Question

What happens when n-butyl chloride reacts with alcoholic KOH?

Hint:

The reaction between alkyl halides and alcoholic KOH is a typical elimination reaction, often leading to the formation of alkenes.

Answer 1

Step 1: Understand the reaction.
When n-butyl chloride (C\(_4\)H\(_9\)Cl) reacts with alcoholic potassium hydroxide (KOH), a dehydrohalogenation reaction occurs. In this reaction, the hydroxyl group from KOH removes a hydrogen atom from the carbon adjacent to the carbon bonded to the chlorine atom, resulting in the elimination of hydrogen chloride (HCl).
Step 2: Mechanism of the reaction.
Alcoholic KOH induces an elimination (E2) reaction, where the chlorine atom is replaced by a double bond, forming butene. The reaction can proceed as follows: \[ C_4H_9Cl \xrightarrow{KOH (alcoholic)} C_4H_8 + HCl \] In this case, n-butyl chloride undergoes elimination to form butene, which is an alkene.
Step 3: Conclusion.
Thus, when n-butyl chloride reacts with alcoholic KOH, it undergoes an elimination reaction to form butene and hydrogen chloride.
Final Answer: When n-butyl chloride reacts with alcoholic KOH, butene is formed through an elimination reaction, releasing HCl.
Final Answer: \[ \boxed{\text{n-Butyl chloride reacts with alcoholic KOH to form butene through an elimination reaction, releasing HCl.}} \]