One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methyl
butane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures?
1 answer:
Answer:
Structures are given below.
Explanation:
- Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
- H atoms adjacent to Br will be eliminated.
- 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
- Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
- Structure of alkenes are given below.
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86 percent is the percent yield for this experiment if he expected to produce 5g of product.
Explanation:
Given that:
mass of test tube = 5 grams
mass of test tube + reactant is 12.5 grams
mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)
mass of reactant = 12.5 -5
= 7.5 grams
when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.
so mass of product formed = 9.3 - 5
= 4. 3 grams of product is formed (actual yield)
However, he expected the product to be 5 grams (theoretical yield)
Percent yield = x 100
putting the values in the formula:
percent yield = x 100
= 86 %
86 percent is the percent yield.