Propanone is treated with iodine and sodium hydroxide to yield a yellow precipitate and the sodium salt of an organic acid. This salt is acidified to give acid (A). Acid (A) is then reacted with one equivalent of chlorine gas in the presence of a small amount of red phosphorus, followed by hydrolysis, to form compound (B). What is the IUPAC name of compound (B)?

Step-by-step solution

Propanone ( CH ₃- CO - CH ₃ ) contains a methyl ketone group. On treatment with I ₂ and NaOH, it undergoes the haloform reaction to give iodoform ( CHI ₃ , a yellow precipitate) and sodium ethanoate ( CH ₃ COONa ). Note that the carbon chain is cleaved, reducing the number of carbon atoms in the acid salt by one compared to the original ketone. Acidification of sodium ethanoate yields ethanoic acid ( CH ₃ COOH ), which is acid (A). Ethanoic acid has -hydrogen atoms. Reaction with one equivalent of Cl ₂ and red phos