kutarbear
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Reaction of amine and bromine water
I did an experiment on the reactions of amine. Here is the experiment:
1. Into a test tube, place 5 drops of phenylamine and carefully add concentrated hydrochloric acid, dropwise, to obtain a clear solution.
2 Drop by drop, add about 1 cm3 of bromine water.
3. Repeat step 1,2 using butylamine.
4. In a test tube, dissolve a spatula measure of ammonium chloride in about 3cm3 of water. Add bromine water.
I did this experiment and got white precipitate for phenylamine, but clear yellow solution for butylamine and ammonia. I can't explain why bromine
only discolour in aniline, but not the other two?
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a123x
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The amine group is very strongly activating to the aromatic ring in aniline. The bromine adds itself to the ring easily because of the activating
group. Bromine does not undergo electrophillic addition with the other two substances.
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Sandmeyer
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You should know some theory behind the experiments you want to carry out. What do you wish to make? Comparing aniline and aliphatic amine is
meaningless as these two classes display entirely different reactivities towards electrophiles. Browse www.orgsyn.org - you should find some interesting stuff, that database covers all the fundamental reactions in organic chemistry. For examples
browse http://www.orgsyn.org/orgsyn/RxnTypes/section.asp?section=11... (bromination)
[Edited on 2-1-2006 by Sandmeyer]
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Madandcrazy
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I know the experiments and different bindings in very strongly amine bindings and aromatic rings are different reactions with not strongly acids.
by the phenylamine involving a chlorine or bromine atom to the aromatic ring, more than ever with more than one chemical.
I would guessing by the phenylamine is the reaction with bromine
C6H5-CH-NH2
-- >
C6H5-CBr-NH2
and by the aniline, i think the same but the aromatic ring
are stable more within double bounds, i don`t know reaction easy ?
[Edited on 2-1-2006 by Madandcrazy]
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Darkblade48
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| Quote: | Originally posted by Madandcrazy
I would guessing by the phenylamine is the reaction with bromine
C6H5-CH-NH2
-- >
C6H5-CBr-NH2
and by the aniline, i think the same but the aromatic ring
are stable more within double bounds, i don`t know reaction easy ?
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Yes, essentially, you are replacing one of the hydrogens on the aromatic ring with the bromine. In the case of phenylamine (trivial name: aniline,
though nobody uses the IUPAC name  ), the NH2 group is highly activating and
allows the bromine to add in the ortho and para positions to the amine group.
The mechanism itself is an electrophilic substitution, where the attacking Br will attack n the o/p position, breaking the benzene ring temporarily,
while the positive charge is "passed around" the ring via resonance. A base (in this case, the Br anion) would come in, take off the hydrogen and
electrons would reform the ring, restoring the aromaticity of the ring.
Edit: Just noticed that you have the following:
| Quote: |
C6H5-CH-NH2
-- >
C6H5-CBr-NH2
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It should be C5H5-C-NH2 --> C5H4-C-Br, assuming that it is monobrominated (most likely not in this case, since the amine group is relatively highly
activating)
[Edited on 2/1/06 by Darkblade48]
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Madandcrazy
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Thanks for the information about my mistake by writing the
formula of the forgotten hydrogen atom.
I meant if it possible break a strong aromatic ring with the bromine or it`s a substitution only with and Br <--> H ion.
I think the amine is stable when attacked with bromine,
more the sometimes funny namings of chemicals, phenylamine is anilin (it sholud be resist and stable)
I meant:
benzylamine
C6H5-CH2-NH2 --> C6H5-CHBr-NH2
I don`t know the aniline is easy broken permanetly when attacked with bromin. I thought it`s stable  too in strong acids.
[Edited on 4-1-2006 by Madandcrazy]
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Darkblade48
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| Quote: | Originally posted by Madandcrazy
Thanks for the information about my mistake by writing the
formula of the forgotten hydrogen atom.
I meant if it possible break a strong aromatic ring with the bromine or it`s a substitution only with and Br <--> H ion.
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No, it's not possible to break the aromatic ring (at least not permanently) with the Br ion. The ring is only temporarily broken when you are doing
the electrophilic substitution.
| Quote: | Originally posted by Madandcrazy
I meant:
benzylamine
C6H5-CH2-NH2 --> C6H5-CHBr-NH2
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I'm not sure that this reaction occurs. You are trying to make bromo(phenyl)methanamine from benzylamine through addition of bromine only. Perhaps
adding NBS would work, since you are trying to brominate in the benzyllic position.
| Quote: | Originally posted by Madandcrazy
I don`t know the aniline is easy broken permanetly when attacked with bromin. I thought it`s stable too in strong acids.
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Not too sure what you mean by this, the aniline is not "broken" in any way when you brominate, you just form a brominated aniline compound.
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Madandcrazy
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I mean the bromine attacks easy more points on the carbon atom which
have no double binding (benzylamine) and can not broken easy the aromatic ring.
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