Aromatic Aldehydes Ketones And Acids Result Question 3
3. Compound $A\left(\mathrm{C} _9 \mathrm{H} _{10} \mathrm{O}\right)$ shows positive iodoform test. Oxidation of $A$ with $\mathrm{KMnO} _4 / \mathrm{KOH}$ gives acid $B\left(\mathrm{C} _8 \mathrm{H} _6 \mathrm{O} _4\right)$. Anhydride of $B$ is used for the preparation of phenolphthalein. Compound $A$ is
(2019 Main, 10 April II)
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Answer:
Correct Answer: 3. ( c )
Solution:
- (i)
(ii) $\mathrm{C}9 \mathrm{H}{10} \mathrm{O}$ on strong oxidation $\left(\mathrm{KMnO}_4 / \mathrm{KOH}\right)$, gives acid $\left(\mathrm{C}_8 \mathrm{H}_6 \mathrm{O}_4\right)$, indicating it can be a dicarboxylic acid. So, ’ $A$ ’ contains $-\mathrm{COCH}_3$ and one $-\mathrm{CH}_3$ group which get oxidised into -COOH and -COOH respectively.
(iii) In the preparation of phenolphthalein from phenol, phthalic anhydride is used. So, ’ $B$ ’ can be phthalic acid (benzene- $1,2$ dicarboxylic acid) which readily forms anhydride.
Thus, the reaction sequence is as follows :
BETA
