Question

When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain this observation and draw a mechanism that shows how racemization occurs. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.

Answer 1

See explanation below

Step-by-step explanation:

In this case, we need to see which is the structure of this compound. Now, racemization occurs basically because we are in an aqueous basic medium, and the ketone can reacts again with water in the medium to form the starting reagent.

First, the base will take out the Alpha hydrogen from the ketone, then, the negative charge goes down and opens up the carbonile group, forming a double bond in there. Later, with the water of the medium, it reacts and substract a proton, and then, with keto enolic equilibrium, forms again the ketone, but this ketone is different from the start, it will be the R isomer which is not optically active.

See picture below for mechanism