Compound `(C_(9)H_(10)O)` shows positive iodoform test. Oxidation of A with `KMnO_(4)//KOH` gives acid `B(C_(8)H_(6)O_(4))` anhydride of B is used for the preparation of phenolphthalein compound is

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, we will analyze the information provided in the question and use our knowledge of organic chemistry.

### Step 1: Identify the structure of compound A (C₉H₁₀O)
Given that compound A has the molecular formula C₉H₁₀O and shows a positive iodoform test, we can conclude that it must contain a methyl ketone group (–C(=O)CH₃). This is because the iodoform test is positive for compounds that have a methyl ketone or a secondary alcohol that can be oxidized to a methyl ketone.

### Step 2: Determine the possible structures for compound A
The presence of a methyl ketone suggests that compound A could be a compound like acetophenone or similar structures. However, since it has a total of 9 carbons, we can consider structures like ortho- or para-substituted phenyl groups with a methyl ketone.

### Step 3: Oxidation of A to form compound B (C₈H₆O₄)
The oxidation of compound A with KMnO₄ in KOH leads to the formation of compound B, which has the molecular formula C₈H₆O₄. This indicates that compound B is likely a dicarboxylic acid. Given the context, it is reasonable to assume that compound B is phthalic acid (1,2-benzenedicarboxylic acid).

### Step 4: Identify the anhydride of compound B
The anhydride of phthalic acid is phthalic anhydride. Phthalic anhydride is used in the preparation of phenolphthalein, which is a well-known pH indicator.

### Step 5: Conclusion
Based on the above deductions, we can conclude that compound A is likely ortho- or para-substituted with a methyl ketone, leading to the formation of phthalic acid (compound B) upon oxidation. The correct answer to the question is compound D, which fits the criteria.

### Final Answer
The compound is **D**.

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