Answer to Question #133524 in Organic Chemistry for Matthew Immordino

Question

when a compound (C2H4)2) with a pK1 of 4.8 reacts with 1. NaHCO3 and 2. C3H7Br, a product forms. What is it?

Answer 1

C₂H₄O₂             PK_A = 4.8 acid

D.B.E      = x+1 - y/2

               = 2+1 - 4/2

               = 1 CH₃COOH

1-  CH₃COOH + NaHCO₃ à CH₃COO^-Na⁺ +H₂O+CO₂

2- CH₃COO + C₃H₇Br à CH₃COO C₃H₇ +HBr