An aromatic compound $A$ (Molecular formula $C_{8}H_{8}O$) gives a positive $2,4-DNP$ test. It gives a yellow precipitate of compound $B$ on treatment with iodine and sodium hydroxide solution. Compound $A$ does not give Tollen's or Fehling's test. On drastic oxidation with potassium permanganate,it forms a carboxylic acid $C$ (Molecular formula $C_{7}H_{6}O_{2}$),which is also formed along with the yellow compound in the above reaction. Identify $A, B$ and $C$ and write all the reactions involved.

(A) $1$. Compound $A$ has the molecular formula $C_{8}H_{8}O$. Since it gives a positive $2,4-DNP$ test,it contains a carbonyl group (aldehyde or ketone).
$2$. It does not give Tollen's or Fehling's test,which confirms that $A$ is a ketone.
$3$. It gives a yellow precipitate with $I_{2}/NaOH$ (iodoform test),indicating the presence of a $CH_{3}CO-$ group.
$4$. The compound $A$ is acetophenone $(C_{6}H_{5}COCH_{3})$.
$5$. Reaction with $I_{2}/NaOH$ gives iodoform ($CHI_{3}$,compound $B$) and sodium benzoate,which upon acidification gives benzoic acid ($C_{6}H_{5}COOH$,compound $C$).
$6$. Oxidation of acetophenone with $KMnO_{4}$ also yields benzoic acid $(C)$.
Reactions:
- $2,4-DNP$ test: $C_{6}H_{5}COCH_{3} + 2,4-(NO_{2})_{2}C_{6}H_{3}NHNH_{2}$ $\rightarrow C_{6}H_{5}C(CH_{3})=NNHC_{6}H_{3}(NO_{2})_{2} + H_{2}O$
- Iodoform test: $C_{6}H_{5}COCH_{3} + 3I_{2} + 4NaOH \rightarrow C_{6}H_{5}COONa + CHI_{3} + 3NaI + 3H_{2}O$. Acidification of $C_{6}H_{5}COONa$ gives $C_{6}H_{5}COOH$ $(C)$.
- Oxidation: $C_{6}H_{5}COCH_{3} + [O] \xrightarrow{KMnO_{4}} C_{6}H_{5}COOH$ $(C)$.