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The Third SEAr

Attaching three groups to a benzene ring by consecutive SEAr reactions is not an easy task. One must start with the least deactivating groups because, otherwise, the last reaction would be very difficult or impossible.

How do you explain this result?

Nitration of o-nitrotoluene: CH₃, activating and ortho-para-directing, and NO₂, deactivating and meta-directing, direct to the same positions; 2,6- and 2,4-dinitrotoluene are formed

Piece of cake! Both groups direct to the same position.

What about this one?

Nitration of p-cresol with dilute HNO₃: the OH, a stronger activator than CH₃, imposes its orientation and gives 4-methyl-2-nitrophenol

If two substituents conflict with their directing trends, the activator or the weaker deactivator beats the other.

And what about this negative result?

m-Dinitrobenzene does not react with Br₂/FeBr₃: two deactivating groups prevent the third substitution

If the benzene already has two deactivating groups, the third SEAr can be very difficult or even impossible.

Last but not least, how can one explain this result?

Bromination of m-hydroxybenzaldehyde: the OH directs, but the position between the two groups is not substituted; 4-bromo-3-hydroxybenzaldehyde and 2-bromo-5-hydroxybenzaldehyde (major) are formed

If two groups are meta to each other, the third substitution does not take place at the position between them, which is the most sterically hindered.