`CH_(2)-CH_(2)-CH_(2)-CH_(3)overset(Br_(2)//hv)(to)`
Major product in the above reaction is:

Options

A. Racemic mixture
B. Meso
C. Diastereomers
D. Constitutinal isomers

Explanation

To determine the major product of the reaction of butane (CH₃-CH₂-CH₂-CH₃) with bromine (Br₂) in the presence of sunlight, we will follow these steps:

### Step 1: Identify the Reaction Conditions
The reaction involves butane (C₄H₁₀) reacting with bromine (Br₂) under sunlight. The presence of sunlight indicates that a free radical substitution mechanism will occur.

**Hint for Step 1:** Look for conditions that favor free radical reactions, such as heat or light.

### Step 2: Understand the Free Radical Mechanism
In a free radical substitution reaction, the bromine molecule (Br₂) will dissociate into two bromine radicals (Br•) due to the energy provided by sunlight.

**Hint for Step 2:** Remember that free radicals are highly reactive species that can initiate chain reactions.

### Step 3: Initiation of the Reaction
The bromine radical (Br•) will abstract a hydrogen atom from butane, creating a butyl radical (C₄H₉•). The position from which the hydrogen is removed will determine the stability of the resulting radical.

**Hint for Step 3:** Consider the stability of the radicals formed; tertiary radicals are more stable than secondary, which are more stable than primary.

### Step 4: Formation of Radicals
When Br• abstracts a hydrogen atom, two possible radicals can be formed:
1. Primary radical (from the terminal CH₃ group)
2. Secondary radical (from the CH₂ group adjacent to the terminal CH₃)

The secondary radical is more stable than the primary radical.

**Hint for Step 4:** Identify which radical is more stable; this will influence the major product.

### Step 5: Bromination of the More Stable Radical
The more stable secondary radical (CH₃-CH₂-CH•-CH₃) will react with another bromine radical (Br•) to form the major product. The bromine will attach to the carbon that has the radical, resulting in the formation of 2-bromobutane.

**Hint for Step 5:** Focus on the product formed from the more stable radical; this will be your major product.

### Step 6: Consider Stereochemistry
Since the formation of the secondary radical creates a chiral center when bromine adds, there will be a racemic mixture of enantiomers (2-bromobutane).

**Hint for Step 6:** Remember that the presence of a chiral center can lead to the formation of enantiomers.

### Final Product
The major product of the reaction is 2-bromobutane (C₄H₉Br), and due to the chiral center, a racemic mixture of enantiomers will be produced.

**Final Answer:** The major product is 2-bromobutane, which may exist as a racemic mixture of enantiomers.

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