`(CH_(3))_(3)CoNa` on reaction with`CH_(3)Br` will give

Options

A. `(CH_(3))_(3)COC(CH_(3))_(3)`
B. `CH_(3)OCH_(3)`
C. `CH_(3)CH_(2)OCH_(2)CH_(3)`
D. `(CH_(3))_(3)COCH_(3)`

Explanation

To solve the question regarding the reaction of `(CH₃)₃CoNa` with `CH₃Br`, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
1. `(CH₃)₃CoNa` (Trimethylsodium cobaltate)
2. `CH₃Br` (Methyl bromide)

### Step 2: Understand the Structure
The structure of `(CH₃)₃CoNa` can be represented as:
- The `Co` (Cobalt) is bonded to three `CH₃` (methyl) groups and a `Na` (Sodium) ion.
- The `Na` ion is associated with the cobalt complex.

### Step 3: Analyze the Reaction
When `(CH₃)₃CoNa` reacts with `CH₃Br`, we need to consider the nucleophilic substitution reaction. The `Na` ion will not participate directly, but the `Co` complex will.

1. The `Co` in `(CH₃)₃CoNa` has a partial negative charge due to the presence of the sodium ion.
2. The `Br` in `CH₃Br` has a partial negative charge (δ-) and the `C` connected to `Br` has a partial positive charge (δ+).

### Step 4: Nucleophilic Attack
The methyl group from `(CH₃)₃CoNa` acts as a nucleophile and attacks the carbon atom in `CH₃Br`:
- The `Br` atom will leave as a bromide ion (Br⁻).
- The methyl group (CH₃) from `(CH₃)₃CoNa` will attach to the carbon atom of `CH₃Br`.

### Step 5: Write the Products
The products of the reaction will be:
1. `NaBr` (Sodium bromide) from the `Na` ion and `Br` ion.
2. The remaining structure will be `Co(CH₃)₄` (Tetra-methyl cobalt complex).

### Final Reaction
Thus, the overall reaction can be summarized as:
\[ (CH₃)₃CoNa + CH₃Br \rightarrow NaBr + Co(CH₃)₄ \]

### Conclusion
The final product of the reaction is `NaBr` and `Co(CH₃)₄`.

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