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Help! I'm Taking Organic Chemistry!
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27th-Apr-2009 04:02 pm - NMR Spectrum
I've been trying to figure out these NMR readings of an aldol (alpha beta unsaturated carbonyl compound), and I can't seem to get the structure. To make matters more difficult, my teacher didn't supply me with the chemical formula...just that the aldehyde has no alpha hydrogens and is aromatic.

Here are the readings...and with my interpretations that I've come up with.

7.4 doublet 2H (para substituted benzene)
7.2 doublet 2H (para substituted benzene)
6.9 singlet 1H (double bond)
2.5 singlet 3H (methyl group either next to carbonyl or double bond)
2.1 singlet 3H (methyl group either next to carbonyl or double bond)
1.9 septet 1H (hydrogen in between these two identical methyl groups)
0.9 doublet 6H (two identical methyl groups)

But for some reason, I cant structure and combine it all correctly to make it an actual product.

Your help is greatly appreciated.
23rd-Nov-2008 05:01 pm - Fish-scented alcohol?
Does anyone know of that alcohol that smells EXACTLY like a fish market/fish?
The scent is so strong; it diffuses very readily.

density about 0.5653 g/mL
Quote/TW/Brain Pretty, Scarecrow
An answer to a question posted by aciel.

The question was whether the nitro group, the amine group, or both would be protonated when you add sulfuric acid.

As you can see from the picture below, m-nitroaniline has two resonance contributors.

The electrons on the nitro group are shared between the two oxygens, making it kind of difficult for the proton to find the electrons. Also, the nitrogen maintains its plus charge, which is balanced by the negative charge on one of the oxygens. If you were to protonate one of the oxygens, the nitrogen would have a plus charge that is not offset by the oxygens.

The amine has an unshared pair of electrons that are concentrated on the nitrogen. Because the electrons are not delocalized, it is much easier for nitrogen to share its electrons than for the nitro group to share its electrons.

Sulfuric acid is one of our strong acids, so the first proton falls off pretty easily. It's much harder to pull off that second proton, and a nitro group that has resonance is certainly not strong enough to do it.

The salt, then, would be NH3+ and HSO4-, which is soluble in your aqueous layer. You actually won't have a precipitate until you add a base to remove the proton from the salt and neutralize the solution. m-Nitroaniline is not soluble in water, so once you take the proton off and return it to its original state, it would precipitate out of solution.

I hope that helps!
8th-Oct-2006 04:40 pm(no subject)
'Allo, there. This community seems pretty empty, but maybe that will change, and the person who created this seems still active.

Entirely selfishly, I have an organic chemistry question, and probably very little I can give in return since I'm more of a physics kind of guy. Thus, your help is greatly and gratefully appreciated.

We're separating a neutral compound (azobenzene) from an organic base (m-nitroaniline) using liquid-liquid extraction. The two compounds are dissolved in ether, and we add aqueous sulfuric acid to cause the salt of the base to precipitate into the aqueous layer.

Now, I get the gist for when we're dealing with hydrochloric acid, but it's only got one free proton. Our organic base has an amine group and a nitrite group, which suggests to me that there are two sites for proton acceptance. Thus, does the sulfuric acid twice deprotonate? And then would the salt be (+)NH3---SO4(2-), or what?

Thanks very much for any help.
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