The major product of the following reaction is :
`{:(CH_(3)CH_(2)CH-CH_(2) ),(" |"" |"),(" "Br" " Br):}` `underset(underset("in liq" NH_(3))(II. NaNH_(2)))overset("I. KOH alc")rarr`

Options

A. `CH_(3)CH_(2)C -= CH`
B. `CH_(3)CH = C = CH_(2)`
C. `{:(CH_(3)CH_(2)CH-CH_(2)),(" |"" |"),(" "NH_(2)" "NH_(2)):}`
D. `CH_(3)CH=CHCH_(2)NH_(2)`

Explanation

To solve the given question, we need to determine the major product of the reaction involving 1,2-dibromobutane when treated with alcoholic KOH followed by NaNH2 in liquid ammonia. Let's break this down step by step:

### Step 1: Identify the Reactant
The reactant is 1,2-dibromobutane. The structure can be represented as follows:

```
Br
|
CH3-CH-CH2-CH3
|
Br
```

### Step 2: First Reaction with Alcoholic KOH
In the first step, we treat the dibromobutane with alcoholic KOH. Alcoholic KOH is a strong base and promotes dehydrohalogenation. It will abstract a proton from the less hindered carbon adjacent to the bromine atom.

1. The base (OH-) abstracts a proton from the carbon that is less sterically hindered (the CH2 next to the Br).
2. The bond between this carbon and the adjacent carbon will form a double bond, and one of the bromine atoms will leave as bromide ion (Br-).

The reaction can be represented as follows:

```
Br
|
CH3-CH-CH2-CH3 + KOH (alc) → CH3-CH=CH-CH3 + HBr
|
Br
```

Now, we have formed butene (specifically, 2-butene) as an intermediate product.

### Step 3: Second Reaction with NaNH2 in Liquid Ammonia
In the second step, we treat the formed butene with NaNH2 in liquid ammonia. NaNH2 is a strong base and will further deprotonate the alkene.

1. The amide ion (NH2-) generated from NaNH2 will abstract a proton from the terminal carbon of the double bond.
2. This leads to the formation of a triple bond (alkyne) as the double bond shifts.

The reaction can be represented as follows:

```
CH3-CH=CH-CH3 + NaNH2 → CH3-C≡C-CH3 + NH3
```

### Step 4: Identify the Major Product
The final product after both reactions is 1-butyne (or but-1-yne), which can be represented as:

```
CH3
|
C≡C
|
CH2
|
CH3
```

### Conclusion
The major product of the reaction is **but-1-yne**.