A compound `C_5H_8` which gives white ppt. with ammonical `AgNO_3`.A give `(CH_3)_2CHCOOH` with hot alcoholic KOH then compound is

Options

A. `CH_3CH_2-CH_2-CH=CH_2`
B. `CH_3-CH_2-C-=CH`
C. `(CH_3)_2CH-C-=CH`
D. `CH_2=CH-CH_2-CH=CH_2`

Answer

To solve the problem, we need to analyze the information given about the compound \( C_5H_8 \) and the reactions it undergoes.

### Step-by-Step Solution:

1. **Identify the Compound**:
- The molecular formula \( C_5H_8 \) suggests that the compound is an alkene or a cyclic compound since it has fewer hydrogens than alkanes (which would have \( C_nH_{2n+2} \)).
- The presence of a double bond or a ring structure is indicated by the degree of unsaturation.

2. **Reaction with Ammonical AgNO3**:
- The compound gives a white precipitate with ammonical \( AgNO_3 \). This indicates that the compound likely contains a terminal alkyne or an alkene that can form a silver complex.
- Terminal alkynes react with ammonical silver nitrate to form a white precipitate of silver acetylide.

3. **Reaction with Hot Alcoholic KOH**:
- The compound gives \( (CH_3)_2CHCOOH \) (which is 2-methylbutanoic acid) when treated with hot alcoholic KOH. This suggests that the compound undergoes a reaction similar to elimination or rearrangement to form a carboxylic acid.
- The formation of a branched chain carboxylic acid indicates that the original compound has a structure that can rearrange to yield this product.

4. **Structure Deduction**:
- Since \( (CH_3)_2CHCOOH \) has a branched structure, we can deduce that the original compound must have a structure that allows for the formation of this acid upon reaction.
- The simplest structure that fits \( C_5H_8 \) and can lead to \( (CH_3)_2CHCOOH \) is 3-pentyne (or 2-pentyne), which has a terminal alkyne that can react with \( AgNO_3 \).

5. **Final Structure**:
- The compound \( C_5H_8 \) is likely 3-pentyne (or 2-pentyne), which has the formula \( CH_3C \equiv CCH_2CH_3 \) or \( CH_3CH \equiv CCH_3CH_2 \).
- This compound can give the required carboxylic acid upon treatment with hot alcoholic KOH.

### Conclusion:
The compound \( C_5H_8 \) is **3-pentyne** or **2-pentyne**.