laczlacylaci

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Why would B not work? There is a disappearance of one C=O bond (bottom left corner), which indicate a disappearance of a strong peak at 1720 cm^-1. Is it because there are 2 OH groups that disappeared?
 

TheLongGame

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The presence of two carboxylic acid groups will cause a broad and very strong absorption in the precursor mixture due to the O-H absorption. This will obscure your monitoring of the Keto groups. In addition, the carboxylic acid also has a C=O bond, so even if a couple of (or even all) actual Keto groups are reduced from the final product, the 1720 absorption will still be present due to carboxyl to ester conversion.

When presented with an absorption spectra problem, look for the functional groups that have unique absorption characteristics that will either disappear or suddenly appear.

In this example, the O-H bond on the carboxylic groups disappear.

Additionally, I believe that the diagram of the final product is misrepresented as there appears to be the presence of a carbocation on one of the former carboxyl groups...this is not likely to be a very stable compound as it is drawn...some electrons got lost in the mix.


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aldol16

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The presence of two carboxylic acid groups will cause a broad and very strong absorption in the precursor mixture due to the O-H absorption. This will obscure your monitoring of the Keto groups. In addition, the carboxylic acid also has a C=O bond, so even if a couple of (or even all) actual Keto groups are reduced from the final product, the 1720 absorption will still be present due to carboxyl to ester conversion.
How do you think this will obscure your monitoring of the keto groups? The O-H stretches are at completely different wavenumbers than the carbonyl stretches.
 

TheLongGame

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How do you think this will obscure your monitoring of the keto groups? The O-H stretches are at completely different wavenumbers than the carbonyl stretches.
You are absolutely right! My mistake! I was getting wrapped around the disappearance of the O-H stretch and crossed my stream of thought. You will be able to monitor the C=O stretch in both the Carboxyl groups as well as the Keto groups, however, you will be very hard pressed to distinguish the two apart. Complicating this issue even more is the fact that not all of the C=O groups disappear means that even though one of them disappeared, the remaining C=O bonds will absorb at the same range. As the questions states "Absence of IR absorption", the only absorption wavelengths that disappear are associated with the O-H stretch of the carboxylic acids.
Yes there are O-H stretches present due to the hydrated ligand, however, I would discount these for two reasons.

1) MCAT does not expect you to know the minutiae of IR spectroscopy procedure, but rather wants to know that you understand the concepts.
2) Initial set-up of reaction monitored IR spectra removes the absorption spikes of the solvents used. In this case a hydrated metallic reagent either directly added as a solid, or more likely as an aqueous solution that would pull the desired product from the organic layer to the aqueous layer as the reaction proceeded forward.
**by subtracting, that only means its contribution, so if there are O-H stretches present after the normalization process (i.e. in a reagent or product), then the contribution of that absorption will register.

Again, sorry for the mix up!
 
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aldol16

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2) Initial set-up of reaction monitored IR spectra removes the absorption spikes of the solvents used. In this case a hydrated metallic reagent either directly added as a solid, or more likely as an aqueous solution that would pull the desired product from the organic layer to the aqueous layer as the reaction proceeded forward.
Not sure what you're saying here but aqua ligands do not exhibit the same IR stretches as their free solvent correlates. This is because there is a degree of bonding/backbonding present that alters the stretch frequency of the O-H bond(s). In fact, IR is a commonly used technique to discern the degree of backbonding in organometallics. Take several electron-rich metals with CO ligands and you can measure how well each metal backbonds - the C-O stretch will change with the metal and will be very different from free CO.
 

TheLongGame

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Not sure what you're saying here but aqua ligands do not exhibit the same IR stretches as their free solvent correlates. This is because there is a degree of bonding/backbonding present that alters the stretch frequency of the O-H bond(s). In fact, IR is a commonly used technique to discern the degree of backbonding in organometallics. Take several electron-rich metals with CO ligands and you can measure how well each metal backbonds - the C-O stretch will change with the metal and will be very different from free CO.
It does indeed alter the O-H stretch frequency, but you are mistaking the OH stretch of newly backboned elements with those that were already present in the hydrated metallic reagent. Those particular OH stretches we can eliminate along with the other water present. No new O-H stretches are formed in the complex. Yes, the C-O stretches will give you indications the degree to which the metal has complexed with the compound, absolutely no argument, but with respect, that is not what the question asked, nor will the MCAT expect anyone to know.

Taking the IR of the metal hydrate in solution, before adding the organic compound, allows the observer to remove that absorption pattern from future readings.

If you have any other questions or concerns feel free to PM!
 

aldol16

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It does indeed alter the O-H stretch frequency, but you are mistaking the OH stretch of newly backboned elements with those that were already present in the hydrated metallic reagent. Those particular OH stretches we can eliminate along with the other water present. No new O-H stretches are formed in the complex. Yes, the C-O stretches will give you indications the degree to which the metal has complexed with the compound, absolutely no argument, but with respect, that is not what the question asked, nor will the MCAT expect anyone to know.
You're not understanding what I'm trying to say. Be clear - which O-H stretches are you talking about? I use IR frequently as an organometallics PhD student and so I'm a bit interested in what you're trying to say. The IR peaks resulting from the O-H stretches from the aqua ligands on the hydrated metal complex will be slightly different than the IR peaks resulting from the O-H stretches from the aqua ligands on the new metal complex shown in the figure. This is because the local electronic environment about the metal center has been altered by the new ligand field. In other words, the metal will bond/backbond to its ligands differently depending on the ligands surrounding it and thus one would expect the O-H stretch from the aqua ligands to be slightly different when one compares a fully hydrated metal with the version shown in the figure (with two aqua ligands).

In fact, the local ligand environment is known to alter the electronic environment about metals so much that the pKas of aqua ligands bound to the metal have been shown to change drastically - in some cases, what you think is M-OH2 is actually M-OH! I understand that's not the question being asked here but I'm making the point that the energy of the O-H bond is altered by the ligand environment about the metal and therefore the IR peaks will be shifted. When you subtract peaks out, the software does it by wavenumber and not by which class of stretch it belongs to - in other words, it takes all the peaks in one reference spectrum and subtracts them exactly from your spectrum. So if the relevant peak has moved from one spectrum to the next, it won't be subtracted.

If one is doing the subtraction manually, it's not good chemical logic to just remove a peak because it's "in the right region" as your reference. That is, a broad peak at 3210 cm^-1 in one spectrum and a corresponding broad peak at 3230 cm^-1 in another spectrum may correspond to the same O-H bond but that doesn't mean one should immediately discard it. It contains useful information!

The upshot of this is that using just the O-H stretch region, one can actually determine whether the metal has bound or not.
 
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TheLongGame

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Wow! Absolutely no arguments here! That was very good explanation!

To answer your question in your second sentence, I was talking about no new hydroxy groups being formed on the starting molecule.

You know what? I didn't consider the impacts of the ligands electronic environment impacting the stretch of the metal complexed water molecules. You brought up a very good point.

Man! We got into the weeds on this one! I appreciate the information and the confusion on my part!


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