Which yields isopropyl methyl ether with little or no by-products?

Options

A. `(CH_(3))_(2)CHO^(-)Na^(+)+CH_(3)Irarr`
B. `CH_(3)O^(-)Na^(+)+(CH_(3))_(2)CHIrarr`
C. `(CH_(3))_(2)CHOH+CH_(3)OH overset(H_(2)SO_(4))rarr`
D. All fo these

Explanation

To determine which reaction yields isopropyl methyl ether with little or no by-products, we need to consider the method of synthesis and the nature of the reactants involved.

### Step-by-Step Solution:

1. **Understanding Ether Formation**: Ethers can be formed through the reaction of alkyl halides with alkoxides. In this case, we are looking for a method that yields isopropyl methyl ether specifically.

2. **Choosing the Right Alkyl Halide**: For the synthesis of unsymmetrical ethers, primary alkyl halides are preferred because they undergo the SN2 mechanism, which minimizes the formation of by-products.

3. **Identifying the Reactants**: To form isopropyl methyl ether, we can use sodium isopropoxide (derived from isopropyl alcohol) and methyl iodide. The reaction can be represented as:
\[ \text{Na isopropoxide} + \text{Methyl iodide} \rightarrow \text{Isopropyl methyl ether} + \text{NaI} \]

4. **Reaction Mechanism**: The reaction follows the SN2 mechanism, where the nucleophile (isopropoxide ion) attacks the electrophile (methyl iodide). Since methyl iodide is a primary alkyl halide, it reacts efficiently with the isopropoxide ion without forming significant by-products.

5. **Conclusion**: The reaction of sodium isopropoxide with methyl iodide yields isopropyl methyl ether as the major product with little or no by-products. Therefore, this is the preferred method for synthesizing isopropyl methyl ether.

### Final Answer:
The reaction of sodium isopropoxide with methyl iodide yields isopropyl methyl ether with little or no by-products.

---