Aldehydes, Ketones and Carboxylic Acids Part 3

Concept:The Haloform reaction (iodoform test) is given by methyl ketones ($\mathrm{-COCH_3}$ group) or compounds that can be oxidised to a methyl ketone, such as ethanol or isopropyl alcohol. Hypoiodite ($\mathrm{KOI}$) is the oxidising agent formed in situ.Explanation:Statement A is true. Sodium hypochlorite ($\mathrm{NaOCl}$) oxidises iodide ($\mathrm{I}^-$) to hypoiodite: $\mathrm{NaOCl + KI \rightarrow NaCl + KOI}$.Statement B is false. $\mathrm{KOI}$ (hypoiodite) is an oxidising agent in the haloform reaction, not a reducing agent.Statement C is true. An $\alpha, \beta$-unsaturated methyl ketone still contains the $\mathrm{-COCH_3}$ group, so it gives a positive iodoform test.Statement D is false. Isopropyl alcohol ($\mathrm{CH_3CHOHCH_3}$) is oxidised by $\mathrm{I_2/NaOH}$ to acetone ($\mathrm{CH_3COCH_3}$), a methyl ketone, so it gives a positive iodoform test.Statement E is false. Methanoic acid ($\mathrm{HCOOH}$) lacks the $\mathrm{-COCH_3}$ group or $\mathrm{CH_3CH(OH)-}$ group required for the iodoform test.Thus, only statements A and C are true.Answer:The correct option is D (A & C Only).