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To solve the problem, we need to analyze the organic compound 'A' with the formula C9H10O and its reactions with conc. HI, which indicates that it is likely an alcohol or ether. The cleavage of 'A' yields compounds 'B' and 'C', where 'B' gives a yellow precipitate with AgNO3 (indicating a halide) and 'C' tautomerizes to 'D', which gives a positive iodoform test (indicating a methyl ketone or a secondary alcohol).
### Step 1: Identify the Structure of Compound 'A'
Given the molecular formula C9H10O, we can deduce that 'A' could be a phenolic compound or an aromatic alcohol. The presence of a hydroxyl group (-OH) is likely due to the formula indicating a degree of unsaturation.
### Step 2: Reaction with Conc. HI
When 'A' is treated with conc. HI, it undergoes cleavage. This suggests that 'A' might be an ether or an alcohol. The cleavage will lead to the formation of alkyl halides.
### Step 3: Analyze the Products 'B' and 'C'
- **Compound 'B'**: Since 'B' gives a yellow precipitate with AgNO3, it is likely a halide (possibly an alkyl halide).
- **Compound 'C'**: The fact that 'C' tautomerizes to 'D' and gives a positive iodoform test indicates that 'C' is likely a secondary alcohol or a methyl ketone.
### Step 4: Determine the Structure of 'D'
Since 'D' gives a positive iodoform test, it must have the structure of a methyl ketone (RCOCH3) or a secondary alcohol (RCHOHCH3).
### Conclusion
Based on the above deductions, the compound 'A' that fits all the criteria is likely to be 1-phenyl-2-propanol (C9H10O). This compound can undergo cleavage with HI to form a phenyl halide and a ketone, which aligns with the properties of 'B' and 'C'.
Thus, the correct answer is the structure that corresponds to 1-phenyl-2-propanol.