The major product of the following addition reaction is:

Step 1: Identify the starting material and reaction

The starting material is propene ($\ce{CH3-CH=CH2}$) or a similar alkene being treated with $\ce{Cl2/H2O}$ (chlorine water).

Step 2: Reaction of alkene with $\ce{Cl2/H2O}$

$\ce{Cl2}$ in the presence of water undergoes halohydrin formation:

1. $\ce{Cl^+}$ (or $\ce{Cl2}$) adds to the double bond forming a chloronium ion intermediate
2. Water (the nucleophile) attacks the more electrophilic carbon (more substituted = Markovnikov)
3. $\ce{Cl-}$ attacks from the opposite face of the more substituted carbon

Step 3: Product Answer: Option (c)

Key Points to Remember:

• $\ce{Cl2/H2O}$ = chlorine water → gives chlorohydrin
• $\ce{OH}$ goes to the more substituted carbon (Markovnikov position)
• $\ce{Cl}$ goes to the less substituted carbon
• Anti addition: $\ce{OH}$ and $\ce{Cl}$ on opposite faces