Total Synthesis Strychnine - Steps 18 & 19

Our next steps in this synthesis involve the de-functionalization of the double bond we had just created.  At this point, it may be well to compare where we are now, to where we want to be.

Figure 1. We can see that several double bonds must be removed to obtain the final structure

Our first step in removing the functional groups that will no longer serve us include removing the vinyl sulfide that we had just added, followed by removing the double bond it is attached to.  In order to do this, Woodward had utilized Raney nickel, followed by palladium on carbon to perform each step, respectively.  The following scheme illustrates the goal.

Figure 2. Following 2 reductions with deactivated Raney nickel, and palladium on carbon.

Overview:
Raney Nickel
The step with Raney nickel I found particularly interesting, as I was not aware that this finely divided nickel over aluminum can de-sulfurize compounds.  This lends it as an exceptionally useful organic reagent.  The summed up version of what Raney Nickel actually is is described here briefly. If you are really curious more about Raney Nickel, I recommend (of course) the wiki page which can be found here.  However, in short, Raney Nickel is a nickel aluminum alloy with a small amount of a third metal to enhance its activity.  It is ground into a small powder, and then activated with strong NaOH to etch away some of the aluminum to create hydrogen gas, which becomes absorbed onto the surface.  This creates a very strong reducing agent (recall that raney nickel and hydrogen can reduce benzene to cylohexane!).  It is for this reason, that deactivated Raney nickel is used, otherwise, we would also reduce our bezene ring, as well with our pyridone ring.  To avoid this, Woodward deactivates it by boiling in acetone, followed by ethyl acetate, each for 3 hours.  This significantly reduces the activity of the catalyst through reduction of both the acetone and ethyl acetate, depriving the catalyst of its rich hydrogen environment in turn for an environment with a lower concentration of hydrogen.  Raney nickel is usually supplied as a slurry in water to prevent its reduction of oxygen in air which will also deactivate it (as well a potentially cause it to ignite!).

Palladium on Carbon
Palladium on carbon seems to me to be the bread and butter for the reduction of simple double bonds as well a benzyl ethers and benzyl esters.  Palladium on carbon is palladium metal with a high surface area due to its dispersion onto porous carbon (same thing as activated carbon).  We can see from the Figure 2 that simple reductions take place very quickly to yield the desired product.  From my experience with using this chemical (only 1 reaction) the product can "stick" to the palladium in the small pores.  The necessitates the use of often excessive rinsing with hot solvent to remove as much of the material as possible.  This is the case in Woodwards case as well it seems as he washes the palladium with lots of solvent.

Mechanism:
The mechanism for the reductions is actually very complicated.  Unless you are getting your Ph.D. studying hydrogenation, it usually suffices to have the following general scheme in your head.  It's accuracy may not be valid, but the mechanism is actually quite complex, and may involve radicals, which I am by no means capable of tackling.  So instead, I leave you with the following scheme.

Figure 3. Generic scheme for reduction by palladium or other metal catalyst.

In this scheme we see that the palladium (or could be nickel) adsorbs hydrogen atoms to its surface.  Transfer of these atoms in a syn addition to the same side of the double bond which is catalyzed by the environment at the surface of the metal releases the product.  In this case, it is important to notice that due to sterics, we only see the cis isomer formed (with the methyl ester coming towards us in Figure 2) and a small fraction of the trans isomer.  This is due to sterics of the approach of the catalyst surface.  We can see from the following figure that approach from the bottom is more favorable than approach from the top due to the 3D structure of the 5 member ring which hinders approach from the top of the molecule.

Figure 4. Pink atoms represent the alkene which will be hydrogenated. A.) Hindered approach from the top of the molecule (see arrow). B.) Less hindered approach from the bottom.

Workup:

The workup for these reactions are often quite simple.  Since the catalyst is suspended in the reaction solvent (usually an alcohol, as it is not reduced by the catalyst) it is simple to just filter it off either through a little bit of celite or silica.  The alcohol will pull the compound through the silica fast if it is a nonpolar compound so usually one doesn't have to worry about the compound getting retained in the silica.  However, one hazard one has to be aware of is the possibility of ignition of the catalyst.  Since there is still often a lot of hydrogen adsorbed onto the catalyst surface, and oxygen in the air, this can create a reaction that, due to the high surface area of the catalyst, can oxidize much of the hydrogen quicky, and release enough heat to start a fire.  For this reason it is necessary to make sure that the catalyst does not become dry during filtering, always keeping solvent above the top of the filtered catalyst.  Also, when done filtering, make sure to cover the catalyst with water, and properly dispose of the catalyst (or reuse it!).

For the first reaction with deactivated Raney nickel, after deactivation (as described earlier in the post), refluxing for 3h with the catalyst was sufficient to complete the reaction.  Often hydrogenations are quick and efficient due to the large excess of hydrogen, so it is likely that this step is very very efficient.  However, we see that when filtering the catalyst, Woodward washed the 500mg scale reaction with over 250mL of hot ethanol.  This is because often the product can get caught on the catalyst.  The poor yield (I believe) is actually due to the incomplete removal of product from the catalyst.  Rinsing with more solvent will likely just be wasteful rather than improving the yield much.  After concentrating and precipitating from a concentrated solution of acetone by diluting with ether, the product was yielded (no description of crude product).

The next step was performed in a similar manner to the previous.  Hydrogen transfer occurred quickly under 2 minutes at room temperature.  Afterwards, filtration and recrystallization from acetone/ether afforded clusters of colorless needles.

Next time we will look at how we can change the stereochemistry of the methyl ester which will be necessary for proving the structure we have created so far through comparison with breakdown products of strychnine.

Total Synthesis Strychnine - Steps 16 & 17

Now, after just completing the synthesis of another ring of srychnine through the Dieckmann Condensation, it is now time to perform a series of steps which remove the vinyl alcohol group.  This is necessary because there are quiet a few steps to accomplish this, and if we performed it at the end, we must worry about the reactivity of the enol group throughout the synthesis.  The rule in synthesis is, keep the number of reactive groups to a minimum and this eases the chemistry a lot.  So here I will take us through the next two steps which will aid in the removal of the double bond, as well as the vinyl alcohol group.

Scheme 1. Steps 16 and 17 involving the addition of tosyl chloride, followed by hetero michael addition of the alkene.

Overview:
The first step is rather straightforward.  Tosyl chloride is a very common reagent for installing a tosyl protecting group, which is more common than tosylic acid for this transformation.  The reason being that the acid chloride is more activated, thus less vigorous conditions are required.  THe only nucleophile available is the alcohol, as we have previously acetylated the secondary (now tertiary) nitrogen to prevent such side reactions from occurring.  The mechanism is similar to any addition to a carbonyl compound with expulsion of chloride ion.  Note that out favorite base for scavenging chloride ions from forming HCl, pyridine, is again used as the solvent.
The next step involves addition of sodium benzylthiolate to the α,β unsaturated methyl ester.  This is a type of michael addition known generally as a hetero Michael Addition, and more specifically as a sulfa Michael Addition.  One thing I would like to point out is that many people incorrectly use the term "Michael Addition" to refer to any atom attacking the alkene of the α,β unsaturated carbonyl compound.  This is incorrect!!  A Michael Addition or Michael Reaction only refers to a enolate (or a carbon nucleophile) which adds to this position.  The addition of other atoms to this position are related somewhat, but the terminology is different.  Any atom other than carbon which is added to the α,β unsaturated carbonyl compound is known as a hetero Michael Addition.  These can be categorized into the sulfa, aza, and oxa Michael Reactions for sulfur, nitrogen, and oxygen acting as the nuclephile, respectively.

The Chemistry:
One thing I would like to point out about the addition of tosyl chloride is the typical lack of propensity for enols to be reacted with electrophiles.  This is because typically the keto tautomer is preferred, thus the utility of treating the ketone with TsCl and forming the enol is perhaps limited (however silyl chlorides are good reagents for trapping the activated enol, See Mukaiyama Aldol).  The reason for the proclivity of this enol to react with TsCl is because of the electron withdrawing pyridone ring which greatly stabilizes the enol form, leading to great yields.

The next step I will focus on will be the sulfa Michael Addition.  The mechanism is the same as for any type of michael addition, however, the catch is that we have a tosylate group (a very stable leaving group due to lots of conjugation) which will be come displaced.  In some senses, this is kind of like a conjugated version of a transesterification, with the double bond acting as a mediator between the tosyl ester for the substitution by the sulfur group.  The mechanism is below.

Scheme 2. Mechanism for sulfa Michael Addition

One thing I find so astonishing is the preference for leaving of the tosylate ion over the methoxide ion. With a yield of 77%, it is possible that some methoxide may leave, however, the preference is obvious for the tosylate group.  The reason is the higher stability of the tosylate ion with its higher conjugation.  Elimination of an akoxide is always a higher energy process unless there is some thermodynamic sink that traps the product in a very stable state upon elimination of such a group.  In this reaction.  If the methoxide ion were to leave, there is no additional conjugation or other interaction that will stabilize it, and so we find that the tosylate is the preferred ion to leave.  One last thing to mention is the solvent for this addition, methanol!  Isn't methanol a nucleophile that could add to the double bond as well?  Well, yes, it is, and it would be labeled as an oxa Michael Addition, however, comparing the nuclephilicities between the thiolate anion and the alcohol, there is a very big gap in nucleophilicity.  According to the Linear Free Energy Relationships (LFER, a huge topic in chemistry if you haven't heard of it, take a peek in Physical Organic Chemistry text here) the thiolate has a Swain-Scott parameter of 9.92 vs methanol, which is 0.0.  This means that the thiolate is a better nucleophile by nearly 10 billion times!  This explains why methanol as the solvent does not affect the reaction.

Workup:
For the tosyl chloride step, reaction took place with pyridine as the solvent.  After standing for 10 hours, water was added and allowed to stir longer.  I believe this extra time stirring helps to get rid of any pyridine complexes that may have formed.  It greatly helps in its removal as I have tried this.  Then addition of 6M HCl helps to protonate the remaining free pyridine to its HCl salt.  Subsequent extraction of the mixture with chloroform.  Washing the combined organic layers with aq. potassium carbonate, followed by water and brine helps to remove any other salts from the reaction mixture, as well as any unreacted enol, as the potassium carbonate may deprotonate it and allow it to transfer to the aqueous layer.  Evaporation and recrystallization from acetone yielded the product as yellow crystals.

The second reaction used an excess of the thiolate to press the reaction forward.  Allowing these reactants to stir at room temperature under nitrogen (this is to protect the thiolate from moisture in the air, and possibly oxidation to disulfide?) for 3 hours resulted in a precipitate forming from the reaction mixture.  Collection of this precipitate yielded the pure product as yellow needle-like crystals.

Next we will explore the final steps remaining in the removal the vinyl sulfide we just added, as well as the removal of the alkene.

Total Synthesis Strychnine - Step 15

Our next step in this synthesis is to perform the Dieckmann Condensation that we wanted to do previously, however, those attempts were thwarted by that N-toluenesulfonyl group.  Now that we have taken the steps to get rid of that and protect other remaining groups on the molecule we can finally complete this reaction.  Below is the scheme.

Scheme 1.  Dieckmann Condensation of our compound to form the fifth ring of strychnine.

Mechanism:
Let's first look at the general mechanism of a Dieckmann Condensation and how it differs compared to say an aldol condensation, or Knoevenagal Condensation.  I will lay out the differences in a table format, and will not explain as the differences are quite obvious.  I will however, discuss the Dieckmann Condensation after.

_______________________________________________________

Dieckmann Condensation
Molecularity: Intramolecular
Functional group: Diester
Conditions: Base
Workup: Acid
Bond formed: Carbon-Carbon (single bond) - (double bond if enol form predominates)

Scheme 2. Dieckmann Condensation mechanism.

Knoevenagel Condesation
Molecularity: Intermolecular
Functional groups reacted: Ketone/Aldehyde with CH2X2, where X=electron withdrawing group
Conditions: Base
Workup: None
Bond formed: Carbon=Carbon (double bond)

Scheme 3. Knoevenagel Condensation mechanism between benzaldehyde and dinitromethane.

Aldol Condensation
Molecularity: Intermolecular
Function groups reacted: Ketone/Aldehyde with Ketone/Aldehyde
Conditions: Base or Acid
Workup: None
Bond formed: Carbon=Carbon (double bond) - Single bond formed if stopped at aldol addition (this is done by performing reaction at low temperatures such as -78˚C)

Scheme 4. Aldol Condensation between benzil and benzaldehyde

Claisen Condensation
Molecularity: Intermolecular
Function groups reacted: Ester with ester/ketone/aldehyde
Conditions: Base
Workup: Acid
Bond formed: Carbon-Carbon (single)
Mechanism: Similar to aldol, however, rather than leaving of OH, the alkoxide leaves.
_______________________________________________________
The Chemistry:
Ok, so that's a lot of information and I think it's helpful to differentiate between the reactions.  They are all very similar, it's really just how you mix and match the groups.  Let's go back to our Dieckmann Condensation and rationalize what happens.  A few things stand out to me.

1.) The molecule has a few basic protons which I see, it's interesting that only the one by the aromatic ring reacts (red protons below).  See Scheme 5 below.

Scheme 5. Two possible reaction pathways.

It is my belief that the red pathway occurs for three reasons.  They are likely most acidic because the pyridone ring is very electron withdrawing (inductively, not through resonance).  While this is the same case for the N-acetyl group, due to the aromatization of the pyridone group, this effect is great.  Second, there are two red protons and only 1 blue proton.  Kinetically speaking, the red ones have a better chance of being deprotonated and reaction.  Third, the blue proton is a tertiary carbon and is very hindered.  The approach of this carbon to form a quaternary center will very very hindered and the blue pathway is thus rather unlikely.

One extra interesting note is that in order for this Dieckmann Condensation to occur, the blue proton must become deprotonated at some point.  This is because in the current geometry, the methyl ester on the pyrrolidine ring is facing away from the red protons and is unaccessible for attack (it may take some careful looking to see this).  Actually, epimerization (steroechemical switching) at this center must occur in order for the reaction to proceed.  Thus we can see that this proton is accessible to the basic environment, but is a poor nucleophile due to the steric encumbrance of the tertiary center.  Once this proton is deprotonated, an equilibrium exists between both isomers at this carbon, and as the favored conformation reacts, it shifts the equilibrium forward driving the reaction nearly to completion.

2.) The enol form predominates when typically the ketone form is favored.

This is odd because ketone forms are very often more stable than the enol form.  The enol dominates due to the electron inductance provided by the pyridone ring.  This effect is so strong that during the reaction, actually the sodium salt of the enol was isolated.  The molecule could actually be deprotonated in sodium bicarbonate solution (pH~7-8) which is astounding for an alcohol.

Workup:
This reaction is almost like a chemist's dream.  There is really no work up here.  The product precipitates from the solution which also helps to drive the reaction forward, as the precipitation of product keeps the concentration of product in solution low, helping to shift equilibrium to the right.  After a brief 20 minute reflux, and allowing the solution to sit in the cold overnight, the light yellow crystals could be simply filtered (how great!).  And in 88% yield, I'm sure Woodward would be high-fiving everyone in lab.  I know I would after first overcoming the problem with the N-toluenesulfonyl group, then beautifully creating the fifth ring in a simple reaction which also elegantly sets up for the next step in another high yielding reaction which I will cover next time.  Genius.

Resources Every Chemist Should Have

A brief excursion to show you all some of the tools I think that every organic chemist (or chemist) should have.  I will list some textbooks, as well as online resources that are great references.

Textbooks:

1.) Organic Chemistry - Clayden (link)
This is a wonderful textbook that for an introductory organic chemistry book, is very detailed.  By far the best text out there.  Covers a large scope of reactions in sufficient detail so that one can gain a very in depth understanding.
2.) Advanced Organic Chemistry Parts A&B - Carey (Part A) (Part B)
These should be your second books that you read on your trip to understanding organic chemistry.  Part A is a more in depth look at mechanisms, Part B looks into learning a bunch of reactions.  To extend and diversify you reaction knowledge, these books are classics.
3.) Modern Physical Organic Chemistry - Anslyn (link)
This book solidifies your understanding of why reactions happen and give you the most in depth and fundamental understanding to all sorts of reactions.  It emphasizes applying thermodynamics and kinetics to all parts of organic chemistry and helps one to use these fundamentals to explain just about everything.  This book is my all time favorite.  Extremely well written, easy to follow, and it is not riddled with equations like most physical chemistry books are.
4.) March's Advanced Organic Chemistry - March (link)
This book is really more of a general reference more than something you would actually read though.  It's filled with so many references you wouldn't believe!  If you are performing research and are looking to learn more about a particular reaction, this is where you should go.
5.) Greene's Protective Groups in Organic Chemistry - Greene (link)
Another reference book and not something to read through.  But like the title says, if you want to protect something, this is where you go.  There are some great tables in the back that show you around 50 conditions that you could apply to your protecting group, and it will tell you if it is stable or not.  Excellent reference, and another classic.
6.) The Art of Writing Reasonable Organic Reaction Mechanisms - Grossman (link)
If you want to be able to predict mechanisms or brush up on your skills to diversify what mechanisms you can describe, this is a great book.  Like the title says, you will be able to better predict what electron pushing mechanisms are most reasonable to predict what products are most likely to form.

Online Resources:

1.) NMR Chemical Shifts (link)
What a great link this is!  Shows you a wide variety of proton chemical shifts for like a bazillion functional groups with many relevant examples of each.  A gold mine if you are curious about where a new functional group should appear.  Also, clicking the home page link in the upper left hand corner takes you to the home page where you can find even more resources!
2.) pKa Table (link) and for Heterocycles (link)
Here are the pKa's of a variety of different functional compounds from where you can predict how acidic something is.  This is the broadest table I have seen and it has been very useful so far.  The second link specializes for some common heterocycles.
3.) Bond Energies (link)
If you want to get a rough idea of how favorable your reaction is, a very simple way to rationalize things are with bond energies.  This is a table that shows you how strong bonds are.  If your products contain higher bond energies, they have stronger bonds, then they are thermodynamically stable, and likely to form.
4.) Elemental Properties (link)
Here you can find any physical property you want for you molecule.  Constants such as Young's modulus, enthalpy of formation, standard entropies, heat of fusion, etc.  You name it, this site has it.

*Note: The best way to find the property you want, is to click on the element you want to learn about (for bond energies, you can click on either element in the bond) and then to look for the link on the right panel that says "Find a Property".  From here, click on the property you want.  The left column defines the property, but the right hand column shows the actual physical data.
5.) Encyclopedia of Reagents for Organic Synthesis† (link) 
This reference lists properties and common uses of tons of different reagents.  So if you are curious about why a reagent is used, or in what reactions it is used, an excellent reference.  I use it frequently in my posts.
6.) Rules for Predicting UV Absorbance (link)
If you want to predict how UV spectra change upon changes in an aromatic system, this website has the rules.  Woodward actually helped to empirically determine these rules :)
7.) What to Dry Solvents With (link)
If you need a dry solvent but are not sure of reagents that are compatible, or what is the best way to dry it, this website has a table with the popular ways to dry a wide variety of common lab chemicals.
8.) Solubilities (link1) (link2)
If you need to know the solubility of a substance, these two links are good places to look.  They give accurate information about the solubility of a substance.  The first link is most easily searched through the google doc on the first page.  The second link works best when searching by CAS number.  Excellent reference I think, and this data should be more available.
9.) Reaction Finder (link1) (link2)
If you want to know what a particular reaction is or what it involves or even if it has a name, these next two links are useful for identifying, but perhaps not for learning more about it.
10.) Journal Abbreviations (link1)
A search engine where you can search from abbreviation to the journal name, or vice versa.
11.) Functional Group References† (link)
The page hosts links to common chemistries of a variety of functional groups.  Very useful if you want to learn more about chemistry of one particular functional group.
12.) Organic Syntheses (link)
A entire collection of the detailed reaction and workup procedures for a variety of compounds.  For me, I just enjoy browsing through these from time to time to learn more about the workup procedures, and to understand what each step in the work up accomplishes.
13.) Organic Reactions† (link)
It gives great detail about the mechanism, side reactions, etc.  To fully utilize this resource, you can navigate the book from the left toolbar, where you can search by structure, title, or reaction type.
14.) Merck Index† (link)
This index is a classic.  So many properties and references of a variety of compounds including their melting points, descriptions, properties, and often solubilities as well.
15.) Reaxys† (link)
An alternative to the popular Sci-Finder reaction search, you can search a reaction or even a structure to search for actual procedures for the reaction, or procedures to synthesize a compound.  Provides references to the actual papers used, just like Sci-Finder.

† = academic access needed

Total Synthesis Strychnine - Step 13 and 14

The next steps in the synthesis involve preparing the molecule for the desired Dieckman Condensation that was expected to occur previously.  Since we have now removed the offending N-tosyl group, this should be accomplishable.  What we have done in this process, however, was also hydrolyze the esters to their carboxylic acids.  Our next goal will be to re-obtain the esters as such a condensation cannot take place with the acids.  This is because under the basic conditions required deprotonate the carboxylic acids, we form a resonance stabilized carboxylate anion which makes deprotonation of the alpha hydrogen unlikely.  To accomplish this, diazomethane is the gold standard reagent for methylation of carboxylic acids.  Additionally, we will also need to re-protect the nitrogen that previously held the N-tosyl group so it does not interfere with future steps.  While accomplishing both of these tasks may seem like a lot of work, actually these modification are quite simple and both can be accomplished in one day.  Taking a look at Scheme 1 below, here are the steps we are going to accomplish.

Scheme 1. Acetylation of the secondary nitrogen followed by methylation with diazomethane.

Overview:
Acetylation of Nitrogen
The first step is the acetylation of the secondary nitrogen.  Why must we add this group?  Well, I suppose there are two possibilities, which we will see only one of which will be relevant.  The first possibility is that if we did these two steps in reverse (that is, diazomethane first, then acetic anhydride) that we may methylate the secondary nitrogen.  Actually, this is not the case.  While diazomethane is a strong methylating agent, it is really only very active in this role once it has been activated with some sort of acid, whether it be the typical bronsted acid, or a lewis acid.  Amines, amides, and alcohols in this compound are not able to attack the diazomethane because they lack a strongly acidic functionality, and so we are not worried about these side reactions.  While one may conceive that the carboxylic acids may make the solution more acidic which could lead to side reactions, once the diazomethane is activated, it becomes so reactive that it has no time to diffuse away before reacting with the carboxylate ion.  For this reason we do not see diffusion of the activated diazomethane to other regions of the molecule.  The second possibility then, is that this amine may react with future steps in the reaction.  For instance, two steps from now we will be using tosyl chloride again, which would react with the secondary nitrogen.  It is really for this precautionary reason that we are protecting the nitrogen with acetic anhydride.  This reaction has been covered before in Step 9.

Dangers of Diazomethane
For those of you who have never heard of diazomethane, this is not a step in the reaction many of us would be comfortable with.  Why is that?  Well, diazomethane is a toxic, carcinogenic, volatile, and spontaneously explosive gas.  Wow, sign me up right?  It makes sense that for these reasons, you can't go to Sigma and buy diazomethane, it's just too unstable.  So what chemists have to do is make it.  Diazomethane like I said before is a gas, but it is soluble in ether, ethanol, and dichloromethane.  Most typically is is prepared as a dilute solution in ether for safety precautions.  Dilute solutions are less likely to explode, and ether is used because its low boiling point.  Purification of diazomethane after synthesizing it involves distillation.  The closer the boiling point of the solvent is to the diazomethane, the less concentrated the vapors are with diazomethane, which leads to a safer isolation.  So how does one actually prepare diazomethane?  Diazomethane is prepared using specialized glassware that contains no sharp points (such as ground glass joints), as these sharp points can detonate diazomethane (scary).  For this same reason, whenever you want to pipette diazomethane, you must use flame dulled pipette tips to prevent detonation as well.  Two common chemicals used as a precursor to diazomethane are diazald and MNNG.  Under basic conditions, these compound decomposes to give off diazomethane and by products.  If you are still curious about the setup and mechanisms, there is a great resource here.  Typically, however, once the diazomethane is generated, it is co-distilled into the receiving flask with the ether and some small amounts of ethanol (which is the solvent of the decomposition reaction) to create a deep yellow clear solution as shown here.  One can either distill the diazomethane/ether solution right into the reaction to use it up right away, or one can store the solution for a while in a freezer (but this is not recommended).  It's best just to use what you need, but the temptation is to make a lot of it at once so you don't have to make it again.  Lastly, when one makes diazomethane, what safety equipment should be used?  Thick gloves, goggles, lab coat, and most importantly, fume hood as well as extra protection with the use of a blast shield.  There are a lot of stories about diazomethane which you can read about here, and it more than clear why all of this is needed.

Mechanism:
Since I had covered the acetylation previously, I will discuss only the methylation reaction.  The reaction of diazomethane with a carboxylic acid is shown below.  You will see why alcohols and amines are not acidic enough to form the activated species, so these functional groups do not react unless a lewis acid is added.  This is why the reaction is allowed to proceed in methanol as the solvent.

Figure 1. Mechanism of diazomethane with a carboxylic acid.

We can see that with the resonance structures there could be two competing sites of protonation from the carboxylic acid, the carbon or the nitrogen, both of which carry a partial negative charge.  If the terminal nitrogen is protonated, it leads to no stable product and so is likely reversible.  What drives the reaction forward is protonation at the carbon.  This creates a strongly electrophilic carbon which reacts with the poor to moderate carboxylate nucleophile.  This displaces nitrogen gas which bubbles out of the solution and drives the reaction forward.  That is what makes this reaction so efficient is that product leaves immediately after the desired reaction takes place effectively forming a sink into which more and more reactant falls into.  It also makes it easy to determine when the reaction is complete.  Just drop in diazomethane until no more gas is evolved.  At that point one knows that they have added either just enough, or a little too much.  The excess diazomethane in the solution is then destroyed using acetic acid, which undergoes the exact same methylation reaction to form methyl acetate which can be easily evaporated away.

Workup:
Acetylation
The first acetylation step involved mixing the starting material into a solution of pyridine, and slowly adding acetyl chloride.  This is slightly exothermic and so the solution is usually cooled, however, in this case it appears that it was not.  After 1h reaction, water was added, and after half an hour, then the solution was evaporated.  As to why Woodward waited the 30 minutes before evaporating is odd.  I have used pyridine once, and in my attempts to remove it, I did notice that upon letting it stir with water for a while did aid in the removal of the pyridine.  Perhaps the water helps to break up some of the pyridine salts and dissolve them establishing an equilibrium between the protonated and deprotonated pyridine.  Water also forms an azeotrope with pyridine, boiling at 92˚C which may also aid its removal.  The dried material was rinsed with ether to get rid of any oily material, possibly extra acetyl chloride or acetic acid, both of which are soluble in ether.  After this, dissolving the remaining material in hot water (remember, we have a dicarboxylic acid at this point, but the rest of the molecule is hydrophobic, thus it may take hot water to dissolve the compound) and acidifying with acid precipitated the protonated di acid as yellow crystals.

Methylation
The material from the acetylation was dissolved in methanol (which to me is surprising, as I would not expect it to be very soluble in methanol, but anyways) to which a fresh, cooled solution of diazomethane was added dropwise until no nitrogen was seen to be evolved (even Woodward practiced the use of fresh diazomethane and not storing it in the freezer).  He let the solution sit in the cold for an hour to allow complete reaction, and then added acetic acid to quench any remaining diazomethane in the solution.  One thing I would like to point out as well, is that we likely now have excess acetic acid in solution as well if we are sure to destroy any remaining diazomethane.  We use acetic acid for two reasons, one, it reacts with diazomethane to form volatile by-products, and two, it is also slightly volatile, and can be pumped off under high vacuum.  Thus, after quenching with acetic acid, the solvent was vacuumed to dryness.  Then, Woodward dissolved the residue in a small amount of ethyl acetate (a small amount because it is likely very very soluble in acetic acid).  The next steps involve adding diethyl ether, followed by cyclohexane.  Why did Woodward do this?  These are poor solvents for our compound as they are both non-polar, with cyclohexane being the most non-polar of the two.  Essentially Woodward is doing somewhat of a crude recrystallization by adding poor solvent.  This doesn't mean he adds both of that very very fast, but in fact, he likely added these slowly, likely drop-wise.  How does he know when to switch over from ether to cyclohexane?  Once no more crystals appear while adding ether, he probably switched to cyclohexane to get any last remaining product out.  So the real recrystallization mixture is really ethyl acetate/ether.  This is in fact the solvent system that Woodward used afterwards to attempt to gain a more pure product, but in fact, it was identical to the crude precipitation/recrystallization.

Next Steps:
Now our molecule is set up to perform the Dieckmann Condensation that was desired this entire time.  As we will see, this is accomplished by adding base, and allowing the condensation to take place.

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.