i looked at the beginning of the page of the synthesis from
5-methoxytryptamine
he used diisopropylethyl amine
from what i see there no usage in that synthesis
the 2-iodopropane react with the 5-methoxytryptamine
for 2 mole of 2-iodopropane
and 1 mole of 5-methoxytryptamine
to give hydrogen iodide and
5-methoxy-N,N-diisopropyltryptamine
in the synthesis he use 0.0158 moles of 5-methoxytryptamine
and 0.0629 moles of 2-iodopropane (10.7 grams)
since he only need 0.0316 moles of 2-iodopropane
(5.37 grams)
so he doubled the quantity of 2-iodopropane
and wasting it
why?497 - 2-7-2011 at 16:05
Its called using a stoiciometric excess.. So the reaction goes further to completion. If you don't understand this concept, you may want to read some
other books to level up your knowledge a ways before going and working out the details of shulgin's reactions in TIKHAL...garry21 - 2-7-2011 at 17:16
hi
hum i just looked a little bit more and maybe the diisopropylethyl amine is used to protect the NH groupe on the indole
because the 2-iodopropan can react with it
i am not too sure about that
still i don't understan too much about the utility of
diisopropylethyl amine in that synthesis
[Edited on 3-7-2011 by garry21]UnintentionalChaos - 2-7-2011 at 18:29
Diisopropylethylamine, Hunig's base, is a tertiary amine resistant/immune to quaternization, as would happen if you used say, triethylamine along with
the MeI forming triethylmethylammonium iodide. Having a base removes the HI byproduct by reacting with it, otherwise the product ends up protonated,
thus losing it's nucleophilicity and halting reaction. The Hunig's base also protects other amines from quaternization. Without it, some of your
product would undoubtedly be turned into quaternary ammonium salt.
dpt
garry21 - 2-7-2011 at 19:22
in a closely related synthesis DPT Synthesis
he use isopropanol instead of sulfolane
in the first synthesis he mentioned there a large residual of sulfolane soo isopropanol would be better to use as solvent ?
easyer to find and less likely toxic
even if the hydrogen iodide react with it to produce
2-iodopropane
Quote:
The most cited path to alkyl iodides include the typical halogenation using a 57% HI solution and 2-propanol,
its even a product used in the synthesis
so 1 that reduct the free hydrogen iodide in the solution .. 2 its create a buffer like what you talked about stoichiometric excess
we can also noted he used 5.1 g of propyl iodide
instead of the 10.3 gram .. so they perfectioned his method
but he use 1.6 g of tryptamine insted of 3 g
but the yield is better
he used isopropanol as solvent to make DPT
so the isopropanol may form isopropyl iodide and and contaminate the product with DIPT
[Edited on 3-7-2011 by garry21]Nicodem - 3-7-2011 at 00:59
The Hunig's base also protects other amines from quaternization. Without it, some of your product would undoubtedly be turned into quaternary ammonium
salt.
Not in this case where the product, an N,N-diisopropylamine, can not form quaternary ammonium salts.
But otherwise, yes, the pKa value of Hunig's base in most solvents does play a role in partial in situ protection from N-alkylation of
secondary and tertiary amines which are generally slightly more basic than primary amines.
in a closely related synthesis DPT Synthesis
he use isopropanol instead of sulfolane
in the first synthesis he mentioned there a large residual of sulfolane soo isopropanol would be better to use as solvent ?
easyer to find and less likely toxic
I truly don't know what to think when a total beginner, who never bothered investing even a single second into reading about something as basic as the
SN2 mechanism, starts playing smart and even goes as far as comparing n-propylations with i-propylations. Complaining about the excess isopropyl
iodide used, and the use of sulfolane, in the only ever documented diisopropylation of an aliphatic amine is ridiculous! If you think you can do
better, then do so or shut up!
Quote:
even if the hydrogen iodide react with it to produce
2-iodopropane
Quote:
The most cited path to alkyl iodides include the typical halogenation using a 57% HI solution and 2-propanol,
its even a product used in the synthesis
so 1 that reduct the free hydrogen iodide in the solution .. 2 its create a buffer like what you talked about stoichiometric excess
we can also noted he used 5.1 g of propyl iodide
instead of the 10.3 gram .. so they perfectioned his method
but he use 1.6 g of tryptamine insted of 3 g
but the yield is better
he used isopropanol as solvent to make DPT
so the isopropanol may form isopropyl iodide and and contaminate the product with DIPT
Your level of ignorance is only surpassed by your level of arrogance. Both personal characteristics can only combine into self-confidence that
ignorance is a bliss.
In any case, it hurts my eyes when reading your posts.
Take the advice 497 gave you.
[Edited on 3/7/2011 by Nicodem]UnintentionalChaos - 3-7-2011 at 08:21
The Hunig's base also protects other amines from quaternization. Without it, some of your product would undoubtedly be turned into quaternary ammonium
salt.
Not in this case where the product, an N,N-diisopropylamine, can not form quaternary ammonium salts.
Derp. For some reason I thought we were doing N,N-dimethyl. That's what I get for not reading.
