Question
Chemistry Question on Organic Chemistry- Some Basic Principles and Techniques
Ethanoic acid undergoes Hell - Volhard Zelinsky reaction but Methanoic acid does not because of
Absence of α-H atom in Ethanoic Acid
Presence of α-H atom in Methanoic Acid
Higher acidic strength of ethanoic acid than Methanoic Acid
Presence of α-H atom in Ethanoic Acid
Presence of α-H atom in Ethanoic Acid
Solution
The Hell-Volhard-Zelinsky (HVZ) reaction is a method used for the α-bromination of carboxylic acids. In this reaction, the carboxylic acid is treated with phosphorus tribromide (PBr3) and a small amount of red phosphorus (P) as a catalyst. The resulting product is an α-bromo acid.
Methanoic acid (formic acid) does not undergo the Hell-Volhard-Zelinsky reaction, whereas ethanoic acid (acetic acid) does. The reason for this is due to the presence of an α-hydrogen atom in ethanoic acid.
Therefore, the correct answer is (D) Presence of α-H atom in Ethanoic Acid.