Question

A compound (A) with molecular formula \( \text{C}_4\text{H}_9\text{I} \), which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), \( \text{C}_8\text{H}_{18} \), which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of (A), (B), (C) and (D). Write the chemical equation when compound (A) is reacted with alcoholic KOH.

Hint:

When reacting an alkyl halide with alcoholic KOH, it typically leads to the formation of alkenes through an elimination reaction.

Answer 1

Compound (A): \( \text{C}_4\text{H}_9\text{I} \) is butyl iodide (n-butyl iodide). Compound (B): The reaction of butyl iodide with alcoholic KOH gives butene (C\(_4\)H\(_8\)). Compound (C): The reaction of butene with HI gives 2-iodobutane. Compound (D): When n-butyl iodide reacts with sodium in dry ether, butane is formed. The chemical reaction for the formation of (B) is: \[ \text{C}_4\text{H}_9\text{I} + \text{KOH (alc)} \longrightarrow \text{C}_4\text{H}_8 + \text{KI} + \text{H}_2\text{O} \]