Total Synthesis Strychnine - Step 2

Alright, so far we are only 1 step into the synthesis of strychnine.  The next step involves the addition of an amine group to the beta position of the indole ring.  As I mentioned before in the previous post, the alpha position is blocked by a substituent.  Essentially, this carbon is conjugated to the benzene ring in the substituent diffusing the charge, while also providing steric bulk.  The main reason for the shielding of this carbon from an electrophile is likely to be due to sterics.

Mannich Reaction

At first glance I had a difficult time determining how this reaction might actually work.  Dimethylamine is a nucleophile, as is the indole ring.  They key here is the formaldehyde which is a potent electrophile which acts to bridge the two nucleophiles together.  Essentially, the formaldehyde and dimethylamine are activated to form a protonated imine (the term Schiff base is also used in place of imine, they are really the same thing) which is the same as an iminium ion.  The iminium ion is a very potent electrophile and reacts readily with a nucleophile giving the product.  This reaction is variant of the Mannich Reaction with indole as the nucleophile rather than the typical enol tautomer of a ketone.  This reaction can be utilized with most nucleophiles, and also electron rich heterocycles.  Notice that this is essentially an electrophilic aromatic substitution reaction on the C-3 of the indole.  While in intro organic everyone learned that these occur on benzene rings, the indole ring is much much much much more reactive and will react in presence of other aromatics (even the activated di-methoxy benzene).

The Mannich reaction proceeds first by formation of the imminium ion (or protonated schiff base).  In Woodward's synthesis, a mixture of the two in a cooled aqueous solution likely provides an equilibrium mixture of the imminium ion, protonated amine, free amine, formaldehyde, the diol of formaldehyde (interestingly enough this diol of formaldehyde known as methanediol is actually in greater amount than the free formaldehyde), as well as likely some condensation products with methanol as well because methanol is often present in aqueous formaldehyde solutions to prevent oxidation to formic acid.  From here, the starting material is added in a dioxane solution with acetic acid (remember that our starting material was insoluble in water, so dioxane is here just to provide solubility.  It is likely that THF could be used as well).  The reaction takes place rather quickly due to the potent electrophile that the imminium ion is.  Acetic acid is used as the acid as it provides a weak-nucleophilic proton source.  I believe that in using HCl, chlorine could act as a nucleophile and destroy the imminium ion?  I am sure that sulfuric or phosphoric acid would work as well, but I cannot be 100% positive.

As an aside I always find it odd how people often overlook the importance the choice of the acid in a reaction.  Things to be aware of are the concentration, oxidizability, and salts that will be formed with species such as amines (as we will see next).  While 1M HCl might not seem that strong of an acid, keep in mind that it is really a pH = 0 solution (only about 3 mL of conc. HCl in 30 mL water).  So even though the HCl is diluted relative to the concentrated stock, 1M HCl is still a very acidic solution and can really mess with your compound if you are not careful.

Workup

After the reaction is complete, our product is in solution, but, like all amines in either neutral or acidic conditions, we have the acetate salt of the amine (perhaps this is why other acids are not used?).  Believe it or not, this one charged group can actually make the entire molecule soluble in water.  With larger molecules there is a balancing effect with the hydrophilicity imparted by the charge, and the hydrophobicity of the rest of the molecule.  However, knowing that the salt of the amine in acidic solution should be soluble, Woodward noticed that there was some insoluble material present in the solution, and filtered this off.  Woodward attributed this solid to the reaction of 2 equivalents of starting material with formaldehyde to from the molecule below.

Notice that this molecule contains no protonatable amines and so should not be soluble in the aqueous solution.  But wait! Wasn't there dioxane in the solution that should solublize this?  Well, the aqueous solution was diluted with nearly 2 volumes of water to essentially dilute the dioxane.  It was this dilution that precipitated this product, as the solution was clear before this.  Once this solid was removed, it is just a matter of making the aqueous solution basic with a cold solution of KOH to deprotonate the amine and precipitate the product as crystals.  Because neutralization of acid is exothermic this is why cold KOH is used, however, if I were performing it, I would add slowly and stir, as the large volume of water will suck up the heat evolved in neutralization.

Looking Forward

Wow, I can't believe how much there is to write just about one reaction.  It just goes to show how complex chemistry can actually be, and how many little things go unnoticed by the inattentive eye.  The next step of the reaction is a very basic and common practice reaction, in which methyl iodide is used to  methylate the amine and activate it as a good leaving group for another nucleophile.  The goal of these and following steps is to form one of the ring of strychnine with control of the stereochemistry, as we will soon see.

Thanks again, if you find this interesting please comment.  Or if you find other interesting observations I would be happy to hear from you.  Thanks again!