How to Draw a Mechanism for a Robinson Annulation

Visualizing and sketching organic reactions can be rather challenging. Therefore, the purpose of this tutorial is to illustrate the Robinson Annulation mechanism. This tutorial is geared towards graduate or undergraduate organic chemistry students. Prior organic chemistry knowledge is recommended but not necessarily required. A mechanism is a process that shows bond breaking and making, it also shows electron transfer between molecules during a reaction. The Robinson Annulation combines two different carbonyl reactions; Michael Addition and Aldol Condensation. These two reactions are used to infuse and construct a six-membered Carbon ring. The resulting product is an essential organic compound which is used in synthesizing antibiotics and steroids, such as cortisone.

Part 1
Part 1 of 5:

Beginning With Starting Materials and Reactants

  1. 1
    Draw the compounds above with their descriptions below. You start with a ketone (alkanone) compound, which has a carbon atom double bonded to an oxygen atom. The carbon atom can also be bonded to other carbon-containing substrates.
    • For the purpose of this reaction, you will work with a 6-membered carbon ring with a Carbon atom double bonded to an oxygen atom, formally know as, cyclohexanon.
    • Write the Lewis dot structures of hydroxide and water (OH/H2O) that will be your solution/mixer that the reaction will be taking place in.
  2. 2
    Show the deprotonation of the alpha-carbon in a base/water solution. In a base solution, the ketone has a unique configuration which allows the alpha-carbon to act as an electrophile (electron accepting).
    • A alpha-carbon is defined as the carbon directly bonded to the carbon that has a ketone (double bonded oxygen).
    • The base acts as a nucleophile (electron donating), which will deprotonate the alpha-carbon by attacking an alpha-hydrogen atom from the alpha-carbon which breaks the carbon-hydrogen bond.
    • After that bond breaks the alpha-carbon accepts the electrons from the broken bond, which will put a lone pair of electrons on the alpha-carbon making it nucleophilic. This negatively charged alpha-carbon with a lone pair of electrons, is called an enolate.
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  3. 3
    Sketch the resonance of the enolate. As illustrated above, the enolate will shift forms between two configurations; this is called reasonance-stabilized enolate.
    • A resonance is considered to be a certain way electrons can arrange themselves within the compound. Compounds are transforming from one resonance to another continuously.
    • One resonance, the more stable one, is the one you've already formed with the alpha-carbon having a lone pair of electrons and a negative charge.
  4. 4
    Draw the other resonance. The other resonance involves the alpha-carbon donating the lone pair of electrons to the carbon with the ketone for a carbon-carbon double bond.
    • However, carbon cannot have more than four bonds, double boned to the alpha-carbon and oxygen and it originally had a hydrogen bonded to it.
    • So a double bond between carbon and oxygen will turn into a single bond and an extra pair of electrons, from the broken bond, will now be on the oxygen atom giving it a negative charge.
    • The compound is at equilibrium between these two configurations, the resonance form used in this reaction is the one with the negatively charged alpha-carbon, the enolate.
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Part 2
Part 2 of 5:

Drawing the Michael Addition

  1. 1
    Draw your resonance-stabilized enolate with another ketone. The particular ketone you will be using in this reaction involves the alpha-carbon having a double bond with another carbon atom, known as a beta-carbon.
    • The specific ketone compound used in this reaction is called methyl-vinyl ketone.
  2. 2
    Draw the enolate attacking the beta-carbon of the ketone in a base solution
    • The negatively charged pair of electrons of the original enolate, will attack the beta-carbon of the vinyl ketone.
    • The alpha-beta carbon double bond will break, sending the pair of electrons from the bond to the vinyl ketone alpha-carbon.
    • The beta-carbon of the vinyl ketone is now bonded to the original alpha-carbon from the original enolate.
    • Our newly formed enolate has resonance-stability, like the original enolate.
  3. 3
    Draw the protonation of the new enolate. The resonance form that will be used in the reaction is the one with the alpha-carbon that has a lone pair of elections.
    • This new enolate is still in a water/base solution. So it will act attack a hydrogen atom from water, using its lone pair of elections, to break the hydrogen-oxygen bond to form a carbon-hydrogen bond on the alpha-carbon.
    • This completes the Michael Addition step of this reaction. You created a bond between the alpha-carbon, from your cyclohexanone, to the beta-Carbon of the vinyl ketone. You also protonated the enolate, that was the result of the broken double bond between the alpha and beta carbons. Now the beta-carbon is bonded to 2 alpha-carbon that were once both enolates.
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Part 3
Part 3 of 5:

Drawing the Aldol Condensation

  1. 1
    Show the deprotonation of the new alpha-carbon. Our methyl-vinyl ketone compound used its vinyl properties to react with the enolate of the cyclohexanone alpha-carbon. There is still another carbon, the methyl group, that exhibits alpha properties in a water/base solution.
    • A hydroxide molecule(-OH) from the water/base solution will attack an alpha-hydrogen from the alpha-carbon to create a new resonance-stabilized enolate from the methyl group directly bonded to the ketone.
  2. 2
    Sketch the intermolecular mechanism of beta-hydroxy ketone. Intermolecular refers to the forces and interactions between atoms within the compound itself.
    • Our enolate will want to attack and give electrons to form a new bond with the carbon directly double bonded to oxygen in your cyclohexanone.
    • At the same time, The double bond between carbon and oxygen breaks into a single bond and the broken bonds electrons travel to oxygen, giving it an extra lone pair of electrons and a negative charge.
    • The reason we know this occurs is because when carbon and oxygen are bonded, oxygen has a greater intermolecular attraction to electrons in the bond than carbon.
    • Since oxygen attracts more electrons, it becomes partially negative. Subsequently, carbon will have less attraction to the electrons, which will make it partially positive since it's losing negative attraction.
    • This creates a new carbon-carbon bond, that closes the ring. This new compound is named beta-hydroxy ketone, but more commonly known as, an aldol.
  3. 3
    Show the protonation of the negatively charged oxygen. Your aldol is still swimming in a water/base solution.
    • The extra lone pair of elections on oxygen will attack and bond to hydrogen from a water molecule and protonate.
    • You have now successfully closed and created a new cyclohexanone that is bonded to your original 6-membered ring.
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Part 4
Part 4 of 5:

Drawing the New Enolate and Elimination

  1. 1
    Draw formation of the enolate. A base molecule(-OH) from your solution will attack a Hydrogen from the alpha-Carbon of the compound.
    • Our deprotonated alpha-Carbon has become an enolate with a pair of electrons and a negative charge.
  2. 2
    Show the condensation of the aldol. Generally, it is difficult to eliminate a beta-hydroxy in a solution. However, due to the properties of the resonance-stabilized enolate, it can be achieved.
    • At relatively high temperatures, The lone electrons of the enolate will attack the beta-Carbon.
    • This attack eliminates the beta-Carbon-Hydroxy(-OH) bond and creates a double bond with the alpha-Carbon and the beta-carbon.
    • This elimination completes the Aldol Condensation, thus completing the Robinson Annulation.
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Part 5
Part 5 of 5:

Understanding a Specific Example

  1. 1
    Draw a specific reaction of the Robinson Annulation. This will just be (starting material + reactant = product). No mechanism is shown.
    • It is important to note that the overall reaction is done at equilibria throughout. This means the reaction will never go to 100% completion; there will always be other remaining intermediates.

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      References

      1. Jones Jr., Maitland, and Steven A. Fleming. Organic Chemistry. fifth ed., Norton, W. W. & Company, Inc., 2014, pp. 263-97.

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