PP's General Chem Question Thread

[Osmium]

I seem to remember that very fine metal powders (including nickel) can be formed by distilling away Hg from their amalgams. But I doubt that they can be used as catalysts.
I'm not fat just horizontally disproportionate.

[Bwiti]

If making methyllithium, phenyl-Mg-bromide, or some other metal complex, is it generally OK to use potassium as a substitute? Patent GB861350 mentions Na and K in producing phenylcyclohexylamines, and I was just wondering if this can be expanded to amphetamine chemistry. Peace!
Love my country, fear my government.

[PrimoPyro]

Bwiti, I dont really know if potassium is an ok substitute, but it would make sense. It may be more reactive though.

Ok, I need someone to throw me a bone here. What would be a good/best way (in your opinion) to prepare any salt or ester derivative of 2-formyl acrylic acid? I don't mean the process has to work for a wide range of salts and esters, I mean if you know a method/reaction to produce any salt or ester, of 2-formyl acrylic acid, please indulge me, and tell me how.

Don't say by Retro-Diels-Alder reaction of the corresponding 2-norbornene, because that is what this is being used to produce!  

I can think of about 6 or 7 different crazy methods, but they all have a flaw in them that will prevent their success. I'm truly stumped, everywhere I turn, I find a reason why I cant use such and such method.

Any ideas? Thanks....

P.S. ten extra points goes to the person who guesses where the interest in this stems from, i.e. why would someone want this compound?  

                                                   PrimoPyro

[PrimoPyro]

I just had an idea. Would this work?

Acrolein could be halogenated at the second position (how would you substitute the H for an X without saturating the olefin?) and then reacted with sodium cyanide, forming the 2-formyl acrylonitrile. This is subjected to alcoholysis with methanol and aqueous HCl, forming the methyl ester of 2-formyl acrylic acid, which is precisely the compound desired.

Would that work well? How would one halogenate the acrolein at the 2 position without reducing the double bond? Is there a better starting material for this?

P.S. hint for 10 point bonus: Oppolzer is a very elegant man. He has my utmost respects.  

                                                  PrimoPyro

[Rhodium]

Regarding

Post 315835

(PrimoPyro: "Formic Dicarboxylic Half Esters", Chemistry Discourse) - Yes, that would work, but you don't need to use an alkoxide, just react the cyclic anhydride with the alcohol of choice.

[PrimoPyro]

Thanks Rhodium, for the answer here and at the tryptamine forum as well.  

Regarding

Post 317041

(PrimoPyro: "Idea", Chemistry Discourse) Thanks goes to hypo for helping me out there, providing a Chem Berichte reference and also physical data on (and the preparation of) 2-bromoacrolein.

If anyone is interested in the article, one of us can copy it here.

Then the 2-bromoacrolein is reacted with sodium cyanide to give 2-formylacrylonitrile, which is alcoholyzed with methanol and HCl to 2-formylacrylic acid, methyl ester, ready for reaction with cyclopentadiene to form that funky bicyclic.  

                                                  PrimoPyro

[PrimoPyro]

Its time to revive my thread once again with another basic question. I almost made a new thread asking this, until I remembered I had made this thread for these type purposes.

Question: What is the mechanism of the safrole + KOH ---> isosafrole reaction?

I am uncertain how it proceeds, and am curious why this occurs.

I imagine that the base -OH removes a proton from the alkene in the first step, forming an intermediate carbanion. I am *guessing* that the formed carbanion would be in the isopropyl position, because it would be the most stable, as it has the least protons/is the most substituted?

If this happens, yay, "go me" but what happens next is beyond me. Logically some sort of electronic rearrangement would take place, but I am unsure just what happens. Carbanions are not prone to rearrangement like carbocations are, so I don't know why or where any formed carbanion would move to, and what effect this would have on the double bond.

Formation of an isopropyl carbanion wouldn't even affect the alkene, it would just turn R-CH2-CH=CH2 into R-CH2-C(-)=CH2, right? Hmm, I wonder:

This intermediate, if *somehow* could get an electron onto that terminal carbon (I have no idea how, so don't ask) then the double bond would be reduced to a terminal carbanion, and a radical at the isopropyl position, along with the isopropyl carbanion, and that radical might rearrange to a benzyl radical. The isopropyl carbanion could give up its electron, which would also form an intermediate radical, and of course two radicals means bond strengthening, restoring the alkenyl bond in a different location: The benzyllic position that corresponds to isosafrole.

Then the terminal carbanion would react with the water produced in the first step to steal back a proton and regenerate the hydroxide catalyst.

Am I close? Not even? What happens? If I am close, where does this mystery electron come from? It can't come from any sodium atoms because they are all already existing as Na+ and won't yield any more electrons.....

How does this isomerization work? Help please? I'm not asking for pictures in ISIS or anything, just a word description like I have given, and maybe an explanation using the "R-CH2-CH=CH2" diagrams I use.

Any help would be appreciated. Thanks.

PrimoPyro
Firm supporter of the "Purge The Couch!" movement. Vote for the purge today.

[Osmium]

Please be patient with me and don't expect too much, I'm tired and exhausted, but I'll try to answr your question:

The KOH removes an H+ from the BENZYLIC position. That is the most stable carbanion that can form, since there is an electron deficient aromatic ring next to it, and it is also the benzylic position of the alkene. Check pKa values of benzylic and allylic hydrogens. This also means that electron can nicely delocalize all over the molecule.
I'm not sure if the H+ really fully separates and produces H2O with the OH- (which would then be absorbed by the KOH or removed by heat/vacuum) or not. I suspect the former, but it doesn't really matter.

So how is the isoalkene produced?

The (delocalized) anion now grabs itself another H+ from the next molecule of safrole. It ends up as predominantly isosafrole since that is the thermodynamically preferred product. It's kinda like a chain reaction proceeding through the whole reaction, with protons jumping from one molecule to the next one. If water mets such a carbanion it will probably give away its H+ and end up as OH- again. Of course all these steps are equilibriums, and the mixture slowly moves energetically downhill, to the preferred trans-isosafrole. The longer you perform the reaction the more trans you end up with, since the cis-isosafrole lies somewhere in between the safrole and the isosafrole.

Hope that was somewhat understandable.
I dunno if that's the real mechanism, if someone has a better explanation please correct me.
I'm not fat just horizontally disproportionate.

[Protium]

To supplement Osmiums explanation, and please correct the details of this if part is incorrect because i'm not very knowledgable with orbitals and such, this is just an educated guess based on memory and now some checking back.  I would say that the KOH dissolves and the hydroxyl abstracts a protium at the most stable position once the temp. allows for the C-H bond dissociation energy requirement, which will be the lowest for the most stable (benzylic) position, to form a singly occupied orbital, which would then bend over and try to grab onto another protium by overlapping of the p-orbital at the radical center with it's neighboring allylic C-H bond, forming into a more planar-type geometry, and this delocalization would continue almost instantaneously down the line until the delocalized radical finds the most stable position that it can on the alkyl chain, naturally the greater number of substituents the greater the strength of the hyperconjugation (stability), and at that point the hyperconjugation stops, and the double bond has shifted to this position of stability.

And so this reaction would be a free radical reaction by the homolysis of an H+ , the formation of a carbon centered radical at the benzylic position, and a hyperconjugation stabilization (as opposed to a resonance delocalization) of the substituent chain containing the pi bond.

Hold on i'll have a picture in a minute too.

Who say's MS Paint is useless?



I can't pull out too great of an analogy right now as Primo can do so well but basically the 4(benzylic) position of the carbon chain loses an H+ and so it reaches over and grabs one from a more stable neighbor on the chain, and in turn the newly deficient molecule does the same thing to it's neighbor, until you get the guy on the end of the rope (alpha carbon) who has even less hydrogen due to it's p-bond, and so the second to last carbon gets screwed for a hydrogen, and being more stable than the end carbon, is forced to take on the pi-bond. 

I've been awake for a long time now, and there's probably a bunch of mistakes, so if you see a correction please point it out. I'm tired as hell but couldn't sleep if I wanted to. The real answer Primo's question, however, is probably a lot more complex than this, as these types of reactions often have different series' of propagation cycles, different initiation and termination and all that as well, so like I said this is my best guess and I share Primo's curiousity in learning more about this reaction.

And if Primo would be so kind as to allow me to impose a follow up question to his own in his thread, does anyone know the mechanism of the equalibrium which determines the cis/trans isomerism ratio?  For most cooks I guess that it's a non-issue but I am rather curious as to the science of why there are differing isomers, as clearly this could not be simply based on chaotic mathematics and complete randomness if there is an equalibrium.

Pr(+)tium