These three groups are listed in our workbook. I don't know the answer and our teacher isn't sure either:

Three substituents

(each has a free valency through which it is attached to a benzene ring):

  1. Diphenylhalomethyl
  2. Difluoromethyl
  3. Dichloromethyl

Now, I've read and understood this question - How does hyperconjugation lead to the directing properties of alkyl group? Therefore, I think that the groups 2,3 are deactivating (due to -I effect of fluorines and chlorines) yet o/p directing due to hyperconjugation with the H atom. On the other hand, the first group should be meta directing since it has only -I effect and no +/-M or hyperconjugation.

I googled for the complete list of directing groups but I could not find these types of groups.

Is my reasoning - and hence answer - correct? Or is there another answer, and why?

  • $\begingroup$ Is 1 a deactivator meta director? 2 deactivator , o/p? 3 deactivator meta? If any of those is correct let me know. I might be thinking on the rightish track. $\endgroup$ – Avnish Kabaj Mar 9 '18 at 12:30
  • $\begingroup$ @AvnishKabaj I don't know the answer :( $\endgroup$ – Gaurang Tandon Mar 9 '18 at 12:59
  • $\begingroup$ I agree with you, on the first group being a meta directing group. But I believe that the degree of hyperconjugation in the second and the third group is very low (due to a single hydrogen atom). Carbon is quite electrophilic in both the cases and hence they must behave like meta directing groups just like CF3. $\endgroup$ – Parth Chauhan Aug 6 '18 at 16:05

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