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I need some help in a electrophilic substitution reaction.

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What will be the major product formed in this reaction and why?

I know that (1) Activator wins over deactivator. (2) A powerful wins over a weak activator. (3) When a substituent possesses a lone pair, it it usually extends the product determining influence. (4) Steric factor always plays a key role.

$\ce{NO2}$ is a powerful deactivator, so I think the middle ring will not undergo substitution. I can't decide which of the other two rings will undergo substitution and at what position.

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    $\begingroup$ Please see our homework policy. Could you show your own thought process/your own efforts in solving this reaction? $\endgroup$ – ManishEarth Feb 25 '13 at 7:35
  • $\begingroup$ Ok, I will add my work. $\endgroup$ – Shobhit Feb 25 '13 at 7:35
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    $\begingroup$ I can answer this easily enough but it does you little good. Draw the resonance structures which show the delocalization of + charge from the nitro. Go ahead and make a carbanion on one of the rings to accomplish this. When you draw these structures, it will become obvious to you which ring will be attacked. $\endgroup$ – Lighthart Feb 25 '13 at 7:59
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    $\begingroup$ @Lighthart can you elaborate on that procedure and maybe turn it into an answer. $\endgroup$ – Aditya Sriram Feb 25 '13 at 9:53
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I no longer am a practicing chemist, so I have no access to a structure drawing program; I'll do my best with words:

Accept the initial premise that the NO2 is strongly deactivating so only the 'end' rings will be attacked. As we know o,p- substituion have functionalyl equivalent resonance forms, so we only have to consider 4 possible products: p-, m-substitutuion on the 3-ring and similar on the 4-ring.

Drawing all four of these structures and the reasonanc forms indicates that the p-,4-ring susbtitution and the m-,3-ring susbtitution are poor choices because of high energy intermediates.

The final question to resolve is whether or not the m-,4-ring is better than the p-,3-ring. Based on the intermediates, this is a even, but the nitro group is in conjugation with the 4-ring, whle only in cross-conjugation with the 3-ring You can see this by drawing a rsonance form with a carbocation at the para position, and rearranging the bonds via resonance all the way to the nitro group. This can't be done with the with the 3-ring, meaning the nitro group deactivates the 4-ring more than the 3-ring.

Thus the product is likely p-substitution on the 3-ring.

Advanced: The resonance for arguments require all three structures be coplanar. This is not the case; biphenyls have some torsion and twisting because o-hydrogens bump. However, there is still enough 'flatness' and orbital overlap to make some use of the resonance arguments presented above, albeit the magnitude of better and worse are smaller than they would be for more classical EAS problems.

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