n-Butane on monochlorination under photochemical condition gives an optically active compound "P". "P" on further chlorination gives dichloro compounds. The number of dichloro compounds obtained (ignore stereoisomers) is :

Step 1: Identify Compound "P"

$$n$$-Butane has the structure: $$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3$$.

When it undergoes monochlorination ($$\text{Cl}_2 / h\nu$$), it can form two structural isomers:

1. 1-Chlorobutane: $$\text{CH}_2\text{Cl}-\text{CH}_2-\text{CH}_2-\text{CH}_3$$ (Achiral / Optically inactive)
2. 2-Chlorobutane: $$\text{CH}_3-\text{CH*Cl}-\text{CH}_2-\text{CH}_3$$ (Chiral at C2 / Optically active)

Since the problem states that compound "P" is optically active, "P" must be 2-chlorobutane.

Step 2: Further Chlorination of "P"

Now, we add a second chlorine atom to 2-chlorobutane at every possible carbon position to find all the unique structural (constitutional) dichloro compounds.

Let's number the chain: $$\overset{1}{\text{C}}\text{H}_3 - \overset{2}{\text{C}}\text{H(Cl)} - \overset{3}{\text{C}}\text{H}_2 - \overset{4}{\text{C}}\text{H}_3$$

• Substitution at C1:

  $$\text{CH}_2\text{Cl}-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3 \rightarrow \textbf{1,2-dichlorobutane}$$

• Substitution at C2:

  $$\text{CH}_3-\text{C(Cl)}_2-\text{CH}_2-\text{CH}_3 \rightarrow \textbf{2,2-dichlorobutane}$$

• Substitution at C3:

  $$\text{CH}_3-\text{CH(Cl)}-\text{CH(Cl)}-\text{CH}_3 \rightarrow \textbf{2,3-dichlorobutane}$$

• Substitution at C4:

  $$\text{CH}_3-\text{CH(Cl)}-\text{CH}_2-\text{CH}_2\text{Cl} \rightarrow \textbf{1,3-dichlorobutane}$$

Ignoring stereoisomers, we get exactly 4 unique structural dichloro compounds:

1. 1,2-dichlorobutane
2. 2,2-dichlorobutane
3. 2,3-dichlorobutane
4. 1,3-dichlorobutane

Therefore, the correct option is B (4).