Major product $$B$$ of the following reaction has ______ $$\pi$$-bond.

Step 1: Formation of Product (A)

• Reactant: Ethylbenzene ($$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3$$)
• Reagent: $$\text{KMnO}_4\text{ / KOH, }\Delta$$ (Strong oxidizing agent)
• Reaction: Vigorous oxidation of the alkyl side chain containing benzylic hydrogens. Regardless of the chain length, the alkyl group is completely oxidized down to a carboxylic acid group attached to the benzene ring.
• Product (A): Benzoic acid ($$\text{C}_6\text{H}_5\text{COOH}$$)

Step 2: Formation of Major Product (B)

• Reactant: Benzoic acid ($$\text{C}_6\text{H}_5\text{COOH}$$)
• Reagent: $$\text{HNO}_3\text{ / H}_2\text{SO}_4$$ (Nitrating mixture)
• Reaction: Electrophilic aromatic substitution (Nitration). Since the carboxylic acid group ($$-\text{COOH}$$) is a strongly deactivating and meta-directing group, the nitro group ($$-\text{NO}_2$$) attacks the meta position.
• Product (B): m-Nitrobenzoic acid (3-nitrobenzoic acid)

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

1. Benzene Ring: Contains 3 $$\pi$$-bonds (alternate double bonds).
2. Carboxylic Acid Group ($$-\text{COOH}$$): Contains 1 $$\pi$$-bond (from the $$\text{C}=\text{O}$$ double bond).
3. Nitro Group ($$-\text{NO}_2$$): Contains 1 $$\pi$$-bond (resonating within the $$-\text{N}=\text{O}$$ bond system).

$$\text{Total }\pi\text{-bonds} = 3 + 1 + 1 = 5$$