`H-overset(O)overset(||)C-H+CH_(3)-CH=Ooverset("Conc. NaOH")rarr` Find out the products of reaction

Options

A. `CH_(3)CO_(2)Na & CH_(3)OH`
B. `CH_(3)CH_(2)OH+CH_(3)OH`
C. `CH_(3)CH_(2)OH & HCO_(2)Na`
D. `CH_(3)CO_(2)Na+HCO_(2)Na`

Answer

To solve the given reaction step by step, we will analyze the reaction between formaldehyde (HCHO) and acetaldehyde (CH3CHO) in the presence of concentrated NaOH.

### Step 1: Identify the Reactants
- The reactants in this reaction are:
- Formaldehyde (HCHO)
- Acetaldehyde (CH3CHO)

### Step 2: Understand the Reaction Type
- This reaction is a **crossed aldol condensation**. In this type of reaction, two different aldehydes react in the presence of a base (here, concentrated NaOH) to form a β-hydroxy aldehyde, which can further dehydrate to form an α,β-unsaturated aldehyde.

### Step 3: Determine the Presence of Alpha Hydrogen
- Formaldehyde (HCHO) has no alpha hydrogen (it has no carbon adjacent to the carbonyl carbon).
- Acetaldehyde (CH3CHO) has alpha hydrogen atoms (the CH3 group has hydrogens that can be abstracted).

### Step 4: Formation of the Enolate Ion
- In the presence of NaOH, acetaldehyde can lose an alpha hydrogen to form an enolate ion:
\[ CH_3CHO \xrightarrow{NaOH} CH_2=CHO^- \]

### Step 5: Nucleophilic Addition
- The enolate ion formed from acetaldehyde will attack the carbonyl carbon of formaldehyde:
\[ CH_2=CHO^- + HCHO \rightarrow CH_3-CH(OH)-CH_2CHO \]
This results in the formation of a β-hydroxy aldehyde.

### Step 6: Dehydration to Form the Final Product
- The β-hydroxy aldehyde can undergo dehydration (loss of water) to form an α,β-unsaturated aldehyde:
\[ CH_3-CH(OH)-CH_2CHO \rightarrow CH_3-CH=CH-CHO + H_2O \]

### Step 7: Conclusion
- The final product of the reaction is **crotonaldehyde (CH3-CH=CH-CHO)**.

### Summary of the Reaction
The overall reaction can be summarized as:
\[ HCHO + CH_3CHO \xrightarrow{Conc. NaOH} CH_3-CH=CH-CHO + H_2O \]