A) \[{{C}_{6}}{{H}_{6}}+FeC{{l}_{3}}+C{{l}_{2}}\to {{C}_{6}}{{H}_{5}}Cl\]
B) \[{{C}_{6}}{{H}_{5}}CHO+C{{H}_{3}}CHO+KOH\to {{C}_{6}}{{H}_{5}}CH=CH-CHO\]
C) \[{{C}_{6}}{{H}_{6}}+C{{H}_{3}}COCl+AlC{{l}_{3}}\to \overset{\,\,\,O}{\mathop{\overset{\,\,\,\,||}{\mathop{{{C}_{6}}{{H}_{5}}-C-C{{H}_{3}}}}\,}}\,\]
D) \[{{C}_{6}}{{H}_{5}}OH+CHC{{l}_{3}}+KOH\xrightarrow{{}}\text{Salicylaldehyde}\]
Correct Answer: C
Solution :
Friedel-craft?s reaction \[\underset{\text{Acetyl chloride}}{\mathop{C{{H}_{3}}COCl}}\,+\underset{\text{Benzene}}{\mathop{{{C}_{6}}{{H}_{6}}}}\,\xrightarrow{\text{anhydrous }AlC{{l}_{3}}}\underset{\text{Acetophenone}}{\mathop{C{{H}_{3}}CO{{C}_{6}}{{H}_{5}}}}\,+HCl\]
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