Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rutherford on August 10, 2012, 09:21:03 AM
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1.I need to assign the following pka :rarrow: 0.8 and 4.6 to the following compounds: (C6H5)2NH+ and C6H5NH3+, I know that the ring delocalisation affects the electrons in the C-N bond (they are near C), so the N-H bond gets polarized more and the proton can be easier released, but there are two rings in one of the compounds and I am confused now. How could two rings affect the electrons in the C-N bond?
2.Just a quick one that bothers me: How to get this compound http://imageshack.us/photo/my-images/98/97005048.gif/ from butan-1,4-diol?
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1.I need to assign the following pka :rarrow: 0.8 and 4.6 to the following compounds: (C6H5)2NH+ and C6H5NH3+, I know that the ring delocalisation affects the electrons in the C-N bond (they are near C), so the N-H bond gets polarized more and the proton can be easier released, but there are two rings in one of the compounds and I am confused now. How could two rings affect the electrons in the C-N bond?
2.Just a quick one that bothers me: How to get this compound http://imageshack.us/photo/my-images/98/97005048.gif/ from butan-1,4-diol?
Question 2: Protect, oxidise and make lactone.
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For 1. see for example: http://www2.fiu.edu/~herriott/ch20-amines-part1.pdf
If one electron withdrawing group makes the N-H bond weaker, what do you think the effect of adding a second electron withdrawing group would be?
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Question 2: Protect, oxidise and make lactone.
I think you can skip the protection and just oxidize it. If you oxidize the alcohol, the intermediate aldehyde will be in equilibrium with the lactol, that is the hydroxyl group of the tethered alcohol will form a cyclic hemiacetal with the aldehyde. The OH of the hemiacetal, upon oxidation, will give the lactone.
That is the principle of the chromic acid and PCC oxidations. If a primary alcohol is oxidized in the presence of water, it is the hydrate of the aldehyde that becomes oxidized to a carboxylic acid. With a PCC oxidation, water is excluded and the oxidation stops at the aldehyde.
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1.Dan, if I understood well, you mean that 2 rings pull the electrons from N-C more to C than 1 ring does, so the N-H bond gets even weaker? I saw there written alkylaminium ion, is this a mistake or not, I thought that it is "alkylamonium"? I saw that methylamine is less basic than dimethylamine, how so?
2.I drew the following scheme, so tell me if it is correct:
http://imageshack.us/photo/my-images/823/reactionscheme.gif/
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The scheme shows how I would expect the reaction to proceed.
@Raderford, are you a stockholder in imageshack? Certainly, we all receive additional advertising from them and their sponsors. I for one am unappreciative of this gratuitous advertisement. The forum does provide for direct hosting of images which avoid additional linking just to view. Just saying'
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Sorry for that, I am not registered there, I just upload images from time to time. I don't know how to upload images directly here, there is "insert image" command, but what to do then?
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1.Dan, if I understood well, you mean that 2 rings pull the electrons from N-C more to C than 1 ring does, so the N-H bond gets even weaker?
Yes
I saw there written alkylaminium ion, is this a mistake or not, I thought that it is "alkylamonium"?
I am not familiar with this term either, I would use alkylammonium.
I saw that methylamine is less basic than dimethylamine, how so?
Read about inductive effects
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I know the inductive effect. For example, trimethylamine has 3 methyl groups that push away the electrons towards the N atom, which pushes a little the electrons of N-H bond towards H, so it is hard to make a H+ ion, meaning that trimethylamine should be the most basic, but on the site you gave it is listed this way (basic property decreases): dimethylamine>methylamine>trimethylamine, this I don't understand.
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I don't know that I can answer why trimethylamine should be less basic. I don't think it is the only instance in which the acidities do not follow an expected pattern. I have pointed out the (di and tri)fluoroacetic acid(s) do not follow the expected trend. Generally a fluoro substituent is less electron withdrawing than the other halogens, compare benzoic acids, phenols, anilines, haloforms, haloethanols, rate of leaving group, haloacid acidity, etc. I certainly cannot explain why fluoroacetic acid should be more acidic than chloroacetic acid. It isn't that this result cannot be explained, but I don't know the reason.
If you compare ammonia (9.21), ethylamine (10.64), diethylamine (10.98), and triethylamine (10.65), there is an increase in basicity from ammonia to ethylamine. There is a smaller increase in basicity in diethylamine and small decrease in basicity in triethylamine. A similar but smaller decrease in basicity occurs. If you compare other amines, it appears that a methyl group does not give the same result as an ethyl or other alkyl groups. (See http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf.) Could the measurement of the pKa data be more complicated than examining a structure would suggest? Let's look at hydronium ions. Perhaps it also is not as simple as we might have expected, see http://www.lsbu.ac.uk/water/ionish.html.
Re: attachments
In additional attachments
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No simple explanation after all, just by looking at the structure. I knew about the hydration of H3O+, very interesting thing. Thanks for the help.
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I found an old explanation on this forum http://www.chemicalforums.com/index.php?topic=23571.0#msg89592
Similar was mentioned on the link Dan gave, only here it is better explained.
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Someone has an idea why basicity increases from left to right in the picture? What should be applied for heterocyclic systems?
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Rationalise the difference between pyrrole and imidazole first, then explain why the trifluoromethylimidazole goes in the middle.
Hint: Consider the stability of the conjugate acids
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There is another N atom in imidazole, but I am not sure on which atom of imidazole will the H+ be attached to make the conjugate acid and why.
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There is another N atom in imidazole, but I am not sure on which atom of imidazole will the H+ be attached to make the conjugate acid and why.
Surely both the N atoms in the imidazole are equivalent!
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So both can bind a H+ ion from water and make OH- ions, but in pyrrole there is only 1N atom, therefore, pyrrole is less basic, right?
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So both can bind a H+ ion from water and make OH- ions, but in pyrrole there is only 1N atom, therefore, pyrrole is less basic, right?
Yes.
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So both can bind a H+ ion from water and make OH- ions, but in pyrrole there is only 1N atom, therefore, pyrrole is less basic, right?
No, pyridine is more basic than pyrimidine, 5.25 vs 1.1 for pyridinium, pyrimidinium ions.
Hint, Huckel aromaticity rules.
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It aren't pyridine and pyrimidine, in every molecule there is a N atom bounded only with single bonds to other atoms.
Does trifluoromethylimidazole go in the middle because of the big electronegativity of F atoms, they attract the lone pairs from N atoms so they are less available for the H+ ion?
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Pyrimidine has two nitrogen atoms and is much less basic than pyridine with one. If that argument were applied to pyrrole, it should be more basic.
Why isn't pyrrole more basic? The answer does not depend on the number of nitrogens, but the availability of the electrons. If the electrons are readily available, the nitrogen will be more basic, or why is pyridine more basic than pyrrole?
Again, think aromaticity. If you look at models, look at the orientation of the electrons in pyrrole, imidazole, and pyridine. How is imidazole like pyridine?
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Okay, then I use Huckel's rule, pyrrole N atom needs to donate its electron pair to the ring so it will follow the rule, in imidazole to make the same effect, only the electron pair from one N atom is enough to follow the rule, so there is a pair that can be donated to a H+ ion, so imidazole is more basic. The compound in the middle has 3 F atoms that push the N electron pair from it, so it is less available, while the pair in the other N atom follows Huckel's rule.
That is clear I think, only one more thing: I don't understand why is pyrimidine less basic than pyridine. The N atom in both molecules needn't to donate an electron pair.
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… why is pyrimidine less basic than pyridine. The N atom in both molecules needn't to donate an electron pair.
I argue this is an inductive effect. Nitrogen has a larger nuclear charge and pulls electrons away. A similar effect can be found with hydroxyl amine (NH2OH) or hydrazine (NH2NH2), but now there is an sp2-carbon intervening.
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"Nitrogen has a larger nuclear charge and pulls electrons away."
I don't understand, away from what?
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"Nitrogen has a larger nuclear charge and pulls electrons away."
I don't understand, away from what?
If you begin with pyridine and replace a CH with a N, that nitrogen can pull electrons from the basic nitrogen making it less basic.
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How can an atom pull electrons from another atom that has the same electronegativity?
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Which is more acidic of X-OH?
HOH
HOOH
ClOH
The pKas are 16, 12, and 7.5. This seems reasonable to me.
If I compare CH2=CHOH, HN=CHOH, and O=CHOH, I would expect the O=CHOH to me the most acidic.
If I compare CH2=CHNMe2, HN=CHNMe2, and O=CHNMe2, I would expect O=CHNMe2 to be the least basic.
Even if you compare HOCH2CH2NH2 or CH2=CHCH2NH2 with ethyl or propylamine, they are less basic.
**Edit**
I am continuing to research this topic (amine basicity). Here is an interesting exception, guanidine is far more basic does not follow the trend. See http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/amine1.htm#amin2
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How can an atom pull electrons from another atom that has the same electronegativity?
If you are talking, for instance, about the 2 nitrogens in 4-trifluoromethylimidazole, then these are not identical nitrogens. Electronegativity scales are a bit misleading in my opinion as they suggest an absolute value. Perhaps you know for instance that the CH3 protons in methanol, methyl chloride, methyl benzoate and acetone have different δ values (we have touched on this a little in your NMR assignment post), but simplistically they can call be described as Me-(electron withdrawing group), right?
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I was talking about the two N atoms in pyrimidine. Orgopete, the thing that confuses me is that the N atoms in pyrimidine are equivalent (they are in a circle) and have the same effect on the C atom and on themselves, so I don't understand how an inductive effect can appear here.
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I don't mean to resurrect this post on ammonium ion acidities, especially has it contains several irrelevant threads. However, I wanted to write my opinion about the pKa issues, solvation, gas phase, and structural effects. As this is long, I have included it in a blog (http://www.curvedarrowpress.com/wpblog/?p=186).