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Step-by-Step Solution
Step 1: Analyze the given data
• The molecular formula of compound A is $C_9H_{10}O$.
• It responds positively to the iodoform test, indicating the presence of a $COCH_3$ (methyl keto) group or a $CH(OH)CH_3$ (methyl carbinol) group.
• On oxidation with $KMnO_4/KOH$, A gives an acid B with the formula $C_8H_6O_4$.
• The anhydride of B is used in the preparation of phenolphthalein, which strongly suggests that B is phthalic acid, since phthalic anhydride is what reacts with phenol to form phenolphthalein.
Step 2: Infer the identity of acid B
Because the anhydride of B is used to make phenolphthalein, B must be phthalic acid ($C_6H_4(COOH)_2$). Its molecular formula is $C_8H_6O_4$, matching the data provided.
Step 3: Deduce the structural requirements of compound A
For compound A to oxidize into phthalic acid, it must have two adjacent substituents on a benzene ring (an ortho-substituted ring) that ultimately become the two carboxylic acid groups in phthalic acid upon oxidation. Also, the positive iodoform test implies that one of these substituents is a methyl ketone group ($-COCH_3$).
Step 4: Propose the structure for compound A
The simplest structure that meets both criteria (an ortho substituent pattern leading to phthalic acid on oxidation, and a methyl ketone group giving iodoform test) is o-methylacetophenone (also called 2’-methylacetophenone). This can be depicted as a benzene ring with a $-COCH_3$ group and a $-CH_3$ group in the ortho positions:
Correct Answer:
Step 5: Connect to the oxidation product
When o-methylacetophenone is oxidized by $KMnO_4/KOH$, both substituents on the ring are converted into $-COOH$ groups in adjacent positions, giving phthalic acid ($C_8H_6O_4$). Its anhydride, phthalic anhydride, is indeed used to synthesize phenolphthalein.
