The major product of the following reaction is:

Step 1: Nucleophilic Substitution with $$\text{HCl}$$

The aliphatic hydroxyl group ($$-\text{OH}$$) undergoes protonation followed by nucleophilic substitution by the chloride ion ($$\text{Cl}^\ominus$$). This replaces the alcohol with a chlorine atom, forming an alkyl chloride intermediate.

Step 2: Intramolecular Friedel-Crafts Alkylation

Anhydrous $$\text{AlCl}_3$$ abstracts the chloride ion from the intermediate to generate a stable carbocation. This carbocation acts as an electrophile and undergoes a rapid intramolecular electrophilic substitution at the highly activated position ortho to the phenolic hydroxyl group, closing the chain to form a fused five-membered ring (indane core).