Consider the following substances
1. `CH_(3)C-= C CH_(3)`
2. `CH_(3)CH_(2)C-=CH`
3. `CH_(3)CH=CHCH_(3)`
4. `CH_(3)CH_(2)CH=CH_(2)`
Which of the following reagent can be used to distinguish the compound (2) from the rest of the compounds?

Options

A. Bromin/ `"CCl"_(4)`
B. Bromine/`CH_(3)COOH`
C. `Alk.KMnO_(4)`
D. Ammonical `AgNO_(3)` or ammoniacal cuprous chloride

Answer

To distinguish the second compound (CH₃CH₂C≡CH) from the other compounds, we can use ammoniacal cuprous chloride (CuCl). This reagent specifically reacts with terminal alkynes, which is the characteristic of compound 2. Here’s a step-by-step solution:

### Step-by-Step Solution

1. **Identify the Compounds**:
- Compound 1: CH₃C≡CCH₃ (internal alkyne)
- Compound 2: CH₃CH₂C≡CH (terminal alkyne)
- Compound 3: CH₃CH=CHCH₃ (alkene)
- Compound 4: CH₃CH₂CH=CH₂ (alkene)

2. **Determine the Type of Each Compound**:
- Compound 1 is an internal alkyne (triple bond between two carbon atoms that are not at the end).
- Compound 2 is a terminal alkyne (triple bond at the end of the carbon chain).
- Compounds 3 and 4 are alkenes (double bonds present).

3. **Select a Suitable Reagent**:
- Ammoniacal cuprous chloride (CuCl) is known to react specifically with terminal alkynes, producing a precipitate of copper(I) acetylide and releasing ammonia gas.

4. **Perform the Reaction**:
- When compound 2 (CH₃CH₂C≡CH) is treated with ammoniacal cuprous chloride, it will react and form a colored precipitate (red or brown) indicating the presence of a terminal alkyne.
- The other compounds (1, 3, and 4) will not react with this reagent in the same way, as they do not have a terminal alkyne.

5. **Conclusion**:
- The formation of a precipitate when reacting compound 2 with ammoniacal cuprous chloride distinguishes it from the other compounds, which do not produce a similar reaction.