1. Find the two dichloro products of propane

Propane is $\ce{CH3-CH2-CH3}$. On chlorination, the possible dichloro structural isomers are:

• $\ce{1,1-dichloropropane}$: $\ce{CH3-CH2-CHCl2}$
• $\ce{1,2-dichloropropane}$: $\ce{CH3-CHCl-CH2Cl}$
• $\ce{1,3-dichloropropane}$: $\ce{ClCH2-CH2-CH2Cl}$
• $\ce{2,2-dichloropropane}$: $\ce{CH3-CCl2-CH3}$

The question says only two dichloro products, $x$ and $y$, are considered, and among them $x$ is optically active.

Among these, the only optically active dichloro product is:

$\ce{1,2-dichloropropane} = \ce{CH3-CHCl-CH2Cl}$

because the middle carbon is attached to four different groups:

• $\ce{H}$
• $\ce{Cl}$
• $\ce{CH3}$
• $\ce{CH2Cl}$

So,

$x = \ce{1,2-dichloropropane}$

2. Now chlorinate $x$ further to get trichloro products

Starting compound:

$\ce{CH3-CHCl-CH2Cl}$

We replace one more hydrogen by chlorine under photochemical chlorination.

Count distinct hydrogen-bearing positions:

• Carbon 1: $\ce{CH3}$ has 3 H
• Carbon 2: $\ce{CHCl}$ has 1 H
• Carbon 3: $\ce{CH2Cl}$ has 2 H

Now substitute one H from each distinct carbon position.

3. Products formed by substitution at each position

(i) Chlorination at carbon 1

$\ce{CH3-CHCl-CH2Cl -> CH2Cl-CHCl-CH2Cl}$

This is:

$\ce{1,2,3-trichloropropane}$

(ii) Chlorination at carbon 2

$\ce{CH3-CHCl-CH2Cl -> CH3-CCl2-CH2Cl}$

This is:

$\ce{1,2,2-trichloropropane}$

(iii) Chlorination at carbon 3

Replacing one H on $\ce{CH2Cl}$ gives:

$\ce{CH3-CHCl-CHCl2}$

This is:

$\ce{1,1,2-trichloropropane}$

4. Check for any additional structural isomers from stereochemistry

The question says consider only structural isomers, so enantiomers/diastereomers are not counted separately.

Thus the total number of distinct structural trichloro products obtained from $x$ is:

$3$

5. Match with options

Option C: 3

6. Comparison with stored correct answer

Stored correct answer = C

Our derived answer = C

So they agree.