Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: aleksxxx on April 27, 2007, 03:10:38 PM
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Okay -
I have to synthesize 1,2,3-tribromo benzene for a study questions starting w/ benzene and using any needed reagents.
I am pretty lost - the only way i can think to do this is the following (but i dont think it will work becuase the groups are deactivators)
1. Make aniline
2. React that W/ Br2 and seperate out the ortho, leaving a Br para to NH2
3. Oxidize the NH2 to a Nitro group again, leaving a NO2 and Br para
4. I would brominate here, since the Br would direct ortho and the NO2 would direct meta (This is where i dont think it would work because the Br and NO2 are both deactivating, so i dont know if the rxn would proceed), if it did go, it would leave my required product only w/ a nitro group i would have to remove.
Please let me know if that route has any shot to get any product, or any tips on how to get there.
Much appreciated.
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Heres what i think...
http://bp1.blogger.com/_60ljWjzZHA0/RjK4yBDZJUI/AAAAAAAAAWc/Vszfmw6anWU/s1600-h/triBromo-ben.JPG
what do guy think
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Heres what i think...
http://bp1.blogger.com/_60ljWjzZHA0/RjK4yBDZJUI/AAAAAAAAAWc/Vszfmw6anWU/s1600-h/triBromo-ben.JPG
what do guy think
ahhh
that looks good.. way better than what i came up with!
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Br and NO2 are both deactivating,
Br is a deactivator but ortho/para director
I think your way is possible... u can get rid of the nitro group by deamination by diazotation
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Br and NO2 are both deactivating,
Br is a deactivator but ortho/para director
I think your way is possible... u can get rid of the nitro group by deamination by diazotation
thats how i was planning on it, but since the Br and NO2 are both deactivators (even though directing where i want the bromines to go) i didnt think it had a shot in reacting.
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Heres the other way
http://bp2.blogger.com/_60ljWjzZHA0/RjPsMBDZJWI/AAAAAAAAAWs/sMCsARoQlXg/s1600-h/tribromosyn2.JPG
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here is a way i thought about today... lemme know..
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That substitution reaction with PBr3 won't work. Aryl compounds do not readily undergo substitution reactions unless you have a decently strong deactivator.
You may have difficulties using Br as a deactivator.
I don't see why you can't just use three equivalents of Br2 and FeBr3.
You'll obviously get para-directing side reactions, but this method seems considerably easier.
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ok, so just brominating benzene w/ 3 equiv. would work you think? Wouldnt para be preferred to ortho becuase of steric hinderance leaving you w/ 2 bromines para and then some ortho?
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ok, so just brominating benzene w/ 3 equiv. would work you think? Wouldnt para be preferred to ortho becuase of steric hinderance leaving you w/ 2 bromines para and then some ortho?
Such a compound is bound to be quite hindered. Making hindered molecules usually entails lower yields.
I'm sure there's some genius alternate process of which I'm unaware. You have two choices though. Your inital, really long and complicated process, or mine, which takes 3 steps. :P
If it's not a matter of yield and only how to get to the product, then don't worry about the complications, just the path that takes you to the end.
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ok, so just brominating benzene w/ 3 equiv. would work you think? Wouldnt para be preferred to ortho becuase of steric hinderance leaving you w/ 2 bromines para and then some ortho?
Such a compound is bound to be quite hindered. Making hindered molecules usually entails lower yields.
I'm sure there's some genius alternate process of which I'm unaware. You have two choices though. Your inital, really long and complicated process, or mine, which takes 3 steps. :P
If it's not a matter of yield and only how to get to the product, then don't worry about the complications, just the path that takes you to the end.
ok sounds good...
would i still have to block the para position w/ somehting though?
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would i still have to block the para position w/ somehting though?
I don't know of a cleaving reaction that would rid you of your blocker afterwards.
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would sulphonation work in this case? also once the first Br is added to the othro postion wouldnt it direct ortho/para relative to it self?
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also once the first Br is added to the othro postion wouldnt it direct ortho/para relative to it self?
That's the idea, yes.
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so your hoping by chance, they'll direct ortho
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so your hoping by chance, they'll direct ortho
Ignoring sterics, there's 2/3 chance of ortho direction and 1/3 chance para direction for the second bromination. The third bromination does not introduce more hindrance at the respective ortho positions of the disubstituted benzene. However, it introduces more problems on behalf of the 50-50 chance of para or ortho; with sterics added in to the formula, the reaction is bound to follow para-direction.
The primary caveat in the second reaction is to ensure than there is only one equivalent of bromine and ferric bromide added, otherwise you could end up with a tetrasubstituted product. Maybe even pentasubstituted.
I don't know how bad the effects are, so I can't say. But there being two ortho positions, I'm confident in the least of this reaction being plausible.
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The direct bromination does make sense, especially not knowing the exact hinderance.
I still think that even if you get a bromine ortho to the first one, the odds of the third going para are far greater than the third going ortho w/o a blocker in the para position.
Im going to bring this up in class tomorrow too and see what they all say and let ya'll know.
Thanks
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so your hoping by chance, they'll direct ortho
Ignoring sterics, there's 2/3 chance of ortho direction and 1/3 chance para direction for the first bromination. The second bromination does not introduce more hindrance at the respective ortho positions of the disubstituted benzene. However, it introduces more problems on behalf of the 50-50 chance of para or ortho; with sterics added in to the formula, the reaction is bound to follow para-direction.
The primary caveat in the second reaction is to ensure than there is only one equivalent of bromine and ferric bromide added, otherwise you could end up with a tetrasubstituted product. Maybe even pentasubstituted.
I don't know how bad the effects are, so I can't say. But there being two ortho positions, I'm confident in the least of this reaction being plausible.
Just wondering g_english, if you take a look at my 2 proposal above how long would that take to make in lab (i dont have much lab experience)?
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Just wondering g_english, if you take a look at my 2 proposal above how long would that take to make in lab (i dont have much lab experience)?
I edited my post; I meant second ad third bromination, respectively.
Are you talking about your diazotization step? That doesn't take long, if I remember correctly, about a half hour or so.
I don't see any way to do this reaction in which you will not encounter steric problems one way or another. Maybe there's a unique synthesis that was published somewhere that we don't know of.
1. A way you could do this is nitration ----> bromination ----> reduction ----> diazotization ----> Sandmeyer reaction ----> bromination
But this reaction sequence gives you a position to be brominated right smack between to bromine atoms; like trying to squeeze through two fat people in a crowded line.
2. Another way to go about it would be to get the first two bromines ortho relative to each other, then get the final bromine at the third position using a nitro group. You could then reduce the nitro group to an amino group, and it may be possible to remove the azo group with enough heat.
This procedure is quite complicated though. I would prefer to 3. to do 3 successive brominations, with 3 equivalents of Br2 and catalytic ferric bromide. This method really doesn't introduce any more hindrance than does method 2.
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How long would these take me to do in the lab (from start to finish)
http://bp2.blogger.com/_60ljWjzZHA0/RjPsMBDZJWI/AAAAAAAAAWs/sMCsARoQlXg/s1600-h/tribromosyn2.JPG
http://bp1.blogger.com/_60ljWjzZHA0/RjK4yBDZJUI/AAAAAAAAAWc/Vszfmw6anWU/s1600-h/triBromo-ben.JPG
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How long would these take me to do in the lab (from start to finish)
http://bp2.blogger.com/_60ljWjzZHA0/RjPsMBDZJWI/AAAAAAAAAWs/sMCsARoQlXg/s1600-h/tribromosyn2.JPG
http://bp1.blogger.com/_60ljWjzZHA0/RjK4yBDZJUI/AAAAAAAAAWc/Vszfmw6anWU/s1600-h/triBromo-ben.JPG
Well the first method gets rid of the problem with having both para and ortho products. And don't forget that an azo group is a triply bonded nitrogen. But you see your first method introduces no more hindrance than mine (with 3 successibe brominations), but it does give a better yield of the necessary tribromo.
Good job. ;)
The second method looks good other than the acidic workup leading to the loss of your sulfonic acid. Sulfonic acid is a very strong acid, so I don't follow. I prefer hydrogenation or Sn/HCl and OH for reduction of your nitro groups. I'm unfamiliar with SnCl2 for this.
I couldn't tell you how long they would take, although the first method has considerably less steps, and fewer diazotizations, so to me it seems faster.
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for azo group theres resonance structures and the triple bonded Nitrogen would have the postive charge on the first N (it's easier for me to draw the structure with N=N and put the postive charge on the second N like the N=N+)
the acid work up is to desulphonate
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the acid work up is to desulphonate
OK, but this requires heat.
for azo group theres resonance structures and the triple bonded Nitrogen would have the postive charge on the first N (it's easier for me to draw the structure with N=N and put the postive charge on the second N like the N=N+)
No. If you do this the nitrogen would have a negative charge, not positive.
I still don't see the logic in the reaction with tin(II) chloride in the second synthesis. In order to reduce a nitro group, your resulting hydrogens must come from somehwere.
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OK, but this requires heat.
Sorry forgot about heat
No. If you do this the nitrogen would have a negative charge, not positive.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg404.imageshack.us%2Fimg404%2F9013%2Fdiazoresmg3.jpg&hash=bc9b8271bb686174045700def023fe3a2f041e7f)
I still don't see the logic in the reaction with tin(II) chloride in the second synthesis. In order to reduce a nitro group, your resulting hydrogens must come from somehwere.
I forgot SnCl2 is in acidic conditions
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No. If you do this the nitrogen would have a negative charge, not positive.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg404.imageshack.us%2Fimg404%2F9013%2Fdiazoresmg3.jpg&hash=bc9b8271bb686174045700def023fe3a2f041e7f)
According to this, your nitrogen does not have an octet. Nitrogen has three bonds when positively charged, not two.
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the middle azo structure, if i move one bond on to the + charge, i get the structure on the left, i would have four electrons (lone pair and 1 each from each actual bond = 4, which mean missing one electron)
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This is the resonance for the N2 portion of the molecule only. Notice the second one has a terminal N with no octet. The structure can therefore be ignored. This is the structure you're referring too.
You can also draw the very last one, but it is very unstable as well. I wouldn't recommend drawing unstable resonance structures!
This is aside the point. Other than those minor things, your syntheses are very creative and do work. I would prefer them over mine. ;)
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OK -
I asked my prof this morning and he suggested acetalating the aniline - thats all he would tell me so far (we will go over it friday). He said the acetylation would make it easier to just put one bromine para, rather than excessive bromination becuase it would become a more mild activator.
I came up w/ the following then but i dont know if once the bromine was on if i continued brominating if they would go Ortho to the Br already on the ring... and also i do not no how to get the amino function off:
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... and also i do not no how to get the amino function off:
Diazotization followed by H2PO3.
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cool, what about those bromines? would they go where i specified??
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Not sure about that last step. An amido group is a moderate activator. The para position would work, but those two at the meta position...I don't see them working.
This would give you a 1,3,5-tribromo- .
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OK -
Im running outta ideas! How is this??
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This is aside the point. Other than those minor things, your syntheses are very creative and do work. I would prefer them over mine. ;)
Thanks, Im glad we can discuss those minor things, that way i can learn . :)
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Looks good. First step you forgot H2SO4. In the acetylation step, I would use an acyl chloride, but that's just me.
In the diazotization step, use HBr instead of H2SO4, this will allow you to eliminate the need for an additional reagent because your Br- from your diazo salt will come from HBr and will allow for the later Sandmeyer reaction.
You got it. :)