups, which defeats the entire purpose.

So now you understand that the preliminary stable carbocation is not always the final stable carbocation. Like I said, they are very prone to rearrangement. CH2(OH)-C(+)H-CH2(OH) rearranges to CH2(OH)-CH2-C(+)H(OH).

Now what happens is due to quantum mechanics, because of orbitals. It is responsible for the property of C(+) being more electronegative than C neutral. This is true for all positive ions. A positive ion of an atom is always more electronegative (at least slightly) than the same atom in neutral state. And the opposite is also true: All negative ions are at least slightly less electronegative than the same atom in its neutral state.

In nearly all cases, the difference is huge! It is more pronounced in the lighter atomic number elements. So obviously the 1st and 2nd and 3rd period of elements exhibit this effect the strongest. F- does NOT want to become F-2, and Na+ does NOT want to become Na+2. These two examples are poor though, because they do not want to do this for more than one reason, and the reason Im about to state actually plays a small role for them because the main reason is that for them to form those second ions, each would have to have reactions invlolving an entirely new electron shell, which takes some pretty far out reaction conditions that very few chemicals can provide.

But for carbon, C+ wants an electron much more than neutral C does. Why? Because in C+, we have 6 protons for only 5 electrons now. The pull on each electron toward the nucleus has increased, and the ionic radius (the distance of the electrons from the nucleus) decreases as the strengthened pull forces them to orbit closer to the nucleus.

Since the electrons are responsible for the bond orbitals that create molecular bonds, the C-O bond shortens from its normal length, and more of the oxygen's electrons come into contact with the carbon. The resulting shift in polarity is from C(+)-OH to C=O-H(+) followed by extrusion of a proton.

But why would the proton leave? Shouldn't formation of H+ from H have the same effect as it did on the carbon, shortening the bond and strengthening it? Yes, normally it would. But in our current equilibrium, H+ extrusion is FAR more favorable, because remember, by forming our carbocation we have decreased the free proton concentration of the solution, and now we technically have an excess of anionic KSO4- which acts as a base, helping to remove the proton from the alcohol. This is why KSO4- (really exisiting as K+ + SO4-2) is called a conjugate base. All acids have a conjugate base, which is the acid structure minus the acidic proton(s), and all bases have conjugate acids, which are the base structures minus the hydroxyl groups, or whatever group gives it basicity (alkoxyl, electron pair, etc).

So we go from CH2(OH)-CH2-C(+)H(OH) to CH2(OH)-CH2-C(H)=O, which is beta-hydroxypropanal. Our acidic proton that caused the reaction has been recycled back into solution to do more work, god bless his hard working soul. :P

The net reaction is simply loss of water, coming as OH from the iso-carbon, and H from the head carbon, and strengthening of the head alcohol bond to an aldehyde.

The next step is a repeat of the first: Protonation of the alcohol to form the C(OH2+) species. This time it happens on our terminal carbon alcohol, going from CH2(OH)-CH2-CHO to CH2(OH2+)-CH2-CHO, followed by dehydration to give the carbocation:

CH2(OH2+)-CH2-CHO ----> CH2(+)-CH2-CHO + H2O, and the water is removed by the dehydrating power of KHSO4 and H2SO4.

Now the next step is similar to the second step of our reaction, but not identical. The carbocation rearranges to the iso-carbon for sure, because the iso-carbon is much more stable, being more substituted and next to a true electron withdrawing group, the carbonyl group. The carbonyl carbon (the head carbon) will not become a carbocation easily. It can be done, but it requires much stronger conditions than simple KHSO4.

So CH2(+)-CH2-CHO ----> CH3-CH(+)-CHO

Now the last step is of course expulsion of that proton again. You would think that it comes from the hydrogen on the iso-carbon, but it doesn't because this would form a carbene, a highly reactive species, that will react very fast with H+, meaning that the opposite reaction of the expulsion here is much more likely to occur, preventing this from happening.

What happens here is a special case for protons. Because a hydrogen atom has only one electron about its nucleus, the ionic radius of hydrogen is very small. It has nearly no electron cloud around the positive nucleus. This means that the positive force from the carbocation exhibits repulsion against the proton nucleus of the hydrogen atom, because they are close in vicinity. This, relatively speaking, makes the H-C bond of the terminal carbon weaker and the proton more acidic. As I said, CHO does not form carbocations, and the CHO group will not expell a proton to form C(-)=O because of the huge instability of that ion in this system full of free protons and the conjugate base for competition.

So that leaves our tail carbon left, which has it's proton removed the most easily by the conjugate base SO4-2 to form a temporary pseudo-carbanion. Of course then C(-)H2-C(+)H-CHO ----> CH2=CH-CHO, which is acrolein/propenal. The enormous electron-withdrawing potential of the C=O group combined with the C(+) group, both on the same side of the terminal carbon, combined with the small repulsion of H from C(+) and the presence of SO4-2 ions, makes elimination of that proton and then electron transfer for neutralization very easy and thus it happens quickly.

So now you know much more about what goes on in this simple reaction, other than just saying "It loses two moles of water to form acrolein." How boring is that? Just how much interesting (and useful) information does that tell you? How can you apply that to other things without fear of being wrong? Well, you can't do anything like that!

But now you can, because you understand HOW this dehydration occurs. In Chemistry, it is not so much important what happens, but HOW and also WHY and WHY NOT. The what comes first, followed by the how, and then the why or why not. But it is very important to understand the mechanics of what goes on if you want to be a chemist or claim to know chemistry.

If you don't understand HOW and WHY things occur, all you can do is follow recipes given to you by others, and make a few simple things. But if you truly understand the depths of mechanics of reactions, you can do nearly anything.

I hope you enjoyed reading this. If there is positive feedback, I will continue with these threads explaining the details of Organic Chemistry and its reactions.



PrimoPyro - 26-9-2002 at 06:30

The title should of course be Glycerol to Acrolein, not Propionic Acid. I had changed my idea for a thread mid-writing, and forgot to change the title! (embarrassed)


madscientist - 26-9-2002 at 09:06

Excellent post! :D Thank you very much for sharing the mechanism of that reaction in such detail!

Very nifty

Polverone - 26-9-2002 at 10:07

A few questions/suggestions about this very interesting and substantial treatment of what initially appears to be a simple reaction.

The iso-carbon is both more substituted, and has a slightly electron withdrawing group on it:the oxygen atom.

I wonder here what effect (relative to the other carbons) the oxygen has. Don't the other two carbons also have an oxygen atom attached (from the -OH)?

And let me see if I have truly learned something - i.e. can properly generalize from the example you have given. Based on what you have said, I would expect that acidic dehydrating conditions would preferentially cause carbocation formation (and likely subsequent rearrangement depending on the stability of the intermediate) at more highly substituted carbons. I would expect this to happen in any alcohol where there are differences in substitution among the different carbons. Is this a fair generalization, or have I absorbed your lesson improperly?

Would this reaction proceed with any other dehydrating agent and strong acid? For example, calcium chloride and anhydrous HCl gas? I know it's not nearly as easy to handle - I just want to know if it is merely acidic dehydrating conditions that are necessary.

Finally, I want to see your propionic acid piece now. The first article has tied together a lot of different chemistry concepts in an interesting way to show how a useful chemical may be formed, and the mechanisms behind it all.

PrimoPyro - 26-9-2002 at 10:43

I wonder here what effect (relative to the other carbons) the oxygen has. Don't the other two carbons also have an oxygen atom attached (from the -OH)?

Not at that stage. There is confusion regarding which carbons still have what groups, I see.

What I meant was that for glycerol, all three hydroxys are present. The iso-carbon offers a more stable immediate carbocation because it is more substituted than the primary terminal alcohols. This is because of the difference in the formed carbocations.

In the iso-carbocation, we have a secondary carbocation. In the terminal alcohols, we have a primary carbocation, which is less stable. Now the iso-carbocation wants to rearrange because of greater stability offered from the terminal oxygens on the terminal alcohols.

So the terminal carbocation is more stable because it contains an oxygen in this case, but visualize preliminary terminal carbocation formation for a second. To form the carbocation at the terminal position, the terminal carbon must lose its hydroxyl, so we have a big difference between the carbocation formed directly from the terminal alcohol, and the carbocation formed by shift of the iso-alcohol.

One carbocation has an -OH present, the other doesn't. This makes the big difference.

I suppose that CH2(OH)COCH3 is formed in minute quantities, but it would be very small, and would be removed by simple distillation of the product.

And your generalization is correct.

As for the conditions, No, I don't think that anhydrous HCl and CaCl2 would work well. CaCl2 is a drying agent, it has no chemical dehydratory powers like KHSO4/H2SO4. Anhydrous HCl would very likely just chlorinate the alcohols in my opinion, which is not necessarily a bad thing.

Solvent has a very large part to play in these reactions, as carriers of the ions, and other functions.

For example of solvent importance, alkyl chlorides with a length of 2 carbons or longer, when reacted with alkalai hydroxides, form two different products, depending on the solvent.

If the solvent is water, the product is an alcohol, formed by nucleophillic substitution yielding RCl ----> ROH + MCl (M = alkalai metal)

If the solvent is dry alcohol, the product is an alkene, via an elimination mechanism, where RH2-R'H2Cl + NaOH ----> NaCl + RH=R'H + H2O


Oh Yeah

PrimoPyro - 26-9-2002 at 10:51

I keep forgetting stuff.

As for the propionic acid, well I'm going to have to keep thinking on that. I wrote the first post while relaxing after work, and I was not thinking clearly apparantly, because I had thought for some unknown reason that acid hydrolysis of the alkene (acrolein) would yield the saturated acid via hydroxide attack on the aldehyde.

Not true as far as I know, and I realized this as I typed up the original post.

For that to happen, the alkene would have to be protonated in water solvent, and then carbocationic rearrangement all the way to the aldehyde bearing the charge, which I stated would not happen. Then hydroxide attack on the positive aldehyde carbon, forming the saturated acid and reviving the absorbed proton from the water used to make the acid.

H2O ----> OH- + H+

The OH- is used to make the acid hydroxyl, and the proton left over takes the place of the one absorbed by the acrolein.

However, I no longer think this happens, at least not this simply. It might still be possible to convert acrolein to propionic acid, but it will take a little more thinking (perhaps a hot shower is in order, I have my best ideas there nightly.) I'll get back to you on it.

If you need propionic acid, the easiest way is just a common Haloform Reaction on methyl ethyl ketone (2-butanone) with hypochlorite (Ca(OCl)2, NaOCl, etc) which will form the propionate salt of the hypochlorite cation, and chloroform "byproduct" as well.

You can liberate the carboxylic acid by addition of aqueous HCl followed by distillation or extraction.


chemrox - 4-9-2010 at 20:16

This was a lot of fun. The only improvement you could make would be to add graphics. I like curly arrows, charges and partial charges.

DDTea - 5-9-2010 at 00:46

You know, in the final days of my sophomore organic chemistry course, I really wanted to ask the professor, "Wait, so what exactly is a carbo-kayshun? Did we even cover that?"

Never had the guts to do it though.

Anyhow, this seems like a pedagogic bit of writing, which leads me to wonder: why don't you describe the electronics of carbocations/sp2 systems? After a couple years of tutoring O. Chem, I can't imagine giving an effective description of carbocations without mentioning their most obvious physical features: the trigonal planar geometry with that big, empty p orbital.

It may be worthwhile to sidetrack a bit to provide a concise, usable explanation of the electronics involved, which would in turn shed a lot more light on why one substituent increases stability while another would decrease it. Simple language is always best, too. One of the best professors back at my university always said things like, "Nature hates charges. It hates point charges more than charges over a large area, and hates two charges more than one." That explanation explained a lot of trends seen in O. Chem.

Good work!

EDIT: just noticed this--

For that to happen, the alkene would have to be protonated in water solvent, and then carbocationic rearrangement all the way to the aldehyde bearing the charge, which I stated would not happen. Then hydroxide attack on the positive aldehyde carbon, forming the saturated acid and reviving the absorbed proton from the water used to make the acid.

H2O ----> OH- + H+

The OH- is used to make the acid hydroxyl, and the proton left over takes the place of the one absorbed by the acrolein.

Careful. In an acidic environment, there is NO -OH present. Your acid is H3O+ and your base is H2O.

In a basic environment, your acid is H2O and your base is -OH.

That's a very common mistake.

[Edited on 9-5-10 by DDTea]

Eclectic - 5-9-2010 at 02:49

Laboratory Practicum: