# Structural Biochemistry/Organic Chemistry/Mechanisms of Organic Molecules
## Fischer Esterification

The Fischer esterification mechanism is a reversible reaction between an alcohol and a carboxylic acid. Upon acid catalysis, the carboxylic acid is protonated and transforms into an activated electrophile. The alcohol, being the nucleophile, attacks the carbonyl carbon because of its positive charged electronegativity experienced by the withdrawing oxygen atom. The result of this attack is a tetrahedral intermediate. Once the intermediate is formed, the alcohol group from the original carboxylic acid undergoes deprotonation while the hydroxyl group protonates. With a protonated hydroxyl group, the functional group is removed by delocalized electrons from the carboxyl group resulting in the removal of a water molecule, leaving the final product as an ester. Due to equilibrium mechanics, all the steps mentioned in the Fischer esterification mechanism can be reversed, which is not desired in order to achieve a final ester product. Le Chatelier’s principle can be used to manipulate the reaction in order for the creation of ester to be more favorable than the reverse reaction. In order for this to happen, an excess of carboxylic acid or alcohol can be added along with the removal of water, the side product of the reaction found in the forward reaction.
When a chemical reaction is introduced, oftentimes only the reactants and products are shown. In the example shown below, normally the ester would be shown as the starting material, along with its product, the carboxylic acid. It is our job to describe the overall transformation taking place in a reaction, in the example below—the hydrolysis. In the example shown below, the ester was able to hydrolyze to the carboxylic acid through a series of reaction intermediates. With organic mechanisms it is our job to study how and when bonds break and form, and how different changes in substrates might affect the outcome of a reaction.

## Base-Catalyzed Hydrolysis and Transesterification

Step 1: Catalysis: The base attacks the carbonyl carbon, pushing the electrons from the double bond to the oxygen.
Step 2: Leaving Group: The electrons from the oxygen anion reform the double bond, kicking out the old ester group, completing the hydrolysis.
Step 3: Deprotonation: If an OH- group would have been the nucleophile, then the hydrogen would be deprotonated.

## Acid-Catalyzed Hydrolysis and Transesterification

Step 1: Catalysis: The carbonyl oxygen is protonated by the acid catalyst.
Step 2: Nucleophilic Attack: The nucleophile attacks the carbonyl carbon, pushing the electrons from the double bond to the oxygen. The oxygen from H2O serves as the nucleophile during the attack at the electrophillic carbon of the carbonyl. This creates a tetrahedral intermediate.
Step 3: Intramolecular Protonation: The original -OMe group picks up a proton from the newlyadded water group. This protonation converts -OMe into a good leaving group.
Step 4: Hydrolysis: The electrons from the hydroxl group moves down forming a double bond, thus kicking out the old ester group.
Step 5: Deprotonation: The solvent picks up the proton from the newly formed double bonded oxygen. The deprotonation of the carbonyl oxygen leads to regeneration of the acid catalyst. This leads to the final product.
This shows one example of a mechanism used in organic chemistry. A good understanding of the properties of organic molecules and the nature of their bonding allows us to predict mechanisms.

## Hydrolysis without Transesterification

The above images show an ester bond being broken and replaced with another ester group.
Below, you will find the same mechanism, but without the reforming of a new ester bond a.k.a transesterification

## Keto- Enol Equilibria

Keto and enol tautomers are in equilibrium in either acidic or basic conditions. An enol is short for an unsaturated alcohol and keto is short for a ketone. In the process of going from a keto to an enol, an enolate is formed which is the resonance structure with a charge delocalized over the oxygen and the carbon. Both the acid and the base catalyzed reactions occur very quickly and that the keto form is more likely to be found in solution.[1]

## Acid-Catalyzed Enol-Keto Equilibrium

Step 1: Protonation: The alkene attacks a free acidic hydrogen proton in the acidic solution.
Step 2: Relief of charge: The lone pair from the alcohol moves down to relieve the positive charge on the carbon.
Step 3: Deprotonation: The resulting carbonyl compound with is then deprotonated, most likely from the solvent or water.
Note that it is the same mechanism but in reverse to go from a ketone to alcohol in acid conditions where the carbonyl is protonated and then a proton from a carbon is pulled off leaving the electrons to form an alkene.[1]

## Base-Catalyzed Enol-Keto Equilibrium

Step 1: Deprotonation: A base pulls off the hydrogen proton from the alcohol group.
Step 2: Resonance: Resonance structures show that the negative charge can be somewhat stabilized on the carbon.
Step 3: Reprotonation: The negatively charged carbon then pulls in a hydrogen proton, most likely from the base or solvent.
Note that it is the same mechanism but in reverse to go from the ketone to an alcohol in basic conditions where the base would pull of the hydrogen forming the alkoxide intermediate which then takes back the hydrogen from the base.[1]

## Aldol Condensation

Aldol Condensation is the attack of an enolate ion on a carbonyl functional group to form a carbon-carbon bond. The name aldol comes from aldehyde alcohol. Aldol condensation can be done with ketones as well as aldehydes, either the same aldehyde/ketone as the starting enolate or a different aldehyde which would be called a crossed aldol condensation. It is a condensation reaction because it releases water. Also note that it regenerates one equivalent of the base used. [1]
Step 1: Enolate Formation: A base (hydroxide in this example) deprotonates a hydrogen and forms an enolate ion with two resonance structures.
Step 2: Enolate Attack: The carbon with the negative charge attacks a second aldehyde (or ketone) at the carbonyl carbon. Note that the enolate alkoxide is more stable with the negative charge than the carbon so the resonance structure is shown to accurately portray the carbon-carbon bond that will form.
Step 3: Protonation: The resulting alkoxide after the enolate attack is then protonated to give the final product. Note that the base is regenerated.

## Michael Addition

Similar to other nucleophiles, enolates can do conjugate additions to α,β-unsaturated aldehydes or ketones to form carbon-carbon bonds by the Michael addition reaction. As before, an enolate ion is formed and then the carbon attaches to another carbon of a second reactant that has a conjugated double bond with a carbonyl, called a Michael acceptor. It is a similar mechanism reaction to the aldol condensation.[1]
Step 1: Enolate Formation: A base such as hydroxide deprotonates a hydrogen from the carbon and forms an enolate ion.
Step 2: Enolate Attack: The alkoxide pushes down its extra set lone pair of electrons reforming the carbonyl and the carbon attacks the conjugated carbon of the Michael acceptor. In a chain reaction of electron pushing another alkoxide is formed after the two reactants are fused together by a carbon-carbon bond.
Step 3: Protonation: the newly formed alkoxide is then protonated by either the solvent or water.

## References

[1] Vollhardt, Peter, and Neil Schore. Organic Chemistry: Structure and Function. 6th ed. New York: W. H. Freeman and Company, 2011.