`PhCHO+HCHOoverset(aq.NaOH)underset(Delta)to` Products obtained are

Options

A. `PhCH_(2)OHandHCOONa`
B. `PhCOONaandCH_(3)OH`
C. `PhCOONaandHCOONa`
D. `PhCH_(2)OHandCH_(3)OH`

Answer

To solve the problem of the reaction between benzaldehyde (PhCHO) and formaldehyde (HCHO) in the presence of aqueous NaOH, we will follow these steps:

### Step 1: Identify the Reaction Type
This reaction is a **Cross Cannizzaro Reaction**. In this reaction, one aldehyde gets oxidized while the other gets reduced.

**Hint:** Remember that the Cannizzaro reaction occurs with non-enolizable aldehydes in the presence of a strong base.

### Step 2: Write the Reactants
The reactants in this case are:
- Benzaldehyde (PhCHO)
- Formaldehyde (HCHO)

**Hint:** Make sure to recognize the structure of both aldehydes before proceeding.

### Step 3: Mechanism of the Reaction
In the presence of NaOH, the hydroxide ion (OH-) will attack the carbonyl carbon of formaldehyde (HCHO), leading to the formation of a formate ion (HCOO-) and a benzyl alcohol (PhCH2OH).

**Hint:** The hydroxide ion acts as a nucleophile, attacking the carbonyl carbon.

### Step 4: Identify the Products
1. **Oxidation Product**: The formaldehyde (HCHO) will be oxidized to formic acid (HCOOH).
2. **Reduction Product**: The benzaldehyde (PhCHO) will be reduced to benzyl alcohol (PhCH2OH).

**Hint:** Remember that in the Cannizzaro reaction, one aldehyde is oxidized to a carboxylic acid while the other is reduced to an alcohol.

### Step 5: Write the Final Products
The final products of the reaction are:
- Sodium formate (HCOONa) from formic acid (HCOOH) reacting with NaOH.
- Benzyl alcohol (PhCH2OH).

**Hint:** The sodium salt of the acid is formed due to the reaction with the base.

### Step 6: Conclusion
Thus, the products obtained from the reaction of benzaldehyde and formaldehyde in the presence of aqueous NaOH are:
- Sodium formate (HCOONa)
- Benzyl alcohol (PhCH2OH)

**Final Answer:** The products obtained are HCOONa and PhCH2OH.

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