Question

A compound (A) with molecular formula $C_4H_9I$ 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), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH. 

Hint:

Elimination reactions, particularly E2, can be triggered by strong bases like alcoholic KOH, leading to the formation of alkenes.

Answer 1

Upon treatment with alcoholic KOH, A undergoes elimination to form an alkene. The reaction follows an E2 mechanism, where the proton is abstracted and the leaving group is eliminated.
Structures: (A) Alkyl halide (B) Alkene formed after elimination (C) Alcoholic KOH used (D) Possible side product from elimination