The purpose of this experiment was to perform elimination reactions with two different bases: potassium hydroxide and potassium t-butoxide. We explored the principles of dehydrohalogenation, Zaitsev’s rule, Hofmann’s rule, and elimination. Using these strong bases eliminated problems that could arise due to by-products produced by substitution reactions, and also allows a study of effects on the product ratios because one base is stronger than the other and has different steric requirements and molecular structure. In comparing the ratios of the different products produced by potassium hydroxide and potassium t-butoxide, we learned how different solvents and reagents affect the course and outcome of a chemical reaction. Reaction Scheme Theory Reaction Mechanism An elimination reaction is a reaction in organic chemistry with which two substituents are …show more content…
The resulting products were 2-methyl-2-butene and 2-methyl-1-butene. For the first part of the experiment we used 0.945 moles of KOH and 0.02 moles of 2-methyl-2-bromobutane, making 2-methyl-2-bromobutane the limiting reagent. To calculate theoretical yield, I multiplied the moles of the limiting reagent and the molecular weight of the product. The major product for the first part of the experiment was 2-methyl-2-butene, so: (0.02 moles)(70.13 g/mol) = 1.4 grams. The weight of 2-methyl-2-butene recovered was 1.056 grams, so in finding percent yield I divided the two: (1.056)/(1.4) x 100 = 75.4%. For the second part of the experiment we used .20 moles of KOt-Bu and 0.02 moles of 2-methyl-2-bromobutane, again making 2-bromo-2-methylbutane the limiting reagent. In calculating the theoretical yield: (0.02 moles)(70.13 g/mol) = 1.4 grams. The weight of the major product (2-methyl-1-butene) recovered was 1.299 grams, making the percent yield: (1.299)/(1.4) x 100 =