In the reaction
`CH_(3)COOHoverset(LiAlH_(4))rarr(A)overset(I_(2)+NaOH)rarr(B)overset(Ag("Dust"))rarr(C)`, the final product C is:-

Options

A. `C_2H_5I`
B. `C_2H_5OH`
C. `C_2H_2`
D. `CH_3COCH_3`

Answer

To solve the given reaction sequence, we will break down each step and identify the products formed at each stage.

### Step 1: Reduction of Acetic Acid
The first reaction involves acetic acid (CH₃COOH) reacting with lithium aluminum hydride (LiAlH₄), which is a strong reducing agent.

**Reaction:**
\[ \text{CH}_3\text{COOH} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} \]

**Product A:**
The product of this reaction is ethanol (CH₃CH₂OH).

### Step 2: Reaction with Iodine and NaOH
Next, ethanol (A) reacts with iodine (I₂) in the presence of sodium hydroxide (NaOH).

**Reaction:**
1. Iodine reacts with NaOH to form sodium iodide (NaI) and sodium hypoiodite (NaOI).
2. The NaI acts as an oxidizing agent, oxidizing ethanol to acetaldehyde (CH₃CHO).

**Overall Reaction:**
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{I}_2 + \text{NaOH} \rightarrow \text{CH}_3\text{CHO} + \text{NaI} + \text{H}_2\text{O} \]

**Product B:**
The product formed is acetaldehyde (CH₃CHO).

### Step 3: Haloform Reaction
The acetaldehyde (B) then undergoes a haloform reaction in the presence of iodine and sodium hydroxide.

**Reaction:**
\[ \text{CH}_3\text{CHO} + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3 + \text{HCOONa} + 3\text{NaI} + 3\text{H}_2\text{O} \]

**Product B:**
The product of this reaction is iodoform (CHI₃).

### Step 4: Reaction with Silver Dust
Finally, iodoform (B) reacts with silver dust (Ag).

**Reaction:**
\[ 2\text{CHI}_3 + 6\text{Ag} \rightarrow \text{C}_2\text{H}_2 + 6\text{AgI} \]

**Product C:**
The final product formed is acetylene (C₂H₂).

### Final Answer:
The final product C is acetylene (C₂H₂).

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