An organic compound (A) with molecular formula `C_8H_8 O` forms an orange red precipitate with 2,4 -DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide . It neither reduces Tollen's reagent or Fehling's solution , nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formulae `C_7H_6O_2`. Identify the compound (A) and (B) and explain the reactions involved .

Explanation

Step-1. To determine the structures of compounds (A) and (B).
(i) Since compound (A), M.F. `C_(8)H_(8)O` forms a 2,4-DNP, it must be either an aldehyde or a ketone.
(ii) Since, it does not reduce tollens' reagent or Fehling's solution, (A) is must be a ketone.
(iii) Since (A) on treatment with `I_(2)//NaOH`, gives yellow ppt. (iodoform), therefore, it must be a methyl ketone.
(iv) D.B.E. for `C_(8)H_(8)O=(8(4-2)+8(1-2)+1(2-2))/(2)+1=5`
One of the five sites of unsaturation must be due to the keto group, the remaining four may be due to the presence of a benzene ring. the presence of benzene ring is supported by the observation that it neither decolourizes bromine water nor Baeyer's reagent.
(v) Since drastic oxidatio of A with chromic acid gives compound (B) with M.F. `C_(7)H_(6)O_(2)`, therefore, (B) must be benzoic acid. further, the formation of benzoic acid suggests that compound (A) is a mono-substituted benzene derivative.
(vi) Thus, compound (A) has a monosubstituted benzene ring, i.e., a `C_(6)H_(5)` ring `CH_(3)CO` group. both these group account for the complete M.F. `(C_(6)H_(5)+COCH_(3)=C_(8)H_(8)O)` of compound (A), therefore, compound (A) is acetophenone or methyl phenyl ketone or 1-phenylethanone.

Step.2: To explain all the reactions involved in the question.