Total Synthesis Strychnine - Step 11

Let's move forward from our previous step, which I'm sure we'll need a refresher on being as it's been quite some time since my last post.  In our last step we finally got rid of our veratryl functional group, and turned it into a useful functional group, a carboxylic acid (rather, two carboxylic acids).  The veratryl group has been a useful group thus far, but through its oxidation, we have a group which can now condense with a nucleophile (in this case a nitrogen) to form ring III of our strychnine molecule.  How will the feat be achieved? Well, in order to condense one of the carboxylic acids with the nitrogen highlighted below, we will first need to remove the acetyl group on the nitrogen because this is rendering the nitrogen non-nucleophilic through resonance effects with the carbonyl, as well as sterics (a tertiary amine is a relatively poor nucleophile).

Overview: One of the brilliances of this part of the reaction is the selectivity for the reaction of the highlighted carbon above.  After all, why couldn't the nitrogen (once freed from the debilitating acetyl group) react with the other methyl ester?  The reason is, they simply can't reach each other!  Shown in the schemes below.  This is due to the slight asymmetry of the carbon attached to the indole ring and the double bond.  Because we previously had a six membered ring with two ortho groups on it, once they became cleaved, we formed this asymmetry.  Because of this we have one bond which can freely rotate (Scheme 1) and the other is a double bond which is rigid in the trans formation (Scheme 2).  This leads to only one possible conformation.

It's also interesting to note that the ozonolysis preserved the remaining double bonds in the molecule, had this somehow been compromised (I'm not sure how this would happen, but it's just good to note) then this step may not proceed forward very well.  Last observation I have is that the ring that will be formed will be a 6-membered ring, which are very easy to make.  Rings larger than this (even 7 membered rings) are not impossible to make, but are exceedingly less likely than a 6-membered ring formation.  All of these factors favor this reaction proceeding forward alongside with the isomerization which will be discussed later in the post.

Mechanism: The mechanism for this is reasonable simple, and if you are reading this, you may already identify it as an acid catalyzed condensation.  Essentially, the highlighted amine will condense with its only other partner, the highlighted carbon of the methyl ester.  I suppose this would be called an amidation.  I won't bother with drawing out the mechanism since it is the same as those of any other acid catalyzed addition to an ester, with the amine acting as the nucleophile, however I will post this link to the acid catalyzed hydrolysis of an ester.  The mechanism is identical, however, instead of water attacking the activated ester, instead the nitrogen will act as this nucleophile, and the remaining steps are the same.  To wrap the mechanism up I would like to say that you can't do this reaction in an aqueous solution.  If you did, you would instead hydrolyze the ester to the carboxylic acid, which is extremely un-reactive and this will destroy your reaction.  Instead, Woodward conducted this in a methanolic solution of hydrogen chloride.  This is prepared by bubbling HCl gas through methanol, rather than water which is the typical solvent for HCl.  Using an excess of methanol also ensures that even if water were introduced into the system and some carboxylic acids were formed, the excess methanol would react in a fischer esterfication to reproduce the methyl ester which can then react with the amide.  This preserves the integrity of the reaction.  One other question I can see being brought up, is why then is the amide also not cleaved with the large excess of methanol back to the starting material?  The reason this does not occur is for multiple reasons.  The first, is that the nitrogen is found very close to the reacting ester.  Thus, the "effective concentration" of the amine is higher than it normally would be, due to proximity effects.  On average, because the nitrogen is closer, it "sees" it more often, and is more likely to react with it.  The second reason is due to the rearrangement of the double bond which I will discuss next.

Moving from the intermediate in parenthesis in the first figure to the final product, we see that the double bond has rearranged.  Because we are in acidic conditions, once we protonate our alkene (as if we were starting to perform an acid catalyzed hydrolysis of a double bond), we form a resonance structure which can rearrange to an aromatically stabilized ring, which is significantly more stable than the initial arrangement.  This drives the reaction forward toward products creating a thermodynamic sink (a low energy product which pulls reactants toward products), while also making the reverse direction, that is, the methanolysis of the amide much less likely.  Also, the interconversion of alpha-beta unsaturated ketones with their beta-gamma counterparts under acidic conditions is also well-known and studied (although I didn't know if it before seeing this).

While again it may seem odd that this above reaction happens, after all, why don't the other double bonds undergo similar reactions during other stages in the reaction?  Well, most likely they do, however, remember that alkenes are not as nucleophilic as you think.  However, this one is also conjugated to two other double bonds, one from the carbonyl and one from the adjacent alkene.  This makes this double bond even less nucleophilic as it is very delocalized over these bonds, imparting some double bond character to the single bonds connecting this chain of double bonds.  However, because carbonyl compounds are most active at their alpha carbon (due to the acidity of this proton) we would expect that proton to be ever so slightly more nucleophilic (although still a poor nucleophile as the charge is very diffuse).  Essentially, once this carbon does graph a proton, it places a positive charge on a tertiary carbon which is conjugated to another double bond forming an allylic stabilized carbocation.  It is likely this resonance stabilization, as well as the low final thermodynamic free energy obtained through aromatization of the pyridone ring.  Essentially, with the small fraction of the double bonds that do react due to the low nucleophilicity of the double bond, they will quickly form products through the large stabilization.

Workup:

This reaction was performed in methanolic hydrogen chloride.  Just because methanol can't dissolve HCl as well, doesn't mean this isn't a strong acid.  Actually methanol is pretty decent at dissolving salts, although not nearly as good as water.  This reaction proceeds toward products over a period of 10 hours.  This is a longer reaction because we are adjusting a lot of parameters to effectively shift our equilibrium toward a more ordered product in the presence of a solvent which only want to break the molecule apart into smaller fragments.  It takes a while for the reaction to finally reach completion not only for a condensation reaction like a Fischer esterification, but also for the rearrangement of the double bond.  Once the reaction is complete however, evaporation of the methanolic HCl and redissolving in methanol and a small amount of chloroform and freezing precipitated crystals as colorless plates in 75% yield.

One thing people often do not think about during reactions is the solubility of their molecule in the desired solvent.  I doubt that the starting material is very soluble in methanol, and we know that the product is not very soluble in methanol because Woodward later recrystallizes this product from methanol for an analytically pure sample.  The hot temperatures of the reflux conditions is what drives the starting material to dissolve in the methanol.  This is also why a small amount (10%) of chloroform was added to the solution, as methanol may not have dissolved everything.  Chloroform and Dichlormethane are notoriously well known for their ability to dissolve most everything, hence chloroform is used (nowadays DCM is more common as it is less harmful than chloroform, however, DCM is still not something I would like to inhale more than I already do, as it is carcinogenic itself).

As a last though to boil your noodle, why didn't the ethyl ester of this molecule react and turn to the methyl ester during the reflux?  It's not a typo I made.  For some reason this ester is just stubborn to transesterfication.  My hypothesis...Woodward didn't care about characterizing it as it will disappear later, and it may actually be the methyl ester.  Just my opinion.