Alkyl iodide can be prepared by:-

Options

A. `ECH_(2)COOAg +I_(2) overset(C CI_(4))underset(Delta)rarr RCH_(2)I`
B. `RCH_(2)CI +NaI overset("acetone")underset(Delta)rarr RCH_(2)I +NaCI`
C. `R-OH +HI rarr RI +H_(2)O`
D. `CH_(4)+I_(2) rarr CH_(3)I`

Answer

To prepare alkyl iodides, we can analyze the given reactions step by step. Here’s how we can determine which reactions will yield alkyl iodides:

### Step 1: Analyze Reaction 1
**Reaction:** Ethyl acetate (CH3COOEt) with iodine (I2) in the presence of carbon tetrachloride (CCl4) and heat.

- **Explanation:** In this reaction, the ethyl acetate does not react to form an alkyl iodide. The presence of Ag and CCl4 leads to dimerization rather than substitution. Therefore, no alkyl iodide is formed.

### Step 2: Analyze Reaction 2
**Reaction:** RCH2Cl (alkyl chloride) with sodium iodide (NaI) in acetone and heat.

- **Explanation:** This is an example of a halogen exchange reaction, known as the Schwarz reaction. In this reaction, the chloride ion (Cl-) is replaced by the iodide ion (I-), resulting in the formation of an alkyl iodide (RCH2I) and sodium chloride (NaCl) as a byproduct. Thus, this reaction does produce an alkyl iodide.

### Step 3: Analyze Reaction 3
**Reaction:** Alcohol (ROH) with hydroiodic acid (HI).

- **Explanation:** In this reaction, HI dissociates into H+ and I-. The H+ ion can protonate the alcohol, leading to the formation of water and allowing the iodide ion (I-) to attack the alkyl group (R). This results in the formation of an alkyl iodide (RI) and water. Therefore, this reaction also produces an alkyl iodide.

### Step 4: Analyze Reaction 4
**Reaction:** A free radical mechanism involving iodide.

- **Explanation:** In this case, the free radical mechanism is not favorable for the formation of stable free radicals with iodide. Therefore, this reaction does not yield an alkyl iodide.

### Conclusion
From the analysis:
- **Reactions that produce alkyl iodides:**
- Reaction 2 (RCH2Cl + NaI in acetone)
- Reaction 3 (ROH + HI)

- **Reactions that do not produce alkyl iodides:**
- Reaction 1 (CH3COOEt + I2 in CCl4)
- Reaction 4 (Free radical mechanism)

### Final Answer:
Alkyl iodides can be prepared by:
1. RCH2Cl + NaI in acetone (Schwarz reaction)
2. ROH + HI

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