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!