In the reaction , `(CH_(3))_(3)C CHO +HCHO overset(NaOH)underset("heat")toA+B`
the products (A) and (B) are respectively

Options

A. `(CH_(3))_(3)C CH_(2)OH` and `HCOO^(-) Na^(+)`
B. `(CH_(3))_(3)C COONa` and `CH_(3)OH`
C. `(CH_(3))_(3)C CH_(2)OH` and `CH_(3)OH`
D. `(CH_(3))_(3)COONa` and `HCOO^(-) Na^(+)`

Answer

To solve the reaction `(CH₃)₃C CHO + HCHO` in the presence of `NaOH` and heat, we will analyze the reaction step by step to identify the products A and B.

### Step 1: Identify the Reactants
The reactants in the reaction are:
- `(CH₃)₃C CHO` (which is 2-methylbutanal)
- `HCHO` (which is formaldehyde)

### Step 2: Understand the Role of NaOH
Sodium hydroxide (NaOH) is a strong base. In this reaction, it will dissociate into Na⁺ and OH⁻ ions. The hydroxide ion (OH⁻) will act as a nucleophile.

### Step 3: Nucleophilic Attack
The OH⁻ ion will attack the carbonyl carbon (C=O) of the formaldehyde (HCHO). This carbon is electrophilic due to the partial positive charge (δ+) on it.

### Step 4: Formation of Alkoxide Ion
When the OH⁻ attacks the carbonyl carbon, it forms a tetrahedral intermediate, which can be represented as:
- `H₂C(OH)O⁻` (an alkoxide ion)

### Step 5: Proton Transfer
The alkoxide ion can lose a proton (H⁺) to form an alcohol. This step involves the extraction of a proton from the hydroxyl group, leading to the formation of:
- `H₂C=O` (aldehyde) and `H₂O` (water)

### Step 6: Hydride Transfer
The hydride ion (H⁻) from the alkoxide can now attack the carbonyl carbon of the other reactant `(CH₃)₃C CHO`. This is a key step where a hydride transfer occurs.

### Step 7: Formation of Products
After the hydride transfer, we will have:
1. The product A: `C(CH₃)₃CH₂OH` (which is 2-methyl-2-butanol)
2. The product B: `HCOO⁻` (which is the formate ion)

### Final Products
Thus, the products A and B are:
- A: `C(CH₃)₃CH₂OH` (2-methyl-2-butanol)
- B: `HCOO⁻` (formate ion)

### Summary of Products
- A: 2-methyl-2-butanol
- B: Formate ion

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