[mysticmuse]

I usually understand what is more reactive when there is only one substituent, ex: phenol, chlorobenzene but I'm stuck when I'm asked to find what is more reactive when there are 2 substituants.

This is the the question: Place in order of reactivity the following compounds - acetanilide, benzene, chlorobenzene, 4-nitrophenol, phenol, 4-chlorophenol.

This is my guess: phenol>4-chlorophenol>4-nitrophenol>acetanilide>benzene>chlorobenzene

The problem with my reasoning is that I only considered the second substituants 4-nitrophenol and 4-chlorophenol when I compared them to eachother, not with the other compounds. Can someone please explain me how to find the correct order? Thanks.

[karlosshughes]

It really depends what the reaction that is being referred to is.

The best way to answer this is to consider the resonance forms of the carbocation intermediates.

For example, in an electrophilic substitution reaction, benzene can delocalise the positive charge around the ring onto the carbons ortho and para to the carbon being attacked by the electrophile (the ipso carbon).

In phenol, one more resonance form can be drawn where the positive charge is delocalised onto oxygen, forming an oxonium ion.

Chlorobenzene has 3 lone pairs on the chlorine atom. This means it can likewise stabilise the cation by delicalisation of the positive charge, but because there is poor overlap between chlorine's 3p orbitals and carbons 2p orbitals, this doesn't have much effect. Not only that, chlorine is also very electronegative and will withdraw electron density by induction and hence there is less electron density available in the pi system for the electrophile to attack. In the end, this e- withdrawal wins out and halogens are deactivating: chlorobenzene will be slower than benzene.

Acetanilide has a lone pair on nitrogen. However, as this is an amide, the lone pair is conjugated with the carbonyl group so is less avaiable for donation into the pi system. Nevertheless, this can happen and amides are weakly activating (although still less activating than phenol)

The nitro group is the most deactivating of all. However, in disubstituted benzenes, the most activating group wins out.

This means that nitrophenol will still be activating and will react as though it were just phenol (although note that only the ortho product will be produced- the para carbon is taken up by the nitro group!)

Finally, as with nitrophenol, chlorophenol will be activating as the hydroxy group wins out. Likewise, only the ortho product will be made since the para position is taken up by chlorine.

So in summary:

Phenol> acetanilide > 4-chlorophenol > 4- nitrophenol > benzene > chlorobenzene.

Whilst chlorophenol and nitrophenol are activating, the chlorine and nitro groups are both more electron withdrawing than the amide group, and the amide group still has a lone pair.

Hope this helps.