`Na//NH_(3)(l)` converts hex -3- yne to :

Options

A. cis - hex -3- ene
B. trans - hex -3- ene
C. hexan -3- amine
D. n- hexane

Answer

To solve the question of how sodium in liquid ammonia converts hex-3-yne, we will follow these steps:

### Step 1: Identify the structure of hex-3-yne
Hex-3-yne is an alkyne with six carbon atoms and a triple bond between the third and fourth carbon atoms. The structure can be represented as:
\[ \text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_2 - \text{CH}_3 \]

### Step 2: Understand the reaction with sodium in liquid ammonia
Sodium in liquid ammonia is a strong reducing agent. When sodium donates an electron, it can convert the triple bond in hex-3-yne into a double bond, forming a radical intermediate.

### Step 3: Formation of a radical
When sodium donates an electron to one of the carbons in the triple bond, it breaks one of the bonds, leading to the formation of a radical. The structure now looks like this:
\[ \text{CH}_3 - \text{C} \cdot - \text{C} - \text{CH}_2 - \text{CH}_3 \]
Here, the dot represents the unpaired electron (radical).

### Step 4: Reaction with liquid ammonia
The radical then reacts with liquid ammonia (NH₃). The radical carbon will abstract a hydrogen atom from ammonia, resulting in the formation of a negative charge on the carbon:
\[ \text{CH}_3 - \text{C}^- - \text{C} - \text{CH}_2 - \text{CH}_3 + \text{NH}_3 \rightarrow \text{CH}_3 - \text{C} - \text{C} - \text{CH}_2 - \text{CH}_3 + \text{NH}_2^- \]

### Step 5: Further reduction
If another sodium atom donates another electron, the process repeats. The carbon that had the negative charge can again abstract a hydrogen from ammonia, leading to the formation of a double bond:
\[ \text{CH}_3 - \text{C} = \text{C} - \text{CH}_2 - \text{CH}_3 \]

### Step 6: Identify the product
The final product formed after the complete reaction is trans-hex-3-ene. This is an alkene with a double bond between the third and fourth carbon atoms, and it is in the trans configuration due to the positioning of the substituents around the double bond.

### Final Answer
The product formed when hex-3-yne reacts with sodium in liquid ammonia is **trans-hex-3-ene**.

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