Help with acid/base question from AAMC's Practice Question?

[m25]

The question and the corresponding passage is below, but you probably just need up to Reaction 1 for the sake of this question.
I originally thought the answer was either B or C.
1) So for B, you can't dissolve a solid base with base? Because I thought like dissolves like.
2) For C, I thought addition of Na+ into the solution would drive the reaction to the left by Le Chatlier's principle, and the free M2+ would react with OH-? But then I guess this is not measureable?

3) Can you tell me what the experiment in D is doing? According to my understanding, adding strong acid HCl to a solid weak base will dissolve it and release it in its acidic form (since it gets protonated). I understand this much, but I don't undertand "The difference in the number of moles of HCl and NaOH required to reach the endpoint". Wouldn't there actually be no difference because ALL of the HCl added would be neutralized by NaOH?? Say for example we had add 1 mole of HCl, and 0.3 mole of H+ react with the RCO2 releasing it as fatty acid (this means that there are 0.3 moles of fatty acid present the precipitate), and the rest of the 0.7 moles of H+ just hangs out in the solution since there's no more weak base to react with. Now we add NaOH to titrate the solution, and because NaOH is a strong base, it neutralizes both the fatty acid and the H+ that is just hanging out in the solution, and we figure out that we used 1 mole of NaOH. So we put in 1 mole of HCl and added 1 mole of NaOH to neutralize it. There would be no difference that the answer explanation is talking about?? Or am I missing some sort of big picture/concept here?

4) I feel like I've over-thought this question, is there a simpler way to approach this question?

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[Cawolf]

1) Like dissolves like applies to polar vs. non-polar solvents and compounds.

2) Na+ is a spectator ion, it has no effect on the equilibrium.

3) The question gives us a salt of M2+ and 2RCOO-. Treating with strong acid will free the M2+ by protonating the ROO-. Have you taken orgo? This is called an acidic workup. The question is confusing, but just seeing that this is the only acid workup should be enough to answer the question. I am not sure on the specifics of the rest of the procedure, but it really doesn't matter.

4) See point 3.

 

[m25]

1) Like dissolves like applies to polar vs. non-polar solvents and compounds.

2) Na+ is a spectator ion, it has no effect on the equilibrium.

3) The question gives us a salt of M2+ and 2RCOO-. Treating with strong acid will free the M2+ by protonating the ROO-. Have you taken orgo? This is called an acidic workup. The question is confusing, but just seeing that this is the only acid workup should be enough to answer the question. I am not sure on the specifics of the rest of the procedure, but it really doesn't matter.

4) See point 3.

Fantastic answer, thank you! I have actually never heard of the term acidic workup before despite having taken orgo class. Could you please elaborate on this a bit? I've looked it up online but the explanations are so brief I am not sure I understand it.

 

[Cawolf]

It is also known as an aqueous workup.

Many reactions produce anions that conjugate with metal reactants after occuring. This is common with Organometallic reagents such as R-MgBr or R-Li. When the nucleophilic R group uses the electrons to attack a carbonyl, the pi bond pushes up to the oxygen, making it negatively charged. This charge will ionically bond with the remaining +MgBr or Li+. In order to isolate the product, water and acid is used to "quench" the charge on the oxygen (O- to OH) and free the metal so the compound can be extracted. Also with Lewis acid catalysts, etc.

I doubt you have never heard of an aqueous workup. It usually written as H3O+ or H2O, H2SO4 (sub. any strong acid).

 

[m25]

Thanks! So basically you add base to extract acid, and add acid to extract base? And this works when the stuff you want to extract is in either aqueous or solid form?

When the nucleophilic R group uses the electrons to attack a carbonyl, the pi bond pushes up to the oxygen, making it negatively charged. This charge will ionically bond with the remaining +MgBr or Li+.

Are you saying here that the nucleophilic R group attacks its own carbonyl? if so, why?

 

[Cawolf]

The steps you listed are good to solubilize an organic compound into an aqueous phase. It charges the molecules and will extract them from the organic phase. This is a standard laboratory technique called liquid extraction.

No, another carbonyl. Like a Grignard reagent adding to a ketone or aldehyde.