When o-bromoanisole is treated with `NaNH_(2)//liq. NH_(3)` , the main product obtained is

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Explanation

To determine the main product obtained when ortho-bromoanisole is treated with NaNH2 in liquid ammonia, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of ortho-Bromoanisole**:
- The structure consists of a benzene ring with a methoxy group (-OCH3) and a bromine atom (Br) at the ortho position relative to the methoxy group.
- The chemical formula can be represented as:
\[ \text{C}_6\text{H}_4(\text{Br})(\text{OCH}_3) \]

2. **Reaction with NaNH2**:
- Sodium amide (NaNH2) in liquid ammonia acts as a strong base. The NH2^- ion will abstract a hydrogen atom from the ortho position of the benzene ring where the methoxy group is located.
- This results in the formation of a negatively charged intermediate (a benzyne-like structure).

3. **Formation of the Anion**:
- When the hydrogen atom is abstracted, a negative charge is generated on the carbon atom that was bonded to the hydrogen. This results in the formation of a reactive anion.

4. **Elimination of Bromine**:
- The negatively charged carbon can then undergo elimination of the bromine atom (Br^-) from the ortho position. This step is crucial as it leads to the formation of a new double bond in the benzene ring.

5. **Final Product Formation**:
- After the elimination of Br^-, the negative charge can stabilize itself by forming a bond with the methoxy group (-OCH3), resulting in the formation of an amino group (-NH2) at the ortho position.
- The final product is therefore ortho-methoxy aniline, which can be represented as:
\[ \text{C}_6\text{H}_4(\text{OCH}_3)(\text{NH}_2) \]

### Conclusion:
The main product obtained from the reaction of ortho-bromoanisole with NaNH2 in liquid ammonia is ortho-methoxy aniline.

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