In the reaction of 2-bromobutane with KOH in ethanol, what type of mechanism is primarily involved?

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Q: In the reaction of 2-bromobutane with KOH in ethanol, what type of mechanism is primarily involved?

Solution: The reaction proceeds via an E2 mechanism due to the strong base (KOH) and the presence of a good leaving group (Br).

Steps: 6

Step 1: Identify the reactant, which is 2-bromobutane. It has a bromine (Br) atom that can leave.
Step 2: Recognize that KOH is a strong base. Strong bases are good at removing protons (H+).
Step 3: Understand that in this reaction, the base (KOH) will remove a hydrogen atom from the carbon adjacent to the one with the bromine.
Step 4: Note that the bromine (Br) is a good leaving group, meaning it can easily leave the molecule.
Step 5: Realize that the reaction happens in one step where the base removes a hydrogen and the bromine leaves at the same time, which is characteristic of an E2 mechanism.
Step 6: Conclude that since we have a strong base and a good leaving group, the reaction primarily follows the E2 mechanism.