An organic compound containing C, H and O gives following observations:
(i) It exists in two isomeric forms (A) and (B).
(ii) 0.108 g of one of the isomers on combustion gave 0.308 g of `CO_(2)` and 0.072 g of `H_(2)O`
(iii) (A) is insoluble in NaOH and `NaHCO_(3)` while (B) is soluble in NaOH.
(iv) (A) reacts with Hl to give compound (C ) and (D). (C ) can be separated from (D) by ethanolic `AgNO_(3)` solution and (D) is soluble in NaOH.
(v) (B) readily reacts with bromine water to give compound (E) of molecular formula `C_(7)H_(5)OBr_(3)`
Compound (A) is

Options

A. a phenol
B. a symmetric ether
C. a mixed ether
D. a carboxylic acid

Answer

To solve the problem, we will analyze the information given step by step.

### Step 1: Determine the empirical formula from combustion data
- Given that 0.108 g of one isomer on combustion produces 0.308 g of CO₂ and 0.072 g of H₂O.
- From the combustion of CO₂, we can find the amount of carbon:
- Molar mass of CO₂ = 44 g/mol (12 g from C and 32 g from O)
- 0.308 g CO₂ contains:
\[ \text{Mass of C} = \frac{12 \text{ g}}{44 \text{ g}} \times 0.308 \text{ g} = 0.084 \text{ g} \]

- From the combustion of H₂O, we can find the amount of hydrogen:
- Molar mass of H₂O = 18 g/mol (2 g from H and 16 g from O)
- 0.072 g H₂O contains:
\[ \text{Mass of H} = \frac{2 \text{ g}}{18 \text{ g}} \times 0.072 \text{ g} = 0.008 \text{ g} \]

- The mass of oxygen can be calculated as:
\[ \text{Mass of O} = \text{Total mass} - \text{Mass of C} - \text{Mass of H} = 0.108 \text{ g} - 0.084 \text{ g} - 0.008 \text{ g} = 0.016 \text{ g} \]

### Step 2: Calculate the moles of each element
- Moles of Carbon (C):
\[ \text{Moles of C} = \frac{0.084 \text{ g}}{12 \text{ g/mol}} = 0.007 \text{ mol} \]
- Moles of Hydrogen (H):
\[ \text{Moles of H} = \frac{0.008 \text{ g}}{1 \text{ g/mol}} = 0.008 \text{ mol} \]
- Moles of Oxygen (O):
\[ \text{Moles of O} = \frac{0.016 \text{ g}}{16 \text{ g/mol}} = 0.001 \text{ mol} \]

### Step 3: Determine the simplest whole number ratio
- Divide each by the smallest number of moles (0.001):
- C: \( \frac{0.007}{0.001} = 7 \)
- H: \( \frac{0.008}{0.001} = 8 \)
- O: \( \frac{0.001}{0.001} = 1 \)

Thus, the empirical formula is \( C_7H_8O \).

### Step 4: Analyze the properties of isomers A and B
- **Isomer A** is insoluble in NaOH and NaHCO₃, indicating it is likely an ether or a non-acidic compound.
- **Isomer B** is soluble in NaOH, suggesting it is likely a phenolic compound.

### Step 5: Identify compound A
- Given that A reacts with HI to yield C and D, where C is soluble in AgNO₃ and D is soluble in NaOH, we can deduce:
- A is likely an ether (specifically, anisole, which is methyl phenyl ether).
- Upon reacting with HI, anisole yields methyl iodide (C) and phenol (D).

### Step 6: Conclusion
- Based on the analysis, compound A is anisole (methyl phenyl ether).

### Final Answer
**Compound (A) is anisole (C₇H₈O).**

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