nice and efficent route for self 2b4m syntehsis (avg 50 usd a kg)

Dr. MMX

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Hey i ll just type reagents used by wise choice (price/effort) aviable to buy in europe
Así que en primer lugar
PCL5 con azufre elemental > cloruro de tionilo

cloruro de tionilo + ácido propiónico > cloruro de priopionilo

4mp síntesis (aquí es un topicso ill sólo enlace):
Cloruro de propionilo
Tolueno
AlCl3 anhidro

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


algunos pueden tener problemas para bromarlo
ácido sulfúrico que es necesario para obtener Hbr que es un poco caro y no a menudo aviable. En su lugar utilizamos
ácido fosfórico y NaBh para la reacción > obtenemos HBR
H2O2 12% puede ser comprado y destilado a algunos más % como 25-30%

así
4 methylopropione + HBR + h2o2 > 2b4m :)

no he calculado el coste pero es muy pequeño y no hay que esperar el envio ni sorprenderse de la baja calidad del precursor.
 
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chost

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¿Puede explicar con más detalle cómo se hace el HBR?
 

Dr. MMX

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Lo siento mucho, me equivoqué. Es correcto:
Para producir ácido bromhídrico (HBr) a partir de ácido fosfórico (H₃PO₄) y bromuro de sodio (NaBr), se puede utilizar una reacción de doble desplazamiento. La reacción puede representarse de la siguiente manera:


  1. Reactantes: Ácido fosfórico (H₃PO₄) y bromuro de sodio (NaBr).
  2. Productos: Ácido bromhídrico (HBr) y fosfato sódico (Na₃PO₄).

La ecuación química equilibrada para la reacción es:


H3PO4+3NaBr→3HBr+Na3PO4


En esta reacción, el ácido fosfórico reacciona con el bromuro de sodio para producir ácido bromhídrico y fosfato de sodio.
 
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chost

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¿Es necesaria la destilación?
¿Y en cuánta agua debo disolver el bromuro de sodio?
 

Dr. MMX

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¿De qué otra forma piensas obtener algo que básicamente es un gas? :)

Depende de lo que % youwanna tener, max es del 47% que yo recuerde y shouldbe ya de esta manera
 

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 con NaBr para producir HBr? Voy a intentarlo dentro de un momento.

Tengo bastante HBr, pero tengo curiosidad por saber si los dos reaccionan.
 

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sí y Na3PO4
 

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eso es maravilloso
 
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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
 

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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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