# How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?

For example consider the bromination of phenol we get the tri-substituted product (even though bromine is deactivating so I suppose it should deactivate the ring with each addition) while sulphonation essentially gives the mono-substituted product (even at high temperatures).

• The OH group is strongly activating, the bromines are only slightly deactivating. – ron Feb 15 '15 at 17:00
• @ron Okay.Then what about Nitration in phenol? It also gives tri-substituted product. NO2 is strongly deactivating. – Kara Feb 15 '15 at 17:17
• A lot of it is experience, you pick it up as you go along. That the nitration of phenol occurs so readily is a further reflection of just how activating the OH group is. – ron Feb 15 '15 at 17:27
• @ron Take a look at this link chemwiki.ucdavis.edu/Textbook_Maps/… It's a good activator not excellent. – Kara Feb 15 '15 at 17:53
• @DelPate Yes, the hydroxyl group activates the ring enough that 3 successive nitrations can occur. As I mentioned in an earlier comment, even the weakly activating methyl group is strong enough to allow 3 successive nitrations under forcing conditions. – ron Feb 16 '15 at 16:08

It is like others have already noted that phenol is very strongly ortho-para directing, while bromine is only slightly deactivating (or sometimes slightly activating). It is so activating that you do not even need a Lewis acidic catalyst or elevated temperature to let the reaction go fast and in good yield. Moreover, if you want the mono substituted product you will actually need to cool the reaction down to < 5 °C and use the rather dangerously flammable solvent $\ce{CS_2}$, yielding 4-bromophenol in 85% yield (source: Vogel). I assume the para product is formed due to the fact that there is no significant difference in electron density between ortho and para carbon atom while ortho is slightly sterically hindered.