Question

An aromatic compound A$^1$ with molecular formula C$_8$H$_8$O gives positive 2,4-DNP test. It gives yellow precipitate of compound ‘B’ on treatment with sodium hypobromite. Compound ‘A’ does not react with Tollen’s or Fehling’s reagent; on drastic oxidation with KMnO$_4$, it forms a carboxylic acid ‘C’. Elucidate the structures of A, B and C. Also give their IUPAC names.

Hint:

Pay attention to the molecular formula and the reactions to determine functional groups in organic compounds, such as the presence of alkenes, aldehydes, and carboxylic acids.

Answer 1

The given molecular formula (C$_8$H$_8$O) and the results of the reactions suggest that compound A is styrene (C$_6$H$_5$-CH=CH$_2$). Styrene gives a positive 2,4-DNP test, indicating the presence of an aldehyde or ketone group, but it does not react with Tollen’s or Fehling’s reagent, which indicates it is not an aldehyde. Upon oxidation with KMnO$_4$, it forms benzoic acid (C$_6$H$_5$COOH). The yellow precipitate formed with sodium hypobromite suggests that compound B is the corresponding brominated product (Bromostyrene). The IUPAC names are: - A: Styrene (C$_6$H$_5$-CH=CH$_2$) - B: 1-Bromo-2-phenylethene (Bromostyrene) - C: Benzoic acid (C$_6$H$_5$COOH)