Step 1: Understanding the Question:
We need to determine which of the given alkyl free radicals exhibits the highest stability.
Step 2: Key Formula or Approach:
The stability of alkyl free radicals is primarily governed by hyperconjugation (delocalization of C-H $\sigma$-electrons into the partially filled p-orbital) and the electron-donating inductive effect ($+I$ effect) of alkyl groups. The greater the number of $\alpha$-hydrogens, the more hyperconjugative structures can be formed, leading to higher stability. The general order of stability is:
$$\text{Tertiary } (3^{\circ}) > \text{Secondary } (2^{\circ}) > \text{Primary } (1^{\circ}) > \text{Methyl radical}$$
Step 3: Detailed Explanation:
Let us analyze each option based on its class and the number of $\alpha$-hydrogens:
1. $\text{CH}_3\text{CH}_2^{\bullet}$ (Ethyl radical): This is a primary ($1^{\circ}$) radical with 3 $\alpha$-hydrogens.
2. $(\text{CH}_3)_3\text{C}^{\bullet}$ (tert-Butyl radical): This is a tertiary ($3^{\circ}$) radical. The central radical carbon is surrounded by three methyl groups, providing a total of 9 $\alpha$-hydrogens. This allows for maximum hyperconjugative stabilization along with strong $+I$ support from three sides.
3. $(\text{CH}_3)_2\text{CH}^{\bullet}$ (Isopropyl radical): This is a secondary ($2^{\circ}$) radical with 6 $\alpha$-hydrogens.
4. $\text{CH}_3^{\bullet}$ (Methyl radical): This is the least stable radical as it possesses 0 $\alpha$-hydrogens and lacks inductive stabilization.
Comparing the counts, the tert-butyl radical has the highest number of hyperconjugative structures (9 $\alpha$-hydrogens), making it the most stable.
Step 4: Final Answer:
The most stable free radical is $(\text{CH}_3)_3\text{C}^{\bullet}$, which is option (B).