I wonder if you want to deprotect the Cbz-group after the direct arylation? If you use catalytic hydrogenation there could be problem with the catalyst being poisoned by the sulphur in thiophene. Its a mystery to me why your professor dont want you to use the Boc-protected amine, you can buy that an the deprotection is just acid? If you want the put on the Cbz-group you can use two-phase-system like you suggest with water and organic solvent, the aminogroup reacts faster than water.
I wonder if you want to deprotect the Cbz-group after the direct arylation? If you use catalytic hydrogenation there could be problem with the catalyst being poisoned by the sulphur in thiophene. Its a mystery to me why your professor dont want you to use the Boc-protected amine, you can buy that an the deprotection is just acid? If you want the put on the Cbz-group you can use two-phase-system like you suggest with water and organic solvent, the aminogroup reacts faster than water.
I followed the following procedure: To a mixture of the crude SM (~10 g) in THF (~1.5L) was added 2N NaOH (168 mL) and benzyl chloroformate (30.2g, 176.8 mmol). After stirring 1 h, the biphasic mixture was separated and the aq phase was extracted with Ethylacetate(50 mL). The combined organic extracts were dried (MgSO4), filtered, and solvent removed by evaporation togive the alcohol. The material was purified by recrystallization from THF/cyclohexane to provide the pure product aswhite needles. [WO2003061598, page 23]
My professor asked to apply the procedure to the 3-amnithiophene oxalate salt instead of making the free amine first.
I did the following calculations based on the procedure I just mentioned. (1eq amine and 1,33eq Cbz-Cl)
To 0,2g of 3-aminothiophene oxalate salt in THF (30ml) was added 2M NaOH (4ml) and 0,68g Cbz-Cl. Stir for 1h at room temperature. The biphasic mixture was separated and the aq phase was extracted with EtOAc. The combined organics were dried (MgSO4), filtered, and concentrated to provide the crude product.
I applied this procedure 2 times on a small amount of the aminothiophene oxalate salt. The first time I got a white powder and the 2nd time I got brown/purple powder. I am not sure which one is the desired product, in the procedure they mention the final product as 'white needles'. I kind of assumed that the white powder is the desired product (not sure though).
The final product is supposed to dissolve in ethyl acetate because it's extracted with ethyl acetate from aq solution. I tried to do an analysis on the TLC but the final product did not dissolve in any of the following solvent (THF, ethyl acetate, chloroform, acetone).
Do you guys have an idea which colour the final product would have (it's supposed to be Cbz-Cl protected 3-aminothiopehen). And would the procedure even work for the aminothiophene oxalate salt insetad of aminothiophene?
I followed the following procedure: To a mixture of the crude SM (~10 g) in THF (~1.5L) was added 2N NaOH (168 mL) and benzyl chloroformate (30.2g, 176.8 mmol). After stirring 1 h, the biphasic mixture was separated and the aq phase was extracted with Ethylacetate(50 mL). The combined organic extracts were dried (MgSO4), filtered, and solvent removed by evaporation togive the alcohol. The material was purified by recrystallization from THF/cyclohexane to provide the pure product aswhite needles. [WO2003061598, page 23]
My professor asked to apply the procedure to the 3-amnithiophene oxalate salt instead of making the free amine first.
I did the following calculations based on the procedure I just mentioned. (1eq amine and 1,33eq Cbz-Cl)
To 0,2g of 3-aminothiophene oxalate salt in THF (30ml) was added 2M NaOH (4ml) and 0,68g Cbz-Cl. Stir for 1h at room temperature. The biphasic mixture was separated and the aq phase was extracted with EtOAc. The combined organics were dried (MgSO4), filtered, and concentrated to provide the crude product.
I applied this procedure 2 times on a small amount of the aminothiophene oxalate salt. The first time I got a white powder and the 2nd time I got brown/purple powder. I am not sure which one is the desired product, in the procedure they mention the final product as 'white needles'. I kind of assumed that the white powder is the desired product (not sure though).
The final product is supposed to dissolve in ethyl acetate because it's extracted with ethyl acetate from aq solution. I tried to do an analysis on the TLC but the final product did not dissolve in any of the following solvent (THF, ethyl acetate, chloroform, acetone).
Do you guys have an idea which colour the final product would have (it's supposed to be Cbz-Cl protected 3-aminothiopehen). And would the procedure even work for the aminothiophene oxalate salt insetad of aminothiophene?
Cyclodextrin does some funny things by bringing molecules into good molecular conformations to do reactions. If you want to know more, follow the reference chain within that picture.
Cyclodextrin does some funny things by bringing molecules into good molecular conformations to do reactions. If you want to know more, follow the reference chain within that picture.
Sach, I don't mean this in a mean way, but do you have any idea what you are doing? Not breaking this down by mols or equivalents is a pretty serious red flag. As is trying to put a chloroformate in neat alkaline water. These mistakes tell me you would probably be best served observing a mentor work for a bit while you learn the basics. I followed someone around and had them check my stuff for many months before seriously attempting real reactions on my own.
Organic synthesis is dangerous, in often unexpected ways. Would you care to share your background at all?
Sach, I don't mean this in a mean way, but do you have any idea what you are doing? Not breaking this down by mols or equivalents is a pretty serious red flag. As is trying to put a chloroformate in neat alkaline water. These mistakes tell me you would probably be best served observing a mentor work for a bit while you learn the basics. I followed someone around and had them check my stuff for many months before seriously attempting real reactions on my own.
Organic synthesis is dangerous, in often unexpected ways. Would you care to share your background at all?
Thank you everyone for the advice. Using Boc protection was also my initial idea but the professor told me that Boc won't be stable in the conditions of our reaction. I also checked it in the book 'protective groups in organic synthesis' and Boc will not be stable in our reaction conditions. I have done my bachelors in applied engineering sciences. I am relatively new to organic synthesis as I did take a minor chemistry but we never did a lot of lab work during the course. But suggesting grams instead of equivalents is still a big mistake about which I should have thought.
Thank you everyone for the advice. Using Boc protection was also my initial idea but the professor told me that Boc won't be stable in the conditions of our reaction. I also checked it in the book 'protective groups in organic synthesis' and Boc will not be stable in our reaction conditions. I have done my bachelors in applied engineering sciences. I am relatively new to organic synthesis as I did take a minor chemistry but we never did a lot of lab work during the course. But suggesting grams instead of equivalents is still a big mistake about which I should have thought.
The reaction that takes place is the completely normal reaction of an acid chloride with an amine to made an amide (in this case its a chloroformate reacting to make a carbamate, but they are quite similar in reactivity to acid chlorides).
Here is an arrow pushing example (http://www.chemguide.co.uk/mechanisms/addelim/amines.html). Primary amines are quite nucleophilic, so the carbonyl of the Cbz-Cl doesn't need to be protonated first like in a Fischer esterification. In fact, you can't do amidation very well under acidic conditions because the amine will be protonated before the carbonyl, and it won't have a nucleophilic lone pair to attack the carbonyl. In the case of a chloroformate like this, its so reactive you might get some conversion, but under aqueous conditions hydrolysis is more likely.
You'll notice that your reaction conditions are very basic, so the carbonyl can't be protonated under this condition anyways.. I'm actually surprised the Cbz protection is done in aqueous NaOH. Typically acid chloride amidations are done in anhydrous polar solvents with tertiary amines, pyridine derivatives, or carbonate salts as proton scavengers.
Lastly, carbamates are sensitive to acid, thats how you deprotect them, so you wouldn't do this under acidic conditions anyways.
The reaction that takes place is the completely normal reaction of an acid chloride with an amine to made an amide (in this case its a chloroformate reacting to make a carbamate, but they are quite similar in reactivity to acid chlorides).
Here is an arrow pushing example (http://www.chemguide.co.uk/mechanisms/addelim/amines.html). Primary amines are quite nucleophilic, so the carbonyl of the Cbz-Cl doesn't need to be protonated first like in a Fischer esterification. In fact, you can't do amidation very well under acidic conditions because the amine will be protonated before the carbonyl, and it won't have a nucleophilic lone pair to attack the carbonyl. In the case of a chloroformate like this, its so reactive you might get some conversion, but under aqueous conditions hydrolysis is more likely.
You'll notice that your reaction conditions are very basic, so the carbonyl can't be protonated under this condition anyways.. I'm actually surprised the Cbz protection is done in aqueous NaOH. Typically acid chloride amidations are done in anhydrous polar solvents with tertiary amines, pyridine derivatives, or carbonate salts as proton scavengers.
Lastly, carbamates are sensitive to acid, thats how you deprotect them, so you wouldn't do this under acidic conditions anyways.
Well, grams are not suitable for this stuff because the compounds have different molar weights so you need to think in moles insted of grams.
Also I dont think your conditions will work: a) I did try making amides with K2CO3 in refluxing toluene and got barely any product. In my experience you need to run it in organic solvent with amine base like triethylamine. Usualy room temperature or even cooling works realy well b) your starting materials are IMO hardly soluble in water c) benzyl chloropfoormate would most likely decompose in aquaeous alkaline conditions
Well, grams are not suitable for this stuff because the compounds have different molar weights so you need to think in moles insted of grams.
Also I dont think your conditions will work: a) I did try making amides with K2CO3 in refluxing toluene and got barely any product. In my experience you need to run it in organic solvent with amine base like triethylamine. Usualy room temperature or even cooling works realy well b) your starting materials are IMO hardly soluble in water c) benzyl chloropfoormate would most likely decompose in aquaeous alkaline conditions
I found some very useful procedures for Cbz-Cl (benzyl carbamate) protection of amines.(http://www.commonorganicchemistry.com/Rxn_Pages/Cbz_Protection/Cbz_Protection_inorganic_base.htm)
You can also find one procedure in the article in the attachment. What I don't completely understand is the mechanism by which the general Cbz-Cl protection of amines takes place. I was thinking that maybe Cbz-Cl gets protonated and reacts with the amine followed by probably the elimination of HCL. But I am not sure really sure about the mechanism. Can someone explain it a bit. Thank you in advance
I found some very useful procedures for Cbz-Cl (benzyl carbamate) protection of amines.(http://www.commonorganicchemistry.com/Rxn_Pages/Cbz_Protection/Cbz_Protection_inorganic_base.htm)
You can also find one procedure in the article in the attachment. What I don't completely understand is the mechanism by which the general Cbz-Cl protection of amines takes place. I was thinking that maybe Cbz-Cl gets protonated and reacts with the amine followed by probably the elimination of HCL. But I am not sure really sure about the mechanism. Can someone explain it a bit. Thank you in advance
I think you can have problems de-protect with catalytic hydrogenation because of the sulfur in your thiophene poisoning the palladium-catalyst. If so you can try de-protection with boiling acid or base. I have written here about this earlier so I repeat myself. The protection of the amine is the easiest step and this will likely go well.
I think you can have problems de-protect with catalytic hydrogenation because of the sulfur in your thiophene poisoning the palladium-catalyst. If so you can try de-protection with boiling acid or base. I have written here about this earlier so I repeat myself. The protection of the amine is the easiest step and this will likely go well.
In the following article, they dissolve catalytic amount of cyclodextrine in water, followed by the addition of amine. Then they add Cbz-Cl and they seem to have very good yields. I don't completely understand how catalytic amounts of B-cyclodextrine dissolved in water would help.
In the following article, they dissolve catalytic amount of cyclodextrine in water, followed by the addition of amine. Then they add Cbz-Cl and they seem to have very good yields. I don't completely understand how catalytic amounts of B-cyclodextrine dissolved in water would help.
If you want the put on the Cbz-group you can use two-phase-system like you suggest with water and organic solvent, the aminogroup reacts faster than water.
If you want the put on the Cbz-group you can use two-phase-system like you suggest with water and organic solvent, the aminogroup reacts faster than water.
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To a mixture of the crude SM (~10 g) in THF (~1.5L) was added 2N NaOH (168 mL) and benzyl chloroformate (30.2g, 176.8 mmol). After stirring 1 h, the biphasic mixture was separated and the aq phase was extracted with Ethylacetate(50 mL). The combined organic extracts were dried (MgSO4), filtered, and solvent removed by evaporation togive the alcohol. The material was purified by recrystallization from THF/cyclohexane to provide the pure product aswhite needles. [WO2003061598, page 23]
My professor asked to apply the procedure to the 3-amnithiophene oxalate salt instead of making the free amine first.
I did the following calculations based on the procedure I just mentioned. (1eq amine and 1,33eq Cbz-Cl)
To 0,2g of 3-aminothiophene oxalate salt in THF (30ml) was added 2M NaOH (4ml) and 0,68g Cbz-Cl. Stir for 1h at room temperature. The biphasic mixture was separated and the aq phase was extracted with EtOAc. The combined organics were dried (MgSO4), filtered, and concentrated to provide the crude product.
I applied this procedure 2 times on a small amount of the aminothiophene oxalate salt. The first time I got a white powder and the 2nd time I got brown/purple powder. I am not sure which one is the desired product, in the procedure they mention the final product as 'white needles'. I kind of assumed that the white powder is the desired product (not sure though).
The final product is supposed to dissolve in ethyl acetate because it's extracted with ethyl acetate from aq solution. I tried to do an analysis on the TLC but the final product did not dissolve in any of the following solvent (THF, ethyl acetate, chloroform, acetone).
Do you guys have an idea which colour the final product would have (it's supposed to be Cbz-Cl protected 3-aminothiopehen). And would the procedure even work for the aminothiophene oxalate salt insetad of aminothiophene?
Thank you in advance
To a mixture of the crude SM (~10 g) in THF (~1.5L) was added 2N NaOH (168 mL) and benzyl chloroformate (30.2g, 176.8 mmol). After stirring 1 h, the biphasic mixture was separated and the aq phase was extracted with Ethylacetate(50 mL). The combined organic extracts were dried (MgSO4), filtered, and solvent removed by evaporation togive the alcohol. The material was purified by recrystallization from THF/cyclohexane to provide the pure product aswhite needles. [WO2003061598, page 23]
My professor asked to apply the procedure to the 3-amnithiophene oxalate salt instead of making the free amine first.
I did the following calculations based on the procedure I just mentioned. (1eq amine and 1,33eq Cbz-Cl)
To 0,2g of 3-aminothiophene oxalate salt in THF (30ml) was added 2M NaOH (4ml) and 0,68g Cbz-Cl. Stir for 1h at room temperature. The biphasic mixture was separated and the aq phase was extracted with EtOAc. The combined organics were dried (MgSO4), filtered, and concentrated to provide the crude product.
I applied this procedure 2 times on a small amount of the aminothiophene oxalate salt. The first time I got a white powder and the 2nd time I got brown/purple powder. I am not sure which one is the desired product, in the procedure they mention the final product as 'white needles'. I kind of assumed that the white powder is the desired product (not sure though).
The final product is supposed to dissolve in ethyl acetate because it's extracted with ethyl acetate from aq solution. I tried to do an analysis on the TLC but the final product did not dissolve in any of the following solvent (THF, ethyl acetate, chloroform, acetone).
Do you guys have an idea which colour the final product would have (it's supposed to be Cbz-Cl protected 3-aminothiopehen). And would the procedure even work for the aminothiophene oxalate salt insetad of aminothiophene?
Thank you in advance
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Organic synthesis is dangerous, in often unexpected ways. Would you care to share your background at all?
Organic synthesis is dangerous, in often unexpected ways. Would you care to share your background at all?
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Using Boc protection was also my initial idea but the professor told me that Boc won't be stable in the conditions of our reaction. I also checked it in the book 'protective groups in organic synthesis' and Boc will not be stable in our reaction conditions.
I have done my bachelors in applied engineering sciences. I am relatively new to organic synthesis as I did take a minor chemistry but we never did a lot of lab work during the course. But suggesting grams instead of equivalents is still a big mistake about which I should have thought.
Using Boc protection was also my initial idea but the professor told me that Boc won't be stable in the conditions of our reaction. I also checked it in the book 'protective groups in organic synthesis' and Boc will not be stable in our reaction conditions.
I have done my bachelors in applied engineering sciences. I am relatively new to organic synthesis as I did take a minor chemistry but we never did a lot of lab work during the course. But suggesting grams instead of equivalents is still a big mistake about which I should have thought.
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Here is an arrow pushing example (http://www.chemguide.co.uk/mechanisms/addelim/amines.html). Primary amines are quite nucleophilic, so the carbonyl of the Cbz-Cl doesn't need to be protonated first like in a Fischer esterification. In fact, you can't do amidation very well under acidic conditions because the amine will be protonated before the carbonyl, and it won't have a nucleophilic lone pair to attack the carbonyl. In the case of a chloroformate like this, its so reactive you might get some conversion, but under aqueous conditions hydrolysis is more likely.
You'll notice that your reaction conditions are very basic, so the carbonyl can't be protonated under this condition anyways.. I'm actually surprised the Cbz protection is done in aqueous NaOH. Typically acid chloride amidations are done in anhydrous polar solvents with tertiary amines, pyridine derivatives, or carbonate salts as proton scavengers.
Lastly, carbamates are sensitive to acid, thats how you deprotect them, so you wouldn't do this under acidic conditions anyways.
Here is an arrow pushing example (http://www.chemguide.co.uk/mechanisms/addelim/amines.html). Primary amines are quite nucleophilic, so the carbonyl of the Cbz-Cl doesn't need to be protonated first like in a Fischer esterification. In fact, you can't do amidation very well under acidic conditions because the amine will be protonated before the carbonyl, and it won't have a nucleophilic lone pair to attack the carbonyl. In the case of a chloroformate like this, its so reactive you might get some conversion, but under aqueous conditions hydrolysis is more likely.
You'll notice that your reaction conditions are very basic, so the carbonyl can't be protonated under this condition anyways.. I'm actually surprised the Cbz protection is done in aqueous NaOH. Typically acid chloride amidations are done in anhydrous polar solvents with tertiary amines, pyridine derivatives, or carbonate salts as proton scavengers.
Lastly, carbamates are sensitive to acid, thats how you deprotect them, so you wouldn't do this under acidic conditions anyways.
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Also I dont think your conditions will work:
a) I did try making amides with K2CO3 in refluxing toluene and got barely any product. In my experience you need to run it in organic solvent with amine base like triethylamine. Usualy room temperature or even cooling works realy well
b) your starting materials are IMO hardly soluble in water
c) benzyl chloropfoormate would most likely decompose in aquaeous alkaline conditions
Also I dont think your conditions will work:
a) I did try making amides with K2CO3 in refluxing toluene and got barely any product. In my experience you need to run it in organic solvent with amine base like triethylamine. Usualy room temperature or even cooling works realy well
b) your starting materials are IMO hardly soluble in water
c) benzyl chloropfoormate would most likely decompose in aquaeous alkaline conditions
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You can also find one procedure in the article in the attachment. What I don't completely understand is the mechanism by which the general Cbz-Cl protection of amines takes place. I was thinking that maybe Cbz-Cl gets protonated and reacts with the amine followed by probably the elimination of HCL. But I am not sure really sure about the mechanism. Can someone explain it a bit.
Thank you in advance
You can also find one procedure in the article in the attachment. What I don't completely understand is the mechanism by which the general Cbz-Cl protection of amines takes place. I was thinking that maybe Cbz-Cl gets protonated and reacts with the amine followed by probably the elimination of HCL. But I am not sure really sure about the mechanism. Can someone explain it a bit.
Thank you in advance
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I have written here about this earlier so I repeat myself.
The protection of the amine is the easiest step and this will likely go well.
I have written here about this earlier so I repeat myself.
The protection of the amine is the easiest step and this will likely go well.
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Thank you very much
Thank you very much
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