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Chemistry Question 65 – JEE-MAIN 2026

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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 :

Determine the structure of "P" by considering the possible monochlorination products of n-butane and checking for chirality.

Step 1: Identify Product 'P'✦ Active

n-Butane (CH3−CH2−CH2−CH3) undergoes monochlorination under photochemical conditions. The possible monochlorination products are 1-chlorobutane (CH3−CH2−CH2−CH2Cl) and 2-chlorobutane (CH3−CH2−CHCl−CH3). 1-chlorobutane is not optically active as it lacks a chiral center. 2-chlorobutane has a chiral carbon at C2, making it optically active. Therefore, "P" is 2-chlorobutane.

Step 2: Identify Unique Hydrogen Environments in 'P'○ Expand

The structure of "P" (2-chlorobutane) is CH3−CH2−CHCl−CH3. We need to identify all unique hydrogen atoms that can be replaced by a second chlorine atom. Let's label the carbons from left to right as C1, C2, C3, C4:

C1H3−C2HCl−C3H2−C4H3

The unique hydrogen environments are on C1, C2, C3, and C4.

Step 3: Determine Dichloro Compounds (Ignoring Stereoisomers)○ Expand

Replacing a hydrogen atom from each unique environment with a chlorine atom will yield the following distinct dichloro compounds:

1. Replacing H on C1: ClCH2−CH2−CHCl−CH3 (1,2-dichlorobutane)

2. Replacing H on C2: CH3−CH2−CCl2−CH3 (2,2-dichlorobutane)

3. Replacing H on C3: CH3−CHCl−CHCl−CH3 (2,3-dichlorobutane)

4. Replacing H on C4: CH3−CH2−CHCl−CH2Cl (1,3-dichlorobutane)

These are 4 distinct structural isomers. Since stereoisomers are to be ignored, the total number of dichloro compounds obtained is 4.

💡 Teacher's Secret Hint

Ensure to check for symmetry to avoid counting identical compounds multiple times.

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