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I Br 0 +1-0 кон Methand он 2. 1C), H₂O Br H ✓ ↓ Acid-base B B "

I Br 0 +1-0 кон Methand он 2. 1C), H₂O Br H ✓ ↓ Acid-base B B "

Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN:9781559539418
Author:Angelica Stacy
Publisher:Angelica Stacy
ChapterU6: Showtime: Reversible Reactions And Chemical Equilibrium
Section: Chapter Questions
Problem 6RE
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draw the complete mechanism for this double E2 elimination reaction?

  • Starting Material: (2R,3S)-2,3-dibromo-3-phenylpropanoic acid
  • Final Product: Phenylpropiolic acid
  • Steps to Include:
    1. Deprotonation of the carboxylic acid group: Show how the carboxylic acid (-COOH) group is deprotonated by KOH to form the carboxylate salt (-COO⁻).
    2. First E2 elimination: Illustrate the elimination of the first HBr molecule, forming the alkene intermediate.
    3. Second E2 elimination: Show the elimination of the second HBr molecule, forming the alkyne.
    4. Final protonation: Depict how HCl protonates the carboxylate group (-COO⁻) to regenerate the carboxylic acid (-COOH).

Make sure to use proper arrow notation to show the movement of electrons in each step. Label all intermediates, reagents (KOH and HCl), and byproducts (HBr and KBr). I inldued some notes for how it kinda should look like

I Br 0 +1-0 кон Methand он 2. 1C), H₂O Br H ✓ ↓ Acid-base B B "
Transcribed Image Text:I Br 0 +1-0 кон Methand он 2. 1C), H₂O Br H ✓ ↓ Acid-base B B "
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