Which of the following is the best procedure to make isopropyl methyl ether using the Williamson's method ?

Options

A. `CH_(3)OH+(CH_(3))_(2)CHOH+H_(2)SO_(4)`
B. `CH_(3)OH+(CH_(3))_(2)CHCH_(2)+H_(2)SO_(4)`
C. `CH_(2)ON a+(CH_(3))_(2)CHBr`
D. `CH_(3)I+(CH_(3))_(2)CHONa `)

Explanation

To synthesize isopropyl methyl ether (CH3)2CHOCH3 using the Williamson synthesis method, we can follow these steps:

### Step 1: Identify the reactants
In the Williamson synthesis, we need an alkoxide and an alkyl halide. For isopropyl methyl ether, we can identify:
- **Alkoxide (R1O)**: Isopropyl alkoxide, which is derived from isopropanol (CH3)2CHO.
- **Alkyl halide (R2X)**: Methyl iodide (CH3I) is a suitable choice for the alkyl halide.

### Step 2: Prepare the alkoxide
To prepare the isopropyl alkoxide, we can react isopropanol with sodium (Na):
- **Reaction**: (CH3)2CHOH + Na → (CH3)2CHO⁻Na⁺ + ½ H2
This forms the isopropyl alkoxide (CH3)2CHO⁻Na⁺.

### Step 3: React the alkoxide with the alkyl halide
Now, we will react the prepared isopropyl alkoxide with methyl iodide:
- **Reaction**: (CH3)2CHO⁻Na⁺ + CH3I → (CH3)2CHOCH3 + NaI
In this step, the alkoxide attacks the methyl halide, resulting in the formation of isopropyl methyl ether and sodium iodide as a byproduct.

### Step 4: Final product
The final product of the reaction is isopropyl methyl ether (CH3)2CHOCH3.

### Conclusion
The best procedure to make isopropyl methyl ether using the Williamson method involves preparing isopropyl alkoxide from isopropanol and then reacting it with methyl iodide.

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