Draw the product of the E2 reaction shown below. Include the correct stereochemistry. Ignore any inorganic byproducts. DBU Draw the E2 Product @ 68% – Br Submit
The Correct Answer and Explanation is :
To answer this question, let’s consider the key points about the E2 reaction and its stereochemical implications:
E2 Reaction Overview:
- Elimination Mechanism: E2 reactions proceed in a single step where the base abstracts a proton from a β-carbon, and a leaving group (e.g., Br⁻) departs simultaneously. This results in the formation of a double bond.
- Anti-Periplanar Geometry: For the reaction to proceed efficiently, the β-hydrogen and leaving group must be in an anti-periplanar (opposite planes, 180° apart) conformation.
- Stereochemistry: The product’s configuration depends on the starting molecule’s stereochemistry and anti-periplanar arrangement.
Given Reaction:
- Base: DBU (a strong, non-nucleophilic base) favors elimination.
- Substrate: A molecule with a bromine atom (-Br) and β-hydrogens available for elimination.
- Focus: Include correct stereochemistry in the product.
Steps to Determine the Product:
- Identify the β-Hydrogens: Locate the hydrogen(s) on the β-carbon (next to the carbon with -Br).
- Ensure Anti-Periplanar Geometry: Rotate the molecule (if needed) to align the β-hydrogen and -Br anti-periplanar. This geometry ensures the correct stereochemical outcome.
- Form the Double Bond: Remove the β-hydrogen and -Br to create a π-bond between the α and β-carbons.
Answer with Explanation:
The product is an alkene with a trans configuration if the starting material allows it. Here’s why:
- DBU abstracts the β-hydrogen anti-periplanar to -Br.
- The double bond forms between the α and β-carbons, with substituents positioned to minimize steric hindrance, favoring the more stable trans configuration.
I will draw the product with correct stereochemistry for clarity.
It seems I can’t directly render the chemical structure here, but I’ll describe it in detail instead.
Structure of the Product:
Assume a starting substrate like 2-bromobutane:
- The bromine is on the second carbon (C2).
- The β-hydrogens are on the adjacent carbons (C1 and C3).
Upon elimination:
- The double bond forms between C2 and C3.
- Substituents on C2 and C3 adopt a trans configuration (opposite sides of the double bond), yielding trans-2-butene as the product.
Conclusion:
The E2 reaction yields a stereoselective product favoring the trans alkene due to its greater stability. The use of a strong base like DBU ensures efficient elimination with the correct anti-periplanar alignment of the β-hydrogen and bromine.