[Rutherford]

1.I need to assign the following pka 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?

[discodermolide]

1.I need to assign the following pka 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.

[Dan]

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?

[orgopete]

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.

[Rutherford]

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/

[orgopete]

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'

[Rutherford]

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?

[Dan]

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

[Rutherford]

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.

[orgopete]

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

[Rutherford]

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.

[Rutherford]

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.

[Rutherford]

Someone has an idea why basicity increases from left to right in the picture? What should be applied for heterocyclic systems?

[Dan]

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

[Rutherford]

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.