Sciencemadness Discussion Board

Alkenes from alcohols- a problem

Artemisia - 20-1-2009 at 07:57

Im currently attempting the dehydation of a tertiary alcohol by formation of a mesylate, then elimination with NEt3 in situ. DMAP is used (4mol %) as a catalyst. I'm currently getting 1:1 mixtures of the 2 alkenes. Does anyone know of any methods of dehydration which allow a degree of control over the alkene formed. The key part of the molecule is shown (the identity of the R group doesnt really matter as its quite out of the way of where the action is). Mineral acids are a no-no for this, as are inorganic acids on solid supports (like CuSO4/ silica). cheers :)

sparkgap - 20-1-2009 at 08:07

Hmm, cursory searching netted this and this, among other things. What do you think?

sparky (~_~)

P.S. This was what you were doing, right?

[Edited on 21-1-2009 by sparkgap]

Artemisia - 20-1-2009 at 08:27

cheers, I havent come across those papers before- I might give the PPh3 method a go tomorrow. The trouble is that neither alkene has appreciably higher stability. Will report back. Any other suggestions welcome!

DJF90 - 20-1-2009 at 08:37

Which of the two alkenes do you want?

sparkgap - 20-1-2009 at 08:37

I'm still trying to look for other stuff, but nearly everybody and his grandma seems to be on the solid support bandwagon... :o

sparky (~_~)

DJF90 - 20-1-2009 at 08:43

I was thinking, if the left hand molecule is the required one, and there was some way of separating the two isomers then perhaps dibromination of the right hand molecule followed by elimination to give the diene, then selective hydrogenation (as the C=C bon that is in the other position is less substituted and hence less stable). This should then yield the left hand molecule again. However I do not know under what conditions the hydrogenation should be done, and if this is a viable method due to separation of the alkenes in the first place.

Artemisia - 20-1-2009 at 10:38

I want the alkene on the left hand side (the 4,5 isomer). I can seperate the two by preparative chromatography if I can make 3g or so. Your idea is nice DJF90, I'll look into the practicalities of it. The hydrogenation would be the main issue. Pd/C isnt too discriminating.
Most of the dehydrations seem to use 1908 Berichte methods like 'stir substrate in sodium phosphite and ground pumice at 300C'. The methyl group will eventually be deuterated and that sort of brutality can really deplete 2H labels!

Ebao-lu - 20-1-2009 at 13:06

Basic elimination (E2) is usually an anti elimination, so stereochemical configuration in the place of cycles connection(gamma-position to OH group) in comparison to OH group should be essential for obtaining the desired alkene. For example, if OH group is down-directed, the base should take that hydrogen that is looking "up". And if the substitute (-CHR-..) is also up-directed, there should be some steric hindrance for the base. It would be better, if base could take proton from opposite side because there is no -CHR-. So the best steriochemical conditions should be the OH group and CHR- syn-configuration in the molecule. In other case, the ratio of desired/undesired alkene would be less, i suppose..


[Edited on 20-1-2009 by Ebao-lu]

Klute - 20-1-2009 at 13:17

Interesting topic!

Is your substrate alcohol racemic? Or can you start from the favorable enatiomer?

DJF90 - 20-1-2009 at 14:23

Ebao-lu: In the process of forming the diene, when elimination the top bromine there is only one choice of hydrogen that the base can take. For the lower bromine (the one on the same carbon as the methyl group), the proton can be abstracted from either the methyl group, or the carbon adjacent in the ring. The saytsev rule then applies as there is a choice of alkene products, so long as you use an unhindered base (like -OH), and so the target alkene becomes the major product of the elimination, as it is the most substituted of the possible alkenes.

Artemisia - 20-1-2009 at 15:24

Klute- yeah the alcohol is racemic (from a Grignard with methyl Iodide). I had thougt about some way of eliminating taking advantage of attack of NEt3 from the least hindered side (since its a decalin it probably exists as a 'folded up' conformation) but as it racemic I dont think itll help.

[Edited on 20-1-2009 by Artemisia]

Artemisia - 20-1-2009 at 15:29

Also- sparkgap, do you have access to that syn. comm. paper? I'm working from a paper by the same authors who use the methodology in the synthesis of chrysanthemic acid but my rubbish library isnt subscribed to syn comm or synlett.

UnintentionalChaos - 20-1-2009 at 15:45

Is the R-group particularly complex or (I presume) delicate that you need to use this original starting material? Or is this something you are not able/willing to share right now? Is it a functionality that can perhaps be added after this part is made if it happens to be delicate? There may be cleaner ways to make this if this is the case.

Artemisia - 20-1-2009 at 16:11

um, id rather not share, but its got a carboxylic acid group on it, no other functionality which would be affected by anything! I do have to use the starting material stated as I'm quite deep into this particular rabbit-hole... The only thing which I cant use is strong acid as the R-group/ decalin bond epimerises in strong acid (>3M) after about 10 minutes.

[Edited on 20-1-2009 by Artemisia]

sparkgap - 20-1-2009 at 17:04

Quote:
Originally posted by Artemisia
Also- sparkgap, do you have access to that syn. comm. paper? I'm working from a paper by the same authors who use the methodology in the synthesis of chrysanthemic acid but my rubbish library isnt subscribed to syn comm or synlett.


But of course. :)

~~~~~~~~~~~~~~~~~

A Facile Conversion of Tertiary Alcohols to Olefins
J. S. Yadav; Sudha V. Mysorekar
Synthetic Communications, Volume 19, Issue 5 & 6 March 1989 , pages 1057 - 1060


Abstract
An efficient method for the dehydration of tertiary alcohols to obtain olefins in high yield, employing methanesulfonyl chloride-triethylamine and catalytic amount of 4-dimethylaminopyridine in dichloromethane, is described.

:)

sparky (~_~)

P.S. I totally forgot to ask, so you do have a mix of cis and trans decalins, or just one of those two? The topological difference might be exploitable, I think.

P.P.S. Damn... it looks like I forgot to attach! :o Duly corrected.

[Edited on 21-1-2009 by sparkgap]

Attachment: tertelim.pdf (138kB)
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Artemisia - 21-1-2009 at 02:40

its 95% cis decalin (by NMR)

Ebao-lu - 21-1-2009 at 11:31

Could some amberlyst resins catalyse the dehydration of such alsohols? Or decaline is intolarable? Is it possible, that weak COO- base, that is in the R radical(if i understood correctly) is sterically favorable to abstract the desired proton? I know, Li2CO3 can be used as a base, so why carboxylic acid salt can't?

Artemisia - 21-1-2009 at 12:12

ebao-lu, amberlyst resin would probably de-deuterate the product (I plan on deuterating that methyl by using CD3I in the Grignard once I have a good methodology for eliminating the alcohol). I was wondering about the role of the carboxylate- I'll have to model it but I think it may well be sterically good. I plan on starting experiments tomorrow as I have about 500mg of the decalin.
sparkgap, thanks very much for that paper. I'm going to run 2 reactions tomorrow- one using PPh3/ I2, one using the mesylate elimination. I'll see if there's any difference between the 2. I'll post results if people are interested?

I really hate to use drug chemistry talk, but I've come across olefin isomerisation using KOH. I dont know if this would be a viable method to isomerise my undesired alkene after seperation by preparative chromatography?

Ebao-lu - 22-1-2009 at 04:20

Isomerisation with KOH (safrole to isosafrole?) is possible due to some acidity of benzyl hydrogen, that it can be abstracted by base, and due to stability of double bond conjugated with benzene ring. I'm sure, it will not work on usual alkenes
Of course, we are interested! This looks like real organic chemistry, not druggie or pyromaniac one, as for me, i'm interested.

Artemisia - 22-1-2009 at 08:31

yeah, everyone I mentioned the safrole isomerisation to has laughed. I totally disregarded the benzylic stabilisation- a proper undergrad style error. :(
NMRs to come on monday (a weekend 3200 scan 13C for confirmation) hopefully so I can report back then. in the mean time, have a good weekend all and thanks for your suggestions...

Sandmeyer - 19-2-2009 at 14:58

Why don't you show what R is? I have an idea but I don't want to share it since you want to selfishly exploit the forum.

PS. What do you meen by "I hate to use drug chemistry talk". Seems like a rather unscientific attitude.


[Edited on 19-2-2009 by Sandmeyer]

Nicodem - 20-2-2009 at 00:31

Artemisia, up to now you only tried various methods for E2 elimination, but that can not be selective for the two isomers you depicted (except for always forming an endocyclic double bond), because both hydrogens anti to the -OH group have practically identical electronic influences. You should try an elimination based on the ene reaction mechanism, like for example the Chugaev reaction which I think is the most commonly used elimination of this type. This reaction does not depend so much on the electronic properties of the beta-hydrogens, as much as their geometric configuration in relation to the -OCSSR group made from the -OH (it is a syn elimination). All beta-hydrogens in your case differ a lot in this regard. However, I have no idea what isomer you would preferentially get, not even if you would get the exocyclic or endocyclic product (pericyclic reactions are not my field). And besides, since you never told what R is, I have no idea if you would have to protect it or how thermally labile is it.

Sandmeyer - 20-2-2009 at 09:21

I don't see why Chugaev would be selective for that particular isomer. Anyways, the two isomers (50:50) can be saparated as already said. Then, oxymercuration-reduction can be done on the unwanted alkene and the elimination repeated to increase the yield of the wanted isomer. How many times this is repeated depends of course on the reaction scale and how valuble the product is. Already the first time gives 3:1 excess in the favour of the wanted product. But, all this provided that R dosen't contain a functionality that can be oxymercurated of course...

Nicodem - 20-2-2009 at 10:06

Quote:
Originally posted by Sandmeyer
I don't see why Chugaev would be selective for that particular isomer.

According to JACS, 81 (1959), 228–231, a paper I only skimmed trough, 1-alkylcyclohexanols do give mostly the product(s) with the endocyclic double bond, so it should be appropriate for this case as well. Yet given the mechanism is different and geometry dependent the ratio of two decalines should be different from 1 : 1 as it is with E2 eliminations. Whether would it be beneficial toward the desired isomer, well... I have no idea.

Edit: OK, I guess it would give a bit more of the unwanted isomer (the one on the right in the scheme), if steric influences have a role (at all) in the transition state. I only now found out that Artemisia actually pointed out which one was the desired somewhere in the middle of the thread.

[Edited on 20/2/2009 by Nicodem]