` CH_3CH_2CHO overset (NaOH, Delta) underset((aldol)) (rarr)A,A` is.

Options

A. ` CH_3CH_2CH= CHCH_2CHO`
B.
C.
D.

Explanation

To solve the given question regarding the aldol reaction of CH₃CH₂CHO (acetaldehyde) with NaOH, we will follow these steps:

### Step 1: Identify the Reactants
The reactant is CH₃CH₂CHO, which is acetaldehyde. In the aldol reaction, two molecules of the same aldehyde will react in the presence of a base (NaOH) to form a β-hydroxy aldehyde.

**Hint:** Look for the structure of the aldehyde and remember that two molecules will participate in the reaction.

### Step 2: Formation of Enolate Ion
In the presence of NaOH, one molecule of acetaldehyde will lose a proton (H⁺) from the alpha carbon (the carbon next to the carbonyl group) to form an enolate ion.

**Hint:** The enolate ion is formed by deprotonation of the alpha carbon of the aldehyde.

### Step 3: Nucleophilic Attack
The enolate ion will then attack the carbonyl carbon of another molecule of acetaldehyde. This forms a tetrahedral intermediate.

**Hint:** Remember that the nucleophile (enolate ion) attacks the electrophile (carbonyl carbon).

### Step 4: Protonation
The tetrahedral intermediate will then collapse, and a water molecule (H₂O) will be eliminated. This results in the formation of a β-hydroxy aldehyde.

**Hint:** During this step, a molecule of water is eliminated, so look for the formation of a double bond.

### Step 5: Final Product
The final product after the aldol condensation will be 3-hydroxybutanal (CH₃CH(OH)CH₂CHO). If dehydration occurs, it can further lead to the formation of crotonaldehyde (CH₃CH=CHCHO).

**Hint:** Check if dehydration occurs to get the unsaturated product.

### Conclusion
The aldol reaction of two molecules of acetaldehyde (CH₃CH₂CHO) in the presence of NaOH leads to the formation of 3-hydroxybutanal (β-hydroxy aldehyde) and potentially crotonaldehyde (if dehydration occurs).

**Final Answer:** The products A and A are 3-hydroxybutanal and crotonaldehyde.