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Topic: Amine and Heterocyclic reaction  (Read 12119 times)

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Offline victor

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Amine and Heterocyclic reaction
« on: July 16, 2006, 11:53:45 PM »
CH3CH2NH2 + CH3CO-O-COCH3 --> CH3CH2-NH-COCH3 + CH3COOH

Question: explain why only one amine hydrogen which is get substituted by acetyl group eventhough acetate anhydride is used excessively in the reaction??

My answer: I can think that, when the product is formed (N-ethylethanoylamide), its resonance structure prevent it from getting further substituted with another acetyl group...what do you think about that?

Oh, one more question..when aldehyde is reacted with primary amine, it will result in imine compound. But, when carboxylic acid is reacted with primary amine, amide will occur. Why? because both of them has carbonyl group and I think that the nucleophilic addition run by amine is nearly the same for both aldehyde and carboxylic acid, then why both compounds not result the same product??  ???
« Last Edit: July 19, 2006, 02:37:12 AM by victor »
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Offline wereworm73

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Re: amine reaction
« Reply #1 on: July 17, 2006, 01:17:27 AM »
Quote
Oh, one more question..when aldehyde is reacted with primary amine, it will result in imine compound. But, when carboxylic acid is reacted with primary amine, amide will occur. Why? because both of them has carbonyl group and I think that the nucleophilic addition run by amine is nearly the same for both aldehyde and carboxylic acid, then why both compounds not result the same product??

Unlike aldehydes, carboxylic acids can instead protonate the primary amine to form an ammonium carboxylate salt.  When you heat this salt, you get an amide and water.   
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Offline AWK

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Re: amine reaction
« Reply #2 on: July 17, 2006, 04:55:03 AM »
Concerning resonance - you are right. But you can also get diacyl amines through the sequence of reactions
R-NH2 + CH3COOC2H5 => RNH-COCH3
Amide treated with NaH or NaNH2 froms [RNCOCH3]-Na+
which after reaction with CH3COCl produces RN(COCH3)2
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Offline wereworm73

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Re: amine reaction
« Reply #3 on: July 17, 2006, 04:05:05 PM »
Also, if you mix aniline with nonaqueous acetic anhydride, the nitrogen often ends up with two acetyl groups instead of one.
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Offline victor

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Re: amine reaction
« Reply #4 on: July 17, 2006, 11:42:03 PM »
Ok, thanks for the answers..it's the question from my book (about amine and acetate anhydride)...so, it's stated that only one acetyl group which is attached to amine (considering there's no other catalysts).
Oh, another question...it's about metamphetamine synthesis...I really can't figure out the mechanism..  :(
it's said that you can synthesize metamphetamine by reacting 1-phenyl-2-propanone and methyl amine. the molecular formula for metamphetamine is:
[Ar-CH2-CH(CH3)NH2+CH3]Cl- --> (sorry, there's no chem structure drawer in this internet cafe..I hope there'll be one who have it.. ;D )
1-phenyl-2-propanone --> Ar-CH2COCH3
Methyl amine --> CH3NH2

What I can think that protonation on carbonyl group (in 1-phenyl-2-propanone) is done by N atome in Methyl amine and, because this is carbonyl group, it will form imine compound. Then, how to make the N atom become positively charged with Cl- attached in form of amonium salt??  ???
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Offline AWK

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Re: amine reaction
« Reply #5 on: July 18, 2006, 07:31:41 AM »
We do not discuss a synthesis of drugs on this forum!
From chemical point of view an aldehyde or ketone forms Schiff base with methylamine, then a double bond is reduced to a single one.
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Offline victor

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Re: Amine and Heterocyclic reaction
« Reply #6 on: July 19, 2006, 02:54:10 AM »
I don't mean to synthesize this kind of drug...I just got it from the question in organic chemistry textbook (if you want. I can give you the page of the question in that textbook) and I just try to answer it.. ;D
Yup, as I thought, it woul result in imine, but, reduction to a single bond really requires Cl- compound??because the product is like this..
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Offline AWK

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Re: Amine and Heterocyclic reaction
« Reply #7 on: July 19, 2006, 03:09:21 AM »
Reduction does not require chloride. Hydrochloride is formed after reduction followed by neutralization with HCl.
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Offline victor

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Re: Amine and Heterocyclic reaction
« Reply #8 on: July 19, 2006, 03:18:58 AM »
Hmm..ok, I got it..thanks.  ;D
Um, next question come from heterocyclics..it's about the structure of Oxazole and Pirole. Does Oxazole get more basic than pirole? I think it's because of there are more pairs of free electrons in Oxazole (in N and O atoms)..what do you think about that?
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Offline AWK

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Re: Amine and Heterocyclic reaction
« Reply #9 on: July 21, 2006, 06:45:29 AM »
Quote from: victor on July 19, 2006, 03:18:58 AM
Hmm..ok, I got it..thanks.  ;D
Um, next question come from heterocyclics..it's about the structure of Oxazole and Pirole. Does Oxazole get more basic than pirole? I think it's because of there are more pairs of free electrons in Oxazole (in N and O atoms)..what do you think about that?
In pirole, a nitrogen elctron pair is involved in aromaticity, hence cannot show basic properties

http://www.scripps.edu/chem/baran/heterocycles/lectures/EssentialsOfHeterocyclesI.pdf
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Offline victor

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Re: Amine and Heterocyclic reaction
« Reply #10 on: July 22, 2006, 01:06:42 AM »
thank you, thank you, thank you...you folks so kind..  ;D
next question, ( I think I posted too many questions here, hope none of you will get upset with this..  ;D ) 2-hydroxypyridineis usually in keto form, can you explain why? I get confuse because I get it compared with phenol which is mostly in enol form...and I can't find any proper explainations for this 2-hydroxypyridine..anyone can help me??

And then, about morphine isolation question..it's said that morphine solubility is 0.2 g/L but, the solubility of Morphine hydrochloride is 57 g/L. Then it asked how to extract and isolate morphine from opium by using the information stated above??
I really have no idea about this question and all the things that I can think of is to solute all of the opium in the water and add some HCl. after that,  use the fractination process and last, we can move the Cl to get morphine from it's water solution...but I need the reaction equation... ???
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Offline victor

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Re: Amine and Heterocyclic reaction
« Reply #11 on: July 23, 2006, 08:01:36 AM »
Those are questions from the book..they're really from my chemistry book (Organic Chemistry; a short course by Harold Hart and friends; 11th edition; chapter 13-heterocyclic compounds). I just wanna answer it, not really synthesize it...so please help me...  :'(
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Offline AWK

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Re: Amine and Heterocyclic reaction
« Reply #12 on: July 24, 2006, 09:24:59 AM »
Quote
2-hydroxypyridineis usually in keto form, can you explain why? I get confuse because I get it compared with phenol which is mostly in enol form...and I can't find any proper explainations for this 2-hydroxypyridine..anyone can help me??
http://en.wikipedia.org/wiki/User:Stone/2-Pyridone

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Offline victor

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Re: Amine and Heterocyclic reaction
« Reply #13 on: August 04, 2006, 12:21:17 AM »
After reding from it (wikipedia) I can conclude that 2-hydroxypyridine tends to be in keto form (2-pyridone) because carbonyl group in 2-pyridone is more stable while the ring resonance also keep maintained because of the free electron pairs in the O atom...
is that right?
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Offline Will

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Re: Amine and Heterocyclic reaction
« Reply #14 on: August 04, 2006, 11:20:31 AM »
Quote from: victor on August 04, 2006, 12:21:17 AM
After reading from it (wikipedia) I can conclude that 2-hydroxypyridine tends to be in keto form (2-pyridone) because carbonyl group in 2-pyridone is more stable while the ring resonance also keep maintained because of the free electron pairs in the O atom...
is that right?

Correct, delocalisation can lead to a resonance structure in which pyridinium-2-olate is aromatic.
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