An organic compound $(A)$ with molecular formula $C_8H_8O$ forms an orange-red precipitate with $2,4-DNP$ reagent and gives a yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehlings' reagent,nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid,it gives a carboxylic acid $(B)$ having molecular formula $C_7H_6O_2$. Identify the compounds $(A)$ and $(B)$ and explain the reactions involved.

(A) forms a $2,4-DNP$ derivative,which indicates it is an aldehyde or a ketone. Since it does not reduce Tollens' or Fehling's reagent,$(A)$ must be a ketone.
$(A)$ responds to the iodoform test,which confirms it is a methyl ketone.
The molecular formula of $(A)$ indicates a high degree of unsaturation,yet it does not decolourise bromine water or Baeyer's reagent,suggesting the unsaturation is due to an aromatic ring.
Compound $(B)$,being an oxidation product of a ketone,is a carboxylic acid. The molecular formula of $(B)$ corresponds to benzoic acid $(C_6H_5COOH)$.
Therefore,$(A)$ is phenyl methyl ketone (acetophenone,$C_6H_5COCH_3$).
The reactions are as follows:
$1$. Formation of $2,4-DNP$ derivative: $C_6H_5COCH_3 + H_2NNHC_6H_3(NO_2)_2 \rightarrow C_6H_5C(CH_3)=NNHC_6H_3(NO_2)_2 + H_2O$
$2$. Iodoform test: $C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 + 3NaI + 3H_2O$
$3$. Oxidation: $C_6H_5COCH_3 \xrightarrow{H_2CrO_4} C_6H_5COOH$