`"Acetaldehyde"overset(HCN)toXoverset(H_(2)O //H^(+))toYoverset("Heat")toZ`
In the above sequence, the end product Z is

Options

A. Prop-2-enoic acid
B. But-2-enoic acid
C. Lactic acid
D. Tartaric acid

Answer

To solve the problem, we will go through the reactions step by step:

### Step 1: Reaction of Acetaldehyde with HCN
- **Starting Compound:** Acetaldehyde (CH₃CHO)
- **Reagent:** HCN (Hydrogen cyanide)

In this step, the cyanide ion (CN⁻) acts as a nucleophile and attacks the electrophilic carbon of the carbonyl group in acetaldehyde. This results in the formation of a cyanohydrin.

**Reaction:**
\[ \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{C(OH)(CN)} \]
- **Product X:** CH₃C(OH)(CN)

### Step 2: Hydrolysis of the Cyanohydrin (X)
- **Starting Compound:** Cyanohydrin (CH₃C(OH)(CN))
- **Reagent:** Water (H₂O) in the presence of an acid (H⁺)

During hydrolysis, the cyanide group (CN) is converted into a carboxylic acid (COOH), while the hydroxyl group remains. Ammonia (NH₃) is released as a byproduct.

**Reaction:**
\[ \text{CH}_3\text{C(OH)(CN)} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{C(OH)(COOH)} + \text{NH}_3 \]
- **Product Y:** CH₃C(OH)(COOH)

### Step 3: Heating the Product (Y)
- **Starting Compound:** Hydroxy acid (CH₃C(OH)(COOH))

When heated, the hydroxy group (OH) and a hydrogen atom (H) from the carboxylic acid group (COOH) are eliminated as water (H₂O), leading to the formation of an α,β-unsaturated carboxylic acid.

**Reaction:**
\[ \text{CH}_3\text{C(OH)(COOH)} \xrightarrow{\text{Heat}} \text{CH}_2=CH\text{COOH} + \text{H}_2\text{O} \]
- **Product Z:** CH₂=CHCOOH

### Final Product
The final product Z is **CH₂=CHCOOH**, which is known as **acrylic acid** or **prop-2-enoic acid**.

### Summary of Products
- **X:** CH₃C(OH)(CN) (Cyanohydrin)
- **Y:** CH₃C(OH)(COOH) (Hydroxy acid)
- **Z:** CH₂=CHCOOH (Acrylic acid)

### Answer
The end product Z is **acrylic acid (CH₂=CHCOOH)**.

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