Consider the following resonance structures of Nitro-benzene:

enter image description here

It is written in the book that because the positive charge is always at $-o$ or $-p$ positions, Nitro group in benzene has meta directing effect. Which means it is easier to attach substituent electrophilic groups at $-m$ position than $-o$ and $-p$ positions.

The reason for the above conclusion is given that because of positive charge at $-o$ and $-p$,$-m$ is relatively negatively charged.

But then look at resonance structures below (I think I drew them properly), enter image description here

These have a negative charge on $-p$ and $-o$ group. This simply conflicts with everything said in the last paragraph.

Also, these structures are as prominent as others in the book because of having the same number of charges.

  • What separates the structures in the book with these ones? why structures in the book predict the reaction mechanism of Nitrobenzene while the structures that have been drawn by me don't.
  • $\begingroup$ You have surely been given some rules as to predict the major resonance contributors. Number of charges is one thing. What are the others? Anything about electronegativity? $\endgroup$ – orthocresol Mar 4 '17 at 14:05
  • $\begingroup$ @orthocresol The ones I can remember are (1) no charge on carbon, (2) -ve charge on electonegative atom, (3) number of charges. I think i am missing one. $\endgroup$ – A---B Mar 4 '17 at 14:13
  • $\begingroup$ So, the ones you drew have a negative charge on carbon, but the ones at the top have a negative charge on oxygen. $\endgroup$ – orthocresol Mar 4 '17 at 14:15
  • $\begingroup$ @orthocresol Ok I did not think that much. My structures are not most prominent but would still have some value say like 50% of book's structures So my point still holds right ? $\endgroup$ – A---B Mar 4 '17 at 14:28
  • $\begingroup$ I wouldn't think it's anywhere near 50%. In any case, the exact number doesn't matter. As long as they are smaller contributors than the ones in the book, you get more reactivity at the meta position than the ortho/para. $\endgroup$ – orthocresol Mar 4 '17 at 14:30

Your diagram is wrong. Have a look at my diagram:

Let me know if you do not understand the curved arrow representation.

In your diagram, in the first picture, you have drawn $+$ and $-$ charges which do not make any sense because both are carbon atoms and the $+$ and $-$ charges will form a $\pi$ bond.

| improve this answer | |
  • $\begingroup$ I did not draw the curved arrows because they are too difficult to draw. Why does not + and - make sense ? I just jumped a pi bond to form a lone pair on a carbon. $\endgroup$ – A---B Mar 4 '17 at 13:36
  • $\begingroup$ Can you tell how you drew these images? $\endgroup$ – A---B Mar 4 '17 at 13:37
  • $\begingroup$ @A---B lol,that's in my book :D $\endgroup$ – Fawad Mar 4 '17 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.