rota agradável e eficiente para síntese de 2b4m (média de 50 dólares por kg)

Dr. MMX

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Ei, vou digitar os reagentes usados pela wise choice (preço/esforço) que podem ser comprados na Europa
Então, primeiro
PCL5 com enxofre elementar > cloreto de tionila

cloreto de tionila + ácido propiônico > cloreto de priopionila

Síntese de 4mp (aqui está um tópico, portanto, vou apenas colocar o link):
Cloreto de propionila
Tolueno
AlCl3 anidro

http://bbzzzsvqcrqtki6umym6itiixfhni37ybtt7mkbjyxn2pgllzxf2qgyd.onion/en/threads/synthesis-of-4-methylpropiophenone-from-toluene-by-friedel-crafts-reaction-1kg-scale.209/


alguns podem ter problemas para bromá-lo
O ácido sulfúrico é necessário para obter Hbr, que é um pouco caro e nem sempre está disponível. Em vez disso, usamos
ácido fosfórico e NaBh para a reação > obtemos HBR
H2O2 12% pode ser comprado e destilado para uma porcentagem maior, como 25-30%.

Assim
4 metilopropiona + HBR + h2o2 > 2b4m :)

Não calculei o custo, mas é muito pequeno e não há necessidade de esperar pelo envio ou de se surpreender com um precursor de baixa qualidade.
 
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chost

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Você pode explicar com mais detalhes como fazer HBR?
 

Dr. MMX

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Sinto muito por ter cometido um erro. Está correto:
Para produzir ácido bromídrico (HBr) a partir de ácido fosfórico (H₃PO₄) e brometo de sódio (NaBr), é possível usar uma reação de deslocamento duplo. A reação pode ser representada da seguinte forma:


  1. Reagentes: Ácido fosfórico (H₃PO₄) e brometo de sódio (NaBr).
  2. Produtos: Ácido hidrobrômico (HBr) e fosfato de sódio (Na₃PO₄).

A equação química balanceada para a reação é:


H3PO4+3NaBr→3HBr+Na3PO4


Nessa reação, o ácido fosfórico reage com o brometo de sódio para produzir ácido bromídrico e fosfato de sódio.
 
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chost

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A destilação é necessária?
E em quanta água devo dissolver o brometo de sódio?
 

Dr. MMX

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De que outra forma você planeja obter algo que é basicamente um gás? :)

Depende da porcentagem que você quer ter, o máximo é 47%, pelo que me lembro, e já deveria ser assim
 

Benz88

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is it Safe to Distille the h2o2 to maybe 30-35% from 12%?
 

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TheVacuumGuy

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H3po4 com NaBr para produzir HBr? Vou tentar fazer isso em um minuto.

Tenho bastante HBr, mas estou curioso para saber se os dois reagem.
 

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sim e Na3PO4
 

Dr. MMX

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isso é maravilhoso
 
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TheVacuumGuy

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Didn't work. I think it would be easier to add sulfuric acid to a solution of KBr (or NaBr) , which would readily yield HBr at 125c
 

Dr. MMX

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I tried to override having sulfuric acid becasue here is troubles to get it. It was very extreme attempt
if there is access to sulfuric acid its easy job
 

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you can get elemental bromine from potasium bromide, dont need to use acids.
it was quite cheap as i remember
 

Toddler

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Hbr2 48% can work ?
(CAS.No: 10035-10-6)

how much ration use for bromination
Hbr2
H2O2
for 100g of 4MPP
 
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w2x3f5

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Instead of propionyl chloride, can use propionic anhydride (obtain phosphorus oxide from propionic acid), can also use propionic acid directly with polyphosphoric acid (there are many options for this reaction, you can choose the best one). If propionyl chloride is used with anhydrous aluminum chloride (can be replaced with cheaper and simpler options). Then it is best to carry out a bromination reaction before hydrolysis in Fridel-Crafts (the complex of 2b4m and aluminum chloride is very active for bromination with a high yield).
 

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In chemistry trade buissness in europe propionic acid is considered with "silent restriction" with reason it can be used as biological weapon, so should be avoided there.

very interesting choice of propionic anhydride
 

Osmosis Vanderwaal

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@TheVacuumGuy you gave him bad figures and he used them and probably ruined 100G 2B4-MP. I dont know what he paid, but I got $200 all day for 10g if it's domestic . In th rules there are a few gold nuggets. "Use the fucking search engine." That's a rule. Another one is " a good chemist i an accurate chemist." Make sure that you clearly state you are guessing when you don't know. You could say nothing at all, you could show your math or you could tell him he needs to figure it himself, but please, dont just throw out random numbers, with no disclaimer.
 
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