Question

Which of the following is the major product of the reaction between 1-bromobutane and potassium hydroxide in ethanol?

• A. Butan-1-ol
• B. But-2-ene
• C. Butan-2-ol
• D. Butane

Hint:

In a reaction with KOH in ethanol, the elimination (E2) mechanism typically leads to the formation of an alkene.

Answer 1

The Correct Option is B

The reaction between 1-bromobutane (an alkyl halide) and potassium hydroxide (KOH) in ethanol is a characteristic E2 elimination reaction. In this mechanism, the hydroxide ion (OH⠻) abstracts a proton from the carbon adjacent to the carbon bonded to the leaving group (Br⠻), resulting in the formation of a double bond. Step 1: Determine the reaction type Ethanol, a polar protic solvent, favors an E2 elimination mechanism. Step 2: Identify the product - An alkene is produced via the E2 mechanism. - Hydrogen is removed from the carbon adjacent to the carbon bearing bromine (the β-carbon), forming But-2-ene. Answer: The principal product of this reaction is But-2-ene. Therefore, option (2) is the correct choice.