The two products formed in above reaction are:

Step 1: Draw 2-hexene

2-Hexene: $\ce{CH3-CH=CH-CH2-CH2-CH3}$

The double bond is between C2 and C3.

Step 2: Apply ozonolysis conditions

The reagents are: (i) $\ce{O3}$ (ii) $\ce{H2O}$

$\ce{H2O}$ workup is oxidative (no $\ce{Zn}$ added) → this is oxidative ozonolysis. Oxidative workup conditions ($\ce{O3/H2O}$ or $\ce{O3/H2O2}$) convert:

• $\ce{R-CH=}$ → $\ce{R-COOH}$ (carboxylic acid)
• $\ce{R-CR'=}$ → $\ce{R-CO-R'}$ (ketone, if no H on carbon)

Step 3: Identify the products

Cleavage of $\ce{CH3-CH=CH-CH2CH2CH3}$ at C2=C3:

• C2 fragment ($\ce{CH3-CH=}$) → $\ce{CH3COOH}$ (acetic acid)
• C3 fragment ($\ce{=CH-CH2CH2CH3}$) → $\ce{CH3CH2CH2COOH}$ (butanoic acid)

Answer: Option (a) — Butanoic acid and acetic acid

Key Points to Remember:

• Oxidative ozonolysis ($\ce{O3}$ + $\ce{H2O}$ or $\ce{H2O2}$): $\ce{-CH=}$ → $\ce{-COOH}$
• Reductive ozonolysis ($\ce{O3}$ + $\ce{Zn/H2O}$): $\ce{-CH=}$ → $\ce{-CHO}$
• 2-Hexene cleaves between C2 and C3 giving $\ce{C2}$ acid (acetic) + $\ce{C4}$ acid (butanoic)