What is the major product of the reaction of 2-methyl-2-bromopropane with KOH?

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Q: What is the major product of the reaction of 2-methyl-2-bromopropane with KOH?

Solution: The reaction favors elimination due to the bulky substrate, leading to the formation of 2-methylpropene.

Steps: 6

Step 1: Identify the starting material, which is 2-methyl-2-bromopropane. This compound has a bromine atom attached to a carbon that is also connected to three other carbon atoms.
Step 2: Recognize that KOH (potassium hydroxide) is a strong base. Strong bases can either cause substitution reactions or elimination reactions.
Step 3: Determine the structure of 2-methyl-2-bromopropane. It is a bulky molecule because of the three methyl groups attached to the central carbon.
Step 4: Understand that bulky substrates tend to favor elimination reactions over substitution reactions. This is because the base can more easily remove a hydrogen atom than replace the bromine atom.
Step 5: In an elimination reaction, KOH will remove a hydrogen atom from a carbon adjacent to the carbon with the bromine atom, leading to the formation of a double bond.
Step 6: The major product of this elimination reaction will be 2-methylpropene, which is formed by the removal of HBr from 2-methyl-2-bromopropane.