The reaction of toluene (methylbenzene) with HBr leads to the alkyl group (methyl) being substituted with a bromine atom at the benzylic position, forming 1-bromo-2-methylbenzene as the major product.
The structure of the product is: \[ \text{C}_6\text{H}_4\text{CH}_3\text{Br} \]
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.