The major product of the following reaction is:

Step 1: Identify the substrate and reaction type

The starting material has:

• An aromatic ring with $\ce{-OCH3}$ (methoxy) substituent
• A side chain ending in $\ce{-CH2Cl}$ (alkyl chloride) — 3 carbons from the ring

With $\ce{AlCl3}$ (Lewis acid), this undergoes intramolecular Friedel-Crafts alkylation.

Many books mark option d as correct answer but in my opinion that's incorrect. Primary carbocations ($1^\circ$) are too unstable to exist as free intermediates. In a Friedel-Crafts reaction with a primary alkyl halide and $AlCl_3$, the reaction proceeds via an $\text{S}_\text{N}2$-like displacement or a tight ion-pair.

By the time the $AlCl_3$ pulls the chloride away enough to allow for a hydride shift, the benzene ring (which is part of the same molecule) is already positioned to attack.

Ring Strain: A 6-membered cyclohexane-like ring has almost zero ring strain, whereas a 5-membered cyclopentane-like ring has approximately $26 \text{ kJ/mol}$ of strain.

The Distance Factor: While 5-membered rings often form faster (entropy), 6-membered rings are thermodynamically more stable. In Friedel-Crafts alkylations, which are often reversible or governed by the stability of the transitioning complex, the 6-membered path is highly competitive.

Key Points to Remember:

• Intramolecular FC alkylation prefers 6-membered ring formation
• $\ce{-OCH3}$ activates ortho/para positions for electrophilic attack
• The most stable carbocation determines the ring closure site