Author Topic: Mannich Reaction On Indole?  (Read 855 times)

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  • Guest
Mannich Reaction On Indole?
« on: June 04, 2002, 07:40:00 AM »
Hi, I have an indole related question, so I will make my first ever tryptamine forum thread.  :)

First of all, what is a Mannich salt? Is it the same as a Mannih base? I cannot locate any data on Google with the term "mannich salt" and it isnt in any chemistry dictionaries I have, nor in March's 5th under that term. But The Chemistry of Mind Altering Drugs, pg. 276 refers to a compound as a mannich "salt" and I dont know what that is.

Basically, I don't have a real question, but rather a statement that I am confused. I don't quite understand how this works or what is going on. here is the excerpt:

"In 1981, Oppolzer reported an elegant synthesis of lysergic acid, in which the C and D rings are formed in a single intramolecular process. The phosphonium bromide [6-20] could be made in near quantitative yield, and bicycloheptene [6-21] in about 60% yield from readily available materials. A Wittig reaction between these two provided [6-22] after removal of the tosyl protecting group, in 62% yield."

That's fine and dandy, I understand all of that just fine, its the next part:

"Condensing the Mannich salt of [6-22] with nitromethane and reduction with TiCl3 gave [6-23] in an overal yield of about 35%...."

It goes on further but I understand all of that too.

What I do not get at all is what the hell is happening in this last part? The whole bicyclo and ylid portion of the molecule is irrelevant, so dont worry about that, This step involves another portion of the molecule seperate from all that.

What they show in the diagram is the 3-position of indole getting alkylated, and having a R-CH2-CH=N-O-CH3 form off the 3-position in indole, where R = indole. What I do not get is where the oxygen comes from, where this second carbon comes from, and how the hell the imine is formed. Basically I just cant put together one single part of this step with the picture provided.

I looked up the mannich reaction and it involves addition of ammonium salts to formaldehyde, forming H2N-CH2-OH, which reacts with compounds containing "an active hydrogen" and the example is with acetic esters at the terminal carbon.

Well I assume that the 3-position of indole serves as the "active hydrogen" spot, but where does the second carbon come from? H2N-CH2-OH should react with indole, forming R-CH2-NH2, not the R-CH2-CH system shown. I dont understand.

Is the oxygen present in the =N-O-CH3 a typo? I wouldnt think so, but it makes no mention kater in the excerpt of removing this methoxy group and methylating the nitrogen to make lysergic acid. I dont see how the reduction with TiCl3 fits in, and how this imine is formed.

Pleeeeeease help?  ::)

anyone who can discuss this is very much welcome to, and I especially invite answers from Rhodium (because I know you have this book and can look up the pictures) from Osmium because you seem to have experience with every damn procedure under the sun, and Lilienthal because I know you know a lot about tryptamine chemistry, and hopefully are familiar with this synthesis. Any help would mean a lot to me.  :)

This is the one and only portion of this synthesis that I do not understand as far as I can tell. I struggled a bit trying to figure out how the hell they created the bicyclo/norbornene compound with a carboxyl and a formyl off the same carbon, but the answer was obvious after I got stuck and admitted that I didnt have the skill to synthesize bicyclics in my head: The opposite of what happens next: the retro-diels-alder that extrudes cyclopentadiene. Obviously reacting methyl 2-formylacrylate with cyclopentadiene would form this compound, as it is the opposite of the "retro reaction" that produces the equivalent to this.

The only other mystery is how they prepared the 4-bromomethyl derivative of indole to make the ylide for the Wittig reaction, but Im sure I'll find out all in due time. I wonder about simply bromomethylating indole, but I doubt its this simple, and I cant predict where the bromomethyl would add. Arg. All in due time all in due time.......

I can assure you that I have searched google for a variety of things related to these compounds and for Oppolzers synthesis as well, Google doesnt know much about it. I should try Sirus next.

Please help with the Mannich part. Thanks again.  :-[



  • Guest
Mannich + SN2
« Reply #1 on: June 04, 2002, 06:56:00 PM »

In the first step, the (substituted) indole undergoes the Mannich reaction to form a 3-(dialkylaminomethyl)-indole (if R=Me the product is gramine) and this is a very standard reaction. The 3-positioned indole hydrogen is definitely active.

In acid solution, the formed mannich base is protonated, and is thus a mannich salt. The mannich salt can undergo SN2 substitution with nucleophiles, expelling R2NH and adding the nucleophile of choice - in this case nitromethane, forming 3-(2-nitroethyl)-indole.

The last step I'm not sure of, but I guess the TiCl3 reduction is performed in methanol, and that the intermediate oxime is somehow etherified with the solvent in the process. I suppose more on this can be looked up in the actual reference.

Halomethylation of indole would preferentially add the substituent in the 3-position, followed by the 5-position. Practically none in the 4-position. The starting material has to be prepared another way.


  • Guest
4% over 11 steps
« Reply #2 on: June 07, 2002, 11:12:00 PM »
Today I found some time to look up that article in my 'Lysergic Acid Total Syntheses' folder.
It's: W. Oppolzer, E. Francotte, K. Bättig; Helv. Chim. Acta 64, 478 1981.

Do you still have questions? Rhodium pretty much answered everything. He is also right about the TiCl3 reduction of the sodium nitronate, it's done with TiCl 3/NH4OAc in MeOH/H2O 3:1 at 20° for 2 h.

The overall yield from 4-hydroxymethyl-1-tosyl-indole is 4% over 11 steps  ::) . But first you have to prepare that strange  starting material...


  • Guest
Starting Material....
« Reply #3 on: June 08, 2002, 05:16:00 PM »
Thanks Rhodium and Lili.  :)

I wonder why he went that route? I look at it now and see an alternate possibility that I'd like to bring up:

At the stage after the Wittig reaction with the bicyclic, but before the Mannich reaction, where one has a 3-unsubstituted indole still, why not chloromethylate the 3-position?

After chloromethylation, react with sodium cyanide, to give the indole-acetonitrile. Then use the Stephen Aldehyde Synthesis (

) to change the nitrile to the aldehyde. Condense with methylamine and remove the water to have the imine.

Now you have the exact same compound they used for the retro-diels-alder, minus the oxygen between the nitrogen and methyl groups.  :)  Eliminate the cyclopentadiene, and Diels-Alder to the lysergo-skeleton cousin we are familiar with, but minus that oxygen.  :)

I like this series better, because it involves simpler reagents than ones like TiCl3 and disubstituted formamides, etc. But of course a rebuttal to that statement could be, "If simple acquisition of chemicals is an issue at this point, you have no chance of succeeding in this anyway."

Well, I agree, but it doesn't change the fact that simpler is better.  :)

You still need the bicyclic for these steps, to protect the double bond and prevent jumping the gun on the diels-alder, as it could occur as soon as the nitrile is formed, and the aldehyde as well.

Another idea, just do the diels'alder on the indole-acetonitrile? Why not? It would cyclize in the traditional manner, but giving the cyclic imine, with no function off the nitrogen. Could the imine be reduced without reducing the olefin as well? Arent imines a lot easier to reduce than C=C olefins?

The worst case scenario would be cyclization at the aldehyde stage, giving an inert ether.

If one chloromethylated either of the following compounds, would the chloromethyl function go into the "correct" 4-position of the indole ring (once the compound is turned into indole that is) ?

1.o-toluidine (or the N-protected toluidine) --> 1-chloromethyl-2-methyl-3-aminobenzene ??? (yes or not likely.)

2.o-nitrotoluene --> 1-chloromethyl-2-methyl-3-nitrobenzene. ?

3.2-amino-phenylethanol --> 1-chloromethyl-3-amino-phenyl-2-ethanol (bad IUPAC I know) ??

The first two compounds can of these compounds can of course be used to synthesize indole rings.

Also, (sorry for so many things in one post) imagine an indole ring, but as an imine. The double bond on the indole ring isnt between the 2 and 3 carbons, but at the 1-nitrogen and the 2-carbon. Would this automatically aromatize to indole? (Indole would be more stable, so I would think so.) I am just curious if 2-amino-phenylacetaldehyde --> indole + H2O.

Thanks for everything so far.  :)

Lili, you actually have a "Lysergic Acid Total Synthesis" folder? That's funny.  ;D  :)



  • Guest
The 3-position of indole is so reactive I believe ...
« Reply #4 on: June 08, 2002, 08:33:00 PM »
The 3-position of indole is so reactive I believe that attempted chloromethylation of it would generate the di-3-indolyl-methane.