`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 question, we need to analyze the given reaction sequence step by step.

### Step 1: Identify the starting compound
The starting compound is CH₃COC₂H₅, which is ethyl methyl ketone. This compound contains a methyl group adjacent to a carbonyl group (C=O), making it a methyl ketone.

**Hint:** Look for functional groups in the compound to determine its reactivity.

### Step 2: Understand the reaction conditions
The reaction involves NaOH and I₂. This combination is indicative of the iodoform test, which is used to identify methyl ketones.

**Hint:** Recall that the iodoform test is specific for compounds that have a methyl group adjacent to a carbonyl group.

### Step 3: Perform the iodoform reaction
When ethyl methyl ketone (CH₃COC₂H₅) reacts with NaOH and I₂, it undergoes halogenation. The methyl group (CH₃) is converted into iodoform (CHI₃), which precipitates as a yellow solid.

The reaction can be summarized as:
\[ \text{CH}_3COC_2H_5 + I_2 + NaOH \rightarrow \text{CHI}_3 (yellow ppt) + \text{CH}_3COO^- \]

**Hint:** Remember that during the iodoform test, the methyl group is transformed into iodoform, which is a yellow precipitate.

### Step 4: Identify the yellow precipitate (X)
The yellow precipitate formed (X) is iodoform (CHI₃).

**Hint:** The yellow precipitate is a key indicator of the presence of a methyl ketone.

### Step 5: React iodoform with silver
Next, we treat the iodoform (CHI₃) with silver (Ag). This reaction leads to the formation of acetylene (ethyne, C₂H₂) and silver iodide (AgI).

The reaction can be represented as:
\[ 2 \text{CHI}_3 + 6 \text{Ag} \rightarrow \text{C}_2\text{H}_2 (g) + 6 \text{AgI} \]

**Hint:** When iodoform is treated with silver, it results in the formation of a gaseous product.

### Step 6: Identify the gaseous product (Y)
The gaseous product Y formed from this reaction is ethyne (C₂H₂).

**Hint:** Look for the products of the reaction involving iodoform and silver to determine the gaseous product.

### Conclusion
Thus, the gaseous product Y is ethyne (C₂H₂).

### Final Answer
Y is ethyne (C₂H₂).

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