Can anyone please show me the rule on how a halogen like bromine reacts with benzene, such as when bromine replaces hydrogen? I'm wondering what happens after the first hydrogen is replaced? Does it replace the hydrogen in the 2- or 4-position after that or something else?

  • 1
    $\begingroup$ SEAr $\endgroup$
    – ParaH2
    Mar 9, 2017 at 12:22

1 Answer 1


The most common reaction of aromatic compounds is electrophilic aromatic substitution, in which an electrophile (E1) reacts with an aromatic ring and substitutes for one of the hydrogens.

In an unsubstituted benzene ring, the halogen will likely take any position (essentially any position is the same - just depends on how you draw the structure) since the ring is only attached to hydrogen atoms with similar effect in the ring system.

The real question is what would happen if we were to carry out a reaction on an aromatic ring that already has a substituent?

A substituent already present on the ring has two effects:

  • Substituents affect the reactivity of the aromatic ring. Some substituents activate the ring, making it more reactive than benzene, and some deactivate the ring, making it less reactive than benzene. In aromatic nitration, for instance, an -OH substituent makes the ring 1000 times more reactive than benzene, while an - NO2 substituent makes the ring more than 10 million times less reactive.
  • Substituents affect the orientation of the reaction. The three possible disubstituted products—ortho, meta, and para—are usually not formed in equal amounts.

Instead, the nature of the substituent already present on the benzene ring determines the position of the second substitution. An -OH group directs substitution toward the ortho and para positions, for instance, while a carbonyl group such as -CHO directs substitution primarily toward the meta position

All meta-directing groups are strongly deactivating, and most ortho- and para-directing groups are activating. The halogens are unique in being ortho- and para-directing but weakly deactivating.

See screenshot below

Classification of substituent effects in electrophilic aromatic substitution.

Taken from Organic Chemistry (Mc Murry) 8th ED

  • $\begingroup$ Does this also apply alkyl benzene? $\endgroup$
    – user42296
    Mar 9, 2017 at 14:39
  • $\begingroup$ Yes, the positioning of the electrophile will depend on the substituent, the substituent can be an alkyl group, alkene, aryl. What matters is whether the substituent group exterts an electronic withdrawal or donating effect. From the illustration above you can notice the alkyl group has an electron donating effect, conversely, the common characteristic of deactivating groups is that they are electron withdrawing. $\endgroup$ Mar 9, 2017 at 14:54
  • $\begingroup$ So Bromine is ortho and para directing but which one is more likely to happen? $\endgroup$
    – user42296
    Mar 13, 2017 at 13:30
  • $\begingroup$ Both -ortho and -para intermediates are stable, however a higher ratio of the para substituted product is formed to enhance the stability of the Π system (since this configuration results in the substituents bring furthest away) $\endgroup$ Mar 13, 2017 at 14:53
  • $\begingroup$ Meant being furthest* $\endgroup$ Mar 13, 2017 at 15:04

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