Acetylene is treated with excess sodium in liquid ammonia. The product is reacted with excess of methyl iodide. The final product is……………….

Explanation

To solve the problem step by step, we will analyze the reactions that acetylene undergoes when treated with sodium in liquid ammonia, followed by the reaction with methyl iodide.

### Step 1: Identify the starting material
The starting material is acetylene, which has the chemical formula C₂H₂. It can be represented as:
\[ \text{HC} \equiv \text{CH} \]

### Step 2: Reaction with sodium in liquid ammonia
When acetylene is treated with excess sodium in liquid ammonia (Na + NH₃), a reduction occurs where sodium donates electrons to the acetylene. This process leads to the formation of a sodium acetylide intermediate. The acidic protons (hydrogens) on the acetylene are abstracted, resulting in the formation of the acetylide ion:
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