A)
B)
C)
D)
Correct Answer: B
Solution :
\[K=\frac{1}{2}\,m{{u}^{2}}\,{{\cos }^{2}}\theta \] is electron donating in nature that is why strongly activating whereas CF is all over electron withdrawing by \[\therefore \] effect so weakly deactivating. Thus, the incoming group achieve a position ortho to \[\frac{1}{2}\,m{{u}^{2}}{{\cos }^{2}}\,\theta =\frac{1}{2}\,\left( \frac{1}{2}\,m{{u}^{2}} \right)\] group as it is o/p directing.
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