Complete the following reaction sequence :
`CH_(3)-overset(O)overset(||)C-CH_(3)overset((i)CH_(3)MgBr)underset((ii)H_(2)O)rarr(A)overset("Na metal")underset("ether")rarr(B)overset(CH_(3)-Br)rarr(C)`

Options

A. Identical
B. positional isomers
C. functional isomers
D. optical isomers

Explanation

To complete the reaction sequence provided in the question, we will analyze each step carefully and identify the products A, B, and C.

### Step 1: Reaction of Propanone with Grignard Reagent
The starting compound is propanone (CH₃C(=O)CH₃). When it reacts with the Grignard reagent (CH₃MgBr), the nucleophilic carbon of the Grignard reagent attacks the electrophilic carbon of the carbonyl group (C=O).

**Reaction:**
1. **Nucleophilic attack:** The CH₃ group from CH₃MgBr attacks the carbonyl carbon, leading to the formation of a tetrahedral intermediate.
2. **Formation of Alkoxide:** The oxygen of the carbonyl group becomes negatively charged (alkoxide), and the magnesium bromide ion (MgBr) associates with it.

**Product A:**
The product after the first step is an alkoxide:
\[ \text{CH}_3C(OH)(CH_3)MgBr \]
This compound is 2-methylpropan-2-ol (also known as tert-butanol) after hydrolysis with water.

### Step 2: Reaction with Water
Next, we add water (H₂O) to the reaction mixture, which protonates the alkoxide oxygen, leading to the formation of the alcohol.

**Reaction:**
1. **Protonation:** The alkoxide oxygen (O⁻) attracts a proton (H⁺) from water, resulting in the formation of the alcohol and the release of MgBrOH.

**Product A (Final):**
The final product after this step is:
\[ \text{CH}_3C(OH)(CH_3) \]
This is 2-methylpropan-2-ol.

### Step 3: Reaction with Sodium Metal
The next step involves the reaction of product A with sodium metal in ether. Sodium metal reduces the alcohol to form a sodium alkoxide.

**Reaction:**
1. **Reduction:** Sodium donates an electron to the alcohol, leading to the formation of a sodium alkoxide.
2. **Formation of Product B:** The hydrogen from the hydroxyl group is replaced by sodium.

**Product B:**
The product formed is:
\[ \text{CH}_3C(O^{-}Na)(CH_3) \]
This is sodium 2-methylpropan-2-olate.

### Step 4: Reaction with Methyl Bromide
Finally, product B reacts with methyl bromide (CH₃Br). The sodium alkoxide acts as a nucleophile and attacks the electrophilic carbon in methyl bromide.

**Reaction:**
1. **Nucleophilic substitution:** The sodium ion (Na⁺) is replaced by the methyl group from methyl bromide.
2. **Formation of Product C:** The product formed is an ether.

**Product C:**
The final product is:
\[ \text{CH}_3C(OCH_3)(CH_3) \]
This compound is 2-methoxy-2-methylpropane.

### Summary of Products:
- **Product A:** 2-methylpropan-2-ol
- **Product B:** Sodium 2-methylpropan-2-olate
- **Product C:** 2-methoxy-2-methylpropane

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