A) COOH
B) CN
C) \[{{\operatorname{COCH}}_{3}}\]
D) \[{{\operatorname{NHCOCH}}_{3}}\]
Correct Answer: D
Solution :
\[\mathbf{NHCOC}{{\mathbf{H}}_{3}}\] It is o, p-directing group, i.e those substituents which direct the incoming group to ortho and para positions. For example \[-C{{H}_{3}},\,-C{{H}_{2}}C{{H}_{3}},-{{C}_{6}}{{H}_{5}},\,-Cl,\,-\operatorname{Br},-\operatorname{I}\] \[-\operatorname{OH},\,-{{\operatorname{OCH}}_{3}},-\operatorname{N}{{\operatorname{H}}_{2}},\,-{{\operatorname{NHCH}}_{3}}\] \[-N{{\left( C{{H}_{3}} \right)}_{2}} -NHCOC{{H}_{3}} etc.\] In general, all electron-donating groups are o,p-directing. -COOH & -CN are m-directing.
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