# Bromination of acetanilide

Why does acetanilide gives exclusively para isomer. I know that -I of nitrogen must decrease the yield of ortho product, but still it should be made in accountable amounts. Where am I going wrong. I am inserting the question and solution screenshots for reference. The question is from IIT JEE 2016 exam, chemistry section of paper 1

Question

Solution

• It probably does give some of the ortho product. Do you have a link or a specific example in mind? – Zhe Oct 17 '16 at 14:28
• Agree - I can certainly imagine that there is less ortho product compared to, say, aniline, but 0% ortho doesn't seem plausible. – orthocresol Oct 17 '16 at 14:30
• @Zhe I uploaded the source. Here you go! – user340743 Oct 17 '16 at 15:10
• But one serious problem with those highschool questions remains: Not giving equivalents, temperatures, solvents, reaction times etc. really makes it impossible to answer those questions without knowing the solution already : ) They could have at least provided the number of equivalents during the bromination. – logical x 2 Oct 17 '16 at 17:17
• @ketbra Yeah, indeed. The truth though is that most people only care about such things when they actually have to go into the lab and do it. – orthocresol Oct 17 '16 at 17:19

One reason: 1,3-allylic strain of the amide group hindering attack at the ortho position. However, I still find it amazing that there is exclusively no byproduct formed. Most aromatic brominations (mostly with NBS) I did required cooling to -78°C and then slowly warming up to room temperature to prevent extensive formation of multiply brominated side products. By the way, the $\mathrm{BrO_3}/\mathrm{HBr}$ combination is used to generate $\mathrm{Br_2}$ in situ thereby leaving the concentration of elemental bromine at a bare minimum. This prevents the formation of polybrominated site products. A similar protocol is $\mathrm{KBr}$/oxone. However, the question is clearly ill-defined as one should have provided the fact that 1 eq of $\mathrm{BrO3}/\mathrm{HBr}$ is used. If you would have used more equivalents, chances are good you would have gotten (d) as well.

Also, I find the description given in (http://websites.rcc.edu/grey/files/2012/02/Bromination-of-Acetanilide.pdf) questionable, I think that the main point for the high regioselectivity is the use of $\mathrm{BrO3}/\mathrm{HBr}$ and not the steric hindrance of the amide. I did bromination on aromatic amides as well and regioselectivities were not that much higher at all, in my experience.

• Please explain how did the use of KBrO3+HBr enhanced regioselectivity. – user340743 Oct 17 '16 at 16:23
• KBrO3 oxidizes HBr -> Br2 is formed. Br2 does electrophilic aromatic substitution. But because the oxidation is relatively slow, low concentration of Br2 is formed at a time. Hence, there is not too much bromine there at a given moment and multiple bromination is prevented. – logical x 2 Oct 17 '16 at 17:14

Turning the amino group into an amide serves two purposes:

1. We are introducing a sterically bulky group. We expect the amide bond to be as complanar as possible with respect to the benzene ring to maximise electronic interactions. That should block one of the two ortho positions sterically.

2. We are turning an electron-rich, $+M$ aromatic system into an electron-poor, $-I$ one.

Especially the second transformation will greatly reduce the reaction rate. Furthermore, a bromine atom is mildly deactivating, too, due to its $-I$ effect (the weak $+M$ effect is neglegible and only responsible for the ortho/para directing abilities). Thus, it is very likely that we can selectively stop the reaction after monobromination.

The question still arises why we only obtain the para-product. Obviously the amide behaves much like a bromide and displays a weak $+M$ effect in spite of the amide resonance. Yet we can imagine the the proximity to the bulky amide group inhibits substitution at both ortho-protons sufficiently to allow isolation of the para-product.