síntese simples de anfetamina

[strannik]

Boa tarde! Peço desculpas antecipadamente pelo meu inglês. Até o momento, surgiu um novo precursor que permite a produção de anfetamina em apenas 1 hora. Esse precursor é a alfa-metilprepanamida. CAS: 7499-19-6

Hidróxido de sódio (álcali) 3 kg Bromo 1,5 kg (0,5 l) Éter de petróleo 1 l Acetona 5 l Ácido sulfúrico hcl 1 l Tornassol Pegamos 3 kg de álcali (hidróxido de sódio), dissolvemos em 3 litros de água, a reação é esotérmica, não despejamos tudo de uma vez, mas gradualmente, a água pode ferver. Em seguida, esfriamos a solução alcalina. Atenção!!! Antes da reação, é necessário resfriar a solução alcalina a uma temperatura de aproximadamente 0 graus Celsius! É necessário trabalhar com agitação - reator, agitador, etc.! Despeje lentamente 0,5 litro de bromo em uma solução alcalina resfriada a 0 graus, enquanto a solução deve necessariamente permanecer incolor ou amarelada, depois de adicionar todo o volume de bromo, despeje propanamida na solução e aqueça-a lentamente enquanto mexe. A uma temperatura de 80-90 graus, ocorre uma reação e a solução fica turva devido à base liberada. Mexa por mais 5 a 7 minutos, deixe esfriar até a temperatura ambiente, despeje 1 litro de éter de petróleo e continue mexendo. Separe a camada etérea superior. Em seguida, é padrão acidificar com ácido sulfúrico, ou seja, despejar um pouco ou pingar ácido sulfúrico até que o pH fique entre 5,5 e 5; se engrossar, você pode diluir com acetona. Depois disso, esprememos a mistura resultante por meio de um vácuo e enxaguamos o pó com acetona.

O bromo pode ser substituído por hipocloreto de sódio

 

[strannik]

Correção
CAS 7499-19-6 Benzeno Propanamida (2-metil-3-fenilpropanamida)

2NaOH + Br2 → NaBr + NaBrO + H2O
O hipobrometo de sódio pode ser substituído por hipocloreto de sódio

 

[OrgUnikum]

Quase. É Benzenepropanamide, a-methyl- você estaria perdendo o grupo metil na cadeia com Benzene Propanamide, CAS e 2-methyl-3-phenylpropanamide está correto. Parece muito com a fenilalanina...

 

[OrgUnikum]

Para mim, isso parece a amida de Helional, é claro. Se a degradação do Hoffman não fosse tão incômoda...

 

[花谢花开]

Olá, você tem alguma receita para fazer o 7499-19-6?

 

[MacondoRC]

I am also interested in this information. This reagent is currently only available in Russia. China does not produce it and it is not available. Custom synthesis will be expensive, but it is possible.

 

[MacondoRC]

Synthesis of amphetamine from alpha-methylprepanamide. CAS: 7499-19-6

 

[MacondoRC]

This is an amateur video. So don't scold me too harshly. There is a synthesis instruction in the format PDF, But I can't send the file, it won't work ((

 

[MacondoRC]

The yield in this video is 85%, but the reagent itself has a purity of 90%, so I think the yield is very good.

 

[MacondoRC]

Prescription for 10g of amphetamine:

Reagents:
- Amide CAS: 7499-19-6 (2-methyl-3-phenylpropanamide, alpha-methylprepanamide, nano-propen) - 10g
- Alkali (NaOH) - 25g
- Bromine - 5ml
- Petroleum ether 40-70°C - 150ml
- Isopropanol (IPA, isopropyl alcohol) - 150ml
- Sodium sulfate (Na₂SO₄) - 2g
- Orthophosphoric acid 85% - 4ml

Synthesis:
We prepare a solution of alkali and water in a measuring cup, approximately 100-120 ml, mix it, and place it in the freezer for about 20 minutes. After taking it out of the freezer, we place it on a magnetic stirrer. We add bromine in small portions. Stir until the solution turns amber in color. Continue stirring continuously and add the amide (one of the names in Russia: nano-propene) in small portions (stir for about 5 minutes). As a result of the reaction, a transparent-murky liquid with white precipitate in the form of lumps is obtained. Filter the liquid to remove the lumps. Discard the lumps; only the filtered liquid is needed. Pour the liquid back into the measuring cup and place it on the stirrer, turn on the heating to 60-70 degrees Celsius. After about 1-10 minutes, a transparent yellowish oil should appear. Continue stirring and add petroleum ether in portions. Stir for about a minute. Remove the measuring cup with the liquid and pour it into a separatory funnel (you can use a syringe to remove the top layer (transparent yellowish oil)). From the separatory funnel, discard the lower cloudy layer (it is not needed), and pour the transparent yellowish oil back into a clean measuring cup. Add sodium sulfate, which absorbs excess water, and stir on the stirrer without heating. Pour the liquid into another clean measuring cup. Discard the sodium sulfate. Place the liquid on the stirrer and stir while adding acetone (about half the volume). After this, begin acidifying with sulfuric acid, adding 5 drops at a time in several steps until the pH reaches 7-8 (do not over-acidify). Filter and wash with acetone.

 

[w2x3f5]

Brom or sodium bromate can be replaced with sodium hypochlorite

 

[MacondoRC]

No, elementary bromine and a highly alkaline environment (pH=14) are needed for the rearrangement.

 

[w2x3f5]

See what is the Hofmann rearrangement. Primary and secondary amids react with Naocl, Naobr and sodium \ potassium hydroxide enter into the regrouping of the Hoffmann, while removing the CO2 molecule and giving Amin, having one carbon atom less than the original amide. Brom in this case is used only to obtain a Naobr.

 

[MacondoRC]

Probably, this is why a cooled alkali solution is used to maximally slow down the reaction 2NaOH + Br2 → NaBr + NaBrO + H2O.

Method:
Synthesis of primary amines by the action of bromine and alkali on carboxylic acid amides. (A.W. Hofmann)
The resulting amines contain one less carbon atom than the original amide:

RCONH2 + Br2 + OH- → RNH2 + CO2 + H2O + Br-.

As a result of this reaction, aliphatic, fatty-aromatic, aromatic, and heterocyclic amines can be obtained. Carrying out the Hofmann reaction in an alcoholic medium leads to the formation of urethanes.

The goal of the reaction is to remove one carbon atom from the amide, for which bromine, an alkaline medium, and an amide are required.

 

[w2x3f5]

But as I said earlier, bromine can be replaced with NaOCl or NaOBr, where was I wrong? Hypochlorite is a widely available precursor, which cannot be said about bromine.

 

[MacondoRC]

Sorry, you are right. Thanks for the comments.

 

[MacondoRC]

Yes, the Hofmann reaction can be carried out using sodium hypobromite (NaOBr) instead of bromine (Br₂) and alkali (NaOH). Sodium hypobromite serves as a source of active bromine and an alkaline environment, making it a convenient reagent for this reaction.

Mechanism of the reaction with NaOBr:

1. Formation of N-bromoamide:
Sodium hypobromite (NaOBr) reacts with the amide of a carboxylic acid (R-CONH₂), forming N-bromoamide (R-CONHBr). This occurs due to the substitution of a hydrogen atom in the amide group with bromine.
R-CONH₂ + NaOBr → R-CONHBr + NaOH

2. Formation of isocyanate:
The N-bromoamide undergoes deprotonation under the action of alkali (NaOH), leading to the formation of an anion. This anion then rearranges, releasing a bromide ion (Br⁻) and forming an isocyanate (R-N=C=O).
R-CONHBr + NaOH → R-N=C=O + NaBr + H₂O

3. Hydrolysis of isocyanate:
The isocyanate reacts with water (which is present in the reaction mixture), forming an unstable intermediate product—carbamic acid (R-NH-COOH). Carbamic acid rapidly decomposes, releasing carbon dioxide (CO₂) and forming a primary amine (R-NH₂).
R-N=C=O + H₂O → R-NH-COOH → R-NH₂ + CO₂

Overall reaction equation with NaOBr:
R-CONH₂ + NaOBr + 2NaOH → R-NH₂ + Na₂CO₃ + NaBr + H₂O

Advantages of using NaOBr:
- Sodium hypobromite (NaOBr) is a more convenient and safer source of bromine compared to the use of gaseous bromine (Br₂).
- The reaction proceeds in a single step, as NaOBr already contains both bromine and alkali.

Thus, the use of sodium hypobromite (NaOBr) is an efficient and convenient method for carrying out the Hofmann reaction.