Synthesis of Amphetamine from P2NP via SnCl2 and Zn

[TucoSalamanca.]

Why 200gr sodium carbonate does not dissolve in 200ml water

 

[GDC]

Why 200gr sodium carbonate does not dissolve in 200ml water

TucoSalamanca.Because Solubility of Sodium Carbonate: 21.4g/100ml H2O and Hydroxide not defined @ 20°C

 

[TucoSalamanca.]

Because Solubility of Sodium Carbonate: 21.4g/100ml H2O and Hydroxide not defined @ 20°C

GDCCan you write the calculations of the 1st and 2nd stages? sodium carbonate

 

[TucoSalamanca.]

Because Solubility of Sodium Carbonate: 21.4g/100ml H2O and Hydroxide not defined @ 20°C

GDCThere is very little p2np left, can you write the full list of ingredients required for 7.5 g p2np, I have sodium carbonate but no sodium hydroxide?

 

[GDC]

Na2CO3
70g/L (0 °C)
164.0g/L (15 °C)
340.70g/L (27.8 °C)
pH=14+log(2×0.09989)=13.3006 inside 1000ml H2O
0,1 Mol Na3CO3 == ~10,6g * 50 = 530g / 245g =

2,16 L H2O with 530g Na2CO3 @ 20°C

NaOH
418 g/L (0 °C)
1000 g/L (25 °C)[
pH=14+log(0.1)=13 inside 1000ml H2O
0,1 Mol NaOH == ~4g NaOH * 50 = 200g

Sodium bicarbonate (baking soda)	0.1 N	8.4
Sodium carbonate (washing soda)	0.1 N	11.6

.

Sodium hydroxide (caustic soda)	N	14.0
Sodium hydroxide	0.1 N	13.0
Sodium hydroxide	0.01 N	12.0

I'm not sure if i'am right xD its to late and you have to check it with someone other

You want the half of the 15g P2NP:

100g NaOH 100ml H2O or 265g NaCO3 and 1,08 L H2O

But caution !!!! I think you need the double of NaCO3

 

[GDC]

Na2CO3
70g/L (0 °C)
164.0g/L (15 °C)
340.70g/L (27.8 °C)
pH=14+log(2×0.09989)=13.3006 inside 1000ml H2O
0,1 Mol Na3CO3 == ~10,6g * 50 = 530g / 245g =

2,16 L H2O with 530g Na2CO3 @ 20°C

NaOH
418 g/L (0 °C)
1000 g/L (25 °C)[
pH=14+log(0.1)=13 inside 1000ml H2O
0,1 Mol NaOH == ~4g NaOH * 50 = 200g

Sodium bicarbonate (baking soda)	0.1 N	8.4
Sodium carbonate (washing soda)	0.1 N	11.6

.

Sodium hydroxide (caustic soda)	N	14.0
Sodium hydroxide	0.1 N	13.0
Sodium hydroxide	0.01 N	12.0

I'm not sure if i'am right xD its to late and you have to check it with someone other

You want the half of the 15g P2NP:

100g NaOH 100ml H2O or 265g NaCO3 and 1,08 L H2O

But caution !!!! I think you need the double of NaCO3

GDCBullshit
I correct it later: https://bbgate.com/threads/synthesis-of-amphetamine-from-p2np-via-sncl2-and-zn.9878/post-53908

 

[TucoSalamanca.]

How can I tell if it's amphetamines?

 

[GDC]

How can I tell if it's amphetamines?

TucoSalamanca.Drugtest, heat some of it to the desired temperature (it will melt and burn)

 

[TucoSalamanca.]

Drugtest, heat some of it to the desired temperature (it will melt and burn)

GDChow can I use it

 

[GDC]

how can I use it

TucoSalamanca.Amphetamine:

• Formula C9H13N
• Molar mass 135.210 g/mol
• Density 0.936 g/cm3 at 25 °C
• Melting point 11.3 °C (52.3 °F)
• Boiling point 200-203 °C (397 °F) at 760 mmHg

What are you mean?

 

[TucoSalamanca.]

Amphetamine:

• Formula C9H13N
• Molar mass 135.210 g/mol
• Density 0.936 g/cm3 at 25 °C
• Melting point 11.3 °C (52.3 °F)
• Boiling point 200-203 °C (397 °F) at 760 mmHg

What are you mean?

GDCCan you write all calculations 7.5g p2np

 

[TucoSalamanca.]

Amphetamine:

• Formula C9H13N
• Molar mass 135.210 g/mol
• Density 0.936 g/cm3 at 25 °C
• Melting point 11.3 °C (52.3 °F)
• Boiling point 200-203 °C (397 °F) at 760 mmHg

What are you mean?

GDCI have some sodium carbonate left, should I sell it, please write the correct calculations

 

[GDC]

I have some sodium carbonate left, should I sell it, please write the correct calculations

TucoSalamanca.Yes you should, sodium carbonate produces a lot of impurity.
NaOH is much better.

7,5 g of P2NP in ~25 ml of anhydrous acetone
~25ml acetone and 21 g of tin(II) chloride dihydrate
75 ml of cold water and 2 ml 36% hydrochloric acid
sodium hydroxide (100g per 100ml of water)
3 x 25 ml portions of ethyl acetate
30 ml glacial acetic acid
5 g of zinc dust.
25 g NaOH in 50 ml water
25 ml dichloromethane
3x 10 ml DCM.
anhydrous magnesium sulfate ~1/20 of the final solution
25 ml of anhydrous acetone

 

[TucoSalamanca.]

Yes you should, sodium carbonate produces a lot of impurity.
NaOH is much better.

7,5 g of P2NP in ~25 ml of anhydrous acetone
~25ml acetone and 21 g of tin(II) chloride dihydrate
75 ml of cold water and 2 ml 36% hydrochloric acid
sodium hydroxide (100g per 100ml of water)
3 x 25 ml portions of ethyl acetate
30 ml glacial acetic acid
5 g of zinc dust.
25 g NaOH in 50 ml water
25 ml dichloromethane
3x 10 ml DCM.
anhydrous magnesium sulfate ~1/20 of the final solution
25 ml of anhydrous acetone

GDCsodium carbonate instead of sodium hydroxide

 

[TucoSalamanca.]

Yes you should, sodium carbonate produces a lot of impurity.
NaOH is much better.

7,5 g of P2NP in ~25 ml of anhydrous acetone
~25ml acetone and 21 g of tin(II) chloride dihydrate
75 ml of cold water and 2 ml 36% hydrochloric acid
sodium hydroxide (100g per 100ml of water)
3 x 25 ml portions of ethyl acetate
30 ml glacial acetic acid
5 g of zinc dust.
25 g NaOH in 50 ml water
25 ml dichloromethane
3x 10 ml DCM.
anhydrous magnesium sulfate ~1/20 of the final solution
25 ml of anhydrous acetone

GDChow much sodium carbonate how much water should be used

 

[GDC]

sodium carbonate instead of sodium hydroxide

TucoSalamanca.So sry for the last confusing post, but i will correct it now:

For the 25g NaOH inside 50ml H2O you need:

We need to calculate the molarity of the NaOH solution.
Molarity (M) = (mass of solute in grams) / (molar mass of solute) / (volume of solution in liters)
Molarity = 25 g / (39.997 g/mol) / (50 ml / 1000 ml/L)
Molarity = 0.625 M

Calculate moles of sodium hydroxide: Moles of NaOH = Molarity x Volume (L) = 0.625 M x 0.050 L = 0.03125 moles of NaOH

Calculate moles of sodium carbonate required to replace NaOH: 1 mole of NaOH requires 1.22 moles of Na2CO3.
Moles of Na2CO3 = 1.22 x Moles of NaOH = 1.22 x 0.03125 moles = 0.038125 moles of Na2CO3

Calculate mass of sodium carbonate required:
Mass of Na2CO3 = Moles of Na2CO3 x Molecular mass of Na2CO3 = 0.038125 moles x 105.9888 g/mol = 4.039 g

So, you will need 4.039 g of sodium carbonate to replace the 25 g of sodium hydroxide. You can dissolve this amount of sodium carbonate in 50 ml of water to achieve a similar concentration as the original NaOH solution.

For the 100g NaOH inside 100ml H2O you need:

Calculate molarity of NaOH solution:
Molarity (M) = (mass of solute in grams) / (molar mass of solute) / (volume of solution in liters)
Molarity = 100 g / (39.997 g/mol) / (100 ml / 1000 ml/L)
Molarity = 2.501 M

Calculate moles of sodium hydroxide:
Moles of NaOH = Molarity x Volume (L) = 2.501 M x 0.100 L = 0.250 moles of NaOH

Calculate moles of sodium carbonate required to replace NaOH:
1 mole of NaOH requires 1.22 moles of Na2CO3.
Moles of Na2CO3 = 1.22 x Moles of NaOH = 1.22 x 0.250 moles = 0.305 moles of Na2CO3

Calculate mass of sodium carbonate required:
Mass of Na2CO3 = Moles of Na2CO3 x Molecular mass of Na2CO3 = 0.305 moles x 105.9888 g/mol = 32.30 g

So, you will need 32.30 g of sodium carbonate to replace the 100 g of sodium hydroxide. You can dissolve this amount of sodium carbonate in 100 ml of water to achieve a similar concentration as the original NaOH solution.

I hope you enjoyed my answer an LIKE it

 

[GDC]

So sry for the last confusing post, but i will correct it now:

For the 25g NaOH inside 50ml H2O you need:

We need to calculate the molarity of the NaOH solution.
Molarity (M) = (mass of solute in grams) / (molar mass of solute) / (volume of solution in liters)
Molarity = 25 g / (39.997 g/mol) / (50 ml / 1000 ml/L)
Molarity = 0.625 M

Calculate moles of sodium hydroxide: Moles of NaOH = Molarity x Volume (L) = 0.625 M x 0.050 L = 0.03125 moles of NaOH

Calculate moles of sodium carbonate required to replace NaOH: 1 mole of NaOH requires 1.22 moles of Na2CO3.
Moles of Na2CO3 = 1.22 x Moles of NaOH = 1.22 x 0.03125 moles = 0.038125 moles of Na2CO3

Calculate mass of sodium carbonate required:
Mass of Na2CO3 = Moles of Na2CO3 x Molecular mass of Na2CO3 = 0.038125 moles x 105.9888 g/mol = 4.039 g

So, you will need 4.039 g of sodium carbonate to replace the 25 g of sodium hydroxide. You can dissolve this amount of sodium carbonate in 50 ml of water to achieve a similar concentration as the original NaOH solution.

For the 100g NaOH inside 100ml H2O you need:

Calculate molarity of NaOH solution:
Molarity (M) = (mass of solute in grams) / (molar mass of solute) / (volume of solution in liters)
Molarity = 100 g / (39.997 g/mol) / (100 ml / 1000 ml/L)
Molarity = 2.501 M

Calculate moles of sodium hydroxide:
Moles of NaOH = Molarity x Volume (L) = 2.501 M x 0.100 L = 0.250 moles of NaOH

Calculate moles of sodium carbonate required to replace NaOH:
1 mole of NaOH requires 1.22 moles of Na2CO3.
Moles of Na2CO3 = 1.22 x Moles of NaOH = 1.22 x 0.250 moles = 0.305 moles of Na2CO3

Calculate mass of sodium carbonate required:
Mass of Na2CO3 = Moles of Na2CO3 x Molecular mass of Na2CO3 = 0.305 moles x 105.9888 g/mol = 32.30 g

So, you will need 32.30 g of sodium carbonate to replace the 100 g of sodium hydroxide. You can dissolve this amount of sodium carbonate in 100 ml of water to achieve a similar concentration as the original NaOH solution.

I hope you enjoyed my answer an LIKE it

GDCSodium hydroxide (NaOH) is a stronger base than sodium carbonate (Na2CO3).
This is because sodium hydroxide completely dissociates in water, releasing hydroxide ions (OH-), which are responsible for its basic properties:
NaOH → Na+ + OH-
On the other hand, sodium carbonate also dissociates in water but forms a weak acid (carbonic acid, H2CO3) and a weaker base (hydrogen carbonate ion, HCO3-) compared to NaOH:
Na2CO3 + H2O → H2CO3 + 2Na+ + 2OH-
H2CO3 → H+ + HCO3-
The weaker basicity of sodium carbonate compared to sodium hydroxide means that solutions of sodium carbonate will have a lower pH and milder basic properties than solutions of sodium hydroxide at the same concentration. Therefore, sodium hydroxide is considered a stronger base than sodium carbonate.

 

[TucoSalamanca.]

Sodium hydroxide (NaOH) is a stronger base than sodium carbonate (Na2CO3).
This is because sodium hydroxide completely dissociates in water, releasing hydroxide ions (OH-), which are responsible for its basic properties:
NaOH → Na+ + OH-
On the other hand, sodium carbonate also dissociates in water but forms a weak acid (carbonic acid, H2CO3) and a weaker base (hydrogen carbonate ion, HCO3-) compared to NaOH:
Na2CO3 + H2O → H2CO3 + 2Na+ + 2OH-
H2CO3 → H+ + HCO3-
The weaker basicity of sodium carbonate compared to sodium hydroxide means that solutions of sodium carbonate will have a lower pH and milder basic properties than solutions of sodium hydroxide at the same concentration. Therefore, sodium hydroxide is considered a stronger base than sodium carbonate.

GDCI don't know molar calculation, can you tell me directly how many grams of sodium carbonate will dissolve in how many ml of water? 2 alkali calculation

 

[TucoSalamanca.]

Sodium hydroxide (NaOH) is a stronger base than sodium carbonate (Na2CO3).
This is because sodium hydroxide completely dissociates in water, releasing hydroxide ions (OH-), which are responsible for its basic properties:
NaOH → Na+ + OH-
On the other hand, sodium carbonate also dissociates in water but forms a weak acid (carbonic acid, H2CO3) and a weaker base (hydrogen carbonate ion, HCO3-) compared to NaOH:
Na2CO3 + H2O → H2CO3 + 2Na+ + 2OH-
H2CO3 → H+ + HCO3-
The weaker basicity of sodium carbonate compared to sodium hydroxide means that solutions of sodium carbonate will have a lower pH and milder basic properties than solutions of sodium hydroxide at the same concentration. Therefore, sodium hydroxide is considered a stronger base than sodium carbonate.

GDCI can't afford sodium hydroxide.

 

[GDC]

I don't know molar calculation, can you tell me directly how many grams of sodium carbonate will dissolve in how many ml of water? 2 alkali calculation

TucoSalamanca.4.039 g of sodium carbonate to replace the 25 g of sodium hydroxide. You can dissolve this amount of sodium carbonate in 50 ml of water to achieve a similar concentration as the original NaOH solution.

32.30 g of sodium carbonate to replace the 100 g of sodium hydroxide. You can dissolve this amount of sodium carbonate in 100 ml of water to achieve a similar concentration as the original NaOH solution.

So:
7,5 g of P2NP in ~25 ml of anhydrous acetone
~25ml acetone and 21 g of tin(II) chloride dihydrate
75 ml of cold water and 2 ml 36% hydrochloric acid
Na2CO3 (32,30g per 100ml of water)
3 x 25 ml portions of ethyl acetate
30 ml glacial acetic acid
5 g of zinc dust.
4,039 g Na2CO3 in 50 ml water
25 ml dichloromethane
3x 10 ml DCM.
anhydrous magnesium sulfate ~1/20 of the final solution
25 ml of anhydrous acetone

 

[hacke8]

oh sry you mean my point 5
You can use ether after zinc reaction

GDCThank you for your answer. This damn translator always makes us wrong. What kind of reagent is GAA? Glacial acetic acid?

 

[G.Patton]

Thank you for your reply. I know that DCM is dichloromethane. I don't know whether ethyl acetate is a reactant or an extractant. If it is a reactant, it may not be replaceable. If the extraction agent is replaced with petroleum ether or xylene, will it affect the subsequent reaction?

hacke8

I don't know whether ethyl acetate is a reactant or an extractant.

Hello, it is said that there is an extraction procedure. EA here is used as extractant (solvent) for extraction of oxime.