Total Synthesis Strychnine - Step 12b

So to return from were we were with the last post, now I will focus on the actual chemistry of the reduction, rather than the rational.  Again, here is the current step (Scheme 1)
we are going to accomplish, and we need to remove the tosylate group because it is causing problems with side reactions leading to undesirable products.

Scheme 1. Reduction of tosylate group in addition to hydrolysis of esters to carboxylic acids.

Overview:
As you can imagine, a N-toluenesulfonyl group (derivative of tosyl) is quite stable.  It is a sulfonamide of the tosyl group forming the structure seen in Scheme 2.  According to Greene's Protecting Groups (An excellent text concerning the types of protecting groups, for functional groups, as well as how to add and remove them, an organic chemists must have), the N-toluenesulfonyl protecting group is really only unstable to extremely acidic conditions (pH < 1, 100˚C), strong reducing conditions (Hydrogen gas, Raney Ni), and strong organometallic reagents (grignard, organolithium).  The reducing conditions will cleave the sulfonamide into from what I can only assume will be a sulfinic acid and the free amine.

Scheme 2. Proposed products of reduction of sulfonamide group forming a sulfinic acid and secondary amine.

As we will see, the hydriodic acid and red phosphorus will react to produce hydrogen gas under very hot conditions to produce a reducing environment that is rather strong.  Under these conditions, we will also see that the esters are cleaved into the acids, and as well the 2-pyridone rings remains stable.  The esters will be cleaved for different reasons than the reducing environment.  These will be cleaved due to the acidic environment, essentially acid catalyzed hydrolysis (remember, these are also extremely acidic conditions with pH <1, and near 100˚C).  The 2-pyridone amide does not cleave because is has a resonance structure that makes is more aromatic in nature, which makes it very stable under these acidic conditions which should hydrolyze the amide (reduction of the amide group is not as common).  One could also envisage using Raney Nickel to reduce the sulfonamide, however, the Raney Nickel would also reduce all of the esters to primary alcohols, and would also reduce all the aromatic groups to hydrocarbons.  A great choice of reducing agent seems to be red phosphorus and hydriodic acid.

Lastly, just some facts about red phosphorus and hydriodic acid.  Firstly, Red Phosphorus is just an allotrope or different crystalline arrangement of elemental phosphorus.  There are actually several allotropes (white phosphorus, red phosphorus, violet phosphorus, and black phosphorus) which are all named after the color in which they appear.  White phosphorus is very reactive and unstable under atmospheric conditions and can spontaneously ignite when reacting with oxygen in the air (check out this video).  Remarkably, this white phosphorus does not react with water, and is actually stored under water to prevent its reaction with oxygen.  Red phosphorus however, is a much more stable allotrope of carbon commonly found on the ends of match sticks.  It is essentially a polymerized form of white phosphorus, and us thus amorphous.

Hydriodic acid is an extremely strong acid (pKa = -9).  It is also a mild reducing agent and in the presence of the oxygen in the air, it can spontaneously oxidize to molecular iodine, which immediately reacts with hydriodic acid to form hydrogentrioidide which is a brown color.  Thus, according to wikipedia, hydriodic acid solutions turn brown over time (I have personally never worked with HI(aq.) so I have not seen the phenomenon).

As a preview for the oxidation reduction chemistry occurring later in this post I would like to introduce some standard reduction potential reactions here.  If it's been a while since you last covered this, consult an analytical or general chemistry textbook to refresh.  Essentially what is important to note, is that the reduction potential that is more positive is more likely to become reduced, and one that is more negative is morel likely to become oxidized.  The difference between these two is related to the free energy of the reaction and thus the amount of energy released.  Here's a simple example of the oxidation of hydriodic acid by oxygen (I just copy pasted these from the wiki page).

O₂(g) + 4 H⁺ + 4 e⁻  2 H₂O     E = + 1.229
I₂(s) + 2 e⁻  2 I^{−                    E = + 0.54}

We can see that because the reduction potential of the oxygen is higher, it is more likely to become reduced (accept electrons from the iodide ion).  Thus the actual reaction taking place looks something like this.

4 I^-+ O₂ + 4H⁺  2 H2O + 2 I2

Thus this reaction is thermodynamically favorable, however, the kinetics or speed of the reaction are unknown.  Likely this takes place slowly.

The Chemistry:

Now for the actual chemistry that will take place for this reaction.  It's actually quite complex, but I will try to be concise yet detailed.  The active species doing the reducing will be the hydrogen gas produced, along with the hydriodic acid itself.  The role of the red phosphorus is to essentially maintain the reducing environment and enhance it.  Reductions with hydriodic acid alone can reduce double bonds under reflux conditions, however, other groups may not be reduced under just these conditions.  Addition of the red phosphorus helps to regenerate the reductive power of the HI, as well as provides a more favorable environment for reducing.  It acts as a catalyst to speed up the reaction as well as to alter the energy terrain of the reaction, lowering the energy required to reduce other more difficult to reduce chemical groups.  So let's take a look at the catalytic cycle (Scheme 3) occurring in aqueous solution.  Recall that the conditions are reflux at 120˚C (BP of HI soln.) in conc. HI with red phosphorus.

Scheme 3. Catalytic cycle for Red phosphorus/HI reaction.  Positive Ecell shows that reaction is favorable.  Reduction potentials gathered from Albouy, D., et. al. J. Organomet. Chem. (1997) 529, pp. 295-299.

From examination of the catalytic cycle, we can see that most importantly, at the top of Scheme 3, there exists an equilibrium between iodine and hydrogen gas at such high temperatures.  The hydrogen gas (and likely the high concentration of HI as well) acts as the reducing agent reducing the desired compound.  Once the hydrogen is used up, there exists molecular iodine in solution.  This is what reacts with the red phosphorus in the aqueous solution to regenerate the HI, while producing phosphinate.  We can track the oxidation states during this redox reaction and see that the iodine is reduced, while the phosphorus atom is oxidized.  The potentials at the bottom of the diagram show that once we subtract the reduction potentials (same as adding the oxidation potential to the reduction potential, which is what I did) we can see that the cell potential is positive, meaning that this is a spontaneous reactions with a negative ∆G.  I have left out other more complicated parts of the cycle however.  The phosphinate can be further reduced in the system to phosphorous acid, and finally to phosphoric acid, each time reducing its oxidation state further, in order to regenerate the hydroiodic acid.  These are covered in more details in this paper if you are interested, however, the general principles are the same.

The Workup:

This reaction was done under vigorous reflux for 3.5 hours, in a solution of 50/50 HI (47%)/Acetic acid, with 1/3 wt% of red phosphorus per compound mass (i.e. in this case, 750mg compound required 250mg red phosphorus).  Once the reaction cooled, the red phosphorus could simply be filtered off.  It is to note that usually filtering iodine solutions tends to oxidize the filter paper turning it a nasty brown, however, these are strongly reducing conditions, so all of the iodine is present as the iodide anion, so no need to worry about any oxidation at room temperatures.  Once the phosphorus is filtered off we have an acidic solution which was evaporated under vacuum.  The residue remaining was dissolved in acetic acid and pumped off a few more times to allow crystallization to set in.  These crystals were then suspended in acetone and stirred to remove any byproducts soluble in acetone (this is known as trituration), followed by filtering to give the final product.

In the next post, we will cover both the acetylation reaction (as we have covered this basic reaction before in Step 9), along with the next step that is the use of diazomethane to regenerate the esters that we destroyed in this step.  Hope you guys enjoyed the post!

Total Synthesis Strychnine - Step 12a

This next step in the total synthesis of Strychnine is a complex yet interesting one indeed.  I was not so familiar with the reaction myself until coming across it, so I took some time to learn as much as I could about it to make this post worth the read.  Below is where we are at in the synthetic scheme (Scheme 1).

Scheme 1. Reaction we will cover today involving reduction of the tosylate group, as well as the esters.

The whole point of this reaction is to remove the tosylate group, but why?  Well, even the great mind of Woodward can't predict everything that can happen during a synthesis.  Actually, this step was not part of the original synthetic scheme, but rather, needed to be included because the tosylate group was interfering with their original synthetic plan.  So let's take a moment to look at what Woodward was really trying to do in Scheme 2 below.

Scheme 2. Original reaction Woodward hoped would happen, but in actuality, this reaction does not occur.  Rather, Scheme 3 occurs.

Woodward desired to perform a Dieckman Condensation, which is essentially just an aldol condensation modified for the enolate of an ester, condensing with another ester.  A strong base such as sodium methoxide is used to deprotonate the enolate (actually, both alpha carbons of the esters can be deprotonated since they are both acidic, however, the less hindered carbon is more likely to be the attacking nucleophile which is shown in Scheme 2).  However elegant this reaction appears to be, actually it does not happen.  The reaction that does occur under the reflux with sodium ethoxide is quite unexpected, but understandable.  Let's take a look at the reaction that actually occurred in Scheme 3.

Scheme 3. One possible mechanism explaining the final confirmed molecule.  Other mechanisms exist, this is one of two proposed originally in the paper.

Wow! What a crazy mechanism, no wonder Woodward didn't predict this.  To verbally explain this mechanism, the key thing is to note we are under very very strong basic conditions (sodium methoxide under reflux, 65˚C).  The pKa of methanol is 15.5, which is over 100X stronger than sodium hydroxide (pKa = 13).  Under these conditions, lots of things can go wrong if you are not careful (obviously).  The first step is deprotonation of the acidic alpha hydrogens, followed by leaving of the stable tosylate anion.  Once this step occurs, it is likely irreversible as the tosyl anion is very stable (lots of resonance structures).  After this, a base induced rearrangement of the alpha-beta unsaturated ester to the beta-gamma isomer (yes there is such a thing!  I had no idea until I saw this however.)  Another deprotonation of the now moderately acidic aldimine leads to cleavage of the C-C bond.  Normally aldimines are not very acidic at the alpha hydrogen, however, because deprotonation here leads to a structure with a high amount of resonance (conjugation through the double bond into the ethyl ester), this can actually have a pKa on the order of 22 (source).  Now I know that a jump between pKa of 15.5 and 22 seems like a large jump, but higher temperatures may help bridge this gap.  Also, aldol condensations typically work with large differences between base strength and pKa.  This is an example of thermodynamic control, where once deprotonated, the enolate reacts quickly to form a stable product, pushing the reaction forward.  This is realized during the last step where formation of the enolate of the methyl ester attacks the newly formed aldimine (which reacts in a similar fashion to a ketone).  Proton transfers catalyzed by the basic conditions pushes the product toward the aromatic product which is the final product.  The large stability gained by this product is what pushes the reaction forward through the last step, elimination of the nitrogen anion which is rather unfavorable.

All of this is what we didn't want to happen.  So to get back to the main point of the post, why do we want to remove this tosylate group with the hydroiodic acid and red phosphorus?  Well, its the leaving of this group that started the whole cascade of reactions that led to the undesired product.  Conditions for removing the tosylate group however, without also hydrolyzing the esters are not realizable, and so hydrolysis of these esters are collateral damage.  We will see in ensuing steps that these esters are reformed to prevent undesired reactions from occurring.

Well despite the whole point of the post, I've been writing this for almost 2 hours now, and I have other things that require my attention.  I hope to cover the more interesting part of the reaction this weekend, which is the actual reduction of the tosylate group.  Stay tuned.

An Interesting Period in Lab

I wanted to break up some of the monotony of the total synthesis to present a conundrum in my current research.  As a brief background, I am currently a first year graduate student at UIUC doing research related to drug delivery.  For a part of my project I want to initiate a polymerization of poly(trimethylene carbonate) with an anticancer drug, camptothecin, so that one of the end groups is this compound.  It has been previously reported that this molecule can undergo such a polymerization in the following paper.  However, my molecule I want to initiate is different.  My synthetic scheme is shown below.

Briefly, the mechanism for this reaction (if you are unfamiliar with polymer chemistry) is actually quite simple.  It all revolves around this zinc complex shown above which is named BDI-II.  The zinc metal is the coordinated metal in the center which is the active catalytic center for the polymerization.  Most polymerizations of this monomer (trimethylene carbonate or TMC) are done at high temperatures in excess of 120˚C.  With the zinc catalyst however, room temperature can start the polymerization (as I can see the solution become more viscous).  However, usually a temperature of 50-60˚C is sufficient with the polymerization being done in under 1 hour.  As I was saying, the mechanism is a coordination-insertion mechanism.  When performing this reaction, my drug molecule is mixed with the catalyst for about 20-30 minutes to form a complex which will initiate the polymerization.  The ligands on the catalyst help to control stereochemistry (none on my polymer) and can selectively activate a less hindered alcohol group over other more hindered alcohols and even amines.  The zinc-drug complex then also coordinates with the carbonyl of the TMC monomer and catalyzes a nucleophilic addition of the alcohol from the drug to the monomer.  In this way, the propagation continues.

The Problem: So here's what I really wanted to share, the rest was just background.  I have done this exact same polymerization before with pyrenemethanol acting as the initiator, and my NMR shows that I have incorporation.  However, when I attempted polymerization with this drug derivative that I made, I could get very little incorporation from the NMR.  The peaks from my drug are almost non-existent.  A further study I did actually showed that my drug disappeared with longer reaction times!  So the longer I ran the reaction, the less drug that was incorporated into my polymer.  This struck me as so strange, since the camptothecin molecule and pyrenemethanol had been shown to undergo polymerization perfectly fine.  Other tests I did to try and understand what happened to my polymer involved MALDI-TOF MS, however, my polymer is difficult to ionize, and some of the data I obtained was very confusing and sometimes even contradictory.  I basically came to the conclusion that certain polymer species were ionizing better than others.  I would be able to solve this problem a little easier if we had a working Gel Permeation Chromatography (GPC) system, however we use a DMF solvent in ours, and the refractive index detector has a difficult time differentiating between my polymer and the DMF because the refractive indexes of them are very close (this leads to a low dn/dc value which is used to calculate molecular weights of polymers).

So anyways, I've been stuck on this for a while until I started to think about it a little more.  What was happening and why was my polymerization not working, and also why was my drug disappearing?  I want to show you two pictures below before I break the answer to you.

After mixing catalyst and drug

After terminating polymerization with 1 drop water

These are images during my polymerization which I must perform with no water present so that I do not destroy my catalyst (water sensitive) or initiate my polymerization with water as the nucleophile instead.  As you can see the picture on the left shows a bright orange and clear solution.  My catalyst is a clear while crystalline solid, and my drug derivative is a fluffy slightly yellow crystalline powder.  This orange color is what I see after mixing these both in THF and letting it stir for 30 minutes (the transition takes around 5-10 minutes to appear).  The initiation with pyrenemethanol showed no color change (only a very light yellow clear solution).  On the other picture after I add 1 drop of water to terminate the polymerization, you can see that now the color is a clear yellow (this color change occurs in 1-2 seconds).  So I began to wonder why I was seeing this very intense color in my polymerization.

The Answer: So after toiling and tinkering and thinking, I came to this conclusion, which I am currently testing.  I believe that I am forming a chelate of my drug with the zinc complex!  How is this so?  Well, the only difference between my molecule and the camptothecin drug that worked fine in a previous paper is my linker that is highlighted below.  With this linker, I believe it is now possible for many of the carbonyl groups and alcohol group of the drug to form a chelate (shown below).

Now I know the picture doesn't look pretty, but remember that this is a 3D molecule.  So my hypothesis explained is this.  The zinc metal forms a very strong chelate with this drug, and essentially kicks of the BDI-II catalyst shown earlier (mixing an organic zinc and my compound also gave an identical clear orange color).  In this way, the chelate is so strong that my compound is essentially rendered unactivated as there is little/no room for my monomer to coordinate with the zinc to initiate the polymerization.  This explains why I have very little drug initiating my polymer, but it does not explain why as I increase polymerization time, my drug initiation efficiency becomes less.  My reasoning for this is that if zinc can catalyze a forward addition, certainly it can also catalyze the reverse (definition of a catalyst).  Because of the strong complex that the zinc forms, it can re-chelate with my drug and essentially rip it off the polymer!  This is an amazing example of thermodynamic control.  As my reaction time increases, the zinc (and my drug) and more happy together achieving a lower energy state.  This hypothesis explains everything down to even the color change too though!  for instance, the orange color change comes from the enhanced coordination of the pyridone carbonyl above.  This enhances the resonance structure with the amine forming a more aromatic structure (sound familiar to this post?).  What this essentially does is enhance the aromaticity of the entire molecule.  When this happens (when something becomes more delocalized) the absorbance and emission spectrums do what is called a red-shift (or bathochromic shift).  This means that everything shifts to a longer wavelength in the photo properties of the molecule.  A very well done explanation of this topic is shown on this site(look at section 4).  This shifts my yellow compound toward an orange color.  After I add my water, the zinc forms very strong coordination complexes with water and kicks off my molecule, returning back to a yellow color.

Lessons learned?  Everything has an explanation no matter how "crazy" things in lab can seem.  I always like to troubleshoot things like these.  It feels really good to figure something out, however, I hate the pressure for results in graduate school which makes expeditions such as this seem like nuisances.  So what am I doing now to prove this complex?  I could crystallize some of this compound and do X-Ray Diffraction and solve the crystal structure.  However, that would be going really above and beyond, wasting lots of time getting trained on the instrument, and the final result would not really be something that is publishable.  However, what I can do is a 2-D NMR technique known as NOEsy.  This essentially can tell me what protons couple to one another through space.  This is contrary to what is taught in undergraduate organic where protons can only couple through bonds.  This technique can show me if my linker arm which should initiate the polymer, is close to the protons near the lactone.  If I see this resonance, it is likely that the pyridone carbonyl is involved and would be indirect proof of my complex.  If I don't see this coupling of these two protons on the linker and the lactone, its not to say that I don't have the complex, but I can't really go much further without wasting time and resources.

Hope you found this interesting.

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.