Question

When optically active (S)-2-methylcyclopentanone is treated with an acid (H3O ), 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

Step-by-step explanation:

When (S)-2-methylcyclopentanone is treated with an acid, the carbonyl oxygen atom is protonated.

This leads to the emergence of a positive charge on the cabonyl oxygen. Then the C=O bond is now transformed into H-C-OH where C is a carbocation which leads to a racemic mixture. Loss of a proton from the adjacent carbon atom completes the mechanism.

In this process, the optically active (S)-2-methylcyclopentanone is converted to a pair of diastereomers thereby loosing its optical activity.