What type of stereochemistry is observed in the product of the reaction between 2-bromobutane and a strong base in an E2 elimination?

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Q: What type of stereochemistry is observed in the product of the reaction between 2-bromobutane and a strong base in an E2 elimination?

Solution: E2 elimination typically leads to the formation of alkenes with trans stereochemistry due to the anti-periplanar requirement.

Steps: 5

Step 1: Understand that E2 elimination is a type of reaction where a strong base removes a hydrogen atom and a leaving group (like bromine) from adjacent carbon atoms.
Step 2: Identify the structure of 2-bromobutane, which has a bromine atom on one carbon and hydrogen atoms on the adjacent carbon.
Step 3: Recognize that for E2 elimination to occur, the hydrogen atom that is removed must be on the carbon adjacent to the carbon with the bromine, and they must be in an anti-periplanar arrangement.
Step 4: Realize that this anti-periplanar arrangement leads to the formation of a double bond between the two carbons, resulting in an alkene.
Step 5: Determine the stereochemistry of the alkene formed. In this case, the product will have trans stereochemistry, meaning the larger groups (like the alkyl chains) are on opposite sides of the double bond.