Knowing What You Are Doing - Stop asking and read

[lutesium]

Though it seems like cooking, chemistry is a combination of brain and labor. No successfu results can be gotten if you treat what you are doing as a cookbook recipe. You have to put your heart in that flask and be a part of the reaction. This means you have to know whats going on there. Its not a mix n bake.

Here I will try to describe what many people dont know. Its not a recipe to make meth. This is an explanation of a basic reaction which is the main reason of the failures.
Learning this is easier than the labor wasted in a failed reaction.

Stop trying and try to understand what you are doing. ITS A VERY SIMPLE REACTION.

R-OH + H-X <-> R-x + H2O

If you are reducing an alcohol (in your case E) with HI you first make the iodo ester (some people dont call it an ester but it is) This ester is what the reaction actually reduces. The formation of the ester is an easy reaction if dry medium is used because HI is an acid which is strong enough to dehydrate the alcohol (the mix is said to liquify quickly in dry reactions this is a sign of the formation of iodoE + H2O) and if theres no H2O in the rxn it will proceed to create more.

But in LWRs (this is the actual method) the rate limiting step is the above mentioned iodination phase due to the high water content. The only chance to drive the reaction right is to increase the amount of HI versus H2O - Thats why the concentration of HI is important.  

If using another solvent the following charachteristics should be sought:
-It should dissolve HI
-It should protonate the PI3 tho create HI
-It should better not be the product of any step (to drive the equations right)

The first solvent coming to mind having all these qualities is acetic acid. It does disolve HI, it is acidic for the protonation of PI3 - HI creation and unlike H2O its not the product of any step thus the iodination of the alcohol(E) will run smoothly without the worry of excess H2O thereby the need of high HI cncentration. Using enough I2 in a dilute acetic acid solution is no problem because the HI is recycled and will always be excess to the H2O which only comes as the product from the iodination of the alcohol - so no worry for concentrated solutions or so.


And the second hydrogenation step:

Think this:
Why would that iodine leave the place it bonded? Why will that hydrogen replace the iodine?

The benzylic carbon (the C that has the -OH group) was very acidic when it was E. HI ionized enough and had a proton (H+) thats more acidic than the above mentioned carbon so the -OH selected to bind it and I formed an ester with the C+ therefore E-I and H2O formed. An acidic ester can not be hydrogenolyzed (breaking the bond with two hydrogens) by the application of an acid. This means that the reaction is not as simple as

E-I + HI -> E-H + I2

The H of the HI can NOT replace the iodine due to the acidity of carbon and H. This suggests that a dissociation is occuring.

A normal ionization is described as follows

HI -> H  + :I

But dissociation is

HI -> H. + .I

(The dots are electrons)

If ionized H has a missing electron, therefore it has a positive charge I has 1 excess electron meaning a negative gharge. But in dissociation both atoms have one electrons. H. wants to give it, so an electron acceptor accepts it (in our case the benzylic carbon of E-I). The carbon now has an octet hole which means that it will strongly pull a H+ (ionized H+) from the solution To share this electron and form a bond. This is the actual reaction and requires energy which means TIME + HEAT.

Not enough time and the electrolytic reduction isnt complete if not using autoclave temps. Not enough heat, you will have to wait longer.

THATS SIMPLE AS SWIM TRIED TO DESCRIBE WHO IS INTERESTED IN REDUCTIONS LIKE THIS. THERE IS NO MAGICAL METHOD TO CREATE A GOOD RESULT IF ENOUGH EFFORT IS GIVEN. ACT ACCORDINGLY

[CharlieBigpotato]

don't expect  much response.
(i'm sure you don't)

understanding things is so slow and boring.

(btw, do you have a recipe for this approach?)

[java]

.....Using enough I2 in a dilute acetic acid solution is no problem because the HI is recycled and will always be excess to the H2O which only comes as the product from the iodination of the alcohol - so no worry for concentrated solutions or so

As my interest lies in a diffferent amino alcohol,  my question was always if plain vinegar could be used in leiu of unavailable  source of acidic acid.  Also  it seems by the equation that I2 is a by product and could be recovered after the reaction. I have no experience with the reaction and my interest is totally academic....java

[lutesium]

No - glacial acetic acid should be used. I meant dilute in in respect to HI concentration in the GAA

[java]

The first solvent coming to mind having all these qualities is acetic acid

Like I said I have no experience in this reaction but you said in your narrative that the" acetic acid had all the qualities" for a good solvent. Would you happen to have any references to go along wih your narrative. I do want to read , and I understand there is an old Assholium post where he uses acidic acid in the reduction of a different Amino alcohol (phenylalaninol) with RP/I  in acedic acid.

[ChemoSabe]

The second recipe given here seems almost relevant. And if I didn't know better I'd say this was coming straight out some form of "high academia".

https://www.thevespiary.org/rhodium/Rhodium/chemistry/meth.hi-p.html

PROCEDURE 2

In a 500 mls round-bottom flask is added 150 mls of glacial acetic acid, 15 grams of red phosphorus, and 33.50 grams of iodine.

[NOTE 6] The mixture is allowed to react for 15 - 20 minutes, until all of the iodine has reacted. Then add 10 mls of distilled water and mix, then add 8.07 grams (0.04 mole) of EPHEDRINE HCl, and mix. A reflux condenser is added to the 500 mls round-bottom flask and the mixture is gently refluxed (gently boiled continuously) for 2 and 1/2 hours. The hot mixture is suction filtered while still hot to remove the excess red phosphorus.[NOTE 7] The hot filtrate is added slowly by pouring into a solution of 20 grams of sodium bisulphite in 1 liter of distilled water. The solution is made basic and solvent extracted as above using two 100 ml portions of ether, washed and dry.The oily residue extracted (after removing the ether) is a mixture of methamphetamine, iodo-ephedrine and minor amount of acetic ester ephedrine.[NOTE 5]

[NOTE 6] Alternatively, 150 mls of hot distilled water, 15 grams (0.48 moles) of red phosphorus, and 33.50 grams (0.132 moles) of iodine in small 3-5 gram portions is added. The mixture is allowed to react for 15-20 min, until all of the iodine has reacted. Then add 8.07 g (0.04 mole) of EPHEDRINE HCl and follow Procedure 1. The yield is 80-88%

[NOTE 7] Have some hot glacial acetic acid on hand to rinse the flask and the filtered red phosphorus.

Calculation of HI to Ephedrine Ratio

In Procedure 1 and 2, two moles of HI react with one mole of Ephedrine giving Methamphamine and I2.

Ephedrine + 2HI ==>> Methamphetamine + I2 + H2O

Therefore the Ephedrine:HI ratio is theoretically 1:2, but to assure that the reaction goes to completion, the actual ratios used above are 1:3.3

This ratios has been calculated as follows:

Procedure 1:

1 ml of 57% HI = 0.99 grams of HI

170 mls of 57% HI = (170 x 0.99) = 168.3 grams of HI = 1.32 moles HI

Since we use 0.4 moles of Eph then (1.32/0.4) = 3.3

Since the reaction of phosphorus, iodine and water is 2 P + 3 I2 + 6 H2O --> 6 HI + 2 H3PO3 the ratio of HI:I2 is 3:3

Procedure 2

For every 0.04 moles of Eph, then (0.04 x 3.3) = 0.132 moles of HI, since the ratio of HI:I2 is 3 : 3 = 1, then we need 0.132 moles of I2 = 33.50 grams of I2.

[java]

I see now , the reaction in Rhodium's page procedure one makes the HI with water then the rest and the solvent for the reaction is the HI and H2O.  Where as in a different amino alcohol (Phenylalaninol) , its solubility in water is slight and hence acetic acid was recomended and later proved by Assolium as was read in an old post by Labrat.

Therefore when I read the original post on this thread and the use of diluted acedic acid my interest was to find out if in fact somone had read and reference the procedure. As I read now the author was talking about procedure two in Rhodium's page.

[Rhodium]

If using another solvent the following charachteristics should be sought: -It should protonate the PI3 tho create HI [...] The first solvent coming to mind having all these qualities is acetic acid. It does disolve HI, it is acidic for the protonation of PI3 - HI creation

The formation of HI from PI₃ is not a "protonation", it is rather a solvolysis. PI₃ can be seen as an anhydride between phosphoric and hydriodic acid, which has to be hydrolyzed to break up into its component acids again...

PI₃ is also only present in fully anhydrous reactions, otherwise other intermediates predominate:

Post 468813

(Rhodium: "Final Word on the HI/RP Mechanism", Stimulants)

The H of the HI can NOT replace the iodine due to the acidity of carbon and H. This suggests that a dissociation is occuring.

A normal ionization is described as follows

HI -> H  + :I

But dissociation is

HI -> H. + .I

(The dots are electrons)

You are confusing the terminology here. What you call "ionization/dissociation" are rather Heterolytic and Homolytic Dissociation.

Homolytic vs heterolytic cleavage
^{http://en.wikibooks.org/wiki/Polar_and_radical_reactions_(Organic_chemistry)}

Two bonded atoms can come apart from each other in one of two ways. Either

• each atom gets away with half of the shared electrons, or
• one of the atoms leaves with more of the shared electrons than the other.

In homolytic cleavage, each atom leaves with one-half of the shared electrons (one electron for a single bond, or two for double bonds).

A—B --> A* + B*

A* and B* represent uncharged radicals. The "*" represents an unbonded, unpaired valence electron.

In heterolytic cleavage, one atom leaves with all of the previously shared electrons and the other atom gets none of them.

A—B --> A^- + B⁺

Homo, or same, indicates that each atom leaves with the same number of electrons from the bond. Hetero, or different, refers to the fact that the atoms each end up with a different number of electrons.

Polar reactions
Polar reactions occur when two bonded atoms come apart, one taking more of the shared electrons than the other. They involve heterolytic cleaveage and result in a positively and negatively charged ion or molecular fragment.

Radical reactions
Radical reactions dont deal with charged particles but with radicals. Radicals are uncharged atoms or molecules with an incomplete octet of valence electrons.

When a molecule comes apart by homolytic cleavage the result is two radicals. Although uncharged, radicals are usually very reactive because the unfilled octet is unstable and the radical can lower its energy by forming a bond in a way that allows it to fill its valence shell while avoiding any electrostatic charge.