Okay, so I need to figure out this NMR and LC-MS question. Let's start by reading the problem carefully.

The question is about a small peptidic compound. The 1H NMR spectrum shows two peaks for the same alpha-proton. These peaks have similar chemical shifts and equal integrals, adding up to 1H. They ruled out spin-spin coupling as the reason. Then, the LC-MS at elevated temperature shows two peaks of equal intensity with the same mass. The mass is consistent with the expected molecule. The question is asking what's the most likely explanation for these observations.

The options are A to D. Let's think through each possibility.

First, the NMR issue: two alpha protons that are equivalent but split into two peaks. The integrals are equal (each about 0.5H). Since they're alpha protons, they're adjacent to the peptide bond. If they're the same, but split into two peaks, maybe they're not actually split. They could be split but with coupling that cancels out? But the question says they ruled out spin-spin coupling. Spin-spin coupling would cause splitting, but if they're not split, maybe another type of coupling?

Wait, but the problem states that they ruled out spin-spin coupling. So spin-spin (like J coupling) isn't happening here. So maybe it's something else. If two protons are equivalent, they should be a single peak. But here, they are two peaks. So why? Maybe if the two protons are in different environments but equivalent in some way, but the coupling is not observed. Or perhaps they are diastereotopic? Wait, but that's about stereoisomerism affecting the signals. If the molecule is a racemic mixture, then the protons would be equivalent, but in different environments, leading to multiplets. But wait, the integrals add to 1H. If it's a racemic mixture, each enantiomer would have one proton, but they would be split into different peaks? Hmm.

Wait, another possibility: if the molecule is a diastereomer mixture. Diastereomers have different physical properties, so their signals might split. But diastereotopic protons can cause splitting. But in this case, the problem states that they ruled out spin-spin coupling. So maybe it's not about coupling between the two protons. If the two protons are in different diastereoisomers, but the molecule is a mixture of diastereomers, then each diastereomer would have its own set of peaks. But if there's a 50:50 mixture, maybe the integrals would average out? Wait, but the problem states that the integrals add up to 1H. If each diastereomer contributes 0.5H each, then combined it's 1H. But if they are diastereomers, the coupling within each diastereomer would cause splitting. But the question says spin-spin coupling is ruled out, so perhaps the diastereomers are not coupled in the same way.

Wait, maybe the issue is with the molecule existing as a mixture of enantiomers. If the molecule is a racemic mixture, then each enantiomer would have a different environment for the alpha protons, leading to different signals. But in that case, the protons would be split into two peaks, each integrating to 0.5H. But then the question says they ruled out spin-spin coupling. Wait, in a racemic mixture, the two enantiomers have different environments. However, the coupling between protons in different molecules (from different enantiomers) isn't really spin-spin coupling; it's more like intermolecular coupling, which usually isn't observed in the NMR. So maybe the splitting isn't happening because of intermolecular interactions. So the two protons would each be split into two peaks, but if the coupling constant is small, maybe they don't split much. Alternatively, perhaps the splitting isn't observed because the coupling is between the two protons and another proton, but that's not the case here.

Alternatively, if the molecule is a diastereomer mixture, then each diastereomer would have its own set of peaks. But if they are diastereomers, they might have different coupling constants. But the problem states that the two peaks have the same chemical shift and the same integral. Wait, but if the diastereomers have different coupling constants, the peaks might be split differently. But the question says that the coupling pattern ruled out spin-spin. So maybe the diastereomers have the same coupling but different shifts. Hmm, this is getting a bit complicated.

Now, the LC-MS part: two peaks of equal intensity, same mass. The mass is consistent with the expected molecule. So the compound is either a mixture of two molecules of the same mass or a single molecule that is fragmented into two products of the same mass. But fragmentation would likely produce different masses, unless there's a specific loss. For example, if the compound is a dipeptide, maybe the cleavage between two amino acids would produce two fragments. If the parent molecule has a mass M, the fragments would have M - (mass of one amino acid). But if the two fragments have the same mass, perhaps the parent is symmetric. For example, if the parent is a dipeptide with two identical amino acids, then cleavage would produce two identical fragments. But the problem states that the crude compound is analyzed, so if there's a mixture, the LC-MS would show two peaks of equal intensity. If the compound is a racemic mixture, the fragments would be the same. Alternatively, if there's a diastereomer mixture, maybe the fragmentation is different. But the problem says the mass is consistent with the expected molecule, so the same mass for both peaks, which would suggest the same molecule. Wait, but if the molecule is a dipeptide, say, with two identical amino acids, then cleavage would give two identical fragments. If the parent is a racemic mixture, the fragments would be the same. But if the parent is a mixture of diastereomers, perhaps the cleavage would give different fragments. However, the problem states that the LC-MS shows two peaks of equal intensity. So perhaps the compound is a racemic mixture, and the two peaks correspond to the same molecule but from different enantiomers. But wait, if the molecule is a racemic mixture, then the two enantiomers would have the same mass, so the LC-MS would integrate into two peaks of equal intensity. But the problem says that the LC-MS shows two peaks with the same mass. So that would fit. But why would the NMR show two peaks for the same alpha-proton?

Alternatively, if the compound is contaminated with a precursor. The LC-MS at elevated temperature shows two peaks. If the precursor is a larger molecule that breaks down into the product. For example, if the precursor is an amino acid and the product is a dipeptide. Then, if the precursor is present, the LC-MS would show the precursor and the dipeptide. But the problem says that the crude compound is analyzed, so maybe the contamination is with a precursor. But the NMR shows two peaks for the same alpha-proton. If the precursor is an amino acid, then the NMR would show signals for the alpha-proton of the amino acid. But in the problem, the NMR is consistent with the expected molecule except for two alpha-proton peaks. If the expected molecule is a dipeptide, then the NMR would show alpha protons from both amino acids. But if one of them is a precursor, maybe a single amino acid, then the NMR would show a single peak. But the problem says two peaks for the same alpha-proton, which would not make sense if it's a single amino acid. Unless the amino acid has two alpha-protons in different environments. Wait, but amino acids have only one alpha-carbon, so only one alpha-proton. So if the crude compound is a mixture of the dipeptide and the amino acid, then the NMR would show two different alpha-protons. But the problem states that the two peaks are for the same alpha-proton, which would mean that the dipeptide has two alpha-protons in the same environment, which isn't possible. So maybe the contamination is with a diastereomeric form. But that would complicate things.

Alternatively, if the LC-MS shows a peak for the dipeptide and another peak for a fragment. For example, if the dipeptide is cleaved into two amino acids, but if there's a dipeptide and the amino acid, but the NMR would show the dipeptide's alpha-protons and the amino acid's. But the problem says the two peaks are for the same alpha-proton. That doesn't add up.

Wait, maybe the two peaks in the NMR are due to diastereotopic protons. Diastereotopic protons are in different chemical environments but equivalent in a certain way. For example, in a chiral environment, two hydrogens that are on the same carbon but in different positions relative to the stereogenic center. In such cases, if the molecule is a racemic mixture, the NMR would show a single peak because the protons are equivalent. But if the molecule is a diastereomer mixture, maybe they have different shifts. But the problem states that the two peaks have similar chemical shifts, so maybe they are diastereotopic protons in the same molecule. Wait, but if the molecule is a racemic mixture, the two enantiomers would have the same diastereotopic protons but in the same environment, leading to a single peak. So that doesn't explain two peaks. Hmm.

Alternatively, the two peaks in the NMR could be due to two different diastereomers. For example, if the compound can form two diastereomers, and the mixture is a racemic mixture of those diastereomers. But that would complicate the NMR. If the two diastereomers have different chemical shifts for the alpha-proton, then the NMR would show two peaks. But the problem states that the two peaks have similar chemical shifts. So that's not likely.

Wait, going back to the LC-MS. The two peaks have the same mass. If the compound is a racemic mixture, then the LC-MS would show two peaks of the same mass (same molecular weight), each integrating to half the sample. That fits. But the NMR shows two peaks for the same alpha-proton. How does that happen?

If the molecule is a racemic mixture, the NMR would average the signals. Wait, no. If the molecule is a racemic mixture, each enantiomer would have the same chemical shift for an equivalent proton. But if the proton is in a chiral environment, then in racemic mixture, the two enantiomers would have the same shift. Wait, no. Wait, in a chiral environment, the two enantiomers would have the same chemical shift for the protons in their respective environments. For example, if the molecule is a chiral center, and the proton is in a chiral environment, then in the racemic mixture, the two enantiomers would have the same chemical shift because the environment is symmetric. Wait, but wait, if the molecule is a racemic mixture and the proton is in a chiral center, the two enantiomers would cause the proton to be split into two peaks, but if they are equivalent, they would average to a single peak. Wait, no. If the molecule is a racemic mixture, the two enantiomers would each have their own signals. But if the molecule is a racemic mixture, then the two enantiomers would have identical environments for the protons. Wait, no. If the molecule is a racemic mixture, the two enantiomers are mirror images. So the protons that are in the same environment in each enantiomer would have the same chemical shift. Therefore, the NMR would show a single peak for equivalent protons. But in this case, the NMR shows two peaks for the same alpha-proton. That suggests that the two protons are not equivalent in the molecule. Wait, but if the molecule is a racemic mixture, the protons would be equivalent in the sense that each enantiomer has a proton in a certain environment, but the two enantiomers would have the same shift. So the NMR would show a single peak. So this contradicts the observation. Therefore, the racemic mixture is not the explanation.

So perhaps the problem is with diastereomers. If the compound is a diastereomer mixture, then each diastereomer would have its own set of signals. If the two diastereomers have the same chemical shift for the alpha-proton, then the NMR would show a single peak. But in this case, the two peaks have similar shifts, but maybe due to different coupling. Wait, the problem states that the coupling pattern ruled out spin-spin coupling. So maybe the two diastereomers have different coupling patterns that are not spin-spin. But if the coupling is something else, like coupling to a different proton, but the problem says that the duplication is not due to spin-spin. So maybe the two diastereomers have different coupling constants, leading to two peaks. But the problem states that the two peaks have similar chemical shifts. If the diastereomers have different coupling constants, the splitting could be different. However, the problem says that the integrals add up to 1H. If the diastereomers are present in equal amounts, each would contribute 0.5H. But if the diastereomers have different coupling patterns, then the peaks would split further. But the problem says that the coupling pattern ruled out spin-spin coupling. So maybe the diastereomers have different coupling, but the problem states that the duplication is not due to spin-spin. So perhaps the diastereomers are present in a 1:1 ratio, leading to two peaks of 0.5H each, and the splitting is not due to spin-spin. But the problem says that spin-spin coupling is ruled out, so maybe the splitting is due to something else, like coupling to another proton.

Alternatively, the two peaks in the NMR could be due to two different diastereomers that have the same chemical shift for the alpha-proton but different coupling. For example, if the diastereomers have different coupling constants, leading to two different splitting patterns, but the same shift. However, the problem states that the integrals add up to 1H, which suggests that the two peaks are of equal area. If the diastereomers are present in a 1:1 ratio, each would contribute 0.5H. But if the coupling is different, the splitting could result in multiple peaks. But the problem says that the two peaks are the only ones observed, each integrating to 0.5H. So perhaps the diastereomers are present as a 1:1 mixture, but they don't split the signal further. Alternatively, each diastereomer has a single peak for the alpha-proton, but they are split into two each, but the integration would be equal. But that seems complicated.

Now, considering the LC-MS. The two peaks have the same mass and equal intensity. If the compound is a racemic mixture, the two enantiomers would have the same mass, so the LC-MS would show two peaks of the same mass (same molecular weight) and equal intensity. That fits. But the NMR shows two peaks for the same alpha-proton. How does that reconcile?

Wait, maybe the racemic mixture is not the issue. Let me think again. The NMR shows two peaks for the same alpha-proton, which suggests that there are two non-equivalent environments for that proton. But the problem says that spin-spin coupling is ruled out. So maybe the two protons are coupled to another proton, but the coupling constant is too small to observe. However, the problem states that the coupling pattern ruled out spin-spin coupling. So perhaps the protons are coupled to a different type of coupling, like dipolar coupling, which is usually not observed in routine NMR experiments. But that's getting into more advanced topics.

Alternatively, maybe the two alpha-protons are part of a conjugated system, like in a peptide bond. Wait, in a peptide bond, the alpha-proton is adjacent to the carbonyl group. The coupling between the alpha-proton and the adjacent protons (like the beta-proton) is possible, but in peptides, such coupling is usually not observed in routine NMR because of the slow rotation around the peptide bond. However, in small molecules, coupling across the peptide bond can sometimes be observed. But the problem states that spin-spin coupling is ruled out, so maybe the coupling is not across the peptide bond. Alternatively, the two alpha-protons are adjacent to each other, but that's unlikely in a dipeptide.

Wait, maybe the two alpha-protons are on different amino acids. For example, if the dipeptide has two alpha-protons from different amino acids, which could be in different environments. But the problem states that the two peaks are for the same alpha-proton, which would imply that the two protons are on the same carbon. But in a dipeptide, each peptide bond connects two amino acids, so each amino acid has its own alpha-proton. So if the dipeptide has two alpha-protons, they are on different carbons. Therefore, they are not the same alpha-proton. So that can't be.

Wait, maybe the two peaks are for diastereotopic protons. Diastereotopic protons are in different environments but on the same carbon. For example, in a chiral center, two protons on the same carbon but in different positions relative to the stereogenic center. If the molecule is a racemic mixture, the two enantiomers would have the same chemical shift for the diastereotopic protons. Therefore, the NMR would show a single peak. But the problem states that there are two peaks. So maybe the racemic mixture is not the case. Alternatively, if the diastereotopic protons are in different diastereomers, but the diastereomers are present in a 1:1 ratio, leading to two peaks of 0.5H each. But the problem states that spin-spin coupling is ruled out, so maybe the diastereotopic protons are coupled to another set of protons, but the coupling isn't observed. Hmm.

Alternatively, the two peaks in the NMR are due to two different diastereomers that have the same chemical shift for the alpha-proton. For example, if the diastereomers have different structures but the alpha-proton is in the same environment. But that seems unlikely unless the substitution pattern is the same. But the problem states that the two peaks have similar chemical shifts, which is possible if they are in similar environments.

But the problem says that the coupling pattern ruled out spin-spin coupling. So maybe the two peaks are due to coupling between the two alpha-protons. But in that case, the coupling constant would cause splitting, but the problem states that spin-spin coupling is ruled out. So perhaps the coupling is through space (dipolar coupling), which is usually not observed in routine NMR. However, if the coupling is through space, then the two protons would split each other's signals. But if they are equivalent, they would average out. But the problem states that there are two peaks, so maybe they are not equivalent. But if they are diastereotopic, perhaps they are not equivalent. Wait, diastereotopic protons are not equivalent, so their coupling would cause splitting. But if the coupling is observed, we would see more peaks. However, the problem says that spin-spin coupling is ruled out, so maybe the coupling is not through bonds (like through space), but the problem states that the coupling pattern ruled out spin-spin coupling, so maybe the coupling is something else.

This is getting complicated. Let's think about the options.

Option A: The crude compound exists as a mixture of enantiomers. If it's a racemic mixture, the NMR would show a single peak for the alpha-proton. But the problem says two peaks. So A is probably not the answer.

Option B: Double coupling has occurred during amide bond forming. That's a bit unclear. Double coupling would mean coupling to two protons. But the problem states that spin-spin coupling is ruled out. So B is unlikely.

Option C: The crude compound exists as a mixture of diastereoisomers. If there are diastereomers, each would have its own set of signals. If the diastereomers have the same chemical shift for the alpha-proton, the NMR would show a single peak. But the problem states that there are two peaks with similar shifts. However, if the diastereomers have different coupling constants, maybe the splitting is different. But the problem says that the coupling pattern ruled out spin-spin coupling, so perhaps the diastereomers have different coupling to another proton. But this is speculative.

Option D: The compound is contaminated with a precursor. If the precursor is a larger molecule that fragments into the product, the LC-MS would show the precursor and the product. However, if the precursor is present in equal amounts, the peak would have a higher intensity. But the problem states that the two peaks have equal intensity. If the precursor is present in equal amounts, then the product would have a peak. But the problem says the product's NMR shows two peaks. So D might not explain the NMR peaks.

Wait, but if the precursor is a larger molecule that breaks down into the product, then the product would have two peaks. But the NMR is of the crude compound. So the crude compound would contain both the precursor and the product. If the precursor is a dipeptide, then the product would be the dipeptide. Wait, but if the precursor is a larger molecule that can fragment into the dipeptide, then the crude compound would have both the precursor and the dipeptide. The LC-MS would show two peaks: one for the precursor and one for the dipeptide. The problem states that the LC-MS shows two peaks of equal intensity. So if the precursor and the dipeptide are present in equal amounts, that would explain the equal intensity. But the NMR shows two peaks for the same alpha-proton. How? If the dipeptide has two alpha-protons, each in different environments, then the NMR would show two peaks. But the problem says the two peaks are for the same alpha-proton. So that doesn't make sense. Unless the precursor has the same alpha-proton as the dipeptide, but that would require the precursor to have the same structure as the dipeptide, which is unlikely.

Alternatively, the precursor is a diastereomer of the dipeptide. For example, if the dipeptide has a certain configuration, the precursor is the opposite configuration. Then, the LC-MS would show two peaks of the same mass (since the diastereomers have the same molecular weight), and the NMR would show two peaks for the alpha-proton. If the diastereomers are present in a 1:1 ratio, each would contribute 0.5H. But the problem states that the NMR shows two peaks for the same alpha-proton. If the diastereomers have different chemical shifts for the alpha-proton, then the NMR would show a single peak. But the problem says the shifts are similar. So this is conflicting.

Alternatively, if the diastereomers have the same shift for the alpha-proton but different coupling constants, leading to two peaks. But the problem states that spin-spin coupling is ruled out, so maybe the coupling is not between the alpha-protons but between them and another set. However, the problem states that the coupling pattern ruled out spin-spin coupling, so perhaps the two peaks are due to coupling between the two alpha-protons, but that's spin-spin coupling, which was ruled out. Therefore, this is not possible.

Wait, maybe the two peaks are due to two different diastereomers, each contributing a peak. If the diastereomers are present in a 1:1 ratio, each would contribute 0.5H. But the problem states that the two peaks have similar chemical shifts. If the diastereomers are present in a 1:1 ratio and have the same shift for the alpha-proton, the NMR would show a single peak. But the problem says two peaks. Therefore, this is not the case.

Hmm. Let's consider another angle. The LC-MS shows two peaks of equal intensity with the same mass. This suggests that the compound is present as a racemic mixture (if the two peaks are enantiomers) or as a diastereomer mixture (if the two peaks are diastereomers of the same mass). If it's a racemic mixture, the NMR would show a single peak. But the problem says two peaks. Therefore, the contamination cannot be a racemic mixture. So Option A is incorrect.

If the compound is a diastereomer mixture, and the two diastereomers have the same mass, then the LC-MS would show two peaks. The NMR would show two peaks if the diastereomers have different shifts for the alpha-proton. However, the problem states that the two peaks have similar chemical shifts. If the diastereomers have different coupling constants, leading to different splitting, but the problem says spin-spin coupling is ruled out. So maybe the diastereomers have different coupling to another proton, but the problem states that coupling is ruled out. Therefore, this is conflicting.

Alternatively, the two peaks are due to two different diastereomers that are present in a 1:1 ratio, but their alpha-protons are equivalent. But that would require the diastereomers to have the same shift for the alpha-proton. If so, the NMR would show a single peak. But the problem says two peaks, so that's not possible.

Wait, maybe the two peaks in the NMR are due to two different diastereomers that have the same shift for the alpha-proton but different coupling to another proton. For example, if the diastereomers have different coupling constants to a third proton, leading to different splitting. However, the problem states that spin-spin coupling is ruled out, so maybe the coupling to the third proton is not spin-spin. But in NMR, coupling to any proton is spin-spin unless it's dipolar coupling, which is usually not observed. Therefore, this seems unlikely.

Another possibility: the two peaks in the NMR are due to the same molecule having two non-equivalent alpha-protons. But in a dipeptide, each amino acid has its own alpha-proton. Unless the two alpha-protons are in different environments, but the problem says they are the same alpha-proton. So that can't be.

Wait, perhaps the two peaks are due to two different molecules in the mixture. For example, the dipeptide and a diastereomeric form. But again, the problem states that the two peaks are for the same alpha-proton. So that's not possible.

I'm getting stuck here. Let me try to think of the options again.

Option D: contamination with a precursor. The LC-MS would show the precursor and the dipeptide. If the precursor is a larger molecule that breaks down into the dipeptide, then the dipeptide would be the main product. The NMR would show the alpha-protons of the dipeptide. But the problem states that the NMR shows two peaks for the same alpha-proton. So unless the dipeptide has two alpha-protons in the same environment, which it doesn't. Each amino acid has its own alpha-proton. Therefore, the NMR would show two peaks for the two amino acids' alpha-protons. But the problem says two peaks for the same alpha-proton. Therefore, this cannot be the case.

Option C: mixture of diastereoisomers. If there are diastereomers, each would have its own set of signals. If the diastereomers have the same shift for the alpha-proton, the NMR would show a single peak. But the problem says two peaks. However, if the diastereomers have different coupling constants, leading to different splitting, but the problem says spin-spin coupling is ruled out. So maybe the diastereomers have the same shift but different coupling via dipolar coupling, leading to two peaks. But the problem states that spin-spin coupling is ruled out. Therefore, this is speculative.

Alternatively, if the diastereomers are present in a 1:1 ratio and their alpha-protons are equivalent, the NMR would show a single peak. But the problem says two peaks. Therefore, this is not possible.

Wait, maybe the diastereomers are not in a 1:1 ratio. If one diastereomer is present in excess, the NMR would show one peak. But the problem states that the two peaks have equal integrals (0.5H each). Therefore, the diastereomers must be present in a 1:1 ratio. But if they are present in a 1:1 ratio and have the same shift, the NMR would show a single peak. Therefore, this is not possible.

Hmm. This is really challenging. Let me think about the problem again.

The key points:

1. NMR shows two peaks for the same alpha-proton, integrated to 1H (each 0.5H). The peaks have similar shifts, ruling out spin-spin coupling.

2. LC-MS shows two peaks of equal intensity with same mass. This suggests either a racemic mixture (if the peaks are enantiomers) or a diastereomer mixture (if the peaks are diastereomers of the same mass). But if the mixture is racemic, the NMR would show one peak, not two. Therefore, the contamination is not a racemic mixture.

3. Therefore, the compound is a mixture of diastereomers. The LC-MS shows two peaks of the same mass (the diastereomers have the same molecular weight). The NMR shows two peaks for the same alpha-proton. How is that possible?

If the diastereomers have the same shift for the alpha-proton, the NMR would show a single peak. But the problem states two peaks. Therefore, the diastereomers must have different shifts. However, the problem states that the two peaks have similar shifts. Therefore, the diastereomers must have slightly different shifts but are close enough to appear as two peaks. If the diastereomers are present in a 1:1 ratio, each contributing 0.5H. But the problem states that the coupling pattern ruled out spin-spin coupling. Therefore, maybe the diastereomers have different coupling patterns leading to different splitting, but the problem states that spin-spin coupling is ruled out.

Alternatively, the diastereomers are interconvertible, but that's not likely in this context.

Wait, maybe the diastereomers are not the same as the parent compound. For example, the parent compound is a dipeptide, and the diastereomers are different diastereotopic forms. But the problem states that the two peaks are for the same alpha-proton, which would not make sense if the diastereomers are diastereotopic.

Alternatively, the diastereomers are not diastereotopic but have different side chain configurations. For example, one diastereomer has an R configuration at the alpha-carbon, and the other has an S. But in a racemic mixture, the NMR would average out. Therefore, the NMR would show a single peak. But the problem states two peaks, so that's not possible.

I'm really stuck here. Let's consider the answer options again.

Option D: contamination with a precursor. If the precursor is a diastereomer of the product, then the LC-MS would show two peaks of the same mass (the diastereomers). The NMR would show two peaks for the alpha-proton. If the diastereomers are present in a 1:1 ratio, each would contribute 0.5H. But the problem states that the two peaks have similar shifts, which would require the diastereomers to have the same shift for the alpha-proton. If that's the case, the NMR would show a single peak. But the problem says two peaks. Therefore, this is not possible.

Wait, but if the diastereomers are not in a 1:1 ratio, but the two peaks have the same integral, then the ratio must be 1:1. Therefore, the diastereomers are present in a 1:1 ratio, but their alpha-protons have different shifts. However, the problem states that the shifts are similar. Therefore, this is conflicting.

Alternatively, maybe the diastereomers are present in a 1:1 ratio, and the coupling of the alpha-proton to another proton is different in each diastereomer. For example, if one diastereomer has coupling to a proton J, and the other has coupling to a proton K, leading to different splitting. But the problem states that spin-spin coupling is ruled out, so this is not possible. Therefore, this is not the case.

Hmm. Maybe the answer is D: contamination with a precursor. Because the LC-MS shows two peaks with same mass. If the crude compound is a mixture of the dipeptide and a precursor that breaks down into the dipeptide, then the LC-MS would show two peaks: one for the dipeptide and one for the precursor. But the problem states that the crude compound is analyzed, so the dipeptide would be the main product. The NMR would show the dipeptide's alpha-protons. But the problem states that the NMR shows two peaks for the same alpha-proton, which would require that the dipeptide has two alpha-protons in the same environment. Which it doesn't. So this can't be.

I think I need to make a decision here. Let's consider the options again.

The most plausible answer seems to be D: contamination with a precursor. Because the LC-MS shows two peaks of equal intensity with the same mass. The diastereomers have the same mass, so they would show up as separate peaks. The NMR would show two peaks for the same alpha-proton if the diastereomers have the same shift for the alpha-proton. However, the problem states that the two peaks have similar shifts, which would require the diastereomers to have the same shift. If they are present in a 1:1 ratio, the NMR would show a single peak. Therefore, this is conflicting.

Alternatively, the diastereomers are present in a 1:1 ratio, and the two peaks in the NMR are due to the fact that the diastereomers are diastereotopic. For example, if the diastereomers have different configurations at the alpha-carbon, causing the alpha-proton to split into two peaks. However, if they are diastereotopic, the splitting would cause more peaks, not just two. But the problem states that the coupling pattern ruled out spin-spin coupling, so perhaps the diastereomers are not diastereotopic. Therefore, the two peaks in the NMR are due to two different diastereomers, each contributing 0.5H, with similar shifts. The LC-MS shows two peaks of the same mass (the diastereomers), each integrating to 0.5H. This would fit the description. However, the problem states that the coupling pattern ruled out spin-spin coupling, which would mean that the diastereomers do not have coupling between the alpha-protons, so the NMR would show two separate peaks. Therefore, this seems plausible.

Therefore, the answer is D: contamination with a precursor. Wait, but the contamination would be the diastereomers. Wait, the contamination is a precursor. If the crude compound contains a precursor and the product, then the LC-MS would show two peaks. The NMR would show the alpha-protons of the product. But the problem states that the NMR shows two peaks for the same alpha-proton, which would not make sense if the product is a dipeptide. Unless the diastereomers are the same as the product. But the problem says the diastereomers have the same mass. Therefore, this is confusing.

Alternatively, perhaps the diastereomers are the same as the product, but present in a 1:1 ratio. The contamination with a precursor would not affect the NMR of the product. Therefore, the NMR would show the product's alpha-protons, which would be two peaks if the product has two alpha-protons in different environments. But the problem states that the NMR is consistent with the expected molecule, which is a dipeptide. Therefore, the diastereomers must be the dipeptide. Therefore, the contamination is not related to the product, but to the diastereomers. Therefore, the contamination would be diastereomers, not a precursor. Therefore, the answer would be C: mixture of diastereoisomers.

But earlier I thought that if the diastereomers are present in a 1:1 ratio, the NMR would show a single peak. However, the problem states that the two peaks have similar shifts, which would require the diastereomers to have the same shift for the alpha-proton. Therefore, the NMR would show a single peak. But the problem says two peaks. Therefore, this is conflicting.

I think I have to make a choice here. After considering all possibilities, the most plausible answer is D: contamination with a precursor. But I'm not entirely sure. Alternatively, the correct answer is C: mixture of diastereoisomers.
</think>

The question involves analyzing a small peptidic compound that has been analyzed using NMR and LC-MS. The NMR spectrum shows two peaks for the same alpha-proton, each integrating to 0.5H, with similar chemical shifts. The coupling pattern ruled out spin-spin coupling. The LC-MS analysis at elevated temperature shows two peaks of equal intensity with the same mass. The task is to determine the most likely explanation for these observations.

### Key Observations and Analysis:
1. **NMR Data**:
   - Two peaks for the same alpha-proton, each integrating to 0.5H, with similar chemical shifts.
   - The coupling pattern ruled out spin-spin coupling. This suggests that the two peaks are not due to coupling between the alpha-protons themselves but rather due to another source of splitting.

2. **LC-MS Data**:
   - Two peaks of equal intensity with the same mass.
   - This indicates the presence of two compounds with the same molecular weight, likely diastereomers or a racemic mixture.

3. **Critical Considerations**:
   - **Option A (Mixture of enantiomers)**: If the compound were a racemic mixture, the NMR would show a single peak for the alpha-proton, as the two enantiomers would average the signals. However, the problem states two distinct peaks, ruling out this option.
   - **Option B (Double coupling during amide bond formation)**: The problem explicitly states that spin-spin coupling is ruled out, making this explanation unlikely.
   - **Option C (Mixture of diastereoisomers)**: Diastereomers can have different physical properties, including different chemical shifts for the same alpha-proton. If the diastereomers are present in a 1:1 ratio, the NMR would show two peaks of equal intensity. However, the problem states that the shifts are similar, which could still be consistent with diastereomers if their chemical environments are slightly different. The LC-MS data supports this explanation, as diastereomers would have the same molecular weight and would appear as two peaks of equal intensity.
   - **Option D (Contamination with a precursor)**: If the crude compound were contaminated with a precursor, the LC-MS would show two peaks: one for the precursor and one for the dipeptide. However, the problem states that the NMR is consistent with the expected molecule (a dipeptide), which implies that the contamination does not involve the product. Additionally, the two peaks in the NMR for the same alpha-proton are not explained by contamination.

### Conclusion:
The most plausible explanation is that the crude compound exists as a mixture of diastereoisomers. Diastereomers can have different physical properties, including different chemical shifts for the same alpha-proton. If the diastereomers are present in a 1:1 ratio, the NMR would show two peaks of equal intensity. The LC-MS data also supports this explanation, as diastereomers would have the same molecular weight and would appear as two peaks of equal intensity.

ANSWER: C