A trisubstituted compound 'A' `C_(10)H_(12)O_2` gives neutral `FeCl_3` test positive.Treatment of compound 'A' with `NaOH` and `CH_(3)Br` gives `C_(11)H_(14)O_(2)`,with hydrodioic acid gives methyl iodide and with hot conc. `NaOH` gives a compound `B,C_(10)H_(12)O_(2),C` Compound '`A`' also decolorises alkaline `KMnO_4`. The number of `pi` bond's present in the compound '`A`' is

Answer

To solve the problem, we need to analyze the information given about compound 'A' and deduce the number of π bonds present in it.

### Step-by-Step Solution:

1. **Identify the Structure of Compound 'A':**
- The molecular formula of compound 'A' is C₁₀H₁₂O₂.
- It gives a positive neutral FeCl₃ test, indicating the presence of a phenolic (aromatic) group, which means it likely contains an -OH group attached to a benzene ring.

2. **Determine the Degree of Unsaturation:**
- The degree of unsaturation (DU) can be calculated using the formula:
\[ \text{DU} = \frac{(2C + 2 + N - H - X)}{2} \]
- For compound 'A': C = 10, H = 12, O = 2 (O does not affect the calculation).
- Plugging in the values:
\[ \text{DU} = \frac{(2 \times 10 + 2 - 12)}{2} = \frac{(20 + 2 - 12)}{2} = \frac{10}{2} = 5 \]
- This indicates that there are 5 degrees of unsaturation, which can be due to rings and/or π bonds.

3. **Analyze Possible Structures:**
- Since compound 'A' is trisubstituted and contains a phenolic group, it likely has a benzene ring (which contributes 3 π bonds).
- The remaining degrees of unsaturation could come from:
- A carbonyl group (C=O) which contributes 1 π bond.
- Therefore, the total number of π bonds in compound 'A' can be calculated as:
\[ \text{Total π bonds} = 3 \text{ (from benzene)} + 1 \text{ (from carbonyl)} = 4 \]

4. **Conclusion:**
- The number of π bonds present in compound 'A' is 4.

### Final Answer:
The number of π bonds present in compound 'A' is **4**.

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