amfetamiinin yksinkertainen synteesi

[strannik]

Hyvää iltapäivää! Pyydän etukäteen anteeksi englanninkielisyyttäni. Tähän mennessä on ilmestynyt uusi esiaste, jonka avulla amfetamiinia voidaan valmistaa vain yhdessä tunnissa. Tämä esiaste on alfa-metyyliprepanamidi. CAS: 7499-19-6

Natriumhydroksidi (alkali) 3kg Bromi 1,5kg(0,5l) Petrolieetteri 1L Asetoni 5L Rikkihappo hcl 1L Lakmus Otetaan 3 kg alkalia (natriumhydroksidi), liuotetaan 3 litraan vettä, reaktio on esoterminen, ei kaadeta kaikkea kerralla vaan vähitellen, vesi voi kiehua. Sitten jäähdytämme alkaliliuoksen. Huomio!!! Ennen reaktiota on tarpeen jäähdyttää alkaliliuos noin 0 asteen lämpötilaan! On välttämätöntä työskennellä sekoittaen - reaktori, sekoitin jne.! Kaada hitaasti 0,5 litraa bromia 0 asteeseen jäähdytettyyn emäksiseen liuokseen, liuoksen on välttämättä pysyttävä värittömänä tai kellertävänä, kun olet lisännyt koko tilavuuden bromia, kaada propanamidi liuokseen ja kuumenna sitä hitaasti sekoittaen. Kun lämpötila on 80-90 astetta, tapahtuu reaktio, liuos muuttuu sameaksi vapautuneen emäksen vuoksi. Sekoitetaan vielä 5-7 minuuttia, jäähdytetään huoneenlämpötilaan, kaadetaan 1 litra petrolieetteriä, jatketaan sekoittamista. Erotetaan ylempi eetterikerros. Seuraavaksi on tavallista happamoittaa rikkihapolla, eli kaada hieman tai tiputtaa rikkihappoa, kunnes pH on 5,5-5, jos se sakeutuu, voit laimentaa asetonilla. Tämän jälkeen puristamme tuloksena olevan seoksen tyhjiön läpi ja huuhtelemme jauheen asetonilla.

Bromi voidaan korvata natriumhypokloridilla.

 

[strannik]

korjaus
CAS 7499-19-6 Bentseeni Propanamidi (2-metyyli-3-fenyylipropanamidi)

2NaOH + Br2 → NaBr + NaBrO + H2O
natriumhypobromidi voidaan korvata natriumhypokloridilla.

 

[OrgUnikum]

Lähes. Se on bentseenipropanamidi, a-metyyli- sinulta puuttuisi metyyliryhmä ketjusta bentseenipropanamidi, CAS ja 2-metyyli-3-fenyylipropanamidi on kuitenkin oikein. Näyttää paljon fenyylialaniinilta...

 

[OrgUnikum]

Minä tyhmä, tämä näyttää tietysti Helionalin Amidilta. Jos Hoffmanin hajoaminen ei olisi niin tuskallista...

 

[花谢花开]

Hei, onko sinulla reseptiä 7499-19-6:n valmistukseen?

 

[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.