Question

Propose a mechanism for this reaction. C=Cc1ccccc1 \(+\mathrm{MeOH}\) COC(C)c1ccccc1 (racemic)

Short answer

Question: Explain the mechanism for the reaction between an alkene compound with a phenyl group and methanol to produce a racemic product. Answer: The mechanism involves a nucleophilic attack of the alkene double bond on the electrophilic oxygen atom in methanol, forming a carbocation intermediate. This carbocation is planar and can be attacked by the oxygen atom from either side, leading to the formation of two stereoisomers (R and S) in equal amounts, creating the racemic mixture. A base then deprotonates the oxonium ion to form the ether compound as the final product.

Step-by-step solution

Identify the Reactive Sites

On the alkene compound, the reactive site is the double bond between the carbon atoms. In methanol, the reactive site is the oxygen atom which has a lone pair of electrons and a hydrogen atom that can be abstracted as a proton.

Determine the Electrophilic and Nucleophilic Sites

The alkene double bond is electron-rich and will act as a nucleophile, while the oxygen atom in methanol can act as an electrophile.

Formation of the Carbocation and Oxonium Ion

The nucleophilic alkene attacks the electrophilic oxygen atom in methanol, forming a bond between the alkene carbon and the methanol oxygen. The other bond in the double bond of the alkene transfers its electrons to the carbon-carbon bond, creating a positive charge on the other carbon atom - this creates a carbocation intermediate. The previously bonded hydrogen in methanol now has a positive charge, creating an oxonium ion.

Deprotonation of the Oxonium Ion

A base (could be another molecule of methanol) abstracts the proton from the oxonium ion, forming a carbon-oxygen bond between the carbocation and oxygen. The molecule has now transformed into the product, which is an ether compound.

Formation of the Racemic Mixture

Since the carbocation intermediate formed in Step 3 is planar, it can be attacked by the nucleophilic oxygen in methanol from either side (above or below the plane). This leads to the formation of two stereoisomers (R and S) in equal amounts, giving the racemic mixture in the final product. Overall, the proposed mechanism involves a nucleophilic attack on the oxygen atom in methanol by the alkene double bond, formation of a carbocation intermediate, deprotonation of the oxonium ion to form the ether compound, and the formation of a racemic mixture due to the planar nature of the carbocation intermediate.