All reagents, `[Cu(NH_3)_2]^(o+)`, `[Ag(NH_3)_2]^(o+)`,
`CH_3MgBr`, and `NaNH_2` react with:

Options

A. (a) Cyclooctyne
B. (b) Pent-1-yne
C. (c) Pent-2-yne
D. (d) Ethyne

Explanation

To solve the question regarding which reagents react with the given compounds, we need to analyze the nature of the reagents and the structures of the compounds mentioned. The reagents are `[Cu(NH_3)_2]^(o+)`, `[Ag(NH_3)_2]^(o+)`, `CH_3MgBr`, and `NaNH_2`.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- The compounds mentioned are cyclooctane, pent-1-yne, pent-2-yne, and ethyne.
- Cyclooctane is a saturated hydrocarbon (not an alkyne).
- Pent-1-yne and ethyne are terminal alkynes.
- Pent-2-yne is an internal alkyne.

2. **Understanding the Reagents**:
- **[Cu(NH₃)₂]⁺**: This is a copper complex that can react with terminal alkynes to form a copper acetylide.
- **[Ag(NH₃)₂]⁺**: This is known as Tollens' reagent, which also reacts with terminal alkynes to form a silver acetylide.
- **CH₃MgBr**: This is a Grignard reagent that reacts with terminal alkynes to form alkynyl magnesium bromides.
- **NaNH₂**: Sodium amide is a strong base that can deprotonate terminal alkynes to form alkynyl anions.

3. **Reactions with Terminal Alkynes**:
- Terminal alkynes (like pent-1-yne and ethyne) can react with all the mentioned reagents.
- Internal alkynes (like pent-2-yne) do not react with these reagents because they lack the acidic hydrogen necessary for deprotonation or complexation.

4. **Conclusion**:
- The reagents `[Cu(NH₃)₂]⁺`, `[Ag(NH₃)₂]⁺`, `CH₃MgBr`, and `NaNH₂` will react with terminal alkynes (pent-1-yne and ethyne).
- Therefore, the correct answer is that these reagents react with **pent-1-yne (B)** and **ethyne (D)**.

### Final Answer:
The reagents `[Cu(NH₃)_2]^(o+)`, `[Ag(NH₃)_2]^(o+)`, `CH_3MgBr`, and `NaNH_2` react with **B (pent-1-yne)** and **D (ethyne)**.

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