Final answer:
A plausible structure for an aromatic hydrocarbon C9H12 that yields two distinct C9H11Cl isomers upon chlorination is 1,3,5-trimethylbenzene. This compound has three methyl groups on a benzene ring, which upon substitution with chlorine at different positions, produces two unique chlorinated isomers.
Step-by-step explanation:
Structure of Aromatic Hydrocarbons C9H12
To propose a structure for an aromatic hydrocarbon C9H12 that can form two C9H11Cl products upon substitution with chlorine, we need to find a molecule that has two types of hydrogens on the aromatic ring which upon chlorination yields two distinct isomers. A suitable starting compound would be 1,3,5-trimethylbenzene. This molecule has three methyl groups attached to a benzene ring, leaving five hydrogens on the ring. When one of these hydrogens is substituted with chlorine, it results in two types of products: chlorination at hydrogen in positions 2 or 4 relative to any methyl group will give one C9H11Cl isomer, while chlorination at position 6 will give another C9H11Cl isomer due to the asymmetry introduced by the three methyl groups.
The presence of the methyl groups and their arrangement in this manner creates positional isomers of the chlorinated products. The two possible products differ in where the chlorine atom is substituted on the aromatic ring.