A) Methyl group
B) Carboxylic acid group
C) Methylene group
D) Bicarbonate
Correct Answer: D
Solution :
\[\underset{\text{Propionic}\,\text{acid}}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH(aq)}}\,\,+\underset{\text{sod}\text{.}\,\text{bicarbonate}}{\mathop{NaHC{{O}_{3}}(aq)}}\,\to \]\[C{{H}_{3}}C{{H}_{2}}COONa+C{{O}_{2}}+{{H}_{2}}O\]You need to login to perform this action.
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