`2CH_(3)CHO overset(OH^(-))rarr X rarr Y overset(H_(2)-Pd//C)rarr Z, Z` is

Options

A. `CH_(3)CH=CHCH_(2)OH`
B. `CH_(3)CH_(2)CH_(2)CH_(2)OH`
C. `CH_(3)CH_(2)CH_(2)CHO`
D. `CH_(3)CH=CHCHO`

Explanation

To solve the problem step by step, let's break down the reactions involved:

### Step 1: Aldol Condensation
We start with two molecules of acetaldehyde (CH₃CHO). In the presence of a base (OH⁻), aldol condensation occurs. The base abstracts an alpha hydrogen from one of the acetaldehyde molecules, leading to the formation of an enolate ion.

1. **Formation of Enolate Ion:**
\[ \text{CH}_3\text{CHO} + \text{OH}^- \rightarrow \text{CH}_2\text{CHO}^- + \text{H}_2\text{O} \]

2. **Nucleophilic Attack:**
The enolate ion then attacks the carbonyl carbon of another acetaldehyde molecule, resulting in the formation of a β-hydroxyaldehyde (X).
\[ \text{CH}_2\text{CHO}^- + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \quad (X) \]

### Step 2: Dehydration
Next, we heat the β-hydroxyaldehyde (X) to eliminate a water molecule, forming an α,β-unsaturated aldehyde (Y).

3. **Dehydration Reaction:**
\[ \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\text{heat}} \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O} \quad (Y) \]

### Step 3: Hydrogenation
Finally, we reduce the α,β-unsaturated aldehyde (Y) using hydrogen in the presence of palladium on carbon (H₂/Pd/C), which will reduce the double bond between the carbon atoms.

4. **Hydrogenation Reaction:**
\[ \text{CH}_3\text{CH=CHCHO} + \text{H}_2 \xrightarrow{\text{H}_2/\text{Pd/C}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \quad (Z) \]

### Final Product
The final product Z is butanal (CH₃(CH₂)₂CHO).

### Summary of Steps:
1. **Aldol condensation** forms a β-hydroxyaldehyde (X).
2. **Dehydration** of X leads to an α,β-unsaturated aldehyde (Y).
3. **Hydrogenation** of Y results in butanal (Z).