1. Tardigrade
  2. Question
  3. Chemistry
  4. Compound A ( 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 phenophthalein. Compound A is:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Compound $A \left( C _9 H _{10} O \right)$ shows positive iodoform test. Oxidation of $A$ with $KMnO _4 / KOH$ gives acid $B \left( C _8 H _6 O _4\right)$. Anhydride of $B$ is used for the preparation of phenophthalein. Compound $A$ is:

NTA AbhyasNTA Abhyas 2020Aldehydes Ketones and Carboxylic Acids

Solution:

$\underset{\text{shows} \, \text{iodoform} \, \text{test}}{\underset{\text{C}_{\text{9}} \text{H}_{\text{10}} \text{O}}{\text{A}}} \overset{\text{KMnO}_{\text{4}} \text{/KOH}}{ \rightarrow } \underset{\text{Acid}}{\underset{\text{C}_{\text{8}} \text{H}_{\text{6}} \text{O}_{\text{4}}}{\text{B}}}$

Note : Anhydride of B is used in preparation of phenolphthalein.

$\Rightarrow $ B must be phthalic acid as phthalic anhydride is used in preparation of phenolphthalein.

Solution

$\Rightarrow $ Option 3 is wrong because it doesn't give iodoform test