Starting with 1 -propanol, 2 -hydroxypropanoate can be prepared by treating with the reagents, in the order (a) \(\mathrm{Cl}_{2} / \mathrm{P}, \mathrm{KMnO}_{4}\) and aqueous \(\mathrm{KOH}\) (b) \(\mathrm{Cl}_{2} / \mathrm{P}, \mathrm{KMnO}_{4}\) and alcoholic \(\mathrm{KOH}\) (c) \(\mathrm{KMnO}_{4}, \mathrm{Cl}_{2} / \mathrm{P}\) and aqueous \(\mathrm{KOH}\) (d) \(\mathrm{KMnO}_{4}, \mathrm{Cl}_{2} / \mathrm{P}\) and alcoholic \(\mathrm{KOH}\)

1-propanol is an alcohol with the chemical formula \(\mathrm{C}_{3}\mathrm{H}_{8}\mathrm{O}\). It has the following structure: H H H | | | H--C-C-C--OH | | | H H H

(a) \(\mathrm{Cl}_{2} / \mathrm{P}\): This reagent is involved in the conversion of alcohols to alkyl halides. (b) \(\mathrm{KMnO}_{4}\): Potassium permanganate is a strong oxidizing agent. It can be used to oxidize primary alcohols to carboxylic acids and secondary alcohols to ketones. (c) Aqueous \(\mathrm{KOH}\): Aqueous potassium hydroxide is a strong base and can be used for hydrolysis of esters to produce carboxylic acids and alcohols. (d) Alcoholic \(\mathrm{KOH}\): Alcoholic potassium hydroxide is also a strong base, but it is used for elimination reactions, such as converting alkyl halides to alkenes.

(a) \(\mathrm{Cl}_{2} / \mathrm{P}, \mathrm{KMnO}_{4}\) and aqueous \(\mathrm{KOH}\): First, 1-propanol would be converted into propyl chloride by Cl2/P. Then, KMnO4 would oxidize propyl chloride into propanoic acid, but not 2-hydroxypropanoate. Aqueous KOH would have no effect in this case. (b) \(\mathrm{Cl}_{2} / \mathrm{P}, \mathrm{KMnO}_{4}\) and alcoholic \(\mathrm{KOH}\): As in (a), Cl2/P and KMnO4 would result in propanoic acid, not 2-hydroxypropanoate. Alcoholic KOH wouldn't have any effect either. (c) \(\mathrm{KMnO}_{4}, \mathrm{Cl}_{2} / \mathrm{P}\), and aqueous \(\mathrm{KOH}\): First, KMnO4 would oxidize 1-propanol into propanoic acid. Then, Cl2/P would have no effect on propanoic acid. Finally, aqueous KOH would convert propanoic acid to potassium propanoate. A further reaction with HCl would yield 2-hydroxypropanoate, so this is the correct sequence. (d) \(\mathrm{KMnO}_{4}, \mathrm{Cl}_{2} / \mathrm{P}\), and alcoholic \(\mathrm{KOH}\): As in (c), KMnO4 would oxidize 1-propanol into propanoic acid, but neither Cl2/P nor alcoholic KOH would produce the desired product. The correct sequence is (c) \(\mathrm{KMnO}_{4}, \mathrm{Cl}_{2} / \mathrm{P}\), and aqueous \(\mathrm{KOH}\).