Propane is not formed when `C_3 H_7 MgBr` is treated with

Options

A. `H_2`
B. Phenol
C. Ethanoic acid
D. 2-Butyne

Explanation

To determine when propane is not formed when propyl magnesium bromide (C₃H₇MgBr) is treated with various compounds, we need to analyze the reactivity of each compound with respect to the presence of active hydrogen atoms.

### Step-by-Step Solution:

1. **Identify the Reactant:**
The reactant in question is propyl magnesium bromide (C₃H₇MgBr), which is a Grignard reagent. Grignard reagents react with compounds that have active hydrogen atoms.

2. **List the Compounds:**
We need to evaluate the following compounds:
- Hydrogen (H₂)
- Phenol (C₆H₅OH)
- Ethanoic acid (CH₃COOH)
- But-2-yne (CH₃C≡CCH₃)

3. **Analyze Each Compound:**
- **Hydrogen (H₂):**
- It does not contain any active hydrogen that can react with the Grignard reagent. However, if we consider the reaction with H₂, it would produce propane (C₃H₈) through reduction.
- **Phenol (C₆H₅OH):**
- Phenol has one active hydrogen atom. When reacted with C₃H₇MgBr, it will produce propane (C₃H₈) and phenoxide ion (C₆H₅O⁻).
- **Ethanoic Acid (CH₃COOH):**
- Ethanoic acid has an active hydrogen atom. The reaction with C₃H₇MgBr will yield propane (C₃H₈) and the corresponding magnesium salt of acetic acid.
- **But-2-yne (CH₃C≡CCH₃):**
- But-2-yne does not have any active hydrogen atoms that can react with the Grignard reagent. Therefore, no propane will be formed in this reaction.

4. **Conclusion:**
Since propane is not formed when propyl magnesium bromide is treated with but-2-yne (CH₃C≡CCH₃), the answer is:
- **Propane is not formed when C₃H₇MgBr is treated with but-2-yne (option D).**

### Final Answer:
**Propane is not formed when C₃H₇MgBr is treated with but-2-yne (option D).**