The hydrocarbon which can react with sodium in liquid ammonia is :
1.`CH_3CH_2CH_2C -= "CCH"_2CH_2CH_3`
2. `CH_3CH_2C -=CH`
3.`CH_3CH = CHCH_3`
4.`CH_3CH_2C -= "CCH"_2CH_3`

Options

A. `CH_3CH_2CH_2C -= "CCH"_2CH_2CH_3`
B. `CH_3CH_2C -=CH`
C. `CH_3CH = CHCH_3`
D. `CH_3CH_2C -= "CCH"_2CH_3`

Answer

To solve the question, we need to identify which of the given hydrocarbons can react with sodium in liquid ammonia. This reaction typically occurs with terminal alkynes, which possess an acidic hydrogen atom that can react with the sodium.

Let's analyze each option step by step:

1. **Option 1: `CH_3CH_2CH_2C ≡ CCH_2CH_2CH_3`**
- This compound is a terminal alkyne because it has a triple bond (≡) at one end of the carbon chain. However, it does not have an acidic hydrogen attached to the carbon that is part of the triple bond. Therefore, it cannot react with sodium in liquid ammonia.

2. **Option 2: `CH_3CH_2C ≡ CH`**
- This compound is also a terminal alkyne. It has a triple bond between the last two carbon atoms, and importantly, it has an acidic hydrogen atom (the hydrogen attached to the carbon that is part of the triple bond). This means it can react with sodium in liquid ammonia, making this option a valid choice.

3. **Option 3: `CH_3CH = CHCH_3`**
- This compound is an alkene, not an alkyne. It has a double bond (instead of a triple bond) and does not have an acidic hydrogen that can react with sodium in liquid ammonia. Therefore, this option is not correct.

4. **Option 4: `CH_3CH_2C ≡ CCH_2CH_3`**
- Similar to option 1, this compound is also a terminal alkyne due to the presence of a triple bond. However, it does not have an acidic hydrogen attached to the carbon involved in the triple bond, so it cannot react with sodium in liquid ammonia.

Based on this analysis, the only hydrocarbon that can react with sodium in liquid ammonia is **Option 2: `CH_3CH_2C ≡ CH`**.

### Final Answer:
The hydrocarbon which can react with sodium in liquid ammonia is **Option 2: `CH_3CH_2C ≡ CH`**.