[impulse29]

For my organic chemistry class, I had a question on my previous test that asked which reagent I would use to separate mixtures. It gave a list of some reagents (I didn't get the question, so I thought I'd memorize the components). THe reagents to choose from were water, CH3CH2OCH2CH3 (diethyl ether), 1M NaOH, and 1M HCl. 

I assume that if I want to separate the mixture, and _keep_ the componds, they must be insoluble?
Because I am not good at making structures on the computer, I will give the name for them:
1. benzoic acid
2. Benzyl-aminochloride +1 (benzene ring with a NH3Cl+ added)
3. Benzanol
4. The same thing as 2. except it is not aromatic. just a cyclohexane.

Thanks for any help. I was really stuck. There's another question, but I can't think of it right now.

[mike]

You need to check the solubilities of the different compounds in water and diethyl ether. Polar compounds will dissolve in water and can be separated from non-polar compounds which will dissolve well in diethyl ether. One example of how to use the NaOH is benzoic acid, which is soluble in ether, however when reacted with NaOH it forms a benzoate salt which will be soluble in water... get where I am going with this?

[impulse29]

Okay, I kinda get it. But should I try reacting them with ones that they are soluble in, or ones that they are not soluble in? The reason would be because we want to isolate them from the mixture. I want to say that they have to be insoluble so that they are left behind, but I remember doing a lab where I added a reactant that was soluble in my desired product, and I was able to filter the other 'stuff' and then crystallize, isolating what my product.

Is this the approach I should take?

I somewhat remember the other question I had. It was concerning electrons jumping to antibonding orbitals when it absorbs light. It asked why this could cause dissociation of a hydrogen molecule into two atoms. Would it be because the bonding and antibonding orbitals have equal number of electrons in them, which would form a node, and break the bond? Now that I wrote out the question, I think I understand it a bit more. I always tend to overthink things on tests

[mike]

You are basically correct. Say for example you wanted to separate a mixture of two compounds, and both of these compounds are non-polar. This means they will both dissolve well in ether (or another organic solvent) so there is nothing to differentiate them yet. What if, however, one of them is an organic acid? You could react the mixture with NaOH, and in this case only the organic acid would react to form a salt of the acid (which is water soluble). Now you could use a separating funnel, diethyl ether and water to separate the two. The bottom layer in the separating funnel would be the water containing the salt of the acid while the other compound would be in the top ether layer. Now you can put each layer in a different flask and you have separated the two compounds.

To isolate the product from the ether you could simply evaporate off the ether so you are left with your product. To isolate the other compound from the water, you could add an acid to convert the salt back into the original organic acid and then filter off the now insoluble organic acid.

[impulse29]

Thanks, you're great *delete me*

Where would I go about finding if a compound is soluble in ___? I have checked some sites (chemfinder and others), but they only give the solubility in water. I have tried the CNC website, but it takes forever (and I don't have a book here).

[Yggdrasil]

In general, you can guess at whether compounds will be soluble in a certain solvent by the principle that "like disolves like."  For the purposes of your problem, you can assume that all organic molecules in their uncharged form are soluble in ether and insoluble in water, and that they are soluble in water and insoluble in ether when they are salts/charged.

As for your other question, the bond order is defined as:

(b - a)/2

where b = the number of electrons in bonding orbitals and a = the number of electrons in antibonding orbitals.  A bond order of 1 is equivalent to a single bond, a bond order of 2 is equivalent to a double bond, etc.  So, when you promote one electron from a bonding orbital in H₂ to an antibonding orbital, what does that do to the bond order?

[impulse29]

That helps. What about getting the last one out. The only difference between it and another compound is that it is not aromatic. Do you recommend I make a table saying what dissolves in what, and then remove the first compound, and try to eliminate one at each step? I am confused because not only is the whole solubility thing confusing to me, but I also think that more than one are soluble in one reagent.

Or should I take the approach of splitting the mixture in half, and then working to isolate them like that?

This stuff really hasn't been covered (although we're probably assumed to connect the dots with what we already know), and I would really hate not knowing how to do this for the next test.

Thanks for all the help, it's greatly appreciated.

[Yggdrasil]

To separate the phenol and the benzoic acid, you definitely need to exploit the differences in pKas between the two species (i.e. in a certain range of pH values, the benzoic acid will be deprotonated and charged, but the phenol will protonated and uncharged).  You may be able to use the same strategy to separate the aminocyclohexane from the aniline.