`CH_(3)CH_(2)COOHCl_(2)overset("RedP")(to)Aunderset(Delta)overset(NaOH//CaO)(to)Bunderset(Delta)overset(Alc.KOH)(to)C 'C'` is

Options

A. `CH_(3)-underset(OH)underset(|)(CH)-CH_(3)`
B. `CH_(3)-CH=CH_(2)`
C. `CH_(3)-CH_(2)-underset(O)underset(||)(C)-OR`
D. `CH_(2)=CH_(2)`

Answer

To solve the given question step by step, we will analyze the reactions involved and identify the products formed at each stage.

### Step 1: Reaction of Propanoic Acid with Red Phosphorus and Chlorine
The starting compound is propanoic acid, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

When propanoic acid reacts with red phosphorus (Red P) and chlorine (Cl₂), it undergoes the HVZ (Hale-Volhard-Zelinsky) reaction. In this reaction, the alpha hydrogen (the hydrogen atom attached to the carbon adjacent to the carboxylic group) is replaced by a chlorine atom.

**Product A:**
\[ \text{CH}_3\text{CHCl}\text{COOH} \]

### Step 2: Reaction of Product A with Sodium Hydroxide and Calcium Oxide
Next, product A (CH₃CHClCOOH) is treated with sodium hydroxide (NaOH) and calcium oxide (CaO), which is known as soda lime. This reaction leads to decarboxylation, where the carboxyl group (COOH) is removed as carbon dioxide (CO₂).

**Decarboxylation Reaction:**
The removal of CO₂ from product A results in:
\[ \text{CH}_3\text{CHCl} \]

**Product B:**
\[ \text{CH}_3\text{CH}_2\text{Cl} \]

### Step 3: Reaction of Product B with Alcoholic KOH
Finally, product B (CH₃CH₂Cl) is treated with alcoholic KOH and heated. This reaction facilitates beta elimination, where a hydrogen atom from the beta carbon (the carbon adjacent to the carbon with the halogen) is removed along with the halogen (Cl) from the alpha carbon.

**Beta Elimination Reaction:**
- Remove Cl from the alpha carbon.
- Remove H from the beta carbon.

This results in the formation of a double bond between the two carbons.

**Final Product C:**
\[ \text{C} = \text{CH}=\text{CH}_2 \] (which is ethylene or ethene)

### Conclusion
The final product C (C dash) formed from the series of reactions is:
\[ \text{C dash} = \text{C}_2\text{H}_4 \] (Ethylene)

### Answer
Thus, the final answer is option 4, which corresponds to the product C dash.

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