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Author: Subject: Organic halides with piperdine
chemrox
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[*] posted on 12-12-2014 at 18:36
Organic halides with piperdine


Would the yield of the reaction of benzyl-Br with piperidine be improved if run with base? I ask because with a primary amine HBr would be released but I don't see this happening with a secondary amine. I was thinking of running it over KCO3 but this might be an unnecessary complication.

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[Edited on 13-12-2014 by chemrox]




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[*] posted on 15-12-2014 at 13:32


Dunno. The authors seem to be indicating that the results are wildly successful without modifications.
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DJF90
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[*] posted on 15-12-2014 at 14:28


Reaction of benzyl bromide with piperidine will give the N-benzylpiperidinium bromide salt. However, in reality, the piperidine is (probably?) a stronger base, giving piperidinium bromide and N-benzylpiperidine, which (being the charge neutral species) will underdo further alkylation to give dibenzylpiperidinium bromide (quaternary ammonium salt). In reality, this is probably not a preparative method. Use of K2CO3 will likely do very little to change the distribution as piperidine is a stronger base. Use of,excess piperidine may help, but in reality you're much better off going with reductive amination using benzaldehyde. A slightly more einded approach would be to form the benzamide (by any which means) and reduce with an appropriate reagent (LiAlH4 works, but BH3 is better for amide reduction due to better matching its natural electronics).
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[*] posted on 16-12-2014 at 02:50


Quote: Originally posted by DJF90  
However, in reality, the piperidine is (probably?) a stronger base

it is
pKb of piperdine is 2.89 whereas K2CO3 is 3.66
Quote:
Use of,excess piperidine may help

what you could do is take a beaker full of piperidine and drip in benzyl bromide very slowly
this method is used while performing a cross aldol between two aldehydes,one of which does not have alpha hydrogen, as you dont want the aldehyde with the alpha hydrogen to self aldol

or maybe you could do this :)

make an unsaturated piperidine ring ,kind of like pyridine but only one double bond ,so that it is a tertiary amine

you could make this ,starting from lysine
do strecker degradation on lysine with bleach to get 5-aminopentanal
heat this to cyclise the straight chain into an unsaturated heterocyclic ring

http://www.amazon.com/BulkSupplements-Pure-L-Lysine-Powder-g...

strecker degradation http://en.wikipedia.org/wiki/Strecker_degradation

then treat this with benzylbromide to get a quat salt
then reduce using formic acid
http://en.wikipedia.org/wiki/Eschweiler%E2%80%93Clarke_react...

but the cyclic imine might hydrolyse :(

[Edited on 16-12-2014 by CuReUS]
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DJF90
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[*] posted on 16-12-2014 at 04:25


Thanks for the pKb values; I was on my mobile device when I wrote that and did not have the appropriate data to hand.
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Dr.Bob
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[*] posted on 17-12-2014 at 07:20


I have alkylated piperidines successfully before. It is too sterically hindered to form the dibenzyl quat salt, for the most part. If you just dissolve it in acetone, put it in one equil. of K2CO3, and then a 1.0 - 1.1 eq of the benzylbromide, you can isolate mostly the desired product, although for that type of reaction, chromatography is often required. If you were using MeI, you might get significant dialkylation, but not as much with a larger group.

But for amines, the best way to make them is to find a different route, like DJF posted that avoids alkylation, or alkylated an amide or similar group. I like to alkylate a Boc protected amine, as those are easy to control, then remove the Boc to get the free amine. But amine chemistry is highly dependent on the size of the alkyl groups. I have even seen cases where TEA reacted with reagents, so even bulk can be overcome with enough force (high temps, harsh reagents, long times). But for simple benzylpiperidine, a simple alkylation would likely suffice.
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[*] posted on 17-12-2014 at 08:40


Quote: Originally posted by Dr.Bob  
piperidines... It is too sterically hindered to form the dibenzyl quat salt, for the most part

i didn't say that dibenzyl quat salt would form.the amine is already tertiary ,so (if the reaction works at all) benzyl quat salt would form
Quote:
If you just dissolve it in acetone, put it in one equil. of K2CO3, and then a 1.0 - 1.1 eq of the benzylbromide, you can isolate mostly the desired product

as DFJ90 said,putting the carbonate will be of no use as piperidine is a stronger base than carbonate
Quote:
although for that type of reaction, chromatography is often required.

please tell more about this technique,it sounds interesting
Quote:
I have even seen cases where TEA reacted with reagents, so even bulk can be overcome with enough force (high temps, harsh reagents, long times).

IIRC TEA is used in the place for pyridine to remove any acid formed
if your reagent has a higher pKb than TEA,then TEA would definitely react :o

could someone tell the pKb of TEA
different sources have different values ,maybe its due to the solvent in which the measurement was done :(

http://en.wikipedia.org/wiki/Triethylamine wiki-3.25

http://employees.oneonta.edu/knauerbr/226lects/226_amines_le... -2.99

https://answers.yahoo.com/question/index?qid=20140105085740A... -11.01 :o
maybe he confused it with pKa

[Edited on 17-12-2014 by CuReUS]
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zed
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[*] posted on 17-12-2014 at 11:19


I'm thinking Dr. Bob might be on the money. Good chance steric hindrance is your friend.

Could theoretically form your quaternary salt, but it is already pretty crowded in the neighborhood before that would have an opportunity to happen.
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chemrox
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[*] posted on 17-12-2014 at 13:41


Quote: Originally posted by zed  
Dunno. The authors seem to be indicating that the results are wildly successful without modifications.

True: will try as published first. Small quantities anyway. 14/20 apparatus




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[*] posted on 17-12-2014 at 19:11


"IIRC TEA is used in the place for pyridine to remove any acid formed"

TEA, pyridine or other bases will work fine. No base is needed for the reaction, most people use base to try to keep the HBr and salts formed from causing solubility issues, and to speed it up and/or drive it to completion. Also easier to extract the final product into organics if not protonated.

"as DFJ90 said,putting the carbonate will be of no use as piperidine is a stronger base than carbonate"

You guys think too fast and not long and slowly. The carbonate and piperidine and their protonated forms are all in a complex equilibrium. But when HBr and K2CO3 are in solution together, they will eventually form KBr and CO2 gas, which will bubble out of the flask. So given enough time, the K2CO3 WILL react with the HBr to form KBr and remove the acid from the solution for the most part, leaving the un-protonated benzylpiperidine. That is why so many reactions use carbonate as a base, it is eventually a non-reversible reaction as the CO2 goes away, driving the equilibrium to the product and KBr. Similar to removing a Boc group with acid, once the CO2 is released, the reaction will NOT go backwards, thus driving the reaction forward. That is why using only the acidity or basicity will not always tell you the correct answer. I certainly agree that the piperidine can react with Benzyl bromide itself, but unless excess piperidine is used (as a base) the reaction will become very slow and not go to completion nearly as quick, as much of the piperidine will be protonated, which will slow down the reaction rate. If you have lots of piperidine, that is a fine way to run the reaction and reduce the dialkylation rate.




[Edited on 18-12-2014 by Dr.Bob]
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[*] posted on 18-12-2014 at 00:25


Quote: Originally posted by zed  
I'm thinking Dr. Bob might be on the money. Good chance steric hindrance is your friend.

Could theoretically form your quaternary salt, but it is already pretty crowded in the neighborhood before that would have an opportunity to happen.


please tell how it is crowded,or what is crowding it
i think you might have misunderstood what i am saying
how do i post the structures of the reagents ,and the product that i am talking about.will chemsketch work :(

see these syntheses ;)

1.minoxidil http://drugsynthesis.blogspot.in/2012/04/laboratory-synthesi...

2.resperidole
https://www.google.co.in/search?q=risperidone+synthesis&...

3.thioridazine
http://www.reactionoftheday.com/2012/07/752012-synthesis-of-...

in the first drug,they treat the ether with piperdine to get the drug.So could benzylbromide be treated with phenol to remove all the HBr formed and then treated with piperdine

in the second drug ,they use Na2CO3/KI in the last step
but i looked up the solubilities and here is what i found

benzylbromide-insoluble in water,soluble in ethanol,benzene
piperdine-miscible in water
k2CO3-soluble in water,insoluble in ethanol

doesnt that mean that if you put all three in a beaker ,benzylbromide and piperdine would be in one layer and the carbonate will be happily enjoying in its own layer ?

for the 3rd drug,they have used soda amide to deprotonate the N
is it really necessary ?

[Edited on 18-12-2014 by CuReUS] in reply to forgotten password

very sorry ,yes he is right :)
i thought he was talking about the synthesis i proposed

[Edited on 18-12-2014 by CuReUS]
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[*] posted on 18-12-2014 at 00:35


He's saying that the nitrogen in N-benzylpiperidine is crowded to the extent that it will not react with benzyl bromide to give the quaternary salt. And he's right.
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[*] posted on 18-12-2014 at 11:55


Yes, exactly. Piperidine can react once with benzyl bromide to form N-benzylpiperidine. That product will be largely stable and not react further with benzyl bromide.

In contrast, if you took piperidine (or almost any small chain alkyl substituted secondary amine) and reacted it with methyl iodide, you are likely to get a mixture containing a lot of N,N-dimethylpiperidine, as the bromide salt (N,N-dimethylpiperidinium bromide), which is known as a quaternary salt. The exact mix is dependent on how much excess alkylating reagent is used, the solvent, temperture, reaction time, etc. But with 2+ equivalents, it is the main product.

If you clean your bathroom with Lysol Tub and Tile cleaner, the main ingredient is a quat salt which consists of a nitrogen with 2 long chain (12-14 carbon atoms) alkyl groups, and two methyl groups (can also be ethyl or benzyls, but harder to make those). The alkyl groups come from stearic acid (from beef fat, lard, grease, etc) or similar sources, which is why C12-14 is the most common one, as it is the main constituent of most animal fats.
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[*] posted on 19-12-2014 at 05:46


Quote: Originally posted by Dr.Bob  

If you clean your bathroom with Lysol Tub and Tile cleaner, the main ingredient is a quat salt which consists of a nitrogen with 2 long chain (12-14 carbon atoms) alkyl groups, and two methyl groups (can also be ethyl or benzyls, but harder to make those). The alkyl groups come from stearic acid (from beef fat, lard, grease, etc) or similar sources, which is why C12-14 is the most common one, as it is the main constituent of most animal fats.


its called benzalkonium chloride
http://en.wikipedia.org/wiki/Benzalkonium_chloride

it is found in most of the toilet cleaners like harpic,lysol etc
they use is as an alternative to HCl ,because if you carelessly mixed harpic with,say bleach,it would form Cl2 which could kill you

but the main drawback of benzalkonium chloride is that it can't kill pseudomonas :(
http://en.wikipedia.org/wiki/Pseudomonas_aeruginosa
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Chemosynthesis
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[*] posted on 19-12-2014 at 11:35


Quat salts including benzyalkonium chloride (alkyl dimethyl benzyl ammonium chloride/ADBAC) do/can kill P. aeruginosa, though it can become resistant. There is evidence that resistance decreases virulence.

Appl Environ Microbiol. Aug 1989; 55(8): 2036–2040.
J Clin Microbiol. Feb 1975; 1(2): 175–179.
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[*] posted on 23-12-2014 at 20:49


I'm planning a reaction of a substrate with piperidine:



I really don't want a quat salt forming, but that would be very difficult, right? The carbon on which the reaction is going on looks quite clustered to me.

Furthermore, I thought that with the given conditions, an SN2 mechanism would predominantly occur. I purposefully made MeCN a solvent because it is a polar aprotic solvent that is not too hard to remove or evaporate. However, a grad student at the lab I intern in said that in practice SN2 reactions with acetonitrile as a solvent are very slow. Is this true? At first I assumed that diisopropylethylamine (Hunig's base) would suck up the tosylate anion to form the corresponding tosylate salt... But I just looked up the pKb value (it's 3.02) which means that piperidine is actually a stronger base. Will this mean that my product will be dissolved as a tosylate salt in acetonitrile?
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[*] posted on 24-12-2014 at 12:38


Do you have to start from that compound? It would be a simple reductive am starting from 2-phenylacetophenone which you would do in toluene with a dean-stark or in DCM with 4A sieves.



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