Final answer:
The reaction involving sodium methoxide (NaOMe) could yield a variety of products, depending on the specific organic compound it reacts with. The given reaction involves an intermediate reacting with sodium methoxide to yield an epoxide.
Step-by-step explanation:
The reaction in question is an organic chemistry transformation involving the use of sodium methoxide (NaOMe) and methanol (MeOH). Sodium methoxide acts as a base and allows the removal of a proton from a suitable molecule, typically inducing a reaction.
The major product of the reaction would depend on the specific organic compound reacting with the sodium methoxide. For instance, in reactions with organic compounds that contain carbonyl groups, methoxide ions (CH₃O-) can act as nucleophile, attacking the carbon of the carbonyl group, leading to a variety of products like alcohols, ethers, etc. It's also worth noting that NaOMe and MeOH are commonly used in methoxymercuration reactions, where they help in creating methoxy groups.
In your example provided, the intermediate 4',6'-O-benzylidene-2',3'-ditosyl is reacted with sodium methoxide. The result is an epoxide with a yield of 99%. The formation of the epoxide is the direct consequence of the reaction with sodium methoxide.
Learn more about Organic Chemistry Reaction