Carboxylic Acids Question 52
Question: When propionic acid is treated with aqueous sodium bicarbonate $ CO_2 $ is liberated. The ‘C’ of $ CO_2 $ comes from
[IIT-JEE (Screening) 1999]
Options:
A) Methyl group
B) Carboxylic acid group
C) Methylene group
D) Bicarbonate
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{Propionicacid}{\mathop{CH_3CH_2COOH(aq)}}+\underset{sod\text{.}bicarbonate}{\mathop{NaHCO_3(aq)}}\to $
$ CH_3CH_2COONa+CO_2+H_2O $
BETA
