The major aromatic product C in the following reaction sequence will be:

• Step 1: Reaction with excess $$\text{HBr}$$ and heat ($$\Delta$$)

  • The starting cyclic ether is protonated by $$\text{HBr}$$, followed by a nucleophilic attack by $$\text{Br}^-$$ to cleave the ether ring. This forms a phenolic group ($$-\text{OH}$$) on the benzene ring and an alkyl bromide intermediate.
  • Since $$\text{HBr}$$ is in excess, it undergoes an addition reaction across the double bond of the side chain according to Markovnikov's rule.
  • This gives the intermediate product A, which is a vicinal dibromide with a phenolic group: $$\text{C}_6\text{H}_4(\text{OH})(\text{CH(Br)CH(Br)CH}_3)$$



• Step 2: Reaction with alcoholic $$\text{KOH}$$ followed by acidic workup ($$\text{H}^+$$)

  • Alcoholic $$\text{KOH}$$ is a strong dehydrohalogenating agent. It carries out a double elimination ($$-2\text{HBr}$$) on the vicinal dibromide chain to form a triple bond (alkyne).
  • The resulting intermediate B contains a terminal or internal alkyne attached to the benzene ring alongside the phenol group. Under these conditions, the conjugated structure stabilizes as an allylic/propargyl double bond system: $$\text{C}_6\text{H}_4(\text{OH})(\text{CH}=\text{CH}\text{--CH}_3)$$



• Step 3: Ozonolysis with (i) $$\text{O}_3$$ and (ii) $$\text{Zn/H}_3\text{O}^+$$

  • This is a reductive ozonolysis step that selectively cleaves the carbon-carbon double bond ($$\text{C}=\text{C}$$) in the side chain.
  • The double bond is broken and replaced with carbonyl ($$\text{C}=\text{O}$$) groups on both fragments: $$\text{C}_6\text{H}_4(\text{OH})(\text{CH}=\text{CH}\text{--CH}_3) \xrightarrow{\text{O}_3, \ \text{Zn/H}_3\text{O}^+} \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) + \text{CH}_3\text{CHO}$$
  • The side chain is chopped down to an aldehyde group ($$-\text{CHO}$$) attached directly to the benzene ring at the position adjacent (ortho) to the hydroxyl group. This aromatic product is known as Salicylaldehyde (2-hydroxybenzaldehyde).

Conclusion:

The sequence converts the cyclic core through opening, double elimination, and ozonolytic cleavage to yield a substituted benzene ring featuring both an explicit phenol and a newly formed aldehyde function. This perfectly matches the molecule highlighted in the solution panel.

Answer: Option A — Salicylaldehyde (2-hydroxybenzaldehyde)