For preparing `R–C-=CH`, from Grignard’s reagent we take -

Options

A. `CH-=CH, CH_(3)MgBr & RI`
B. `CH_(3)–C-=CH + RMg Br + CH_(3)I`
C. `CH-=CH, CH_(3)MgBr + R’I`
D. none of these

Answer

To prepare the compound \( R-C \equiv CH \) from Grignard's reagent, we need to follow a series of steps involving the reaction of Grignard reagents with alkynes. Let's break down the solution step by step.

### Step-by-Step Solution

1. **Understanding Grignard Reagents**:
- Grignard reagents have the general formula \( R-MgX \), where \( R \) is an alkyl or aryl group, \( Mg \) is magnesium, and \( X \) is a halogen (like Cl, Br, or I).
- They are nucleophilic, meaning the carbon attached to magnesium has a partial negative charge, making it reactive towards electrophiles.

2. **Identify the Target Compound**:
- We want to synthesize \( R-C \equiv CH \), which is a terminal alkyne. The terminal hydrogen is acidic, which is important for the next steps.

3. **Choosing the Right Reaction**:
- To form \( R-C \equiv CH \), we can start with a terminal alkyne, such as \( CH \equiv CH \) (ethyne or acetylene).
- The reaction involves the addition of a Grignard reagent to this terminal alkyne.

4. **Reacting Grignard Reagent with Terminal Alkyne**:
- When we react \( CH \equiv CH \) with a Grignard reagent, such as \( CH_3MgBr \), the nucleophilic \( CH_3^- \) from the Grignard reagent will attack the terminal hydrogen of the alkyne.
- This results in an acid-base reaction where the terminal hydrogen is removed, forming \( CH_3C \equiv C^- \) (the anion) and releasing methane \( (CH_4) \).

5. **Introducing Alkyl Halide**:
- Now, we need to introduce an alkyl halide \( R-I \) (where \( R \) is any alkyl group) to the anion \( CH_3C \equiv C^- \).
- The negative charge on the carbon will attack the carbon of the alkyl halide, displacing the iodine and forming the desired terminal alkyne \( R-C \equiv CH \).

6. **Final Product**:
- The final product after the reaction will be \( R-C \equiv CH \), which is the target compound.

### Conclusion:
Thus, the correct option for preparing \( R-C \equiv CH \) from Grignard's reagent is to use a terminal alkyne like \( CH \equiv CH \) and react it with \( R-I \) after the initial reaction with the Grignard reagent.