Total Synthesis Strychnine - Step 9

So it's been such a busy week that I haven't had time to post anything in a while!  But now, it's the Sunday of the Oscars, and I'll try to squeeze in a post on a very common type of reaction, acetylation.

Overview:

Acetylations are so widespread throughout the organic world due to the efficiency and speed of the reactions.  In this case what we are doing is creating an amide from an anhydride.  Before we get into the reaction itself, we should ask ourselves, why?!  Why are we turning this nitrogen into an amide?  The next step is an ozonolysis how should that affect anything?  Well, the ozonolysis in the next step involves reacting the electrophilic ozone molecule (O3) with the nucleophilic veratry group (the di-methoxy benzene).  This works (although the yield is poor) because this group is much more nucleophilic than a normal benzene.  However, we notice that without acetylating the amine, actually it is possible that the benzene group could be cleaved too because the amine is also very strongly electron donating, making the benzene ring a better nucleophile.  It is for this reason that we are turning the amine into an amide, which is a non-electron donating group.

The reaction for this is quite simple and the generic reaction can be found below.

As seen in the mechanism above, there is really an equilibrium of pyridine and acetic acid, however, the equilibrium favors the protonation of the pyridine and the deprotonation of the acetic acid.  A question that may arise is, why do we have the pyridine?  After all, without the pyridine the acetic acid could be released as the acid and all would be well right?  Not so.  Remember that acetic acid wants to be deprotonated because it is an acid, and all amines (including the amine we are trying to acetylate!) like to take away these protons.  So our pyridine is our sacrificial amine essentially that sucks up the protons so that our reactant can still react with the acetic anhydride.  Why then does pyridine not react with acetic anhydride to form an acetylated pyridine?  It's not that nucleophilic, is the answer.  Compared to our secondary amine (which is attached to an aromatic ring which in addition to the sterics is likely why this reaction is heated.  Typically these reactions go at room temp. quite nicely in an hour or two) the pyridine is not nucleophilic and is a tertiary amine.  These are much less likely to react, although its' always probably that a small percentage will.  Other than pyridine, triethylamine (TEA) could also be used for purposes like this.

Again, like I said before, reactions like this usually go at RT and are finished in hours to minutes for simple substrates.  In this case, reflux is used likely to speed up reaction, and perhaps to push reaction forward, as the secondary amine is somewhat hindered, and since the nitrogen electrons are conjugated with the aromatic ring (just like in aniline) it is likely not as nucleophilic.  In addition to amines (the most common substrate) alcohols and thiols can also be acetylated as they are good nucleophiles.

Workup:

I feel like I keep repeating myself with Woodward here.  It's another precipitation into water that produces our product.  Upon cooling for 2 hours and adding a little bit of methanol (another non-solvent) Woodward was able to isolate the crystal as well as remove the pesky pyridine from the reaction.  Additionally the acetic acid can be difficult to remove if you just use a vacuum pump, but in this case, this also goes into the aqueous phase upon filtering and so product plus some minor impurities are obtained.  The product description for this was not given (in terms of its appearance) likely because reactions like this are so commonplace (even though I myself have never done one).  Recrystallization from chloroform/methanol yielded a more pure product.  Also to note are the high yields for these reactions (98.5%).  This is due to the pyridine shifting the equilibrium to the products side by absorbing the proton.  I am not sure what the yield would be without, but I am sure it would be around 50%, as this is when half of the amine has reacted, leaving the other remaining half protonated.

Tune in for the ozonolysis in the next step of the synthesis which actually was a big setback in the synthesis due to the poor yield.  But nevertheless, a poor yield does not necessarily mean a failure of an experiment, although it is certainly a hinderance.  Even in my own research I have one step which is 50% yield for removal of a protecting group.  This means that for every 150mg of material I have, I can only get 60mg out of it.  It's a little depressing because it is a lot more wasteful, but in this case, sheer large mass of material gets the job done.

Total Synthesis Strychnine - Step 8

Let's take a look at where we want to go from here now that we have one ring down (ring V).

Our next goal is to create rings III and IV which are directly attached to our current molecule.  Highlighted in red is are the atoms from our molecule thus far.  In blue is also the ethyl ester carbon from our current molecule, and the purple bond will be one created from the veratryl group (the di-methoxy benzene substituent) and the ethyl ester.  Rings VI and VII will come after this.

Below are listed the next few (very general) steps of this synthesis to create ring III.

Overview:

The focus today will be on the borohydride reduction.  While similar to the lithium aluminum hydride covered in a previous post, this chemical is an extremely common reducing group and is weaker than the aluminum hydride.  The choice if reagent is simple, we want to reduce the double bond of the imine, but not other functional groups (such as the ester!).  Use of LiAlH4 would certainly reduce the ester (as that is what it is used for) or the amide (hidden in the tosylate protecting group).  Interestingly enough, sodium borohydride can actually be stored in water (crazy huh?).  Even though it does form some hydrogen gas initially, a basic solution quickly forms, which is stable.  You can purchase it as a basic solution if you really want to.  Solvents for reduction of sodium borohydride are typically done in methanol or ethanol which have weakly acidic protons (pka ~ 20-25, so they are to no extent deprotonated in aqueous solutions near pH=7).  If you are really into the specifics, there is a lot of work done on these reactions which can be nicely summed up in this article.

The hard part about all of these reducing agents is knowing the strength, which honestly seems greatly ambiguous in undergraduate organic.  Actually, as an interesting aside, sodium borohydride can reduce esters under particular conditions.  Using a lewis acid (such as CaCl2) or by heating to 65-80 C.  In reality, esters can be reduced at room temperature if you let the reaction stand long enough, but typically these reactions are performed quickly (< 1 hour) and so reduction of the ester is not a problem.  

Last thing to note about reductions using these hydride reagents, the cation really matters!!  If you are hurrying to get a reaction done and you are all out of sodium borohydride, but you happen to have lithium borohyride, you should really just wait to get the correct reagent.  This is because the cation that coordinates to the carbonyl group plays a key role in the activation of the carbonyl toward nucleophilic attack by the hydride.  This is why adding such lewis acids like CaCl2 or LiBr (of which the Ca2+ or the Li+ ion coordinates with the carbonyl) leads to increased activity.

How does this mechanism look like though?  After all, we form an alcohol, but we really don't have any spare protons lying around.  The actual reaction is shown below, and the protons to form the amine (or alcohol if you are reducing a ketone) come after the aqueous work up, which always follows these reactions.  There is actually a lot going on in solution.  Since we are usually using methanol or ethanol, these can react slowly with the borohydride forming alkyl borates (alkoxy groups on a boron).

Workup:

I'm always so surprised at how easy Woodward makes chemistry look.  Keep in mind that I'm sure a lot of these procedures took lots of time to figure out, but once you figure it out, it makes the synthesis seem so elegant sometimes.  Anyway, let's see what Woodward did.  First of all, he dissolved the compound in ethanol.  But wait, didn't he crystallize the previous compound from methanol?!  Yup he did, so chances are when he put this compound into ethanol, it didn't dissolve either.  Since ethanol is more non-polar (but still very polar), what is the average chemist to do?  What most chemists tend to do...use brute force (i.e. heat the darn thing).  After all, it must be soluble in hot solvent since it was recrystallized from hot solvent.  Will we worry about reducing the ester?  It's a little more of a worry, but still, the sodium borohydride reacts much faster with the imine than the ester that unless we let this reflux this reaction for a long period of time with a large excess of borohydride, we really don't have to worry about it.  So after slowly adding the borohydride (exothermic) and heating for 1 hour, a clear solution had formed.  Doing what Woodward does best, precipitating product while quenching reaction  with the water to donate the final proton to the now amine, gave a solid (yellow and impure) product.  Notice how all the borate salts formed will have been washed away in the water.  However, any remaining unreacted starting material (or the very very small amount of reduced ester) has not been separated from the product (hence the yellow color).  Recrystallization from chloroform (good solvent) and methanol (poor solvent) was able to purify the compound to give colorless crystals.

See you next time where we delve into the ever-so-common acetylation reactions.

Total Synthesis Strychnine - Step 7

It's a late night at lab and I'm getting bored, so it's time to finish up the V ring of Strychnine (a.k.a. the first step with some actual complexity).  What we will try to accomplish today will be the formation of a 5 membered ring with concomitant formation of a stereocenter.  While doing this we will also be breaking the aromaticity on the indole ring which will actually cost us quite a bit of energy to do so.  Below is what we are trying to accomplish.

General:

It is easy to observe that this transformation could be made to happen with acid (well, maybe not easy, but at least it can be understood).  This occurs through protonation of the imine which allows the nucleophilic beta-carbon of the indole to arrange for good trajectory for attack, and formation of the ring.  The trouble is, while Woodward attempted these conditions, actually he was unable to attain a useful result (so it seems is the case with most experiments).  Other papers at the time had shown actually the reverse reaction of the above to occur under acidic conditions due to the aromaticity gained and higher stability in the reverse direction.  Actually for them to get the reaction to go forward, they had to somehow lower the energy of the product molecule to stabilize it.  The very crude energy diagram may help to show what is happening in this step.  The question is, well, how do the pyridine and tosyl amide stabilize the product.  First of all, deprotonation of the indole hydrogen by pyridine is not a likely thing to happen.  The pKa of the indole hydrogen is nearly 20 (in DMSO) and requires very strong bases.  So why then is such a weak base like pyridine used?  Well, in this case, only a weak base is needed, due to the activation of the electrophile by the work of tosyl chloride on the imine nitrogen.  This creates a potent and reactive electrophile which should react with the already nuclephilic indole beta carbon.  The pyridine is simply here to remove the proton once this attack happens so as to push the reaction forward.

The purpose of the tosyl chloride also serves in this case to stop the reverse reaction from happening.  once the amide forms (with the tosyl group) the electron pair of the nitrogen is no longer a good donating group, and so the reverse reaction is essentially hindered, driving the reaction forward.

During this process the one thing still yet untouched has been the stereochemistry.  Woodward was able to determine (I am not sure how) that only one diastereomer (because 2 chiral centers were formed) was obtained.  In this case we created a stereocenter on the indole ring, as well as on the carbon alpha to the ethyl ester.  As is in most cases, this is driven by almost solely sterics.  When Woodward carried out this reaction he had predicted that this product was likely to be formed, but actually he also stated that it was very possible that the other diastereomer of the carbon alpha to the ethyl ether could have been formed.  We can now rationalize that the ester has steric clashes with the veratryl (the fancy name for the di-methoxy benzene ring) group.  As for the stereochemistry of the indole ring, I had tried to find a convincing argument as to why only one stereocenter was formed, but I came up short.  Seems that this may be the most favorable conformation of the ring, but to me it seems that the opposite stereochemistry would be equally likely.  I'll have to think on this a bit more.

Workup:

The solvent in this reaction is actually the pyridine (as is the case in most reactions where pyridine is the base.  Usually seen in acetylations with acetyl chloride).  Letting the reaction stir essentially overnight (18h) completed the reaction.  Now, how can we isolate the product.  Precipitating in water seems to be something Woodward enjoys doing, and se he employs this here.  Upon adding an equivalent amount of water as solvent, a precipitate formed, while at the same time, excess p-toluenesulfonyl chloride was destroyed.  Collecting these crystals and then washing with methanol and water yielded a fairly pure product.  Further recrystallization from hot methanol was able to give an even more pure product.

Again we see that precipitation of the product and filtering is the easiest way to not only get all of the material out of the reaction, but also to wash away that pesky pyridine which can often be so tricky to remove.  What I am surprised about is how the starting material did not precipitate in this reaction.  I am curious to know if it is soluble in methanol which would lead to the purification of the material from the starting compound.

Last thing to note.  It is interesting to note how only the intramolecular product was formed, and how one indole ring did not attack the activated imine on another molecule.  The reason for this suppression is due to the close proximity of the imine on the same molecule.  The term "effective concentration" is often used, where most of the time, the imine on the same molecule is in higher concentration because of its close proximity.  For this reason, 5 and 6 membered rings are often favored as products rather than dimers or other oligomers.

Total Synthesis Strychnine - Step 6

We are almost to the good stuff in this total synthesis.  And by good stuff, I am talking about the very important stereocenter forming step.  But before we get there, you will notice we are still a couple carbons short of making one of our rings of strychnine.  As we stand in our synthesis so far, we have been planning along the way, that the beta carbon of indole ring would act as a nucleophile, reacting with an electrophile to form a new carbon-carbon bond with the proper stereochemistry.  It seems to me that these types of carbon bonds for chiral centers, usually take place with some sort of carbonyl derivative.  With this in mind, by reacting our primary amine with ethyl glyoxalate, we can set up our reaction for a nucleophilic attack on an electrophilic imine.  Below is the scheme with the nucleophile in red, and the electrophile in blue, which will be reacted in the next step.

Overview:

What we are doing is this step is a condensation between the primary amine and the aldehyde.  From Organic II, you likely learned about the reactivity of aldehydes vs. ketones and other carbonyl derivatives.  The aldehydes are significantly more reactive than esters for two main reasons.  As is always, these two reasons come from electronics, and sterics.  The steric argument comes from the hydrogen rather than another alkyl or heteroatom.  The hydrogen is much smaller and so allows approach of the nucleophile with much ease.  Compared to the ester, which has a resonance structure with the oxygen, it is also more bulky.  For these reasons, actually the ester is much less reactive than the aldehyde.  so much so that we see attack of the aldehyde moiety in place of the ester, even at reflux temperatures.  One must really understand the depth that these seemingly small differences have on the reactivity of the compound.  Even a difference in reactivity of only 2 orders of magnitude can really allow selectivity of one group over another.  Again, it's not that it's impossible for an amide to form, but it's less likely.  Another reason why, the leaving group would be an ethoxy anion, which, in benzene which is a non-polar solvent, may not solvate the anion well, and it's actually a poor leaving group in the first place.  The mechanism for this reaction is shown below.

While it may be possible that the leaving of the hydroxide group may occur concomitantly with the removal of the last proton, I put them as two separate steps for clarity.  In fact, because benzene is a non-polar solvent, it is more likely that water is the leaving group.

Workup:

One important consideration in these types of condensation reactions is the production of water.  It can be difficult to push the equilibrium toward the products if the water is not removed.  To remove this water and aid the equilibrium, there are a couple of tricks a chemist can use.  One, is to place molecular sieves in the reaction flask.  Molecular sieves are one such option.  These are simply some microporous material which has small pores to let small molecules fit in, but not larger molecules.  In this way, water will work its way into the sieves and essentially be removed from the reaction.  Woodward, however chose to use another, more popular technique.  This involves using a Dean-Stark trap.  This trap collects vapors during the reflux reaction, which contain benzene and water, and condenses them in the typical reflux condensor, however, the mixture is not dripped back directly into the reaction flask, but rather, into a "trap" where the water will sink to the bottom, and displace benzene which will fall back into the reaction.  It's a little hard to explain, but here is an image which should help.  

Lastly, the workup.  Refluxing for 5 hours is necessary to collect the water, and to help drive the reaction forward.  Refluxing is very common in condensation reactions.  To work up the material it is actually quite easy.  The material precipitates out of solution in the refluxing benzene as yellow crystals.  However, this often doesn't mean all of the material has been precipitated.  To ensure this, Woodward had cooled the flask overnight to aid in precipitating all of the material.  Even so, Woodward was clever enough to know that there was likely more material in the flask, even after filtering the solid away.  Nowadays we would use TLC to see if more material remained in the organic layer.  To get the rest of the material out of the benzene, he added lots of diethyl ether (which our desired compound should not be soluble in, as diethyl ether is pretty non-polar).  This helped to precipitate actually the largest amount of material out of the flask.  Concentrating the mixture either by boiling or by rotary evaporation, and then re-diluting with ether, again, gave more product (because some of the benzene was removed) and then, adding petroleum ether (really just a mixture of alkanes) to the mixture, precipitated another smaller batch of material from the solution.  A final recrystallization from benzene yielded the yellow crystalline powder in good yield (92 %).

Overall, a very simple and common reaction, but as I always say, there's a lot that goes on behind the scenes.  See you guys next time.

Total Synthesis Strychnine - Step 5

Time for another rather simple and common chemical reaction, which is treatment of the nitrile group with lithium aluminum hydride (LiAlH4), a very strong reducing agent.  Under most circumstances, we will likely avoid this reagent, not because of its hazards, but rather due to its reactivity.  LiAlH4 reduces a myriad of different functional groups.  From the common ester, to nitro, azide, oxime, to the often difficult to reduce amides and, in this case, nitriles.  Notice that in the reaction below, that the only reactive group is the nitrile.  Typically there will be other groups one has to worry about, but when there are not, lithium aluminum hydride is a great reagent because of the high yields and speed of the reaction.  Below is the step we will focus on today.

General:
In general, these LiAlH4 is run in ethereal solvents.  These include tetrahydrofuran, THF, and diethyl ether.  THF is more commonly used because of supposed impurities in diethyl ether, despite the higher solubility of the salt in ether (weird for a salt to be soluble in ether, a non-polar solvent).  These solvents should be dried using an SDS (solvent dispensing system) or other methods (drying over anhydrous salts, or some other common methods are here).  This is because of the reactivity of the reagent.  This reagent is much more reactive than sodium borohydride, for two reasons.  One, the lithium ion is smaller than sodium, this ion is involved in coordination with the group to be reduced (in this reaction is is the nitrogen of the nitrile).  Second, the aluminum is less electronegative than boron, and so the electrons are held on looser, and thus the hydrides have more "hydride character" than in the borohydride.  A reagent that is often omitted from undergraduates is lithium borohydride, which I'm sure you can guess has properties in-between the two.  For this reaction, only lithium aluminum hydride is strong enough to reduce the nitrile, and since there are no other ketones, aldehydes, amides, etc., we will only get reduction at our nitrile.  Lastly, when you see hydride reagents such as those just mentioned, you should always think nucleophile.  These all behave similarly in that the hydride ion behaves as a nucleophile, attacking electrophilic centers.  In this reaction, the carbon of the nitrile is the electrophile as it is triple bonded to the nitrogen, highly polarizing the carbon.  The mechanism for this reaction (general scheme) is below.  It is interesting to note that in fact the hydride adds twice to the nitrile to isolate the amine.  During the reaction, the remainder of the aluminum complex acts as basically a giant proton (more specifically, a lewis acid).

The aqueous work up does not occur until after the reaction is complete.  Otherwise we would destroy all of our hydride and turn it to hydrogen gas and aluminum hydroxide immediately (in fact, LiAlH4 reacts very violently with water).

Workup:
There are many ways to work up a reaction after it has reacted with the hydride.  One of the most common is to do a basic hydrolysis.  This is essentially the aqueous work up in the mechanism above.  The base acts to slow the rate of hydrogen gas evolution (low H+ concentration) and also to help hydrolyze the aluminum to form aluminum hydroxide as the final byproduct.  The aluminum hydroxide is not very soluble in THF, nor is it in water.  Because of this what we will see is a lot of precipitate starting to form.  This makes the workup quite simple, and all we need to do is filter off the salts, and our product should be soluble in the THF, so we can then just collect the liquid after rinsing away any product that adsorbed to our salt with a little solvent.  Another method which Woodward uses in this synthesis, is to used a saturated aqueous solution of sodium sulfate.  Now what does this accomplish?  Well, sodium sulfate first of all is an aqueous solution, this is really the most important part.  Secondly, the sulfate is a flocculant.  A flocculant is a compound that aids in precipitating things from solution, in this case, we want to precipitate the aluminum hydroxide.  The sulfate helps in doing this, giving us a again, a precipitate, from which we can collect the liquid from.  After this, Woodward had added some chloroform, likely to solubilize all of the organics, but not the salts.  After this, he removed the solvent leaving a goop behind.  Trituration from ether (essentially just adding ether, stirring to dissolve impurity, then removing ether leaving solid behind) gave the desired product which were light yellow crystals.  He recrystallized from benzene to obtain a purer product for analysis.

Notice that woodward did not use the other type of workup that is possible for this reaction, and that is an acidic workup.  In workups such as these, usually aqueous ammonium chloride is added to the solution to gently (relatively) protonate the amine, as well as degrade the remaining aluminum hydride.  This was not done however, because we are reducing the nitrile to a primary amine.  The primary amine is a very basic group, and so we would be left with a charged group, making isolation more difficult as it will not be very soluble in organic solutions, and separation from the lithium salts can be difficult (after all, the protonated form of the amine is really just a salt itself once it is dried).

The last thing I would like to mention is that this reaction was not done under nitrogen.  But why?  The hydride is very reactive to moisture in the air!  Well, in reactions like these, usually 3-5 equivalents of hydride is added.  So any losses to water in the air is not going to greatly affect the reaction.  If you were carrying out this experiment though, should you keep it under nitrogen?  Well, probably.  The thing is you can't add the hydride too fast or you will create too much exotherm.  But how to you add a solid reagent to a flask without exposing it to the atmosphere?  Unless you cary this out in a glovebox (not a good idea), you can't really add the solid.  So, what is sometimes done, is that the ethereal hydride solution is placed into a flask, and the compound we want to reduce is slowly added, after dissolving in solvent, to our solution, via syring, or addition funnel capped with a drying tube (calcium chloride tube on top of the addition funnel).

Well, that's all for today.  Please leave a comment if you have any other questions or if you like my blog.  It will be good to have some feedback.  Thanks!

Total Synthesis Strychnine - Step 4

Hey there.  Today our goal is to look at the next transformation in the total synthesis of strychnine.  Yesterday we looked into the preparation of the tetraalkyl ammoium group.  The next goal is to perform an SN2 displacement of this group with a nucleophile, the nucleophile in this case being cyanide ion.  Below is the general scheme.

Overview:

We will see that the point of this step is to actually just extend the carbon chain by one.  This problem seems like it could have easily been circumvented, but remember that the carbon alpha to the amine was donated by the formaldehyde from the Mannich reaction.  There is no way to add two carbons in this way, and so we are left with the somewhat roundabout way of extending the carbon chain.

One thing to note first of all is the solvent, DMF.  DMF (if you are not familiar) is a very useful solvent.  Like DMSO, it dissolves a very wide range of compounds and is quite polar (you can view the resonance structures on the wiki page).  However, because of its high polarity, it has a very high boiling point (153 ºC).  Solvents with BP's this high are very difficult to remove completely from you compound, and so you must always be careful when using DMF, that you can remove it from your reaction.  For example, don't use DMF for a solvent if your final compound is a liquid/oil.  Chances are you are going to have a hard time separating the two if you don't have a good distillation column (actually our lab doesn't even have a distillation column, not sure if most labs nowadays do).  So, why is DMF used then!?  Two reasons.  First, DMF is a polar aprotic solvent.  You may remember from Organic I, that these types of solvent are polar, but have no hydrogens available for hydrogen bonding. What these solvents effectively do for SN2 reactions is to stabilize the nucleophile and the leaving group.  This brings down the activation energy of the reaction and dramatically (and I mean dramatically) increase the rate.  One example I found briefly online showed that performing an SN2 in methanol vs DMSO, took 20h for ~70%, or 20 minutes for ~90% conversion, respectively.  That is a huugee difference!

The second important part of the DMF, is that it is a high boiling point solvent, and so it allows an increase in the temperature up to 150 ºC, which should help drive the reaction forward.  Other factors driving the reaction forward include the excess amount of sodium cyanide, as well as the leaving group.  The tetraalkyl ammonium group is positively charge, and as it coverts to a neutral species in the transition state, it becomes actually not such a great nucleophile (tertiary amines are significantly worse nucleophiles than primary amines).  I could go on and on and on about SN2 reactions, and maybe I will make a post about them in the future.  In the meantime, if you are still curious about all the possibilities I recommend 2 sources.  This page, and "Modern Physical Organic Chemistry" by Anslyn and Dougherty (Chapter 11).

Workup:

For this reaction, simply mixing the two compounds (In a fume hood because of the sodium cyanide!) and heating for 1 h completes the reaction.  As I said before, DMF is a great solvent for SN2 reactions. However, removing the DMF can usually be problematic.  However, in this case, isolation was actually very easy.  Diluting the solution of DMF (70 mL) into 350 mL of water precipitated out the product (again, our product is pretty non-polar, and DMF dissolves just about everything).  Isolating the solid on a vacuum funnel and washing with water should remove all the DMF.  Further purity of the sample was obtained by recrystallizing from ethanol.

Here is one thing that I would like to say that I believe is so so so often left out of descriptions, and can actually be very tricky in the laboratory.  How do you pick a solvent to recrystallize from?  All the advice I have recieved, is just "try a lot of different solvents".  Not the best advice... So here is where I would like to give some insight into the choice of ethanol.  Ethanol is usually a great choice for recrystalliztion of aromatic compounds.  It's a trend I have noticed through some of my literature searching, as well as brief experience.  If you come across an aromatic compound, chances are you can recrystallize it from ethanol or often isopropyl alcohol too.  Ethanol is a polar solvent and usually aromatics are not that polar.  Thus, by heating, the aromatics will usually dissolve (as ethanol can be heated up to ~70 ºC).  Also of importance in this reaction was the fact that it was performed under nitrogen atmosphere.  But why?  Well, water getting into the reaction can be bad because it can destroy some of the sodium cyanide.  The pka of the cyanide ion is 9.1, meaning that if you put the sodium cyanide into a water solution at pH 7, actually most of the sodium cyanide (about 99%) would turn to hydrogen cyanide (the poisonous gas you are scared about) and so it is important to keep the sodium cyanide in its ionic form for the sake of the reaction, and for yourself.

Lastly, how does one deal with the aqueous waste leftover?  After all, there are likely still cyanide ions leftover in the water solution.  As with many things, there are lots of ways, but most commonly you will see some sort of oxidant used.  This can be KMnO4, hydrogen peroxide, bleach (sodium hypochlorite).  By making sure that the solution stays basic (around pH 9-10) ensures a quick reaction.  Using ~100mM hydrogen peroxide at pH 10, the reaction was essentially complete in 9 minutes (Link).

Looking Ahead:

Anyways, there is always so much to talk about and I really end up writing much more than I ever though I would.  At this point we are still attempting to create a ring (specifically this ring is known as ring V) of the strychnine molecule.  Our next step will be to reduce the cyanide group into a primary amine.  To do this, we will need a strong reducing agent (LiAlH4).

Total Synthesis Strychnine - Step 3

So to recap, so far we have just added an amine group to our compound.  The end goal of this and to following transformations is to synthesize one of the rings of strychnine (see below in red).

With this goal in mind we can see that in the following steps (shown below), that this creates a stereogenic center which leads to the question, what stereospecific reaction will be necessary to form this center, and how can we arrange our molecule to set us up for this reaction.  In present day chemistry, this is often where some transition metal catalyst from some far out land comes into play, and quite honestly I can't stand when people do this.  After all, it's not like I have that catalyst just lying around the lab.  What I love about Woodward, was in ingenuity in forming these stereocenters.  In this reaction we will see that a common base (pyridine) is used to form this stereocenter.

I will only cover the first step today (1 to 2), the treatment with methyl iodide.

Overview:

Let's first start with what methyl iodide is, and how it acts.  Methyl iodide is a very potent electrophile.  The iodine (as you may remember from electronegativity trends) is pretty electronegative and pulls lots of the electron density from the carbon.  This polarization makes it a good leaving group for attack by a nucleophile.  If we look at our reaction scheme above the numbered compound 1 we see only one nucleophilic nitrogen.  Recall that the nitrogen on the indole ring is not nucleophilic as its electron pair is involved in the conjugated aromatic system between both fused rings (it is also conjugated to the dimethoxy benzene substituent too!).  One thing to note about nitrogens is that they are very nucleophilic.  I remember first learning organic chemistry and noting that nitrogens and oxygens were both nucleophilic, and just accepting it as so.  In some of my current research, actually I have come to learn that nitrogens can displace a halide in an SN2 reaction in the presence of an alcohol group.  I was surprised as well to see some acetylation reactions of nitrogens using acetic anhydride done with methanol as the solvent with good yields.  It just goes to show that Orgo I and II are really just scraping the surface of the actual practice of organic chemistry.

Back to methyl iodide.  Since the tertiary nitrogen is the only nitrogen allowed to attack the methyl iodide we see displacement of the iodine to form the tetraalkyl amine salt.  You may be wondering why a tertiary and thus hindered amine is allowed to form such a salt, well, it's because the methyl iodide is is such a great electrophile, and also because even tertiary amines are good nucleophiles for a reaction like this.  I don't believe that doing this reaction in an alcoholic solvent would be a good idea as I am sure that the alcohols would react to form methyl ethers (unlike the anhydride reactions).  Since our compound is now positively charged, what we will actually isolate will be its salt.  Since the reaction is performed in benzene (nowadays we would not use such a solvent, but perhaps toluene, or THF would be better) the only counteranion is the iodine leftover from the methyl iodide.

Another thing you may or may not be asking yourself is why methyl bromide or methyl chloride couldn't be used.  In fact, they could be used, however, methyl iodide is a liquid, while the bromide and chloride are gases.  Not only are gases more difficult to handle, they are also potent electrophiles which in a chemists mind should instantly hint at carcinogens.  They will react with logs of nucleophiles in the body and cause a ton of dammage to tissues and the like.  In gaseous form, these are not fun to deal with.

Workup:

The work up for this reaction is actually quite easy.  Because the reaction is performed in a non-polar solvent at ice bath temperatures (0C).  What is nice about reactions like this, where the starting material is non-polar (relatively) and the products are a salt, is that often they will just precipitate from the reaction.  This is the case with this reaction.  The crystals will slowly precipitate from the reaction, and can simply be vacuum filtered away.  Washing with a volatile non-polar solvent (diethyl ether) washes away any benzene or unreacted material away leaving the pure, pink, crystals in this case in great yield (92%)!  This is because methyl iodide is such a great electrophile, the reaction goes well toward the right, giving excellent yields, which are typical of these methylations with methyl iodide.

Looking Ahead:

The next step is to actually remove this group we just made to add another carbon into the chain giving a primary amine.  The positively charged tetraalkyl ammonium ion is a great leaving group for the cyanide ion.  We will cover this step (and the precautions to take while using cyanide) in the next post.

Take care!

Total Synthesis Strychnine - Step 2

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

Mannich Reaction

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

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

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

Workup

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

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

Looking Forward

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

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

Total Synthesis Strychnine - Step 1

To start off, I love total synthesis.  Even though I don't myself perform it, I appreciate the complexity and ingenuity that goes into figuring out synthetic schemes of these complex molecules.  So every so often I like to take some time to read "Classics in Total Synthesis", of which the first entry is the total synthesis of Strychnine by B.R. Woodward.  I will try to go over each step every day or two, and really break down the importance of every chemical involved, possible side reactions, and planning.

Here is the structure of Strychnine.  While not as complex as some other molecules, there is still quite a bit of somewhat hidden complexities, most notably the number of chiral centers, and the odd ring structure.  I'm still always amazed that what may look like a straightforward molecule, in practice, can be extremely difficult to synthesize.

If you are interested in the synthesis, it's reported synthesis by Woodward was published in Tetrahedron in 1963.  It can also be found on the SynArchive website.


To start off, the first step of the synthesis will involved the least labile groups of the molecule.  In this case, this would involve the indole ring on the left side of the molecule.  The benzene is quite unreactive and should maintain its integrity throughout the remainder of the synthesis.  So here is the first step of the synthesis.

Overview:

In this first step, the reaction that takes place is the Fischer Indole Synthesis.  It is one of the oldest ways of making an indole ring.  First, the hydrazine moiety is condensed with the ketone of the di-methoxy substituted benzene to form a hydrazone (the hydrazine equivalent of a ketone) with a concomittant loss of water.  The use of polyphosphoric acid in the reaction is twofold.  First of all, polyphosphoric acid is simply a very very concentrated phosphoric acid solution.  So concentrated that it forms essentially a polymer of phosphates.  Its first use is that water will hydrolyze these bonds to form regular concentrated phosphoric acid.  Well wouldn't you know that water is evolved from this reaction in the first step of the condensation of the hydrazine to form the hydrazone.  Secondly, the fischer indole synthesis is catalyzed by acid throughout the duration of the reaction.

Workup: I find it interesting how quickly this reaction proceeds, finishing in nearly 10-15 minutes.  It is to note that this reaction is highly exothermic due to the additional stabilization attained by the aromatic indole product.  For the same reason, this is the driving force of the reaction bringing it forward, creating a sink to bring products toward reactants.  It is to note that the entire molecule is conjugated and relatively hydrophobic.  The amine in the indole is not very basic (it takes a pH of 3.6 to protonate half of these sites) as the electron pair is involved in the aromatic system, again, making this molecule relatively non-polar.  Isolation of this compound involved precipitating the reaction melt into water, then extracting the compound into chloroform (a good solvent for most compounds, non-polar compound such as this should be soluble).  Using a chloroform/methanol mixture the compound was recrystallized in a 54.4% yield.  It is to note that the more polar phenylhydrazine is likely to be soluble in the methanol and as the desired compound precipitates out as the more polar solvent is added, the phenylhydrazine should remain in the liquid layer, effectively purifying the compound.

Looking Ahead: One note about indoles is that they are quite nucleophilic at the alpha and beta nitrogen carbons, leaving them reactive with electrophiles (see figure below).  The introduction of the di-methoxy benzene moiety on the alpha carbon effectively blocks this carbon and helps to activate the other for attack in the next step via the indole modified Mannich Reaction.

Important resonance forms of Indole

Welcome

Hello everyone!

This blog is a way for me to organize my thoughts about chemistry, and also take a more in depth analysis into some experiments I'm performing, or to go into more of the chemical logic that goes into running and working up a reaction, as well as explaining some of the logic and techniques that are often times left out of literature and textbooks.  The goal is to develop a sense of chemical logic and awareness to not only improve my abilities in this field, but I hope that you also will take away some valuable information.  Enjoy!

Ryan