Which one of the following compounds undergoes predominantly `S_N^2` reaction with aqueous NaOH in a polar aprotic solvent ?

Options

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B.
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Answer

To determine which compound undergoes predominantly an \( S_N^2 \) reaction with aqueous NaOH in a polar aprotic solvent, we need to analyze the electrophilicity of the carbon atom bonded to the leaving group in each compound. The \( S_N^2 \) mechanism involves a nucleophile attacking the electrophilic carbon, leading to the formation of a transition state and ultimately resulting in the substitution of the leaving group.

### Step-by-Step Solution:

1. **Identify the Compounds**: We have four compounds to analyze. Let's denote them as:
- Compound 1: Methoxybenzene (C6H5OCH3)
- Compound 2: Nitrobenzene (C6H5NO2)
- Compound 3: Benzyl bromide (C6H5CH2Br)
- Compound 4: Aniline (C6H5NH2)

2. **Understand the \( S_N^2 \) Mechanism**: The \( S_N^2 \) reaction requires a good electrophile (carbon attached to a good leaving group) and a strong nucleophile. The rate of the reaction is influenced by the electrophilicity of the carbon atom bonded to the leaving group.

3. **Evaluate the Electrophilicity of Each Compound**:
- **Compound 1 (Methoxybenzene)**: The methoxy group is an electron-donating group (+R effect), which increases electron density on the benzene ring. This decreases the electrophilicity of the carbon attached to the leaving group, making it less favorable for \( S_N^2 \).

- **Compound 2 (Nitrobenzene)**: The nitro group is an electron-withdrawing group (-R effect), which decreases electron density on the benzene ring, increasing the electrophilicity of the carbon attached to the leaving group. This makes it more favorable for \( S_N^2 \) reactions.

- **Compound 3 (Benzyl bromide)**: The benzyl group can stabilize the transition state through resonance, but the carbon is not as electrophilic as in nitrobenzene. However, bromine is a good leaving group, which is favorable for \( S_N^2 \).

- **Compound 4 (Aniline)**: The amino group is also an electron-donating group (+R effect), which increases electron density on the benzene ring and decreases the electrophilicity of the carbon attached to the leaving group.

4. **Rank the Compounds Based on Electrophilicity**:
- Nitrobenzene (Compound 2) has the highest electrophilicity due to the electron-withdrawing nitro group.
- Benzyl bromide (Compound 3) is next due to the good leaving group (Br) and some resonance stabilization.
- Methoxybenzene (Compound 1) and Aniline (Compound 4) are less favorable due to their electron-donating effects.

5. **Conclusion**: The compound that undergoes predominantly \( S_N^2 \) reaction with aqueous NaOH in a polar aprotic solvent is **Compound 2 (Nitrobenzene)**.

### Final Answer:
**Compound 2 (Nitrobenzene)** undergoes predominantly \( S_N^2 \) reaction with aqueous NaOH in a polar aprotic solvent.