Identify the hydrocarbon having molecular formula `C_(5)H_(6)` which gives white ppt with ammonical `AgNO_(3)` ?

Options

A.
B.
C.
D.

Answer

To identify the hydrocarbon with the molecular formula \( C_5H_6 \) that gives a white precipitate with ammonical \( AgNO_3 \), we can follow these steps:

### Step 1: Understand the molecular formula
The molecular formula \( C_5H_6 \) suggests that the compound is an unsaturated hydrocarbon. The degree of unsaturation can be calculated using the formula:
\[ \text{Degree of Unsaturation} = \frac{(2C + 2 - H)}{2} \]
For \( C_5H_6 \):
\[ \text{Degree of Unsaturation} = \frac{(2 \times 5 + 2 - 6)}{2} = \frac{(10 + 2 - 6)}{2} = \frac{6}{2} = 3 \]
This indicates that the compound has three degrees of unsaturation, which could be due to multiple bonds (double or triple bonds) or rings.

### Step 2: Identify possible structures
Given that the compound has three degrees of unsaturation, it could be an alkyne (which has a triple bond) or a combination of double bonds and rings. However, since the question specifies that it gives a white precipitate with ammonical \( AgNO_3 \), we focus on terminal alkynes, as they react with ammonical silver nitrate to form a white precipitate of silver acetylide.

### Step 3: List possible terminal alkynes
The terminal alkynes that fit the formula \( C_5H_6 \) are:
1. 1-Pentyne (CH≡C-CH2-CH2-CH3)
2. 2-Pentyne (not terminal, so it won't give a precipitate)

### Step 4: Analyze the options
Among the possible structures:
- **1-Pentyne** has a terminal triple bond and fits the formula \( C_5H_6 \).
- **2-Pentyne** does not have a terminal triple bond and will not react with ammonical \( AgNO_3 \).

### Step 5: Conclusion
Since 1-Pentyne is the only hydrocarbon that meets the criteria of having a terminal triple bond and the molecular formula \( C_5H_6 \), it is the compound that gives a white precipitate with ammonical \( AgNO_3 \).

Thus, the answer is **1-Pentyne**.

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