A) \[PC{{l}_{5}}\]
B) \[C{{l}_{2}}\]
C) \[SOC{{l}_{2}}\]
D) \[PC{{l}_{3}}\]
Correct Answer: B
Solution :
(a) \[C{{H}_{3}}COOH+PC{{l}_{5}}\xrightarrow{{}}\underset{acetylchloride}{\mathop{C{{H}_{3}}COCl}}\,\] \[+POC{{l}_{3}}+HCl\] (b)\[C{{H}_{3}}COOH+C{{l}_{2}}\xrightarrow{{}}C{{H}_{2}}ClCOOH+HCl\] (c) \[C{{H}_{3}}COOH+SOC{{l}_{2}}\xrightarrow{{}}C{{H}_{3}}COCl\] \[+S{{O}_{2}}+HCl\] (d) \[C{{H}_{3}}COOH+PC{{l}_{3}}\xrightarrow{{}}3C{{H}_{3}}COCl+{{H}_{3}}P{{O}_{3}}\] Hence, acetic acid reacts with chlorine gas to give mono chloro acetic acid and it does not give acetyl chloride.
You need to login to perform this action.
You will be redirected in
3 sec
