[english]

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.

[aleksxxx]

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

[refid]

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)?

[english]

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.

[refid]

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

[english]

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.

[refid]

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

[english]

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. 

[refid]

OK, but this requires heat.

Sorry forgot about heat

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. 

I forgot SnCl2 is in acidic conditions

[english]

No.  If you do this the nitrogen would have a negative charge, not positive. 

According to this, your nitrogen does not have an octet.  Nitrogen has three bonds when positively charged, not two.

[refid]

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)

[english]

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.   

[aleksxxx]

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:

[english]

... and also i do not no how to get the amino function off:

Diazotization followed by H₂PO₃.

[aleksxxx]

cool, what about those bromines? would they go where i specified??