Pseudoephedrine ---> Methamphetamine

flunky nassau

Senior member
Feb 17, 2007
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It's getting ridiculous here at the pharmacy.

The federal government is trying to crack down by putting a federal daily & monthly limit of pseudoephedrine that one can purchase from a given pharmacy. However, now there are clans going around buying the daily limit from different pharmacies. It's a huge waste of our time because we have to log in their driver's license.

It's pretty obvious too, because if I tell one person I'm out of stock, a whole line of people leave. It's getting annoying, but I don't want to start profiling people.


These people have no problems paying $20 for 15 tablets, so anyone know the street value?
 

GregGreen

Golden Member
Dec 5, 2000
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I donno why they don't just start getting their cheap meth from the Mexican cartels just like everyone else...
 

flunky nassau

Senior member
Feb 17, 2007
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Ok, let's put it this way.

One box of Claritin-D 24-Hour #15 (their purchase of choice) contains 3.6 grams of pseudoephedrine. They pay ~$20 for this. Once this is converted, how much it worth?

 

IEC

Elite Member
Super Moderator
Jun 10, 2004
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Meth has a street value of $100/g +- $50.

Assuming it's $100/g I'm assuming it's more than worth it to them to buy pseudoephedrine for <$6/g.
 

flunky nassau

Senior member
Feb 17, 2007
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Originally posted by: Spartan Niner
Meth has a street value of $100/g +- $50.

Assuming it's $100/g I'm assuming it's more than worth it to them to buy pseudoephedrine for <$6/g.



But how many grams (or milligrams) of meth is converted from 3.6 grams of pseudoephedrine?
 

aphex

Moderator<br>All Things Apple
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Jul 19, 2001
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thinking of starting a new business venture?
 

destrekor

Lifer
Nov 18, 2005
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Originally posted by: flunky nassau
Originally posted by: Spartan Niner
Meth has a street value of $100/g +- $50.

Assuming it's $100/g I'm assuming it's more than worth it to them to buy pseudoephedrine for <$6/g.



But how many grams (or milligrams) of meth is converted from 3.6 grams of pseudoephedrine?

I've read the final product is up to 80% or so of the original amount of pseudoephedrine. However, likely less as that's probably ideal based on mathematics. Labs are rarely perfect and don't always use red phosphorous (of which is what the article stated should be used. something else can be used too I think.. ether?).
 

IEC

Elite Member
Super Moderator
Jun 10, 2004
14,600
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Originally posted by: flunky nassau
Originally posted by: Spartan Niner
Meth has a street value of $100/g +- $50.

Assuming it's $100/g I'm assuming it's more than worth it to them to buy pseudoephedrine for <$6/g.



But how many grams (or milligrams) of meth is converted from 3.6 grams of pseudoephedrine?

????????
 

Gibsons

Lifer
Aug 14, 2001
12,530
35
91
Originally posted by: destrekor
Originally posted by: flunky nassau
Originally posted by: Spartan Niner
Meth has a street value of $100/g +- $50.

Assuming it's $100/g I'm assuming it's more than worth it to them to buy pseudoephedrine for <$6/g.



But how many grams (or milligrams) of meth is converted from 3.6 grams of pseudoephedrine?

I've read the final product is up to 80% or so of the original amount of pseudoephedrine. However, likely less as that's probably ideal based on mathematics. Labs are rarely perfect and don't always use red phosphorous (of which is what the article stated should be used. something else can be used too I think.. ether?).

They can make up for that by cutting the final product.
 

uclaLabrat

Diamond Member
Aug 2, 2007
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Assuming a decent yield of 80% or so, I'd say in round numbers that 3.6g pseudoepehdrine could yield 3g meth, if done well.

It's only one step, removing a hydroxyl group.

As far as the chemistry, they use the red phosphorous and iodine to reduce off the hydroxyl group, I believe. I'm not sure what the active reducing agent is.
 

FallenHero

Diamond Member
Jan 2, 2006
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Originally posted by: uclaLabrat
Assuming a decent yield of 80% or so, I'd say in round numbers that 3.6g pseudoepehdrine could yield 3g meth, if done well.

It's only one step, removing a hydroxyl group.

As far as the chemistry, they use the red phosphorous and iodine to reduce off the hydroxyl group, I believe. I'm not sure what the active reducing agent is.

Anhydrous ammonia and battery acid.

Edit: Assuming I have my steps in making meth correct.