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.