4-bromoaniline can be prepared by aniline from several steps. Aniline is an ortho para activator compound like phenol. In 4-bromoaniline bromine atom exist in para position related to the amine group (-NH2 group).
Reaction of aniline and bromine liquid will give 2,4,6-Tribromoaniline. Therefore we have to control the reaction (to stop joining three bromine atoms to benzene ring and stop it by joining only one bromine atom at para position). if we need to prepare 4-Bromoaniline.
Question: Synthesis 4-bromoaniline from benzene. In this reaction you can only use benzene and CH3COCl as organic compounds.
Solution - steps of the conversion
Aniline is a ortho para activator. When liquid bromine (Br2(l)) is added to the aniline, it gives 2,4,6-bromoaniline.
But we need only one bromine atom to connect to the para position of aniline. Therefore we reduce the activity of aniline by the reaction of aniline and ethanoyl chloride (CH3COCl). Then we add Br2(l) to the newly formed compound.
Aniline is a strong ortho para activator. Aniline and ethanoyl chloride react to give N-phenylethanamide which is a ortho para Moderate activator. This product is an amide compound.
-NH-CO-CH3 group results a steric impediment around that group. Therefore, substitutions to ortho positions are more difficult than para places.
Two products are given. Due to steric impediment, bromine is substituted more to para position. Therefore major product is para substituted compound, 4-bromoaniline.
We have already discussed how to prepare p-bromoaniline from aniine. Now, we want to prepare aniline by benzene.
We know, electrons density of ortho and para positions are higher than meta positions. So, brominine subsitution should be occurred at ortho or para positions because this reaction is an electrophilic substitution reaction.
Brominum positive +1 ion forms by the reaction of FeBr3 and Br2. Br+ is an electrophile.
Benzene ring of aniline attacks the brominum ion. But due to steric impediment around ortho positions, Br+ cannot reach to the ortho position. Therefore brominium ion goes to the para position.
Reaction between aniline and liquid bromium will give 2,4,6-tribromoaniline. 2,4,6-tribromoaniline is a white precipitate.
Aniline cannot be transformed to 2-bromoaniline directly because aniline is a strong activator. So, as preparation of 4-bromoaniline, we first add ethanoyl chloride with aniline. Then add liquid bromine to it. As major product, 4-bromoaniline will be given But as a by-product, 2-bromoaniline is given and it can be separated from a sepation method of chemicals.
According to the previosly discussed method, o-bromoaniline is prepared as a minor product. So we can distillate mixture of p-bromoaniline and o-bromoaniline to separate both compounds. But receiving o-bromoaniline amount is very low tobecause it is a minor product of the reaction.
We cannot prepare m-bromoaniline in one step from aniline because aniline is a ortho - para activator. In m-bromoaniline, bromine atom is attached to the meta position of aniline. We have to convert aniline to a deactivater and then attach bromine atom to the meta position.
p-bromoaniline | para-bromoaniline
There are three steps in preparation of 4-bromoaniline from aniline.
Phenyl amine is aniline. The conversion of aniline to 4-bromoaniline is explained above.
We know, aniline is an ortho para director. Therefore we cannot connect chlorine into the meta position. Then We use nitrobenzene to connect the chlorine atom to the meta position. Finally nitrobenzene is reduced to aniline.