Organic chemistry

Question

Under what conditions does transesterfication tend to favor the product?

Answer 1

Hi Samantha!

This is a great question. Let's look at a few case studies:

1.) Le Chatlier's Principle and Equilibrium: Transesterification of Cyclohexylacetate:

Cy-O-C(O)CH₃ + MeOH + K₂CO₃ --> Cy-OH + MeO-C(O)CH₃

Cy = cyclohexyl

In this reaction, we have converted the cyclohexyl ester to a methyl ester by reacting the starting material (SM) with a basic methanol solution (MeOH is the solvent, K₂CO₃ is a catalyst). This reaction is favored in the forward direction due to Le Chatlier's principle. Let's say we had 0.1 mol of the SM in 1 L of methanol. The concentration of SM is 0.1 M. The concentration of methanol is very high, as it is the solvent ([MeOH] ~ 25 M at room temperature). If the reaction were to go to completion in the forward direction, the concentration of Cy-OH would be 0.1 M, while [MeOH] wouldn't change much. Let's write the equilibrium constant for this reaction:

Keq = [CyOH] [MeO-C(O)CH₃] / [SM] [MeOH]

Lets assume the equilibrium constant is ~ 1, which is a fair assumption since CyOH and MeOH are very similar substituents (both are aliphatic alcohols). If the reaction were to go to completion, the ratio of MeOH : CyOH would be 250:1, meaning the ratio of MeO-C(O)CH₃ : Cy-O-C(O)CH₃ would be 250:1. The reaction is driven to product in order to balance the equilibrium. Another way to think about this is that there is way more MeOH around to react in the forward direction than there is CyOH to react in the reverse.

These Conditions (K₂CO₃ in MeOH) are often used in total synthesis of natural products to remove acyl protecting groups, most commonly acyl (-C(O)CH₃; -Ac) as a protecting group for a hydroxyl group.

2.) Lactone Formation: intramolecular reaction

HOCH₂-(CH₂)₄-CO₂Me + 0.01 equiv. NaOMe --> δ-valerolactone + MeOH

run in dichloromethane (also known as methylene chloride, CH₂Cl₂)

δ-valerolactone is a 6-membered ring lactone.

The starting material is an ester with a pendant alcohol. This means it can undergo an intramolecular lactonization. This reaction is highly favored under these conditions. This is a product of kinetics of the reaction--because the pendant alcohol is always nearby the ester, its almost like the pendant alcohol is in high concentration. The MeOH will drift away into solution and be diluted.

Note: This reaction is very facile because 6-membered rings have little to no strain. Small rings, like 4-membered lactones (β-lactones), are highly strained, and will therefore favor the ring-opened ester.

3.) Leaving groups: Methanolysis of phenyl acetate

Ph-O-C(O)CH₃ + MeOH + K₂CO₃ --> PhOH + MeO-C(O)CH₃

Ph = C₆H₅–

This reaction is highly favored because of the relative stability of the reactants and products. For an exothermic reaction (products more stable than reactants), Keq > 1, meaning the forward reaction is favored (see Gibbs Free Energy). This reaction results in formation of the Phenolate anion (PhO¯) in the forward direction, and methoxide anion (MeO¯) in the reverse reaction. The pKa of PhOH is ~10 and the pKa of MeOH is 16. This means PhOH is 6 ORDERS OF MAGNITUDE more acidic than methanol, and that the phenolate ion is more stable than the methoxide anion.

Please take these cases as helpful examples of 3 key principles of chemical reactions

1) Reaction Equilibria/Le Chatlier's Principle

2) Reaction Kinetics

3) Reaction Thermodynamics

Other concepts relevant to these examples include resonance (Phenol is more acidic than methanol, because the negative charge of the conjugate base is stabilized by resonance with the aryl ring), strain and sterics. Please let me know if I can help you further with other topics in Organic Chemistry!