Carboxylic Acids Question 43
Question: In the reaction, $ C_6H_5OH\xrightarrow{NaOH}(A) $ $ \underset{140{}^\circ C,\ (4-7atm)}{\mathop{\xrightarrow{\ CO_2\ \ \ \ \ \ \ \ \ \ \ \ }}}(B) $ $ \xrightarrow{HCl}(C), $ the compound is
[Pb. CET 2001]
Options:
A) Benzoic acid
B) Salicylaldehyde
C) Chlorobenzene
D) Salicylic acid
Show Answer
Answer:
Correct Answer: D
Solution:
Treatment of sodium salt of phenol with $ CO_2 $ under pressure bring about substitution of the carbonyl group $ -COOH, $ for the hydrogen of the ring. This is called as Kolbe’s reaction
BETA
