The product (C) in the following sequence of reactions has ______ $$\pi$$ bonds.

Step 1: Formation of Intermediate (A) and (B)

• Reactant: Propylbenzene ($$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_3$$)
• Reagents: 1) $$\text{KMnO}_4\text{ - KOH, }\Delta$$   followed by 2) $$\text{H}_3\text{O}^+$$
• Reaction: Vigorous side-chain oxidation. Any alkyl group with at least one benzylic hydrogen is completely oxidized down to a carboxylic acid group attached directly to the benzene ring.
• Product (B): Benzoic acid ($$\text{C}_6\text{H}_5\text{COOH}$$)

Step 2: Formation of Major Product (C)

• Reactant: Benzoic acid ($$\text{C}_6\text{H}_5\text{COOH}$$)
• Reagents: $$\text{Br}_2\text{ / FeBr}_3$$ (Electrophilic Aromatic Substitution / Bromination)
• Reaction: The carboxylic acid group ($$-\text{COOH}$$) is a strongly deactivating and meta-directing group. Therefore, the bromine atom attacks the meta position.
• Product (C): m-Bromobenzoic acid (3-bromobenzoic acid)

Step 3: Counting the $$\pi$$-bonds in Product (C)

Let's count the $$\pi$$-bonds in the chemical structure of m-bromobenzoic acid:

1. Benzene Ring: Contains 3 $$\pi$$-bonds (alternate double bonds).
2. Carboxylic Acid Group ($$-\text{COOH}$$): Contains 1 $$\pi$$-bond (from the carbon-oxygen double bond, $$\text{C}=\text{O}$$).
3. Bromine Atom ($$-\text{Br}$$): Contains only $$\sigma$$-bonds (no $$\pi$$-bonds).

$$\text{Total }\pi\text{-bonds} = 3 + 1 = 4$$