Tertiary alkyl halides are practically inert to substitution by `S_(N)^(2)` mechanism because of –

Options

A. instability
B. insolubility
C. steric hindrance
D. inductive effect

Answer

To answer the question regarding why tertiary alkyl halides are practically inert to substitution by the \( S_N2 \) mechanism, we can break down the explanation into clear steps:

### Step-by-Step Solution:

1. **Understanding Tertiary Alkyl Halides**:
- Tertiary alkyl halides have the general structure where the carbon atom bonded to the halogen is also bonded to three other carbon atoms (R groups). This structure can be represented as R3C-X, where X is the halogen.

2. **Mechanism of \( S_N2 \) Reactions**:
- The \( S_N2 \) (bimolecular nucleophilic substitution) mechanism involves a single step where the nucleophile attacks the carbon atom from the opposite side of the leaving group (the halogen). This results in the formation of a transition state where both the nucleophile and the leaving group are partially bonded to the carbon.

3. **Steric Hindrance**:
- In the case of tertiary alkyl halides, the carbon atom is surrounded by bulky alkyl groups. This crowding creates steric hindrance, making it difficult for the nucleophile to approach and attack the carbon atom.

4. **Rate of \( S_N2 \) Reactions**:
- The rate of \( S_N2 \) reactions is inversely proportional to steric hindrance. As steric hindrance increases, the rate of the reaction decreases because the nucleophile cannot effectively approach the carbon atom.

5. **Conclusion**:
- Due to the significant steric hindrance in tertiary alkyl halides, the nucleophile cannot effectively attack the carbon atom, making these compounds practically inert to substitution via the \( S_N2 \) mechanism.

### Final Answer:
Tertiary alkyl halides are practically inert to substitution by the \( S_N2 \) mechanism because of steric hindrance, which prevents the nucleophile from effectively approaching and attacking the carbon atom bonded to the halogen.

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