Chapter 5: Problem 61
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 \(\mathrm{KOH}\)
Short Answer
Step by step solution
Key Concepts
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