Propane molecule on chlorination under photochemical condition gives two di-chloro products, "x" and "y". Amongst "x"…

Step 1: Dichlorination of propane

Propane: $\ce{CH3-CH2-CH3}$

Dichlorination gives two products:

• Product x: $\ce{CH3-CHCl-CH2Cl}$ (1,2-dichloropropane) - has a chiral center at C-2, optically active
• Product y: $\ce{CH2Cl-CH2-CH2Cl}$ (1,3-dichloropropane) - no chiral center, not optically active

Since x is optically active, x = 1,2-dichloropropane = $\ce{CH3-CHCl-CH2Cl}$

Step 2: Trichlorination of x

Starting from $\ce{CH3-CHCl-CH2Cl}$, add one more Cl:

Possible positions for the third Cl:

1. On C-1 methyl: $\ce{CH2Cl-CHCl-CH2Cl}$ (1,1,2-trichloropropane)
2. On C-2: $\ce{CH3-CCl2-CH2Cl}$ (1,2,2-trichloropropane)
3. On C-3: $\ce{CH3-CHCl-CHCl2}$ (1,2,2-trichloropropane) - wait, this is C-3, so it's $\ce{CH3-CHCl-CHCl2}$ which is actually 1,2,3-trichloropropane

Let me reconsider. Starting structure: $\ce{CH3-CHCl-CH2Cl}$

Positions for adding 3rd Cl:

• C-1: $\ce{CH2Cl-CHCl-CH2Cl}$ → 1,1,2-trichloropropane
• C-2: $\ce{CH3-CCl2-CH2Cl}$ → 1,2,2-trichloropropane
• C-3: $\ce{CH3-CHCl-CHCl2}$ → 1,2,3-trichloropropane

Step 3: Count structural isomers

Three distinct structural isomers:

1. 1,1,2-trichloropropane
2. 1,2,2-trichloropropane
3. 1,2,3-trichloropropane

Answer: (a) 3

Key Points:

• Optically active dichloro product must have a chiral center
• Free radical chlorination can occur at any C-H position
• Count only structural isomers, not stereoisomers