dormouse (Member)
04-20-00 15:32
No 108407
      LSD synth via Bromocryptine: Piglet? -CZ_74  Bookmark   

Author  Topic:   LSD synth via Bromocryptine: Piglet? 
CZ_74
unregistered   posted 06-25-98 08:37 PM           
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It won't work. The R-Mg-Br part of one bromocryptine molecule reacts with the carbonyl group of another making a teritary alcoholic mess on acidic quenching. Questions?
CZ
 
Piglet
Member   posted 06-26-98 05:25 AM          
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Are we talking about the =O on the amide group here ?
 
Piglet
Member   posted 06-26-98 09:29 AM          
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O.K.: The Grignard reagent forms on a carbonyl group by forming an -OH+ intermediate. This is normally a H from the amide (primary/secondary) but with a tertiary that cannot happen. Failing that, a H could be (possibly) moved from the carbon on the ring. Unfortunatly this carbon is conjugated and doesn't have any hydrogens. That is the 'acidic carbon' I was talking about.
You understand that this is just MY view of the problem because when D342 posted I thought exactly the same as you. Then I thought that this person is USUALLY pretty good so I must bee missing something...
If this is rubbish, say so! I'm keen to learn!
Thanks for your question,
Best wishes,
Piglet 


Piglet
Member   posted 06-26-98 10:08 AM          
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Actually, the N in the indole ring is going to cause problems for the grignard... Protection is possible (I think methylation has little effect on the potency ?) but of course that makes another step.
I think that some other metal complexes can remove the Br without the NH interfering. LiCu meybee... I shall go home and do some hard reading. Suggestions anyone ?
Piglet 


drone 342
Member   posted 06-29-98 05:17 PM          
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Permit me to repeat myself for ther fifth time. Amides do not react well with Grignards. The carbonyl just isn't electrophilic enough. As for the concern with the indole, I'll get some ref's describing Grignard reactions involving indoles to put your minds at ease.
CZ, to put it another way, the R-MgBr will not react with any others, and upon quenching, the R-MgBr turns to RH, just like every other grignard in the presence of water. There is only one carbonyl to worry about, I've asked a PhD about this, and they took the time to convince me that it would work.

Piglet, I'm afraid I don't quite see what you're getting at with what problem you're seeing. If you could, please describe the mechanism you're concerned about using the carbon numbering system, or better yet, is there a way you could get me a picture of what your saying?

Once again, the Grignard leaves the carbonyl ALONE in this case. I'm trying my hardest to think of new ways to explain this, but I'm running out of ideas.

Think about it this way. Imagine a grignard reagent R-MgBr reacting with acetone. Why does it attack the carbonyl? Because it is electron-deficient. The oxygen pulls an electron away from the doulbe bond, leaving its aqdjacent carbon positicvvely charged. The Grignard's respective carbon has more electron richness than it knows what to do with, so it attacks the carbonyl, the MgBr+ is scooped up by the oxygen, and upon quenching, you do get a tertiary alcohol.

Now compare this scenario with that of bromo LSD. The carbonyl you're worried about is bonded directly to a nitrogen. If the oxygen descides to steal man electron, the nitrogen will donate one to the carbon to replace it. This makes it considerably less electron deficient, and makes it a poor candidate for Grignard addition. Period. A good overview for this can be found in The Encyclopedia of Chemical Technology, vol 12.

As for the H+ moving from the "ring", which ring are we talking about here, and from which carbon? What is the mechanism (in grueling, electron-pushing detail)? As previously stated, THERE'S NO CONJUGATION HERE. None. Nada. Take out your Merck index, and take a look at bromocriptine. The carbon you're refering to does have a proton; its a stereocenter, and the proton's position in fact determines biological activity (LSD vs. iso-LSD.) If that carbon you're talking about were doulby-bonded, then yes, it would be in a conjugated system. But it's not, so it isn't.

Now how exactly will the N in the indole ring cause problems? What do you forsee? In any case, this could easily be proected with a little acetic anhydride, which incidentally, would give a final product ALD-52.

-drone #342


drone 342
Member   posted 06-29-98 08:26 PM          
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I did a little thinking, and I think I understand what you meant by the indole nitrogen. You're worried that due to the NH's greater acidity, it will donate a proton to the 2-position. Here's why not to worry about it.
1)That particular nitrogen is less acidic than your average amine (though it is more acidic than the carbon in the 2-position.)

2)Asumming it would do that, its still not a bad thing. If you balance it out, You wind up eleminating the MgBr group (which is something you wanted anyways), but wind up with a MgBr- substituent on the 1-position. This N-MgBr is easily hydrolyzed by water into NH and MgBrOH.

3) Like I said before, even if this is a problem, you can always acetylate that N, and you'll just wind up with ALD-52 -- an alalog that is *alledgedly* smoother and more pleasant than the Colonel's original extra crispy recipe.

Hope this puts your clever little mind at ease.

-drone #342


Piglet
Member   posted 06-30-98 03:47 AM          
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D342: You described my worry better than I was able. I already did the electron-pushing (boring!). I hope you noted that I posted this before we had our little chat. I wasn't registering continued uncertainty.
What I was thinking was that when you form the grignard, the H off the indole N would be acidic enough to move. I also wondered if that double-bond (9-10 I think, from memory) could also move towards the amide, conjugating it. I know that particular bit of the molecule is rather delicate.
I feel more comfortable now, anyway! So don't feel that you have to answer me.

Here is some conjecture: I wonder, I wonder, I wonder what would happen if you added an extra group on the benzene ring of the indole in, say, the 5 position like in that frog-mad-brain-go-bye-bye-drug whose name temporarily eludes me...

Best wishes to ya,
Piglet 


CZ_74
unregistered   posted 07-01-98 01:59 AM           
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Drone342, u are wrong. I spoke to my organic chemistry professor, who argeed with the arguement that Grignards DO add to amides, regardless of the amide nitrogen's substitutents. The amide is stabilised by electron delocalisation between the O=C(R)-N-R' group, but not enough to stop the highly polar >C-Mg-Br from attacking the amide carbon. See 'Organic Chemistry' by McMurry, 4th ed (around QD 251)under 'limitations of Grignards' for details. My original point stands.
CZ
 
CZ_74
unregistered   posted 07-01-98 02:30 AM           
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Piglet:, Grignards don't have an =O-H+ intermediate. The magnesium atom makes the carbon attached to it very nucleophillic & the C-Mg bond is very polar. The mechinism I've been taught involves the nucleophillic carbon-(Mg-Br) forming a bond with the carbonyl carbon, breaking the amide electron delocalisation & pushing 2 electrons from the carbonyl carbon onto the oxygen. This leaves the electrophillic +Mg-Br ion, which then receives two electrons from the nucleophillic oxygen. This organometallic compound (R-C(-O-Mg-Br)-R' is then quenched with H2O to make the teritary alcohol + Br-Mg-OH. Trying to react/form Grignards is impossible
in the presence of H2O, as Mg(s) + H2O -> Mg 2+(aq) + 2OH-(aq) + H2(g).
CZ


drone 342
Member   posted 07-01-98 10:38 PM          
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Piglet, even if the proton on the indole nitrogen were to quench the Grignard (which it just might), this wouldn't be a problem.
As for the 9,10 stability, I think it has a lot more to do with ring strain. It doesn't move. It can't move. The reason for this ring strain is that you got this ring with a double bond conjugated to the indole, which is going to flatten out the molecule a bit to maintain its conjugation. By doing this, it is awefully twisted.

CZ: like I said, I talked to a PhD too, and he told me the *exact opposite*. I got copies of the Grignard reactions overviews from the past 25 years, and I'll look through them to see what's the deal. Besides, look at the stearic hindrance we're looking at in this situation; that alone would keep things from getting too out of hand.

When I checked my o-chem books at home, I found no examples of amides + Grignards. In any case, as long as thing didn't get too hot, my belief is it would work. I think you'd agree though, that an amide, from an electrophilicity standpoint, would be not as reactive as, say, a ketone, an ester, or a carbonyl. I'm curious, and I'll keep looking. I don't know if I have access to that textbook you listed. I'm not fionding much for info on amides and Grignards, so could you supply another ref for me?

-drone #342


CZ_74
unregistered   posted 07-06-98 04:11 AM           
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I can't find any other references }:-( but can post u a verbartim quote of the page. I don't mean to bee painful but sterics don't usually affect Grignards. On the topic of stabillity, amides are the most stable carboxylic acid derivative (acid chlorides < acid anhydrides < esters < amides) but not stable enough to stop agressive Mr Grignard.
If u find a ref contrary to mine, let me know. Also, acetylating the indole N doesn't protect it from Grignard attack!! The Grignard will rip into the ester faster than u can say Oh,fu*k! Forget the whole thing.
It's good to think about these things, but organic synthesis is like chess: u need to know what u can/can't do before commiting!
CZ
 
drone 342
Member   posted 07-06-98 08:50 PM          
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"Chess" yes, but I don't think we've established entirely the question of improbability. First, I do believe stearics have a lot to do with Grignard reactions in this case: look at the size of that molecule! If you take a look at the stearic conformation of LSD, you'll see that the place of attack is VERY out-of the way. Ths point becomes especially significant if you look at the actual mechanism involved here. The intermedate forms a sort of 6-membered ring structure, involving a carbonyl and two Grignard reagent equivalents. I know its wierd, but what happens is the grignard reagent attacks the carbonyl, the oxygen pulls up an electron and attacks a separate magnesium, which releases an electron to its respective R-group, which attacks the magnesium bonded to the original R group attacking a carbonyl. For whatever reason, 6-member-ring-type-intermediate-complex reaction mechanisms are common. In this case because most Grignards actually exist as dimers in solution (if memory serves), which means that you got three separate fat-assed molecules tryng to hunker down and react in a daisy-chain fashion at a spot that's already pretty stearicly hinderd anyways.
Plus, I do believe piglet had a point, and I'm liking it more and more. That proton on the 1-position is acidic. You can't get any closer than right-next-door, so what I'd think would happen now would be a quenching of the R-MgBr on the 2-position by the proton on the 1-postion. This will be considerably less reactive.

Besides, if this were done at, say, -78 deg C, in a dilute solution, the addition could be kept from happening fairly easily ( if it were to happen at all, which I doubt.)

I *too* can offer verbatim quotes from PhD's, but what's needed here are some good ref's. I'll do my best to dig up stuff on this, and I hope you'll be kind and do the same.

This is fun! This s the kind of healthy academic debate that The Hive needs lots of. I'm glad you're taking such interest in this pursuit, and I'll keep you posted on my lit search. It's so nice to have such a stimulating chemistry-related discussion like this with someone who has obviously done some studying.

-drone #342


CZ_74
unregistered   posted 07-07-98 01:40 AM           
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The size of the molecule is generally irrelevant in rxns also (as most are done in liquids or gas tanks, allowing the molecules to move freely as they choose). Given this, the just formed C-Mg-Br group (& position 2)of a certain bromocryptine molecule is attracted to the carbonyl carbon of another bromocryptine molecule elecrtostatically. The
indole part of the molecule is unhindered & the acid diEtamide is free to rotate around the axis of the single (rings)-(acid diethylamide) bond (as all single bonds in chains can rotate the chain into myriad combinations, for a small amount of energy).
Cooling the rxn down slows the movement (decreases kinetic energy) of all the atoms in the system (not necessarily increase the stability of the amide group etc), thus rxn rate slows down proportional to temperature.
The temp would need to lowered to about -200C or lower to stop the rotations around the single bonds, but that isn't to say that the conformation it stops in (of lowest energy) won't be able to add to the Grignard of another molecule). Lets say u could achieve these conditions when adding the Mg(s) to bromocryptine in liquid Argon. The Grignard forming reaction would take weeks & be rediculously expencive.
Now, peforming the quenching of the Grignard selectively is exceptionally difficult, as the water isn't @ a high enough temperature to break the hydrogen bonds with itself to quench the Grignard. Warm it up (slowly or quickly) and the Grignard adds to the carbonyl group (as posted before) making ergocryptine chains of ramdom length (ie a BIG mess) while quenching some of the Grignard. Expect yields of <5% (optimistic).

Hope u haven't hallen asleep...
CZ


CZ_74
unregistered   posted 07-07-98 01:52 AM           
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Above, when I was talking about H2O not having enough energy to break bonds with itself, I was referring to ice, H2O(s).
CZ
 
drone 342
Member   posted 07-07-98 08:36 PM          
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Size *does* matter, and especially here. The Grignard reaction affords enantioselectivity, and as I said before, you're dealing with a reaction involving a cyclohexyl-shaped intermediate -- good indications of the inportance of geometric and steric hindrance.
To put is shortly, you're dealing with a molecule with relatively bulky substituents, and you're proposing three of them huddling around in the same place at the same time. This means steric hindrance. It does matter; quite much. And not just in this reaction, but throught almost all organic synthesis.
In addition, while the amide substituent may rotate, it still has a prefered position relative to the rest of the molecule. Conveniently, this postion makes it hindered. This is certainly the least important factor here in determining the feasability of said reaction, but something to consider anyways.

There's still the issue of the acidic proton on the 1-position -- a prime candidate for quenching the Grignard reagent. Considering the extreme closeness of the 1 and 2 positions, it would seem it would do a fine job quenching it long before tail ends of a couple other molecules got in the way.

Also, the whole issue of Grignard automatically bouncing into the amide. This can be easily minimized by maintaining a low concentration of the grignard reagent, even if such a reaction were to occur. For that matter, I still have yet to find much written on Grignards reaction with amides at all (I'm sure there are a fedw examples, but I really think it is considerably less reactive than other carboxylic acid derivatives -- enough so that it should work.) Besides, the Grignard may not be necessariuly in such a hurry to react with it anyways; remember, this same reaction can be used to produce ketones, and other products.

Besides, if its really that big of a problem, what about using a protecting group on it? Even with this extra step, its still worth a pure source of LSD starting material here in the US. Hell, for that matter, there are other ways of dehalogenating. If memory serves, there's a few reducing reagents that would do the trick, yet leave the amide alone.

Also, water would react just fine at lower temperatures. Just because it will freez solid doesn't mean you can't have it dissolved in solution. Even if this were the case, perhaps methanol could be used in relacement of it.

-drone #342


CZ_74
unregistered   posted 07-07-98 10:41 PM           
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Drone:
The acidic proton on the 1 position would only disolve signifigantly in alkaline protic solution (eg NaOH in H2O/ NaOMe in MeOH). It can be neglected in anhydrous aprotic organic solvents(like Et2O, which Grignards are performed in). Hence the indole proton won't quench the Grignard.
After consulting the Merck index, the bromocryptine molecule has 3 amide groups (2 of them cyclic, called lactams). Both the lactams are on the peptide part of the molecule & are much less hindered strerically than the amide linking the lysergic acid moiety to the peptide moiety. Thus there is an extremely good chance that the Grignard will attack one of them.
Substituting anhydrous MeOH for H2O won't stop the Grignard from adding to the amide, as the addition is done immediately after Grignard formation. All the MeOH does is donate its proton to the newly formed -O-Mg-Br oxygen & make MeO-Mg-Br.
Also, if an Organic Chemistry textbook says that Grignard reagents add to amides & mentions no exceptions, the textbook is assumed correct. There are no ifs or maybes. If u are brave enough to email me your address, I could send you a photocopy of the relevant pages free.
Please post the reaction which dehalogenates the indole without destroying the molecule or making more problems than it solves. I'm interested.
Apart from that, I see no point furthering this discussion. Thanks for participating.
CZ
PS: If you are still convinced that it will work, do the rxn and post the results. You will prove 1 of us wrong & learn heaps...


 
Piglet
Member   posted 07-08-98 04:22 AM          
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CZ: Your point about the amides on the peptide molecule is certainly right but surely we are talking about making bromo-LSD and then cleaving that? There is only 1 carbonyl and 1 NH and 1 Br group to think about ?
Also, my book says that when grignards add to carbonyl groups they form an OH+? Is it a misprint ? I don't know enough to say although all my books do agree about the limitations of the grignard...
Might I add a new spin by suggesting other reagents to remove the Br. I have read of several other metals reacting with halogens such as the zinc/copper couple and organolithium reagents. I'm not saying that this is right, I'm just thinking (duuhh) out loud.
I cannot believe that a simple route does NOT exist to use this handy source.
Best wishes,
Piglet 

P.S. Your postings have been most helpful for helping me learn about these reactions, I thank you!


Piglet
Member   posted 07-08-98 10:59 AM          
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From 'A Guidebook to Mechanism In Organic Chemistry' 6th edition by Peter Sykes
8.4.1

The actual composition/structure of Grignard reagents- commonly written as RMgX- is still a matter of some dispute.
It appears to depend on the nature of R, and also on the solvent in which the reagent is, or has been, dissolved.
Thus, the NMR spectrum of MeMgBr in Et2O indicates that it is present largely as MgMe2+ & MgBr2, while X-ray
measurements on crystals of PhMgBr isolated from Et2O solution, indicate that it has the composition PhMgBr.2Et2O,
with the 4 ligands arranged tetrahedrally round the Mg atom. Whatever the details may be, Grignard reagents may be
regarded as acting as sources of negatively polarised carbon, i.e. as -RMgX+.
There is evidence of complexing of the Mg atom of the Grignard reagent with the carbonyl oxygen atom and it is
found that TWO molecules of RMgX are involved in the addition reaction, in some cases at least, possibly via a
cyclic T.S.

The second molecule of RMgX could be looked upon as a Lewis acid catalyst, increasing the positive polarisation
of the carbonyl carbon atom through complexing with oxygen. It is indeed found in practice that the addition of
Lewis acids, e.g. MgBr2, does speed up the rate of Grignard additions. Reliable details of the mechanism of Grignard
addition to C=O are surprisingly scanty for so well-known a reaction but pathways can be invoked to explain two
important further observations: a) that Grignard reagents having H atoms on their beta carbon atom tend to reduce
C=O --> CHOH, being themselves reduced to alkenes in the process (transfer of H rather than RCH2CH2 taking place).
b) That sterically hindered ketones having H atoms on their alpha carbons tend to be converted to their enols, the
Grignard reagent, RMgX being lost as RH in the process.

Grignards act as strong nucleophiles and the addition reaction is essentially irreversible. The end products of
addition, after aqueous hydrolysis, are alcohols. It is, however, important to emphasise that the utility of Grignard
, and similar, additions to C=O is a general method of joining 2 different carbon atoms together.

In the past, organo-zinc compounds were used in a similar way, being largely displaced by Grignard reagents;
in turn, Grignard reagents tended to be replaced by lithium alkyls and aryls, RLi & ArLi. These reagents tend to give
more of the normal addition product with sterically hindered ketones than do Grignard reagents, and also more 1,2-
and less 1,4- addition with C=C-C=O than do grignard reagents.

I hope that clears it up (!).

Piglet 

P.S. You know, I don't follow that stuff too well...


CZ_74
unregistered   posted 07-09-98 12:06 AM           
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Piglet>After the Grignard has added to the carbonyl & quenched, it creates a secondary or teritary alcohol depending on the R groups. The + bit is a misprint. I'll get back 2 u on the organo-Li rxns. Sorry about the Br-cryptine sidetrack, I forgot. The Grignard rxn in this instance will probably yield <20%.
CZ
 
drone 342
Member   posted 07-09-98 12:43 PM          
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CZ,
As stated previously, we're not just talking about bromocriptine. What I'm proposing is the synthesis of bromo-LSD, followed by halogen cleavage. There's only one amide in this case to worry about, and I'm still waiting for ref's from you on why this won't work.

Grignard reactions are touchy, but nowhere near as touchy as you claim them to be. That R-MgBr group may be reactive, but its not lurking in the shadows, waiting for a chance to react with innocent amide substituents -- or at least the reaction can be set up so that this isn't a problem. They're used to synthesize carboxylic acids as well as aldehydes, fer Pete's sake! If ketones and acids can survive (both of which involve considerably more electron-deficient carbonyls than what we're taking about) then why would amides be a particularily good target? There's a lot of variations on this reaction, and a lot of conditions can be adjusted to make this work.

There's a great deal of reason to further this discussion! I've taken the time to get copies of all the reviews of this reaction in the literature, and as far as I've seen, there really is a lot of reason to believe this would work.

As for the proton on the 1 position. This is a lot more acidic than you think. Check out the pKa for indole, and you'll find that it'll dissociate just fine on its own in good old anhydrous aprotic DMSO, thank you. Why do you think you need basic conditions for such a thing to happen?

Now on to steric hindrance. A direct quote from "Advanced Organic Chemistry", 3rd ed., by Francis Carey and Richard Sundberg:

"Grignard additions are sensetive to steric effects, and with hindered ketones a cometing process involving reduction of the carbonyl group is observed. A cyclic transition state similar to that proposed for the Meerwein-Pondorff-Verley reduction is involved.

The extent of this reaction increases with the steric bulk of the ketone and the Grignard reagent. For example, no addition occurs between diisopropyl ketone and isopropylmagnesium bromide, and the reduction product diisopropilcarbinol is formed in 70% yield*. Competing reduction can be minimized in troublesome cases by using benzene or toluene as the solvent.**"

*JOC 27, 1 (1962)
**Tet Lett 4383 (1979)

You said:

"Also, if an Organic Chemistry textbook says that Grignard reagents add to amides & mentions no exceptions, the textbook is assumed correct. There are no ifs or maybes."

Obviously you haven't been studying organic chemistry for very long! There are "if's" and "maybe's" for almost everything. Not only that, but chemistry is science, not religion; there are *NO* holy books, absolute unquestioned authorities, etc. -- especially text books.

I haven't seen your textbook, though I assume it didn't say "Grignards have a particular affinity to amides that is absolutely merciless and voracious. They will absolutely sodomize any carbonyl that it comes in contact with. It happens instantaneously. There are no exceptions, no ways around it, and this is The Word of God."

There are plenty of examples of carbonyls getting by in the presence of Grignards *without* reacting. If your book says otherwise, its wrong. Period. To interpret whatever it said to mean that a carbonyl as stable as an amide doesn't stand a ghost of a chance is interpolation.

In fact, the aforementioned book of mine, when describing the action of Grignards upon acid derivatives, says that Grignards react with acids, acid chlorides, nitriles, esters, aldehydes, etc. But nowhere are amides mentioned as being particularily reactive. Now, it would be gauche of me to say that because such a nice, well-referenced book that does such a thorough job on the subject gives no heed to it, it mustn't be so. But I still contend that its at least a fairly decent indication.

Here's a short list of a few examples of the synthesis of carboxylic acids using Grignards (something that, according to your arguement, shouldn't happen):

Org Synth III, 553 (1955)
" " I, 353 (1932)
" " 59, 85 (1977)

My e-mail adress is drone342@hotmail.com I'd love to see what you have.

Also, where are you getting these "5% yields" and "20% yield" numbers? What info source are you using to make these estimates? What references?

The reason we're having this discussion is to mash out what would be the most favorable conditions for removing that bromine. I can't do the reaction if I don't know what conditions will work best. If you can come up with a better reaction, please be my guest and propose it in this forum.

If I become privy to lab notes on this reaction, you better believe I'll post them!


 
drone 342
Member   posted 07-09-98 07:27 PM          
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CZ:
LSD's pKa=6.37 (aq. EtOH, Helv. Chim. Acta 37, 1954 2039, 2054)

Now that's plenty acidic for Grignard quenching. Consider the Grignard as a bad-assed base, ready to do some serious deprotonation, given the chance. When I talked about adding MeOH, this was what I was refering to. You said:

"Substituting anhydrous MeOH for H2O won't stop the Grignard from adding to the amide, as the addition is done immediately after Grignard formation. All the MeOH does is donate its proton to the newly formed -O-Mg-Br oxygen & make MeO-Mg-Br. "

What I was refering to was that the MeOH would donate its proton the same way water would to quench the reaction at the end. The only reason I brought it up was that you said this would have to be done pretty cold, and by using MeOH, you'll have an easier time adding it at really cold temperatures than perhaps water. Personally, I think it just might be feasable at 0 deg, but that's only my humble gut feeling. In any case, I'm throwing out MeOH as a viable quenching alternative to burnt hydrogen.

I'm not uneducated, and I didn't just stumble across organic chemistry a few days ago. I've done my share of Grignards, and none of this is new to me. You're not dealing with a neophyte, but you seem to think so.

Piglet:
I finally figured out what you're worried about with the "OH" intermediate, and you're right in thinking that's a little strange, since it doesn't work that way. There is no "OH" intermediate, the oxygen is protonated *after* quenching. Before then, you have a -OMgBr substituent. Have no fear, my fine cloven-hooved friend; I haven't led you wrong yet.

Back to CZ:

So what we're dealing with is a weakly electrophilic carbonyl and some severe steric hindrance, involving 3 hefty molecules configuring in a daisy-chain cyclohexanyl fashion. This reaction is competing with an acidic proton that just so happens to be RIGHT NEXT TO a very strong nucleophile, able to quench it with amazing ease and speed. In fact, there'd really be no way to get that acidic proton closer, and that nitrogen is making that proton significantly less hungry for electrons.

Now considering all of this, which reaction do you think would be more favored? My guess is you're going to say the first one, but I still don't understand why you like that paradigm so much. Skepticism is one thing, but this is something else.

-drone #342


drone 342
Member   posted 07-09-98 08:48 PM          
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This is a picture of the two mechanisms in debate. Hopefully, I got this html stuff working properly, but if not, here's the adress:
"http://www.geocities.com/CapeCanaveral/Hall/1399/mechanisms.gif"
CZ_74
unregistered   posted 07-09-98 11:58 PM           
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Drone:
Ignore my yield estimates. They were gut feelings (proof they're not always right!).I wasn't aware of the cyclic transition of hindered Grignards or the high acidity of LSD. If u can minimise the addition rxn, I'm all for this synth.
 
PS: If I believed u were a twit, I would have said so.


CZ_74
unregistered   posted 07-10-98 02:13 AM           
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Drone:
Saying 'Sterics don't generally effect Grignards' was a stupid assumption. I withdraw that comment. It does. (J.Org.Chem 27,1(1962). I also withdraw the 'textbook is final' comment. Not clever...
The Tet.Lett.4383(1979) ref. was in french. The only english in the whole thing was the summary & said 'The addition of Ar to Et2O in Grignard rxns causes a 70% increase in yield'
Or something? Of what that is I couldn't work out. None of the refs. mention Grignard/amide rxns. Keep @ it.
 
 
Piglet
Member   posted 07-10-98 09:26 AM          
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The =OH+ I was thinking about was from the =O on the amide. IF the Mg did form a bond there, what pathway would be followed? As the text I posted said 'carbonyls with a beta carbon tend to from alcohols. so I was guessing that OMgH would be an intermediate via =OH+. But I also wondered if that 9,10 double-bond would migrate as the H was displaced, thus conjugating the group.
Thats why I thought that it WOULDN'T add. But as it turms out, I was talking rubbish!
Best wishes to both of you,
Piglet 


drone 342
Member   posted 07-10-98 05:22 PM          
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CZ,
Tell, you what. I have a friend who's fluent in French, as well as an avid chemistry student. I'll have him translate if you want. The articles I've listed have little to do directly with amides, but are excellent examples of even less stable carbonyls withstanding Grignard additions -- proof positive that this method has great potential.

I'm glad you've played the skeptic here; that little bit of antagonism was exactly what was needed to get some homework done. I'll keep searching on this one, and I'll keep you posted on my findings w/resp. to RxN conditions.

Piglet,

If the Mg were to add to the ketone, it would be the second reaction I pictured. The mechanism is there, but if you need a picture of the final product, just give a holler.

There is no =OH+. Period. What happens here is that the oxygen pulls up from the double bond, leaving the carbon positively charged, and leaving itself positively charged. This attracts the electrophilic MgBr, which is then attacked by the oxygen. At this point, you have essentially a tertiary alcohol, but where the hydroxy is deprotonated, and the MgBr takes the proton's repective place.

Let me try to explain it another way, step-by-step.

1.That electron-greedy oxygen yanks up an electron from the double bond. This makes the oxygen negative, and the carbon positive. This means the oxygen is nucleophilic, and the carbon is electrophilic.

2.The grignard reagent's active carbon, which is FAR more electronegative than Mg, steals an electron from the Mg, cuasing it to break its bond with it, and making it pretty friggin' nucleophilic.

3.The previously mentioned Grignard reagent, which is a strong nulceophile, attacks the aforementioned carbon that's part of the ketone.

4.This leaves the oxygen dangling. Oxygen, a nucleophile, is looking for something to bond with. There's plenty of MgBr+'s floating around, so it latches onto that.

5.Hydrolysis. Up to this point, protons (i.e. H+'s) haven't played a significant role. This changes. Oxygen now hydrolyzes the -OMgBr.

5a.Water dissociates into H+ and OH-.
5b.OH- attacks the -MgBr+, cleaving it off the oxygen in question, and leaving that oxygen negatively charged.
5c.The proton is scooped up by the oxygen.

As for the double bond migrating, it won't. Try drawing the mechanism. It's actually more stable right where it is, anyways (conjugated to the indole.) What you're talking about requires a H+ to be displaced, but there are no H+'s to be seen. Why would a proton in that region of the molecule dissociate?

If you have the technical capabilties where you are, try drawing for me the elctron-pushing diagram for the mechanism you're concerned about. I'm curious now.

If you're refering to the proton on the 1 position, then you have nothing to worry about. THis proton would quench the Grignard, and keep this hellacious reaction from doing anything but dehalogenating - yeah! This is what I believe it'll.

-drone #342


 
Piglet
Member   posted 07-12-98 10:07 AM          
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D342: I have always thought it strange that there is so little info about the grignard reaction of amides. The point I made in the last post was my explanation of why I was wrong! The next stage must bee to obtain a sample and give it a bloody good try.
There does seem to be a number of these compounds in circulation. Maybe I should get some vet supplier catalogues and see what horse-medicine contains this stuff. At least I can make myself useful that way.

Nice work,
Piglet 


 
CZ_74
unregistered   posted 07-14-98 12:32 AM           
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Forget the Organo-Li(Gillman) rxn (unless u can make LiCuH2..). Don't worry Drone, i saw yaw point.
CZ
 
drone 342
Member   posted 07-14-98 12:28 PM          
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The best source is of course the net. Search for "bromocriptine" and "catalog" and whatever else; they're a number of overseas suppliers who will sell it to you without perscription if you ask nice enough.
-drone #342
 
 

 
 
 
 
    bujinkan
(Stranger)
12-11-01 11:27
No 246513
      Re: LSD synth via Bromocryptine: Piglet? -CZ_74  Bookmark   

Interesting. Any comments from new arrivals?
 
 
 
 
    Rhodium
(Chief Bee)
12-11-01 12:07
No 246525
      Re: LSD synth via Bromocryptine: Piglet? -CZ_74  Bookmark   

I still think it would be an interesting thing to try, but you'll never know until someone has successfully performed it in reality.
 
 
 
 
    bujinkan
(Newbee)
12-12-01 17:25
No 246936
      Re: LSD synth via Bromocryptine: Piglet? -CZ_74  Bookmark   

bromocryptine---oh and how nice it would be.