In the chlorination of `(CH_(3))_(2)CH–CH_(2)–CH_(3)` the substitution at -

Options

A. `1^(@)` carbon would be fastest
B. `2^(@)` carbon would be fastest
C. `3^(@)` carbon would be fastest
D. `1^(@), 2^(@), 3^(@)` carbon atoms all will occur at the same rate

Answer

To solve the problem of chlorination of `(CH₃)₂CH–CH₂–CH₃`, we need to analyze the structure of the compound and determine the most stable free radical that can form during the chlorination process. Here’s a step-by-step solution:

### Step 1: Identify the Structure
The given compound `(CH₃)₂CH–CH₂–CH₃` is 2-methylbutane. It has a branched structure with a central carbon atom connected to two methyl groups and two other carbon atoms.

### Step 2: Understand the Chlorination Mechanism
Chlorination of alkanes occurs through a free radical mechanism. This process involves three main steps: initiation, propagation, and termination. The key step we will focus on is the propagation step where the free radicals are formed.

### Step 3: Determine Possible Free Radicals
During chlorination, hydrogen atoms in the alkane can be substituted by chlorine atoms, forming free radicals. The stability of these free radicals will influence the rate of the reaction. The stability order of free radicals is:
- Tertiary (3°) > Secondary (2°) > Primary (1°)

In the case of 2-methylbutane:
- The carbon atom attached to the two methyl groups can form a tertiary free radical.
- The carbon atom attached to the CH₂ group can form a secondary free radical.

### Step 4: Identify the Most Stable Free Radical
The most stable free radical that can form during the chlorination of 2-methylbutane is the tertiary free radical. This is because it is stabilized by the two adjacent methyl groups, which provide hyperconjugation and inductive effects.

### Step 5: Conclusion
Since the tertiary free radical is the most stable, the chlorination will preferentially occur at the carbon that leads to the formation of this tertiary radical. Therefore, the substitution at the carbon that leads to the formation of the tertiary radical will be the fastest reaction.

### Final Answer
The substitution at the tertiary carbon will be the fastest reaction in the chlorination of `(CH₃)₂CH–CH₂–CH₃`.

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