A mixture of `(CH_(3))_(3)C-CHO` and HCHO on heating with aqueous NaOH solution will give:

Options

A. `HCOONa+(CH_(3))_(3)C-CH_(2)OH`
B. `(CH_(3))_(3)C-COONa+CH_(3)OH`
C. `HCOONa+(CH_(3))_(3)C-COONa`
D. `CH_(3)OH+(CH_(3))_(3)C-CH_(2)OH`

Answer

To solve the problem, we need to analyze the reaction between the given aldehydes, `(CH₃)₃C-CHO` (which is 2-methylbutanal) and HCHO (formaldehyde), when heated with aqueous NaOH.

### Step-by-Step Solution:

1. **Identify the Aldehydes**: The two aldehydes in the mixture are:
- `(CH₃)₃C-CHO` (2-methylbutanal)
- HCHO (formaldehyde)

2. **Check for Alpha Hydrogens**:
- 2-methylbutanal has no alpha hydrogen because the carbon adjacent to the carbonyl group is attached to three methyl groups.
- Formaldehyde (HCHO) has alpha hydrogens.

3. **Determine the Reaction Type**:
- Since 2-methylbutanal does not have alpha hydrogens, it cannot undergo an aldol condensation.
- However, formaldehyde can participate in the Cannizzaro reaction, which occurs in the absence of alpha hydrogens.

4. **Cannizzaro Reaction**:
- In the Cannizzaro reaction, one molecule of aldehyde is oxidized to a carboxylic acid, while another molecule is reduced to an alcohol.
- Here, formaldehyde (HCHO) will undergo the Cannizzaro reaction.

5. **Reaction Mechanism**:
- The hydroxide ion (from NaOH) will attack the carbonyl carbon of formaldehyde, leading to the formation of a tetrahedral intermediate.
- This intermediate will then collapse, resulting in the transfer of a hydride ion to another molecule of formaldehyde.
- One molecule of formaldehyde will be oxidized to formic acid (HCOOH), and the other will be reduced to methanol (CH₃OH).

6. **Final Products**:
- The final products of the reaction will be:
- Formic acid (HCOOH) from the oxidation of formaldehyde.
- Methanol (CH₃OH) from the reduction of formaldehyde.

### Conclusion:
The mixture of `(CH₃)₃C-CHO` and HCHO on heating with aqueous NaOH will yield:
- Formic acid (HCOOH)
- Methanol (CH₃OH)