What is the stereochemistry of the product formed in an SN2 reaction involving a haloalkane? Provide an example

The stereochemistry of the product formed in an SN2 reaction involving a haloalkane is inverted compared to the starting material. This is due to the backside attack mechanism of the SN2 reaction.

Here's a step-by-step example:

1. Consider a simple haloalkane, like (R)-2-bromobutane. In this molecule, the bromine atom is attached to a chiral carbon atom, which has four different groups attached to it.

2. In an SN2 reaction, a nucleophile (for example, a hydroxide ion, OH-) will attack the carbon atom attached to the bromine from the side opposite to the bromine atom. This is known as a backside attack.

3. As the nucleophile attacks, it begins to form a bond with the carbon atom. At the same time, the carbon-bromine bond begins to break.

4. At the transition state of the reaction, the carbon atom is attached to both the bromine atom and the nucleophile. The molecule is flat at this point, with the remaining two groups attached to the carbon atom pointing away from the bromine and the nucleophile.

5. Finally, the bromine atom leaves, and the nucleophile forms a full bond with the carbon atom. Because the nucleophile attacked from the opposite side to the bromine atom, the configuration of the carbon atom is inverted compared to the starting material.

6. The product of this reaction is (S)-2-butanol, which has the opposite configuration to the starting material, (R)-2-bromobutane.

So, the stereochemistry of the product formed in an SN2 reaction involving a haloalkane is inverted compared to the starting material.