What is the expected product when 2-bromo-2-methylpropane reacts with a strong base in a bimolecular elimination reaction?

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Q: What is the expected product when 2-bromo-2-methylpropane reacts with a strong base in a bimolecular elimination reaction?

Solution: The strong base facilitates the E2 elimination, resulting in the formation of 2-methylpropene.

Steps: 5

Step 1: Identify the starting compound, which is 2-bromo-2-methylpropane. This compound has a bromine atom attached to a carbon that is also connected to three other carbon atoms.
Step 2: Recognize that a strong base is present. Strong bases can remove protons (H atoms) from molecules.
Step 3: Understand that in a bimolecular elimination reaction (E2), the strong base will remove a hydrogen atom from a carbon adjacent to the carbon with the bromine atom.
Step 4: When the strong base removes the hydrogen atom, the bromine atom leaves, resulting in the formation of a double bond between the two carbon atoms.
Step 5: The product of this reaction is 2-methylpropene, which has a double bond between the second and third carbon atoms.