This page guides you through the mechanism for the nitration of benzene involving an electrophilic substitution reaction between benzene and nitric acid.

Important!  It would help if you first read the page What is electrophilic substitution? before you went on.

The electrophilic substitution reaction between benzene and nitric acid

The formation of the electrophile

If you are going to substitute an -NO2 group into the ring, then the electrophile must be NO2+. This is called the "nitronium ion" or the "nitryl cation", and is formed by reaction between the nitric acid and sulphuric acid.

Note:  If you don't understand why the electrophile has got to be NO2+, then you really should look at What is electrophilic substitution? before you go on.

If you are going to substitute X onto the ring, then the electrophile must be X+. If you are going to insert an -NO2 group onto the ring, then the electrophile must be NO2+.

The equation

Important!  Check past exam papers to see whether you will need to quote this equation in an exam. It's probable that you will, so learn it! There are ways of building it up, but they involve much more effort than learning it in the first place.

The hydrogensulphate ion, HSO4-, will also be involved in the mechanism. The hydroxonium ion, H3O+, isn't involved.

Note:  The hydroxonium ion (also called the hydronium ion or oxonium ion) is simply a hydrogen ion attached to a water molecule - what is often written more simply as H+(aq).

The electrophilic substitution mechanism

Stage one

As the NO2+ ion approaches the delocalised electrons in the benzene, those electrons are strongly attracted towards the positive charge.

Two electrons from the delocalised system are used to form a new bond with the NO2+ ion. Because those two electrons aren't a part of the delocalised system any longer, the delocalisation is partly broken, and in the process the ring gains a positive charge.

The hydrogen shown on the ring is the one which was already attached to that top carbon atom - it's nothing new or subtle! We need to show it there because it has to be removed in the second stage.

Stage two

The second stage involves a hydrogensulphate ion, HSO4-, which was produced at the same time as the NO2+ ion (refer back to the equation showing the formation of the electrophile if you've forgotten).

Note:  Only one of the lone pairs in the hydrogensulphate ion is shown. There are lots more, but those aren't involved in the reaction.

This removes a hydrogen from the ring to form sulphuric acid - the catalyst has therefore been regenerated. The electrons which originally joined the hydrogen to the ring are now used to re-establish the delocalised system.

Stage two - a sloppy way of writing the same thing!

You will often find the second stage of this reaction simplified in many (or even most!) books. The second stage is shown as:

The hydrogen is shown as "falling off" the ring as a hydrogen ion. This is sloppy and unsatisfactory on two counts:

  • Hydrogen ions never exist on their own in a chemical reaction. A hydrogen ion is a raw proton - the most intensely positive thing you can imagine. It will always be attached to something else.

  • By not showing the hydrogensulphate ion, you can't show that the sulphuric acid catalyst has been regenerated. That's simply unsatisfying!

Showing exactly how the hydrogen is removed from the ring isn't difficult - do it properly!

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© Jim Clark 2000