Question

Elimination Reactions Prelab Date 1. Explain why we get a different product when we use bulky bases compared to smal bases. (3 pts) 2. If we start with 25.0 mL of 0.100 M sodium methoxide and 25.0 g of 2-bromobutane, what will be the theoretical yield of your product? (Use dimensional analysis/ Do not break up your steps/Read the beginning of your lab to guide you on how to do this problem). (3 pts) 3. How can one determine which peak represents the Hoffman product and which represens the Zaitsev product? (3 pts) 4. Why do we use an acid in week II in the elimination reaction using a bulky base? (3 pts) 5. Write the reaction mechanism for the reaction of sodium methoxide with 2-bromobutane. (3 pts)

Answer 1

Q1: A bulky base would remove $\beta$ hydrogen from the least hindered carbon atom and less substituted alkene is the major product i.e. zaitsev product.

While a small base would remove $\beta$ hydrogen atom from the more hindered carbon atom because that would result into a more stable product i.e. Hoffman product.

Q2:

Moles of sodium methoxide = 25.0×10^-3L × 0.100mol/L = 2.5×10^-3 mol

Moles of 2-bromobutane = 25.0g/137.02g/mol = 1.825×10^-2

Sodium methoxide is the limiting reagent.

Moles of the product = 2.5×10^-3 mol

Q5: sodium methoxide shows elimination reaction (E₂ mechanism) with 2-bromobutane . The reaction proceed in a single step .

Br (E,) H E Na Octy