Question

What is obtained when (S)-2-butanol is treated with tosyl chloride and pyridine, followed by exposure to bromide? O (R)-2-bromobutane O (S)-2-bromobutane O (S)-2-chlorobutane O (R)-2-butanol

Answer 1

The product obtained is (R)-2-bromobutane.

Step-by-step explanation:

The product obtained when (S)-2-butanol is treated with tosyl chloride and pyridine, followed by exposure to bromide is (R)-2-bromobutane.

In this reaction, the hydroxyl group (-OH) of (S)-2-butanol is converted to a good leaving group (tosylate group) by reacting with tosyl chloride in the presence of pyridine. Then, the tosylate group is replaced by a bromide ion, resulting in the formation of (R)-2-bromobutane.

The stereochemistry of the product is determined by the configuration of the starting (S)-2-butanol. Since the bromine replaces the hydroxyl group on the same side of the molecule, the product is (R)-2-bromobutane.

Answer 2

When (S)-2-butanol is treated with tosyl chloride and pyridine, and then exposed to bromide, it undergoes an SN2 reaction resulting in configuration inversion at the stereogenic center, yielding (R)-2-bromobutane.

Step-by-step explanation:

When (S)-2-butanol is treated with tosyl chloride and pyridine, followed by exposure to bromide, (S)-2-butanol is converted into an alkyl tosylate which is then replaced by bromide to give (S)-2-bromobutane. This reaction proceeds via an SN2 mechanism, which involves a backside attack by the bromide ion on the alkyl tosylate. During this process, the configuration at the stereogenic center is inverted due to the backside attack, and thus the (S)-configuration of the starting alcohol becomes the (R)-configuration in the final product. Therefore, the correct answer is (R)-2-bromobutane.