Seems like not many Methylamine (g) flowed into iPrOH;
1. mixture of 50 g NaOH  and 35 ml 40% Methylamine is contains about 20 ml water, i think 50 g NaOH isn't enough except in fine powder form.
2. the solubility of Methylamine in saturated water-NaOH solution is unknown
Maybe CaO (readily available in fine powder form) will work well.

And worse if Methylamine contained "many" water,  you got 3,4-methylenedioxy-2-propanol  as "a few grams of safrole-smelling oil"

I didn't understand the mecanism you were trying to explain ( I still don't know if you are talking about SN1 or SN2) but here is a trial with methylamine SN on iodosafrole in isopropanol:

Dont mean to disrupt this thread but Im fairly sure the mechanism in this proposal is SN1 as a protic polar solvent is used (i.e IPA). The halide bond dissociates to form a secondary carbocation and the latter is attacked by a methylamine molecule. The rate of the reaction is solely determined by the dissociation step to form the carbocation and is therefore unimolecular.

no there are competing reactions because methylamine is a strong nucleophile it wins out the ipa has it's oxygenated proton sheilded so the sn1 reaction is'nt as predominate.
this is also dependant on concentration.

As I said, its SN1. All the criterions are there. Its because methylamine is a good nucleophile and temperature is dropped to avoid elimination (i.e methylamine dont act as a base to react with the alpha proton). And of course its dependant on concentration of the substrate. Your proposal basically involves an halide swimming in a sea of methylamine so the concentration of nucleophile is not considred in the rate determining step.

no your'e right concentration is'nt affecting the rate determining step but it does affect what kind of reactions take place.
a little water in the mix and now you have oh- ions floating around causing e2 reactions also it blocks the nucleophile quite effectively from attacking the backside of that carbon.
and temp is key  just heating that on running water and you get big tarballs yes that's also true.

Well then youre describing side-reactions and not the actual process of making MDMA. All reactions have undesired reactional path, im sure youll agree with that.

Im just stating that the actual path to MDMA in this route from iodosafrole is via SN1, all other processes are side reactions. I concur they may happen and the purpose is to avoid them, but they have nothing to do with the desired target which is MDMA.

Hello,

I am sorry, I forgot to precise this report was not made by either me nor somebody of my team because our company don't allow us to work with illegal compound within our territory (Belgium). He used methylamine water solution 40% because he didn't have Methylamine HCl. If the reaction was performed in our lab we would used MeNH2 in cylinder but apparently his appartus is limited. The product was not analysed because he has no MS,NMR,...

Seems like not many Methylamine (g) flowed into iPrOH;
1. mixture of 50 g NaOH  and 35 ml 40% Methylamine is contains about 20 ml water, i think 50 g NaOH isn't enough except in fine powder form.
2. the solubility of Methylamine in saturated water-NaOH solution is unknown
Maybe CaO (readily available in fine powder form) will work well.

And worse if Methylamine contained "many" water,  you got 3,4-methylenedioxy-2-propanol  as "a few grams of safrole-smelling oil"

The solubility of methylamine is really poor in water. Only methylammonium hydroxide is soluble in water, therefore adding NaOH will kill ammonium ions.

[/quote]

iodosafrole is'nt green that's bromosafrole, iodosafrole has a purple tint
i remember saying to make a solution of naoh first and add it to solid methylamine because the dissolution  of the naoh kicks off a lot of heat, and that carries water with it.
 did you use a suckback trap?
did you calculate moles bubbled in before procedding?
it looks like you got water in there somehow.
i can't tell your notes are'nt too specific.
***5g of iodosafrole as green needles (melts at RT to give brown oil)***
that's not right.

another thing did you use a gas dispersion device to bubble it in??
i'm am of the above poster's opionion you did'nt buble that much in without a gas dispersion end you only "think" you bubbling methylamine.
and when you drip it onto naoh solid like that it gets real hot and comes over real angry.

I didnt know iodosafrole is purple. The iodo was syntetised upon dropping bromosafrole on a saturated 1.3 eq KI in acetone, then refluxed for 1h, cooled,filtered, solvent boiled off, filtered again to get a brown oil wich cristallise to green crystals upon freezing. Perhaps the anion exchange failed and there is too much less reactive bromosafrole in the oil. The iodo was not tested by MS.

The moles of MeNH2 were not calculated. I said him to add 15+ eq. I suggested to dissolve MeNH2 HCl in 80% NaOH, bubble through anhydrous iPrOH (with a simple glasspipe with no gas dispersion device, due to the high solubility of MeNH2 in alcools I don't think it would be necessary), let the solution for one week over molecular sieve 3A before using, but he apparently has no sieve and no methylamine HCl...

Dont mean to disrupt this thread but Im fairly sure the mechanism in this proposal is SN1 as a protic polar solvent is used (i.e IPA). The halide bond dissociates to form a secondary carbocation and the latter is attacked by a methylamine molecule. The rate of the reaction is solely determined by the dissociation step to form the carbocation and is therefore unimolecular.

Hey it's difficult to say if its SN1 or SN2 here. an SN1 would give much more side products : the intermediate carbocation can rearrenge to the benzylic position and can also react with the solvent. An SN2 is much more preferable here and I think it's an SN2 but IMO the reaction cannot be as fast as you said and it's unprobable to form the tertiary amine here, the MDMA base is sooo bulky and you've got plenty of MeNH2 in excess...

Don't forget : Low temp. favorise the SN2 and high temp the elimination.

Another thing :Can you explain me, jon, why you said me to use iPrOH instead of DMF for the reaction if its an SN2 ?? The only reason can be the solubility of the MeNH2 in iPrOH.

***I didnt know iodosafrole is purple. The iodo was syntetised upon dropping bromosafrole on a saturated 1.3 eq KI in acetone, then refluxed for 1h, cooled,filtered, solvent boiled off, filtered again to get a brown oil wich cristallise to green crystals upon freezing. Perhaps the anion exchange failed and there is too much less reactive bromosafrole in the oil. The iodo was not tested by MS.

The moles of MeNH2 were not calculated. I said him to add 15+ eq. I suggested to dissolve MeNH2 HCl in 80% NaOH, bubble through anhydrous iPrOH (with a simple glasspipe with no gas dispersion device, due to the high solubility of MeNH2 in alcools I don't think it would be necessary), let the solution for one week over molecular sieve 3A before using, but he apparently has no sieve and no methylamine HCl...***


you fucked it when you refluxed it that simple. it decomposed
all you do is add to the ki in acetone and stir it happens pretty fast. ki should be finely grind.
add wasser until phase separation and extract with dcm.
wash dcm, evap dcm, done.

no suckback trap? you could suck back a lot of your solution and throw all you calculations off.

no gas dispersion fitting? probably bubbled the methylamine out of the ipa, believe me i know what i'm talking about.
gas dissolution is a function of *surface area*
there's a pretty good explanation why ammonylisis reactions are done in alcohols. in j. march's "advanced organic chemistry"
first it's a strong nuclephile so crowding by protons is'nt a big issue.
the reason you use dmf for sn2's as there are different kinds is that you have a weak nucleophile and the nucleophile contains an anion.
the anion is energized by the solvation effect and the nuclephile is promoted in energy in this manner.
sn1 sn2? i dunno man i'm just a dumbass.
 
re:

*1. The halohalide dissociates in a protic polar solvent generating a secondary carbocation R1R2HC+ and an iodide ion I-.
2. A methylamine molecule CH3NH2 attacks the carbocation to generate R1R2HC-N+H2CH3
3.A nucleophile such as the iodide ion I- comes along (could be other things) and grab that proton R1R2HC-NHCH3 + H-I

End of story. No basic protons or**

QD, that mechanism may occur if the concentration of the methylamine is low more likely the sn1 reaction will just produce an irreversable ether product.


man you's guys are working my noodle today.

Just quickly QD is right,

Nucleophilic substitution reaction with methylamine, on a secondary alkyl halide. Rate = k[RX] is unimolecular as QD stated S_N1 reaction, they will both be competing as usual but is classified S_N1 or pseudo first order kinetics if a very large excess of methylamine is present. So you just need to ask yourself, do you want fries with that ? (jk)

[edit; I have to expand a little on that answer]

quote from March p 390;

If a large excess of the nucleophile is present- for example, if it is the solvent- the mechanism may still be bimolecular, though the experimentally determined kinetics will be first order;

Rate = k[RX]

As previously mentioned (p.293), such kinetics are called pseudo-first order.

Measuring a second order reaction rate with reactants A and B can be problematic: the concentrations of the two reactants must be followed simultaneously, which is more difficult; or measure one of them and calculate the other as a difference, which is less precise. A common solution for that problem is the pseudo first order approximation
If either (A) or (B) remain constant as the reaction proceeds, then the reaction can be considered pseudo first order because in fact it only depends on the concentration of one reactant. If for example (B) remains constant then:

where k' = k(B)0 (k' or kobs with units s?1) and an expression is obtained identical to the first order expression above.
One way to obtain a pseudo first order reaction is to use a large excess of one of the reactants ((B)>>(A) would work for the previous example) so that, as the reaction progresses only a small amount of the reactant is consumed and its concentration can be considered to stay constant. By collecting k' for many reactions with different (but excess) concentrations of (B); a plot of k' versus (B) gives k (the regular second order rate constant) as the slope.
http://en.wikipedia.org/wiki/Rate_equation#Pseudo_first_order

Nice thread, Dr Methoxy!

well the theory of sn1 vs. sn2 is confirmed for Q.D. as per table 1

http://en.wikipedia.org/wiki/Nucleophilic_substitution

to sum it up on secondary alkyl halides the two different reactions compete.

**Ok, but did you totally dissolve the KI? How much solvent did you use for every g of bromosafrole ?  I know the exchange reaction from bromo to iodo is pretty rapid, even for sec. carbon halides (something like 10 min. IIRC).**

**If the reaction goes , like you said, via SN1 mechanism, perhaps jon made some 1-methylamino-1-phenylpropane as side product because this latter crystalises along with MDMA when gassed with HCl.**

possible indeed  but i did'nt have an nmr of it so i can't say for sure.



KI is soluble only 1g/100ml acetone so that is not a factor as it reacts very fast in solution with bromosafrole.
the only thing that matters is that the mixture can be stirred.
 10 ml of acetone per gram of bromosafrole is sufficient to obtain a mixture that can be stirred.


regarding the expiriment of reacting 5 moles of propylamine with bromo/iodo safrole i may has well have been talking out my arse there because i've found yeilds to be shit unless the ratio is 8-10 moles of methylamine to the alkyl halide but i was'nt taking into consideration the you were using a bulky amine.

i just indicated i did'nt know for lack of an nmr.
what i can say is this: they are competing reactions the extent to which they compete we do not know.
hmm looking at the table both rates are moderate.
for sn1 and sn2.
so assuming there is a 50/50 ratio of intermediates one of which is prone to rearrangements the variable would be to what extent does the rearrangement occur?

m.p. of the acetate was 100C because it was a hydrate.
or a hemihdrate.
i would like to also comment that the commonly used wacker oxidation gives rearrangemant products also.

the o2 system gives aldehydes i think the benzoquinone is less prone to that.
could explain your broad values. another is that it was racemates too.
i guess the guys getting the mdp2p from china have the sharpest melting points so to speak as they use the peracid process very convenient for them.

By simply doing HCl titrating to iPr-OH - MeNH2 solution we can determine how much MeNH2 out from naOH solution.
Another idea, start from HCl salt of MeNH2, and drop saturated KOH - iPr-OH (or MeOH?)  solution into that salt, by this way you can heat it to push MeNH2 flow into IPA without worry about water.

you can add Na to isopropanol, then neutralise it with methylamine HCl. Filter the NaCl and store it over sieve.

Dr.Methoxy:

You have an organic bromide. You want an organic iodide. So just get your bromo, dissolve it in acetone, add your KI (all of it, as is), stir like crazy, KBr precipitates, filter and rinse, evap, and as Gordon Ramsay would say, "Iodosafrole: done." No reflux, no dropping funnels, no bullshit. Just qualitative yield. It really is that simple.

As for the second step, take the others advice: use the hydrochloride salt, and add concentrated base. Using 40% is not adding a shitload of unnecessary water to the system - the Whole Point^TM is to keep the reaction anhydrous - it's also a waste of perfectly good 40% methylamine. You lucky bastard .

And SN₁, SN₂, E₂? Well, it's nice to think about those things - and it makes me feel better, because if you're talking reaction mechanisms, you probably aren't the average happy-go-lucky muppet who is going to blow himself up. But no less than a professor and head researcher of a university organic chemistry department once told me that working that shit out is what physical chemists are for... all she cares about is what works. One can take that the wrong way, of course, but I guess the point is that if you do the practical side of things right, then the molecules are still going to do their thing, whether you understand it or not. It's not bad advice.

V

Would Triethanolamine also work for neutralizing the MeHN2 as it is also a strong organic base? Would the OH groups result in water being formed?

that has been tried it does'nt work do i need to stamp it onto your forehead so when you fuck it up you can go read it in the mirror?
bubble dry methylamine in.

That is what I said...