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.
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