Number of Dichloro Compounds from Photochemical Monochlorination of n Butane

To determine the number of dichloro compounds obtained, we analyze the reaction step-by-step:

Step 1: Identification of Compound "P"

Monochlorination of $n$-butane ($\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3$) gives two structural isomers:

1. 1-chlorobutane ($\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Cl}$): It does not possess any chiral center, making it optically inactive.
2. 2-chlorobutane ($\text{CH}_3-\text{CH}_2-\overset{*}{\text{C}}\text{H(Cl)}-\text{CH}_3$): The second carbon ($\text{C}_2$) is attached to four different groups ($-\text{H}$, $-\text{Cl}$, $-\text{CH}_3$, and $-\text{CH}_2\text{CH}_3$), making it a chiral center and thus optically active.

Since compound "P" is optically active, P must be 2-chlorobutane.

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Step 2: Further Chlorination of Compound "P"

The structure of 2-chlorobutane is: $\overset{4}{\text{C}}\text{H}_3-\overset{3}{\text{C}}\text{H}_2-\overset{2}{\text{C}}\text{H(Cl)}-\overset{1}{\text{C}}\text{H}_3$

Replacing a hydrogen atom with a chlorine atom at each distinct carbon position gives the following dichloro compounds:

1. Substitution at $\text{C}_1$: $\text{CH}_3-\text{CH}_2-\text{CH(Cl)}-\text{CH}_2\text{Cl} \quad \implies \text{1,2-dichlorobutane}$

2. Substitution at $\text{C}_2$: $\text{CH}_3-\text{CH}_2-\text{C(Cl)}_2-\text{CH}_3 \quad \implies \text{2,2-dichlorobutane}$

3. Substitution at $\text{C}_3$: $\text{CH}_3-\text{CH(Cl)}-\text{CH(Cl)}-\text{CH}_3 \quad \implies \text{2,3-dichlorobutane}$

4. Substitution at $\text{C}_4$: $\text{ClCH}_2-\text{CH}_2-\text{CH(Cl)}-\text{CH}_3 \quad \implies \text{1,3-dichlorobutane}$

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Step 3: Total Count

Ignoring stereoisomers, the unique structural isomers formed are:

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

Thus, the total number of dichloro compounds obtained is 4.

Correct Option: B