Question

Compound $A\left(C_9{H}_{10}O\right)$ shows positive iodoform test. Oxidation of $A$ with $KMn{O}_4/KOH$ gives acid $B\left(C_8{H}_6{O}_4\right)$. Anhydride of $B$ is used for the preparation of phenophthalein. Compound $A$ is:

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

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

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.



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