2,3-dibromobutane is treated with sodium amide (excess) in liquid ammonia. The main product is

Options

A. but-1-yne
B. buta-1,2-diene
C. but-2-yne
D. buta-1,3-diene

Explanation

To solve the question regarding the reaction of 2,3-dibromobutane with sodium amide in liquid ammonia, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
We have 2,3-dibromobutane (C4H8Br2) as our starting material. Its structure can be represented as follows:
```
CH3-CH(Br)-CH(Br)-CH3
```
Here, bromine atoms are attached to the second and third carbon atoms of the butane chain.

2. **Understand the Role of Sodium Amide**:
Sodium amide (NaNH2) is a strong base. In this reaction, it will act to deprotonate (remove hydrogen) from the adjacent carbon atoms, facilitating the elimination of bromine atoms.

3. **First Elimination Reaction**:
Sodium amide will first extract a hydrogen atom from the carbon adjacent to one of the bromine atoms.
- The hydrogen can be removed from carbon 1 or carbon 4 (which have three hydrogens each) or from carbon 2 (which has one hydrogen due to the presence of Br).
- However, since carbon 2 is bonded to Br (an electron-withdrawing group), it is less stable for hydrogen to be removed from here.

Let's assume the hydrogen is removed from carbon 1:
```
CH3-CH(Br)-CH(Br)-CH2- → CH2=CH(Br)-CH(Br)-CH3 + NaBr + NH3
```

4. **Second Elimination Reaction**:
The product from the first elimination still contains a bromine atom on carbon 3. Sodium amide will again act to remove a hydrogen atom, this time from carbon 3:
```
CH2=CH(Br)-CH(Br)-CH3 → CH2=CH-C≡C-CH3 + NaBr + NH3
```

5. **Final Product**:
After both elimination reactions, the main product formed is 2-butyne (Bute-2-yne), which can be represented as:
```
CH3-C≡C-CH3
```

### Conclusion:
The main product of the reaction of 2,3-dibromobutane with excess sodium amide in liquid ammonia is **2-butyne (Bute-2-yne)**.