I've destilled some propylene glycol containing anti freeze. Just to get a new compound on my shelf I've mixed some of it with sodium hydroxide and
refluxed it to eliminate it.
What are the products? Is it amyl alcohol CH2=CH-CH2-OH or is it 1 propen-1ol CH3-CH=COH. Th H-Atom in the middle/2-position should not eliminate
because a secondary carboanion is more unstable than a primary carboanion(It's an E-2 reaction H+goes first I think/bad leaving group, relativly
strong base). So there are the the two outer H-atoms, the one of the methyl group and the other close to the OH group. If the one close to the OH
eliminates, the product will be the enol of propanal, not stable I think, so there should only be the amyl alcohol as product but I didn't smell
anything mustard/ethanol like so I wonder if it reacted at all.
[Edited on 16-8-2006 by hinz]not_important - 16-8-2006 at 11:57
Bases aren't good catalysts for elimination of water, acids are better.
1-propen-2-ol is not stable, it would rearrange to acetone.
You left out getting propanal - propionaldehyde. If you made this under basic conditions you would get aldol condensation products of it.
You can get any or all three, allyl (not amyl) alcohol, acetone, or propionaldehyde, depending on conditions. Propylene oxide acts in a similar
fashion,
Ber. 36, 2017 (1903)
German patent 618.972 (spet 19, 1935)
US patent 2.159.507 (May 23, 1939)
US patent 2.426.264 (Aug 23, 1947)
Much of the research was done on the oxide as it is used to make the glycol.
The boiling point, or points - do a distillation, will tell you what you have.
The elimination works better in the vapour phase, run through a heated tube containing the catalyst.
[Edited on 16-8-2006 by not_important]Nicodem - 16-8-2006 at 20:57
Quote:
Originally posted by not_important
Bases aren't good catalysts for elimination of water, acids are better.
Bases are not catalyst for elimination of water from alcohols at all. I don't know where Hinz got the idea of using a base but all he could have
recovered is unchanged 1,2-propandiol.Nick F - 17-8-2006 at 04:33
not_important: "You left out getting propanal - propionaldehyde."
hinz: "or is it 1 propen-1ol CH3-CH=COH."
CH3-CH=CH-OH <- ------> CH3-CH2-CHO
Propene-1-ol converts mainly into propanal in equilibrium.
Base would have been a good choice if you were trying to eliminate a halide, but not so good for OH, as people have said.not_important - 17-8-2006 at 07:27
1-propen-1ol is not propionaldehyde, even though they co-exist. While the enol form is important in a goodly number of reactions, the bulk properties
are governed by the 5 or 6 oders of magnitude greater stability of the aldehyde.
There are base catalysed dehydrations, but I've only seen them talked about in vapour phase reactions and generally with compounds such as
keto-alcohols.hinz - 17-8-2006 at 13:40
"Bases are not catalyst for elimination of water from alcohols at all."
But where are the differences between an E2 elimination at alkyl chlorides and alcohols. The hydrogen at the possition which gives the more stable
carboanion will leave. Then the eliminated anion (halogen or hyroxy) will leave and the double bond forms. But when I thought about it, I realised
that the hyrogen at the alcoholic OH might be more acidic and form a salt with the base. Now its a carbocation that doesn't have an acidic hyrogen at
a carbon close to the now formed R2HC-O(-), so the base can't capture an acidic C-H hyrogen.
S you're right, as always Nicodem.
With acid the OH will be protonated first, leaves then as H2O and a carbocation will form. Then the carbocation will leave a proton, the one which has
more electon donators like CH3 or C6H5 around and a double bond will form.
I'll try it tomorrow with sulfuric acid. HCl will substitude the OH group by Cl I think, so it will give 1,2 dichlorpropan. The idea to eliminate the
compound with a base was an experiment, I didn't knew if it would work. I read it in my organic chemistry book and I just wanted to try it. Always
cooking after a recipe in a book is boring
At the 1 propen-1ol I forget a H atom at C1 sorry, I meant it like Nick F, but you showed it more comprehensible then me.
[Edited on 17-8-2006 by hinz]Nicodem - 18-8-2006 at 00:01
The fact that the OH group would be the first to deprotonate is not the only problem. Especially since the E2 mechanism does not require the beta-H to
deprotonate (it is a concerted transfer of electron pairs so bonds are created synchronously while other are broken). The major problem is that the
hydroxide (OH-) is one of the worst imaginable leaving groups. So nothing happens.
As you already know the protonated OH makes it into H2O as a leaving group and that is many, many magnitudes better than hydroxyde.
Quote:
Always cooking after a recipe in a book is boring
How very true.
But nevertheless, first check the patents Not_important provided.
Substitution of OH's with HCl is unlikely to procceed so easily and without extensive elimination side reaction. Besides the primary OH needs a
catalyst for something like this.Nicodem - 23-8-2006 at 13:53
Hinz, if you are still looking for ways to use your 1,2-propanediol this might be apealing. It seams it might be possible to make 1-chloropropan-2-ol
by bubbling HCl trough 1,2-propanediol at about 110°C. At least by analogy with glycerol as described at Org. Synth. Glycerol can actually even be dichlorinated up to 1,3-dichloropropan-2-ol. There is no reason why this would not work with 1,2-propandiol as well under the same conditions.
1-chloropropan-2-ol is an interesting compound that can be used to prepare a useful reagent, propene oxide.cristiro - 16-4-2008 at 03:01
I hope this research might be of good help (A simple method to obtain propionaldehyde ( propanal ) but also acetaldehyde, acetone, 2-propanol,
1-propanol, allyl alcohol, acetol, and dipropylene glycol )
[Edited on 16-4-2008 by cristiro]
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