GreenD
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Hofmann rearrangement
http://pubs.acs.org/doi/pdf/10.1021/ed076p1717
They do the rearrangement on nitrobenzamide with household bleach (5.25% active).
I'm wondering, with the strong withdrawing groups and conjugation would their yield (80%) be lower for aliphatic amides? For example propamide would
be much less than 80% theoretically - correct, or would a longer reaction time be necessary.
Or would the reverse hold true? I seem to have trouble with this concept.
I tried the search engine - most that came up was simply "Use the hofmann rearrangement you big dummy" with little experimental or sources.
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Rich_Insane
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Semi-on topic: Hoffman Rearrangement of Acetamide
They use Ca(OCl)2. The nifty thing there is that the final product is the HCl salt. Really just a deal-sweetener, since often times people desire the
salts of the amine rather than the free amine. Now this applies to your question, because they reported 55% yield after thorough purification. They
reported 70% yield of the crude product.
I'm no expert, but I think the proximity of the amide group to the ring certainly has some effect. If you look at the mechanism, the reaction is
extremely dependent on the ability of the amide to accept electrons from the base (usually NaOH). Formation of the isocyanate intermediate largely
depends on the ability of the hydrogens on the nitrogen to bounce off to the base (the bromine forms the N-bromine bond before the second
deprotonation). Even after the isocyanate intermediate is formed, the base plays a crucial role in hydrolyzing it. What I , in my limited knowledge
think, is that the carbon in the carbonyl group must be willing to donate the electron to the oxygen (which is easy, because oxygen hoards electrons)
and the ability of the oxygen to give back the electron to the C=NH bond to form the N-bromo amide intermediate. The process is again repeated using
yet more base to cause a migration of the R-group.
So here's what I would ask to answer your question: Is the electron accepting ability of the electrophilic carbon in the C=O bond affected? Is the
electron donating property of the corresponding O- affected? Most importantly (the point in the mechanism where I think your idea might come into most
importance), is the migration of the R-group affected?
Just my $0.02, though it seems that aliphatic amides undergo reactions with lower yields, but theoretically, this may or may not be the case.
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GreenD
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I'm not sure but I think most of what you're saying isn't stated correctly, but you have the idea.
I believe that the proximity of an amide to the aro-ring would cause more acidity (greater ability for the removal of the N-H proton), as is seen with
benzoic acid (pKa = 4.2) and acetic acid (pKa = 4.8) and hexanoic acid (4.9).
The proximity of the group to the benzyl ring will cause the removal of the proton to be easier, allowing for the first step (halogenation) to occur
faster. That is really what I think is the difference between doing this with benzoic acid versus a more aliphatic group. Which is, of course, a
result of the aromaticity stabilizing the anion (conjugation).
The step where the nitrogen migrates, I believe, would be also lowered from that of the benzoic acid-aniline reaction for aliphatics, for the nitrogen
in benzamid could much more easily access the pi orbitals of the ring with some stabilization (might be talking out my ass here). Where as in an
aliphatic system, the nitrogen migration will have to contend with steric hindrance, and no p-orbitals.
At any rate I think the answer to my question is, yes - aniline will have exceptional yields for this reaction, while aliphatic amides will be
lowered.
[Edited on 16-3-2012 by GreenD]
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Hexavalent
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I agree.
Also, how does one insert mechanisms into a post?
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UnintentionalChaos
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It's the image of the mechanism from wikipedia with a transparent background inserted with image tags.
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Hexavalent
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Ah, OK, thanks for that. Might try and use it in a future post
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GreenD
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So my question is, would the use of an even stronger (albeit not OTC) base increase yields here?
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Rich_Insane
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Electrochemical Hoffman
This paper doesn't exactly address your question, but there is some mention of using methanolic NaOCH3 and bromine for the rearrangement. I've also
heard of using NBS and sodium methoxide to achieve the rearrangement, though I can't remember where exactly the source is...
Yea, I just skimmed over a Youtube video and checked out the O Chem portal. But it really does seem that deprotonation is key, this lower pKa may be
crucial in increasing efficiency of the reaction. In another note, how exactly does this migration of the alkyl group work? I've always had a hard
time understanding these migrations, for some reason they end up confusing me.
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GreenD
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I wonder if perhaps sodium methoxide and bleach would be compatible in the rearrangement?
The only problem with using bleach on an aliphatic system is the acidity will be low, and a NaOH/Bleach system will invariably be low (but not
useless). A stronger base could help, but I wonder if there will be compatibility issues with bleach and a very strong base like sodium methoxide.
If not this would be an interesting experiment to see if improved yields can be found by increasing base strength with household bleach.
Sodium methoxide can be made by methanol + sodium metal.
Apparently, due to this source, fatty amides can be made in high yield.
"Fatty amide (0.2 mol) was dissolved in methanol (250 cm3)
with stirring. To it, a freshly prepared aqueous solution
of NaOCl (0.21 mol, 125 cm3) was added immediately
with good stirring. Additional methanol (100 cm3) was
added to the reaction mixture in the case of undecylamide
and lauramide. The temperature rose to about 52-55°C
within 30 min and the solution was then refluxed for 2 h.
Methanol (200 cm3) was recovered from the reaction
mixture by distillation. On cooling the reaction mixture
was diluted with water (200 cm3) and extracted with
dichloromethane (3 x 50 cm3). The combined extracts
were washed with water and dried over Na,S04.
Dichloromethane was recovered by distillation, leaving
carbamate as the residue."
http://onlinelibrary.wiley.com/doi/10.1002/jctb.280590310/pd...
[Edited on 19-3-2012 by GreenD]
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bbartlog
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| Quote: | | Apparently, due to this source, fatty amides can be made in high yield. |
I don't know exactly what's going on in the paper you link to, but they start with amides and synthesize carbamates. It looks like the Hofmann
rearrangement is part of the mechanism but regardless, they don't make amides.
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GreenD
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| Quote: |
I don't know exactly what's going on in the paper you link to, but they start with amides and synthesize carbamates. It looks like the Hofmann
rearrangement is part of the mechanism but regardless, they don't make amides. |
Exactly correct. I think your last word in this you mean amines, correct?
At any rate, they make the carbamates, which can be decarboxylated to give the appropriate amine.
PS if you look at the mechanism above, the last species is the carbamate they make. If you follow that synthesis mechanism the paper will make crystal
clear sense!
[Edited on 19-3-2012 by GreenD]
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