An organic synthesis problem

  1. Hi,

    The problem was to propose a synthetic scheme for making maleic acid (cis-but-2-ene-1,4-dioic acid) from acetylene (as part of a larger synthesis where I would have to use the maleic anhydride in a Diels-Alders cycloaddition with 1,3-cyclopentadiene made from cyclopentanol to get a tricyclic molecule). After several hours of work, I worked out a synthesis that would synthesise a mixture of both maleic acid and fumaric acid (its trans isomer), but I can't achieve the required stereochemistry.

    I figured out that any synthesis will most probably proceed through the diol, which could then be oxidised to the acid. I used two disconnections in my retrosynthesis: first one of the terminal carbons containing OH (synthetic equivalent:Grignard addition) and then the double bonded carbon to which the previous carbon was attached (synthetic equivalent:Wittig reaction).

    Immediately after completing the analysis I felt that stereochemistry had not been ensured. I went back over my scheme trying to identify the steps controlling the stereochemistry. In one of the steps, I had proposed anti-Markownikoff (using peroxide) hydrobromination of a terminal alkyne, so that I could get an alkenyl Grignard reagent. I needed to make sure that the vinylic bromide formed in this step was cis to the hydroxymethyl group, after which the Grignard reaction with formaldehyde (methanal) would preserve this stereochemistry. However, as far as I know the free radical hydrobromination of an alkyne is not exclusively or dominantly syn.

    Can the synthesis be achieved with any minor modification to my scheme, or would I need some entirely different pathway?

    Thanks a lot for your help.

    Molu

    QUERY IN SUMMARY: Propose a synthesis of maleic anhydride from acetylene.
     
  2. jcsd
  3. A bit of help guys?

    Molu
     
  4. chemisttree

    chemisttree 3,723
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    Are you proposing to generate the double bond via a Wittig reaction? If so, why not use the unsaturation already present in the acetylene?

    What reactions are you familiar that yield cis olefins? What would likely starting materials be for these?

    Your grignard approach would work for the first methylol addition but how would you add the second? You might not get past the propargyl alcohol.

    If you did produce propargyl alcohol via Grignard chemistry, how would you add another carbon the the terminal alkyne?

    Have you studied the addition of formaldehyde to acetylene? (Reppe Synthesis)
     
  5. What I did in my original synthesis was to make ethylene bromohydrin (2-bromoethanol) from acetylene (bromine to protect one double bond, stopping the required alcohol from tautomerising to acetaldehyde), oxidise it to 2-bromoethanal and add a carbon by the Wittig reaction. Then I dehydrohalogenated to allene (propadiene), performed hydroboration-oxidation to get the prop-2-en-1-ol and then halogenation-dehydrohalogenation to prop-2-yn-1-ol (propargyl alcohol). Then I hydrobrominated with peroxide propargyl alcohol to get the grignard reagent and then added to formaldehyde, getting (EZ)but-2-ene-1,4-diol and then oxidised it to the diprotic acids. But since the free-radical hydrobromination step is not stereoselective, I would get a mixture of the geometrical isomers.

    The only reaction I know to yield cis isomers is the surface-catalysed hydrogenation of alkynes, but that would need a triple bond instead of the double bond, I can't think of how to introduce it. Halogenation-dehydrohalogenation will favour the diene over the alkyne.

    I've not seen the addition of formaldehyde to acetylene.

    Thanks.

    Molu
     
  6. chemisttree

    chemisttree 3,723
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    I see a few problems with your synthesis. You propose some chemistry that I have never seen applied for the generation of propen-2-ol (prop-2-en-1-ol?).

    This is a very simple molecule most easily generated by adding formaldehyde to a grignard of the acetylene starting material. You don't need all of the exotic bromohydrin/hydroboration(?)/allene/oxidation stuff that probably won't work anyway.

    From propargyl alcohol where do you go? Peroxide propargyl alcohol? I've never heard of it but I'm sure it is not compatible with grignard conditions...

    Once you have propargyl alcohol, where do you go? If you want to elaborate the other end of the acetylene functionality, you will be advised to mask the OH functionality. I would suggest converting it to a trimethyl silyl ether. From that step forward, all reactions must be water-free. Another grignard reaction to add formaldehyde would yield 2-butyne-1,4-diol after an aqueous work-up.

    Try it from there and review the following:

    1. Reppe synthesis (shortcut to everything above!)
    2. Chromium oxidations (don't touch that alkyne functionality in this step!)
    3. Lindlars catalyst
     
  7. How foolish of me! I chose the wrong disconnections.

    What I propose to do with propargyl alcohol is as follows:-

    Perform hydrobromination in presence of peroxides (Khrusch effect), giving the bromide with anti-Markownikoff orientation. This is to get the terminal bromide rather than the internal one I would get by simple hydrobromination. I then put this bromide with Mg in ethoxyethane to get the Grignard reagent. I then add this to formaldehyde, getting but-2-ene-1,4-diol. I don't see any of these reactions interfaring with the hydroxyl. I can then use dichromate to
    oxidise the diol to the dioic acid.

    If I could get but-2-yne-1,4-diol then I could produce fumaric or maleic acid as desired by choosing the hydrogenation method. How are you making it? It seems to me that the grignard addition will yield a molecule with a double rather than a triple bond.

    Thanks.

    Molu
     
  8. chemisttree

    chemisttree 3,723
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    Absolutely not. Review the incompatibilities for grignard conditions...

    You mentioned that you thought that Grignard conditions would convert a triple bond to a double bond? ("It seems to me that the grignard addition will yield a molecule with a double rather than a triple bond.") This is incompatible with your own logic above. Grignard conditions will not reduce triple bonds.

    A most effecient way to generate grignard reagents from terminal alkynes is to treat with butyllithium or some other strong base such as sodium hydride followed by magnesium bromide (MgBr2). Remember that terminal alkynes are acidic and lose the proton very easily. The carbanion is localized on the terminal carbon barring any conjugation elsewhere in the molecule. You can see that these conditions too are incompatible with alcohols since the alcohol proton is orders of magnitude more acidic than the alkyne proton.

    What are you going to do about the propargyl alcohol functionality?

    You must mask it some way to continue down the Grignard pathway...
     
  9. I do not understand, how will the whole process of Grignard reagent formation, its addition to a carbonyl and the subsequent hydrolysis affect a OH group? Why do I need to protect it? Is the alcohol sufficiently acidic to form a Mg salt under Grignard conditions?

    I fail to see how you arrive at a triple bond in the product. I did not mean that the Grignard reaction itself would reduce the triple bond, but to make a Grignard reagent I would have to add a halogen group first and that would reduce the triple bond to a double bond. But you seem to suggest some alternate pathway where the Mg will substitute a terminal hydrogen rather than a halogen. Would you mind explaining this reaction?

    Thanks.

    Molu
     
  10. chemisttree

    chemisttree 3,723
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    The following is a list of incompatible functionalities with the Grignard reagent: carbonyl compounds, alcohols, thiols, primary and secondary amines, nitro, nitroso, anything with an acidic proton. Things with acidic protons include terminal alkynes.

    The answer to your question about the acidity of the alcohol being sufficient to form a magnesium salt under grignard conditions is "yes". You can think of a grignard reagent as a masked carbanion where the magnesium bonds to the grignard. All of the acid/base reactions of carbanions will apply to grignard reagents for the most part. There are some exceptions, however...

    If we consider the grignard reagent as a masked carbanion it is straightforward to rationalize the approach to make the reagent by lithiation followed by treatment of that carbanion with magnesium bromide (MgBr2, not Mg). Traditionally, the haloalkane(-ene or -yne) is treated with zero valent magnesium to produce a reagent that contains a magnesium in a +2 oxidation state. You may also consider the haloalkane as a masked carbonium ion (+ charge) and therefore the oxidation of the zero-valent magnesium donates two electrons to that carbonium ion to yield a masked carbanion.

    You mentioned that generating a grignard would need to go through a halogen-substituted vinyl group. I assume that you are thinking of producing the 3-halo subsituted vinyl propenol by adding HX across the triple bond in an anti-Markovnikov fashion. If you added HX under free radical conditions (with peroxides), that would certainly happen.

    Consider the alternative of first making the anion with butyllithium or sodium hydride and then adding X+ (in the form of X2 or N-halo substituted succinimide as in N-bromo-succinimide). This would result in the haloalkyne.

    Preserving the alkyne functionality will greatly simplify the chemistry.
     
  11. I see now. You are using the acidity of the acetylenic hydrogen to substitute it with the MgBr group. Is it possible to use Sodium Amide as the base in liquid ammonia as solvent? And when I try to add Br+, can I do it in the same solution and would I need some polarising agent like AlCl3 to generate the electrophilic halogen? And will silyl ether be the best protecting group for the OH? Thanks.

    Molu

    Molu
     
    Last edited: Mar 9, 2007
  12. chemisttree

    chemisttree 3,723
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    Sodium amide is indeed a strong base... and liquid ammonia is a good solvent for it. Review the incompatibilities for Grignard reagents.....

    When I have made haloalkenes, I just dripped liquid bromine into a solution of the anion. Alternatively, you could weigh out some NBS (N-bromosuccinimide) and add it as a solution in the same solvent as the Grignard (usually an ether like THF or diethyl ether).
    Trimethylsilyl ether is the first protecting group that comes to mind but there are others.... Search for protecting groups in organic synthesis, alcohols for a more complete list.

    Remember the incompatibilities with Grignard reagents.... you know, like amines (ammonia, for example).
     
  13. So I could do this:

    Put the alkyne with sodium amide in liquid ammonia, treat the sodium acetylide formed with bromine (in alcohol?) to get an alkynyl halide, treat this with Mg in THF to get an akynyl grignard reagent, add it to formaldehyde and then finally hydrolyse to get the alcohol.

    I'm wondering something. When I finally try to oxidise the diol to dioic acid using, for example, acidified potassium dichromate, wouldn't such a strong oxidising agent also oxidise the triple bond? If that was the case, could I protect the double bond using a vic dihalide?

    Also, if I use NBS to brominate the second acetylide (with a methyl alcohol attached to the other end of the triple bond), wouldn't allylic substitution become a competitor against electrophilic addition of bromine? Even if I use pure bromine, each addition of electrophilic bromine will be accompanied by the generation of a nucleophilic bromine. Couldn't these engage in nucleophilic substitution?

    The synthesis is almost complete. Thanks.

    Molu
     
  14. chemisttree

    chemisttree 3,723
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    Sodium amide in liquid ammonia is a strong base and will certainly deprotonate the terminal acetylene. Two equivalents would generate the dianion. You could then add Br+ to generate the halide if you want. You would need to add Mg to the isolated, purified alkynyl halide since the Grignard reaction is incompatible with ammonia. In my experience, adding Br2 to the anion does not add Br2 across the triple bond if Br2 is added carefully and only equimolar amounts are used. One equivalent for the mono ion and two equivalents for the dianion. It is difficult to use and weigh accurately (liquid, volatile, corrosive) and therefore NBS (solid, stable) is usually employed.

    Review the chemistry of chromic acid. Does it oxidize triple bonds?
     
    Last edited: Mar 12, 2007
  15. You are suggesting that I perform both Grignard additions in the same step by substituting both acetylenic hydrogens with bromine? If this works, then I wouldn't need to protect the alcohol?

    Thanks.

    Molu
     
  16. Could this work:

    To acetylene, you add methyl magnesium halide to give you 2-butyne. You reduce this with hydrogen/Pd/BaSo4 to give you cis-2-butene. Then you use 6 moles of NBS to substitute the methyl hydrogen with bromine and then hydrolyse to get maelic acid.

    The only step Im slightly shakey about is the NBS addition. Everything else will work. If that doesnt work, is there some other way to replace the methyl hydrogen with a halogen? Because then we could hydrolise the halogens to give the acid.
     
    Last edited: Mar 13, 2007
  17. chemisttree

    chemisttree 3,723
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    Exactly!

    I wouldn't use sodium amide in ammonia in the bromination step. Remember that the formal charge on bromine in this step is positive (Br+) and ammonia is a base. Use butyllithium instead...

    In response to Chaos' post.... I wouldn't try free radical brominations on a compound that contained any unsaturation, especially a cis double bond.
     
  18. But NBS replaces the hydrogen on the allylic position with bromine. Thats why I didnt use Br2/light. In any case, is there any other way to replace the hydrogens with bromine/chlorine?
     
  19. Using NBS to create a low concentration of bromine is an accepted method of performing allylic substitution. But what is that first reaction? Grignard reagents reacting with acetylene? I've never seen it before. What is it called?

    Molu
     
  20. I dont know what the reaction is called, but grignards can react with a hydrogen on a sp hybridised carbon. It acts as a base for the acidic hydrogen.
     
  21. Ouch. I made a mistake. When you would do that reaction, you would get methane, as the CH3 group would take away the hydrogen on the sp hyb carbon.
     
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