`CH_(3)COC_(2)H_(5)overset(NaOH//I_(2))tounderset("yellow ppt")(X)overset(Ag)toY(g)`.
In this sequence, the gaseous product Y is

Options

A. Ethyne
B. Ethane
C. Ethene
D. Propane

Answer

To solve the problem step by step, let's break down the reaction sequence provided in the question:

### Step 1: Identify the starting compound
The starting compound is CH₃COC₂H₅, which is an ethyl methyl ketone (or 3-pentanone).

### Step 2: Understand the reaction with NaOH and I₂
When CH₃COC₂H₅ reacts with NaOH and I₂, it undergoes the haloform reaction. This reaction specifically occurs with methyl ketones (compounds containing the CH₃CO- group).

### Step 3: Formation of yellow precipitate (X)
During the haloform reaction, the methyl ketone (CH₃CO-) will react with the iodine (I₂) in the presence of NaOH to form iodoform (CHI₃), which is a yellow precipitate. The reaction can be summarized as:
\[ \text{CH}_3COC_2H_5 + \text{NaOH} + \text{I}_2 \rightarrow \text{CHI}_3 (yellow \, ppt) + \text{RCOO}^- \text{(sodium salt of carboxylic acid)} \]
Here, RCOO⁻ is the sodium salt of the carboxylic acid formed from the ethyl group (C₂H₅).

### Step 4: Reaction of yellow precipitate (X) with silver
The yellow precipitate (X) is iodoform (CHI₃). When iodoform reacts with silver (Ag), it produces ethyne (C₂H₂) and silver iodide (AgI). The reaction can be summarized as:
\[ \text{CHI}_3 + 6 \text{Ag} \rightarrow \text{C}_2\text{H}_2 + 6 \text{AgI} \]
Here, C₂H₂ is ethyne, which is a gaseous product.

### Conclusion
Thus, the gaseous product Y formed in this sequence is ethyne (C₂H₂).

### Final Answer
The gaseous product Y is **C₂H₂ (ethyne)**.

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