[stam]

Bear in mind I am a year removed from learning ochem material:

In my ochem lab II, we are synthesizing 4-bromo-2-nitroaniline and the third synthetic step involves:

4-bromo-acetanilide ----HNO3/H2SO4---> 4-bromo-2-nitroacetanilide

The H2SO4 is added to the 4-bromoacetanidilde thereby protonating the amide substituent, rendering it electron deficient, and therefore deactivating and meta-directing. Therefore, since the -Br group is weakly activating, then the nitration SHOULD occur @ #3C, due to both ortho-directing of the -BR and meta-directing of the deactivating protonated amide form. but the lab protocol says that nitration occurs @ #2C.

What is wrong with my logic? does protonation of the amide substituent NOT occur since amides are NOT BASIC, and the lone pair of that nitrogen is actually involved in the resonance stabilization of the amide? therefore the amide is the stronger activator than the weakly deactivating halogens? and therefore the amide directs ortho, so nitration occurs at #2 carbon?

Thank you

[jake.n]

If I remember correctly, bromine is not a very strong director.  For example, toluene when treated with excess bromine and ferric chloride yeilds 2,4,6-tribromotoluene.  Also, aniline is a much weaker base than aliphatic amines because of the extent of delocalization of the lone pair electrons (through the ring and the carbonyl).  Remember that H2SO4 is required to activate the HNO3 and generate an electrophile.

[nox]

That acetanilide is not going to be very basic, especially since nitrogen is conjugated with both the ring and the carbonyl, so you're looking at a pKa of maybe -2 or even lower. At any rate under the reaction conditions, you can safely assume the majority of your acetanilide will be unprotonated.

Bromine is indeed ortho/para directing, but it is a moderate deactivator, and most definitely NOT an activator! However the acetanilide group is mildly activating, and directs ortho/para substitution. Since activators always exert a stronger effect, the acetanilide wins out. Now because your para position is blocked by the bromine, you have no choice but to nitrate at the 2 position.