`CH_3-CH_2-undersetunderset(Cl)|CH CH_3` obtained by chlorination of n butane, will be : -

Options

A. Meso form
B. Racemic mixture
C. d-form
D. l-form

Explanation

To solve the question regarding the product formed by the chlorination of n-butane, we can follow these steps:

### Step 1: Understand the Structure of n-Butane
n-Butane is a straight-chain alkane with the formula \( C_4H_{10} \), which can be represented as:
\[ CH_3-CH_2-CH_2-CH_3 \]

### Step 2: Identify the Chlorination Process
When n-butane undergoes chlorination, chlorine gas (\( Cl_2 \)) reacts with it. Chlorination involves the substitution of hydrogen atoms with chlorine atoms. This reaction typically follows a free radical mechanism.

### Step 3: Determine Possible Products
During the chlorination of n-butane, two main products can be formed:
1. **1-Chlorobutane**: Chlorine attaches to the terminal carbon (first carbon).
\[ CH_3-CH_2-CH_2-CH_2Cl \]
2. **2-Chlorobutane**: Chlorine attaches to the second carbon.
\[ CH_3-CHCl-CH_2-CH_3 \]

### Step 4: Analyze the Product 2-Chlorobutane
The structure of 2-chlorobutane shows that it has a chiral center at the second carbon. This means that 2-chlorobutane can exist in two enantiomeric forms (D and L forms).

### Step 5: Determine the Mixture of Products
Since both 1-chlorobutane and 2-chlorobutane are formed during the chlorination of n-butane, we have a mixture of products. The presence of 2-chlorobutane, which can exist as two enantiomers, leads to the formation of a racemic mixture (equal amounts of both enantiomers).

### Conclusion
The correct answer to the question is that the product obtained by chlorination of n-butane will be a **racemic mixture**.

### Final Answer
The product will be a **racemic mixture** (Option B).

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