Question

G the ratio of elimination to substitution is exactly the same (26% elimination) for 2−bromo−2−methylbutane and 2−iodo−2−methylbutane in 80% ethanol/20% water at 25°c. what two elimination products are formed from each substrate? select all that apply.

Answer 1

Answer:

2-methyl-2-butanol and 2-ethoxy-2-methylbutane

Explanation:

When the elimination occurs, a good leaving group must leave the compound and will be replaced. A good leaving group is a nucleophile, which intends to bond with compounds with positive charges. In this case, the bromine and iodine ions are nucleophiles that will leave the compounds.

In the mixture ethanol and water, two compounds must replace the leaving groups, which are strong nucleophiles: OH⁻ and ⁻OCH₂CH₃ (O is a strong nucleophile, and after the bonding, H⁺ will be lost in the ethanol molecule).

So, the two possible products are 2-methyl-2-butanol and 2-ethoxy-2-methylbutane, as shown below.