`CH_3-C-=C-CH-3 overset(NaNH_2)rarr'X'`. What is X.?

Options

A. `CH_3CH_2CH=CH_2`
B. `CH_3CH_2C-=CH`
C. `CH_3-CH=CH-CH_3`
D. `CH_2=C=CH-CH_3`

Explanation

To solve the problem, we need to determine the product 'X' formed when butyl 2-yne (CH₃-C≡C-CH₃) reacts with sodium amide (NaNH₂).

### Step-by-Step Solution:

1. **Identify the Reactant:**
The reactant is butyl 2-yne, which has the structure:
\[ CH_3-C≡C-CH_3 \]
This is a symmetrical alkyne.

2. **Understand the Role of NaNH₂:**
Sodium amide (NaNH₂) is a strong base. When it reacts with alkynes, it can abstract acidic protons from terminal alkynes, leading to the formation of an anion.

3. **Proton Abstraction:**
In this case, since butyl 2-yne is symmetrical and does not have a terminal hydrogen, the reaction will lead to the abstraction of a proton from one of the carbon atoms adjacent to the triple bond, forming a carbanion:
\[ CH_3-C≡C^- - CH_3 \]

4. **Formation of the Carbanion:**
The carbanion can be represented as:
\[ CH_3-C^-C-CH_3 \]
The negative charge is now on one of the carbon atoms adjacent to the triple bond.

5. **Tautomerization:**
The carbanion can undergo tautomerization, where the negative charge can shift, and a proton can be abstracted from the adjacent carbon. This results in the formation of a double bond:
\[ CH_3-CH=C^-=C-CH_3 \]
This structure can further rearrange to form a terminal alkyne.

6. **Final Product:**
After the tautomerization and rearrangement, the final product is:
\[ CH_3-C≡C-H \]
This is 1-butyne, a terminal alkyne.

### Conclusion:
The product 'X' formed from the reaction of butyl 2-yne with NaNH₂ is **1-butyne (CH₃-C≡C-H)**.