Okay, so I need to figure out which compound is the correct answer based on the given spectral data. Let me start by going through each piece of information step by step.

First, the molecular ion peak is at m/z 156, and there's a peak at 158 with 32% intensity. The molecular formula can be calculated from the molecular ion. The molecular ion (M+) is the most abundant peak, so if the molecular weight is 156, the molecular formula's molecular weight should be 156. The next peak at 158 is probably a fragment, like a loss of a hydrogen atom (since 158 - 156 = 2, but wait, loss of H is usually by 1, but maybe that's a mistake here. Wait, 158 - 156 is 2, which could be a loss of two H atoms, but that's less common. Alternatively, maybe it's a different isotope? But chlorine has a natural abundance of about 75%, so maybe the molecular weight is Cl, which is about 35.5. Wait, 35.5 would make the molecular weight 156 + 35.5 = 191.5, which doesn't match. Hmm, maybe not. Alternatively, maybe the molecular ion is 156, and the next peak is 158, which might be a chlorine substitution. Wait, molecular weight of Cl is around 35, so 35 would make the molecular weight 156 + 35 = 191, but that's not matching. Wait, perhaps the molecular formula is C8H8O2Cl? Let me check. Let's see: 12*8 + 1*8 + 16*2 + 35.5 = 96 +8 +32 +35.5 = 171.5. That's higher than 156. So maybe that's not right. Alternatively, maybe the molecular formula is C7H8O2Cl. Let's calculate that: 12*7 + 1*8 + 16*2 +35.5 = 84 +8 +32 +35.5 = 159.5. Close to 156. Hmm, maybe the molecular ion is 156, so perhaps the molecular formula is C7H6O2Cl. Let's check: 12*7=84, 6*1=6, 16*2=32, 35.5 for Cl. Total 84+6+32+35.5=157.5. Close to 156, but not exact. Maybe another approach. Alternatively, perhaps the molecular formula is C7H6O2Cl. Wait, 12*7=84, 6*1=6, 16*2=32, Cl is 35.5. Total 84+6=90, +32=122, +35.5=157.5. Hmm, not 156. Maybe the molecular formula is C6H5ClO2. Let's calculate: 6*12=72, 5*1=5, Cl=35.5, 2*16=32. Total 72+5=77, +35.5=112.5, +32=144.5. That's too low. Alternatively, maybe the molecular formula is C7H5ClO2. Let's see: 7*12=84, 5*1=5, Cl=35.5, 2*16=32. Total 84+5=89 +35.5=124.5 +32=156.5. Close to 156. So molecular formula could be C7H5ClO2. That would give a molecular weight of approximately 156.5, which rounds to 156. The molecular ion is 156, and the next peak at 158 could be a chlorine isotope? Wait, chlorine has two main isotopes: 35Cl (about 75%) and 37Cl (25%). So a molecular ion at m/z 156 would be for C7H5ClO2 with 35Cl. The 37Cl would be at m/z 158 (since 35Cl is 35, 37Cl is 37, so 156 +2 =158). So the peak at 158 would be chlorine substitution. That makes sense. So the molecular formula is likely C7H5ClO2. Let me confirm: 7 carbons, 5 hydrogens, one chlorine, two oxygens. So the molecular weight would be 7*12 +5*1 +35.5 +2*16 = 84 +5 +35.5 +32 = 156.5. That's spot on. So the molecular formula is C7H5ClO2.

Next, the IR spectrum: a broad peak from 3500-2700 cm^-1. Broad peaks in that region usually indicate O-H stretching, like in phenols or alcohols. Also, there's a strong sharp peak at 1720 cm^-1. A peak around 1720 is typical for a carbonyl group, like a ketone, aldehyde, or maybe an ester. However, the presence of a broad O-H peak suggests a phenol or alcohol. But wait, if it's a carbonyl (like an aldehyde or ketone), then the IR would show that. Let's see the options.

Looking at the options:

A. 3-Chloro-2-hydroxybenzaldehyde: This would have a benzaldehyde group (aldehyde) and a hydroxyl group (phenol). The IR would show a strong aldehyde C=O (around 1720 cm^-1) and the O-H stretch (broad). So that fits the IR data.

B. Phenyl chloroformate: Chloroformate would have a carbonyl group. The structure is Cl-CO-O-Ph. The IR would show a strong carbonyl (around 1720 cm^-1), but phenyl chloroformate has a Cl and an OCH3 (but wait, chloroformate is Cl-CO-O-R? Wait, chloroformate is more like Ar-CO-O-Cl? Wait, no, chloroformate is O=C(O-Cl)Ph. So the carbonyl is present, but the O-Cl might not show a strong peak. Also, the molecular formula would be C7H5ClO2. Let's check: phenyl chloroformate is Ph-C(O)-O-Cl. Ph is C6H5, so total C7H5ClO2. Yes, that matches. So molecular formula is correct. IR would show a carbonyl at around 1720, but chloroformate might have a strong Cl-O peak? Wait, chloroformate (OCH2Cl? No, chloroformate is more like O=C(O-Cl)Ph. The O-Cl group is an O-C-O-Cl structure. The O-Cl stretch is around 600-1000 cm^-1. But the IR here shows a broad peak from 3500-2700, which would be O-H if present. Phenyl chloroformate doesn't have an O-H group. So the IR data for B would have a carbonyl (1720) but no broad O-H. However, option B's IR would not match the given data. Wait, but the given IR has a broad peak from 3500-2700, which is O-H. So if the compound is phenyl chloroformate, there's no O-H, so the IR wouldn't show that. Therefore, B is unlikely.

C. 2-chlorobenzoic acid: This would have a carboxylic acid group (-COOH) and a Cl substituent. The IR would show a broad O-H stretch (from COOH) around 2500-3500, which matches the given data. The carbonyl would be around 1700 cm^-1, but benzoic acid's carbonyl is around 1700, but the given IR has a strong peak at 1720. Benzoic acid's carbonyl is a bit lower. Also, benzoic acid's IR would have a strong O-H (broad) around 2500-3500, which matches. But the molecular formula for 2-chlorobenzoic acid is C7H5ClO2. Let's check: benzoic acid is C7H6O2, replacing one H with Cl gives C7H5ClO2. Yes, molecular formula matches. So molecular weight would be 12*7 +1*5 +35.5 +16*2=84 +5 +35.5 +32=156.5. So molecular formula is correct. The IR would show a broad O-H (matches) and a carbonyl at around 1700, but the given IR has 1720. So maybe the carbonyl is slightly higher. Benzoic acid's carbonyl is a bit lower than 1720. So this could be possible, but the IR here has a stronger peak at 1720. Wait, but the IR given has a strong sharp peak at 1720. Benzoic acid's carbonyl is usually a strong absorption around 1700, but if there's electron-withdrawing groups, it can shift higher. Wait, 2-chlorobenzoic acid would have a Cl substituent adjacent to the COOH group. Chlorine is an electron-withdrawing group, so the carbonyl would be deshielded, shifting the absorption to a higher wavenumber. So 1720 would be plausible. So C could be a possibility.

D. 4-chlorobenzoic acid: Similar to C, but with Cl at the para position. The same reasoning applies. Molecular formula is C7H5ClO2. IR would have a broad O-H (from COOH) and a carbonyl around 1720 (shifted up due to Cl). So this also fits. But the question is which one is correct. Let's look at the NMR data.

The 1H NMR: 11.0 ppm (s, 1H), which is a singlet integrating to 1H. That's probably a carbonyl proton, like in a carbonyl group. But wait, in aldehydes, the proton is around 9-10 ppm. In ketones, it's lower. Wait, 11 ppm is quite high. That's more like a carboxylic acid proton, which is around 12-13 ppm. But in the NMR given, there's a peak at 11.0 ppm. Alternatively, maybe it's an aldehyde proton, but aldehydes are usually around 9-10. Wait, perhaps it's an aromatic aldehyde proton. Wait, but the IR shows a strong peak at 1720, which is more like a carbonyl (maybe aldehyde or ketone), and the O-H (phenol or carboxylic acid). Let's think.

In the NMR, 11.0 ppm (s, 1H): if that's a carbonyl proton, like an aldehyde, then the structure would have an aldehyde group. Then the IR would show a strong aldehyde C=O (around 1720, which matches), and a broad O-H (from a phenol). So that would fit option A (3-chloro-2-hydroxybenzaldehyde). Let's check the NMR for that structure.

In 3-chloro-2-hydroxybenzaldehyde, the aldehyde proton is at around 9-10 ppm. But here it's at 11.0 ppm. That's higher than typical aldehydes. Alternatively, maybe it's a carboxylic acid proton. But carboxylic acid protons are usually around 12-13 ppm. However, the IR for a carboxylic acid would have a broad O-H, but the NMR for a carboxylic acid would have a singlet around 12-13 ppm. Let's see the options again.

Looking at the NMR data: 8.02 ppm (d, 2H) and 7.72 (d, 2H). These are aromatic protons, likely in a para-disubstituted benzene. The coupling constants (d, doublet) suggest that each proton is coupled to another, so maybe two doublets, indicating a para-substituted benzene with two substituents in the para positions. Wait, but if there are two substituents, say Cl and OH in para positions, but the NMR shows two sets of doublets. Wait, maybe the substituents are in the meta positions. Wait, no, in a para-disubstituted benzene, if there are two substituents, they could be in the 1,3 positions (meta) or 1,2 (ortho), but the splitting would differ. Wait, let's think. For a para-disubstituted benzene with substituents in the 2 and 5 positions (meta), the coupling would result in two doublets. Wait, maybe the structure is such that there are two substituents leading to two doublets. Let's consider the 1H NMR data.

The NMR shows two doublets at 8.02 ppm (2H) and 7.72 ppm (2H). Let's think about the substitution pattern. For example, if the substituents are in the 2 and 5 positions (meta), the coupling would cause the aromatic protons to split into two doublets. Alternatively, if the substituents are in the 3 and 5 positions (which is para to each other?), wait, maybe not. Let me draw this out mentally.

Alternatively, perhaps the substituents are in the 2 and 4 positions (ortho), but that might cause a different splitting. Wait, maybe it's a para-substituted benzene with two substituents in the 2 and 5 positions, making them meta to each other. Wait, but that would leave the aromatic protons split into two sets. Let's think: if the benzene ring has substituents at positions 2 and 5, then the protons at positions 3 and 4 would be equivalent, and positions 6 and 1 would be equivalent. But this would lead to two sets of doublets, each with 2H. The chemical shifts would depend on the substituents. For example, a substituent at position 2 (ortho to the aldehyde) and position 5 (meta to the aldehyde) would cause different shifts. Alternatively, maybe the substituents are at positions 2 and 5, making the aldehyde at position 1, with substituents at 2 and 5. Let's think of the structure.

Alternatively, maybe the substituents are at positions 3 and 5. Let me think of the 3-chloro-2-hydroxybenzaldehyde structure. The aldehyde is at position 1 (assuming numbering starts at the aldehyde), then substituents at 2 (OH) and 3 (Cl). Then the aromatic protons would be at positions 4,5,6. Wait, but that might complicate the splitting. Alternatively, perhaps the substituents are at positions 3 and 5, leading to two doublets in the NMR.

Alternatively, maybe the substituents are at positions 4 and 5. Let me try to figure this out.

Wait, the NMR shows two doublets with 2H each. Let's consider the structure of 3-chloro-2-hydroxybenzaldehyde. The aldehyde is at position 1, then substituents at 2 (OH) and 3 (Cl). The aromatic protons would be at positions 4,5,6. Let's see:

- Position 4: adjacent to substituent at 3 (Cl) and 5 (if any). Wait, in this structure, there are substituents at 2 (OH) and 3 (Cl). So the aromatic protons would be at positions 4,5,6. The substituents at 2 and 3 would affect the shifts and splitting. The protons at 4 and 6 would be adjacent to substituents, leading to different shifts. The coupling between protons would depend on their positions. For example, if there are two substituents, the aromatic protons might split into two doublets. Let me think of the coupling constants. The NMR shows two doublets, each integrating to 2H. Maybe the protons at positions 4 and 6 are equivalent (meta to substituents) and the protons at positions 5 and (another position) are equivalent. Wait, perhaps the substituents are in positions 2 and 4, leading to a para-substitution. But I'm getting a bit confused here. Let me try another approach.

The NMR data: 11.0 ppm (s, 1H), 8.02 ppm (d, 2H), 7.72 ppm (d, 2H). The singlet at 11.0 ppm suggests a proton with no neighboring protons, which is unusual for an aromatic proton. Wait, unless it's an aldehyde proton. But aldehyde protons are typically around 9-10 ppm. Here it's at 11.0, which is higher. Alternatively, maybe it's a carboxylic acid proton, which is usually around 12-13 ppm. But the IR shows a broad O-H, which could be from a carboxylic acid. Let's check the options again.

Option C: 2-chlorobenzoic acid. Structure is Cl-C6H4-COOH. The COOH group would have a proton at around 12-13 ppm (singlet if no neighboring protons). The aromatic protons would be in positions 3,4,5,6. If the substituents are at positions 2 (Cl) and the COOH is at position 1, then the aromatic protons would be at positions 3,4,5,6. The coupling would depend on the substituents. For example, the protons adjacent to the Cl (position 3) and the COOH (position 1) would have different shifts. However, the NMR shows two doublets, each integrating to 2H. Let's see: if the substituents are at positions 2 (Cl) and 1 (COOH), then the aromatic protons would be at positions 3,4,5,6. The coupling between protons at positions 3 and 5 (meta) and 4 and 6 (para) might create two doublets. Wait, but if there's a Cl at position 2, then the protons at positions 3 and 6 would be ortho to Cl, and the protons at positions 4 and 5 would be meta. This might lead to two doublets. Let me check the chemical shifts.

In 2-chlorobenzoic acid, the COOH group is at position 1. The substituent at position 2 (Cl) would deshield the aromatic protons. The protons at positions 3 and 6 would be ortho to Cl and meta to COOH. The protons at positions 4 and 5 would be ortho to COOH and meta to Cl. The coupling would cause the protons at positions 3 and 6 to be a doublet, and the protons at positions 4 and 5 to be another doublet. The chemical shifts for the protons ortho to Cl (positions 3 and 6) would be downfield, around 8.5-9 ppm, while the protons ortho to COOH (positions 4 and 5) would be around 7.5-8 ppm. However, the given NMR has peaks at 8.02 and 7.72 ppm, which are within this range. The singlet at 11.0 could correspond to the COOH proton, which is a singlet because it's not adjacent to any other protons (since COOH is at position 1, the adjacent protons would be at position 2 (Cl) and position 6 (aromatic). Wait, but in benzoic acid, the COOH is at position 1, and position 2 has Cl. So the COOH proton is at position 1, and the adjacent aromatic protons are at positions 2 and 6. However, the COOH proton is part of the carboxylic acid group, which is deprotonated in solution, but in the NMR, the carboxylic acid proton would be present if the solution is acidic. Wait, but if the solution is not very acidic, the carboxylic acid proton might not be present. Alternatively, if the sample is in a form where the carboxylic acid is deprotonated, the NMR would not show that peak. But the given NMR has a peak at 11.0 ppm, which is a singlet for 1H. If that's the COOH proton, then the sample is in an acidic solution where the COOH is protonated. So that could fit. The aromatic protons would show two doublets. If the substituents are at positions 2 (Cl) and 1 (COOH), then the aromatic protons would be at positions 3,4,5,6. The coupling between 3 and 6 (ortho to Cl) would be a doublet, and the coupling between 4 and 5 (meta to Cl and para to COOH) would be another doublet. The chemical shifts would be around 8.02 ppm (positions 3 and 6) and 7.72 ppm (positions 4 and 5). The singlet at 11.0 would be the COOH proton. This matches the NMR data. So option C (2-chlorobenzoic acid) seems to fit.

But wait, let's check the other options again. Option A: 3-chloro-2-hydroxybenzaldehyde. Structure: aldehyde at position 1, OH at 2, Cl at 3. The aromatic protons would be at positions 4,5,6. The aldehyde proton is at position 1, so the adjacent protons are at 2 (OH) and 6 (aromatic). The aldehyde proton is expected around 9-10 ppm, but here it's at 11.0. The aromatic protons would be influenced by the substituents at positions 2 (OH) and 3 (Cl). The OH is at position 2, which is ortho to the aldehyde. The Cl is at position 3, which is meta to the aldehyde. The coupling between aromatic protons would lead to a more complex splitting. The NMR shows two doublets, each integrating to 2H. Let's think: if the substituents are at positions 2 (OH) and 3 (Cl), the aromatic protons would be at positions 4,5,6. The coupling between 4 and 5 (ortho) and 5 and 6 (meta) might create a multiplet. But the given NMR shows two doublets, so maybe this structure doesn't fit. Alternatively, perhaps the splitting is due to coupling with the aldehyde proton, but the aldehyde proton is at 11.0, which is a singlet. Wait, but in the NMR, there's a singlet at 11.0 ppm (1H), which would correspond to the aldehyde proton if present. However, in the structure of 3-chloro-2-hydroxybenzaldehyde, the aldehyde group is at position 1, and the adjacent protons are at positions 2 (OH) and 6 (aromatic). However, the aldehyde proton is usually a singlet because there's no coupling with adjacent protons (since the aldehyde is a CHO group, the adjacent carbon is a CH, but the aldehyde proton is usually deshielded and doesn't split with adjacent protons unless there's conjugation. Wait, but in the NMR, the aldehyde proton is a singlet. Wait, the given NMR has a singlet at 11.0 ppm (1H), which could be the aldehyde proton. Then the aromatic protons would be split into two doublets. Let's think: in 3-chloro-2-hydroxybenzaldehyde, the aromatic protons would be at positions 4,5,6. The substituents at 2 (OH) and 3 (Cl) would affect their shifts. The coupling between 4 and 5 (ortho to Cl) and 5 and 6 (meta to Cl and OH) might cause the splitting. But would they result in two doublets? Let's imagine: the protons at 4 and 6 are each adjacent to Cl and OH, respectively. The protons at 5 would be between them. The coupling constants (J values) would determine the splitting. If the coupling between 4 and 5 is J1 and between 5 and 6 is J2, then the splitting could create a doublet of doublets or a multiplet. However, the NMR shows two doublets, each integrating to 2H. This suggests that there are two sets of equivalent protons, each as a doublet. So perhaps the substituents are arranged such that there are two sets of equivalent aromatic protons. For example, if the substituents are in positions that create two sets of para protons. Wait, but in 3-chloro-2-hydroxybenzaldehyde, the substituents are in positions 2 (OH) and 3 (Cl), leaving the aromatic protons at 4,5,6. The protons at 4 and 6 are each ortho to Cl and meta to OH, while the protons at 5 are between them. This might not result in two doublets. Alternatively, if the substituents are in positions that create two doublets. Maybe the substituents are in positions that make the aromatic protons split into two sets. For example, if the substituents are in the para positions, but in this case, the substituents are at 2 and 3, which are meta. Hmm, perhaps this structure doesn't fit the NMR data. Alternatively, if the substituents are in positions that create two sets of equivalent protons. Wait, maybe the substituents are in positions 3 and 5, making the aromatic protons split into two doublets. Let me think: substituents at 3 (Cl) and 5 (OH), then the aromatic protons would be at 2,4,6. But that's if substituents are at 3 and 5. Wait, but in the given structure, substituents are at 2 (OH) and 3 (Cl). So the aromatic protons are at 4,5,6. The coupling between 4 and 5 (J1), and 5 and 6 (J2) might create complex splitting. But the NMR shows two doublets, which suggests two sets of equivalent protons. So perhaps the substituents are arranged such that there are two sets of equivalent aromatic protons. For example, in 4-chlorobenzoic acid (option D), the substituents are at positions 4 (Cl) and 1 (COOH). Then the aromatic protons would be at 2,3,5,6. The substituents at 4 (Cl) and 1 (COOH) would influence the shifts. The protons at 2 and 6 would be ortho to Cl, and the protons at 3 and 5 would be meta to Cl and adjacent to COOH. The coupling would be between 2 and 3 (meta), 3 and 4 (Cl), but this is getting complicated. Alternatively, perhaps the substituents are in positions that create two doublets. But I'm not entirely sure. Let's consider the molecular formula again. Both options C and D have the correct molecular formula. Now, looking at the IR data again: strong peak at 1720. For option C (2-chlorobenzoic acid), the carbonyl is from the COOH group. Benzoic acid's carbonyl is around 1700 cm-1, but with electron-withdrawing groups like Cl, the carbonyl would be shifted higher, around 1720. That matches the IR data. For option D (4-chlorobenzoic acid), the substituents are para, so the effect on the carbonyl might be different. The para-substitution might have a smaller effect on the carbonyl. However, the IR data shows a strong peak at 1720, which suggests a higher carbonyl. So option C is more likely. Now, the 1H NMR: option C would have a COOH proton at around 12-13 ppm, which matches the peak at 11.0? Wait, no. Wait, if the COOH is in position 1, then the adjacent protons are at 2 (Cl) and 6 (aromatic). The COOH proton is in position 1, so it's not adjacent to any aromatic protons. Wait, but COOH is a carboxylic acid group, but in the NMR, the COOH proton would usually appear around 12-13 ppm as a singlet. However, in the given NMR, there's a singlet at 11.0 ppm. That's lower than typical for a COOH proton. Wait, unless the sample is in a basic solution, but that would deprotonate the COOH. Alternatively, perhaps the sample is in an acidic solution where the COOH is protonated, but that's unusual for an IR analysis. Alternatively, maybe the COOH is not present. Wait, but the IR shows a strong peak at 1720, which is consistent with a COOH group. So the COOH proton should be present. But the NMR shows a singlet at 11.0 ppm. The only way this makes sense is if the COOH is not present. Wait, but the IR shows a strong carbonyl peak, which would be present in COOH. So this is conflicting. Alternatively, maybe the COOH is not present. Wait, but the IR shows a broad O-H peak (from 3500-2700), which is present in COOH. So this suggests that the compound has an O-H group, which is present in COOH. Therefore, the COOH proton should be in the NMR. But the given NMR has a singlet at 11.0 ppm, which is lower than expected. Hmm, maybe the COOH is in a different position. Alternatively, perhaps the COOH is not at position 1. Wait, maybe the structure is 3-chlorobenzoic acid. Wait, but that would have the COOH at position 1 and Cl at position 3. Then the aromatic protons would be at 2,4,5,6. The substitution pattern would lead to different splitting. But I'm getting confused. Let me think again. The molecular formula is C7H5ClO2. Let's check the options again:

Option A: 3-Chloro-2-hydroxybenzaldehyde. Molecular formula: C7H5ClO2. Correct.

Option B: Phenyl chloroformate. Molecular formula: C7H5ClO2. Correct.

Option C: 2-chlorobenzoic acid. Molecular formula: C7H5ClO2. Correct.

Option D: 4-chlorobenzoic acid. Molecular formula: C7H5ClO2. Correct.

So all have the same molecular formula. Now, considering the NMR data. The singlet at 11.0 ppm (1H) could be the COOH proton if present, but the expected shift is higher. Alternatively, maybe it's the aldehyde proton, which is higher than typical. But aldehydes usually have protons around 9-10. Let's check the options:

Option A: aldehyde group. So the aldehyde proton would be around 9-10 ppm. But the NMR has a peak at 11.0 ppm. That's possible if the aldehyde is conjugated or in a specific environment. However, the IR data shows a strong carbonyl peak at 1720, which is more in line with a COOH group. So perhaps the aldehyde is not present. Therefore, options A and B (which have aldehyde or ketone groups) might not fit. Wait, option B is phenyl chloroformate, which is Cl-CO-O-Ph. The carbonyl would be around 1720, which matches. However, the IR would show a strong O-H from the phenol? Wait, no, phenyl chloroformate doesn't have an O-H group. So the IR would not show a broad O-H peak. But the given IR does have a broad O-H peak. Therefore, option B is unlikely. So options C and D are possibilities. Now, the NMR: 11.0 ppm is a singlet of 1H, which would be the COOH proton if present. But the expected shift for COOH is higher, maybe around 12-13 ppm. However, the given shift is 11.0. That's lower. Hmm. Alternatively, perhaps the sample is in a basic solution, but that would deprotonate the COOH. Wait, but if the sample is in a basic solution, the COOH would be deprotonated, and the NMR would not show the COOH proton. However, the IR would still show the carbonyl peak. But then the NMR would not have a peak at 11.0 ppm. Therefore, the presence of the peak at 11.0 ppm suggests that the COOH is protonated, but that's unusual. Alternatively, perhaps the COOH is in a different position. Let me think again. If the compound is 4-chlorobenzoic acid, the substituents are at positions 4 (Cl) and 1 (COOH). The aromatic protons would be at 2,3,5,6. The substituents at 4 (Cl) and 1 (COOH) would influence the shifts. The protons at 2 and 6 would be ortho to Cl, and the protons at 3 and 5 would be meta to Cl and adjacent to COOH. The coupling between protons would lead to two doublets. The COOH proton would be at position 1, adjacent to Cl at 4. So the COOH proton would be at position 1, and the adjacent aromatic protons are at 2 and 6. The coupling between 1 and 2, and 1 and 6 would cause splitting. However, the COOH proton is usually a singlet because it's not adjacent to any other protons (since it's a carboxylic acid group). Wait, but in the NMR, the COOH proton would be a singlet. However, the given NMR shows a singlet at 11.0 ppm (1H). If that's the COOH proton, then the peak would be around 12-13 ppm. But here it's at 11.0. So that's conflicting. Alternatively, perhaps the COOH is not present. But the IR data shows a strong peak at 1720, which is more in line with a COOH group. Therefore, this is confusing. Let me think of another approach. The molecular formula is C7H5ClO2. Let's calculate the degrees of unsaturation. Degrees of unsaturation = (2C + 2 - H - X + N)/2. Here, C=7, H=5, X=Cl (1), N=0. So (2*7 +2 -5 -1)/2 = (14+2-5-1)/2 = (10)/2 = 5 degrees of unsaturation. Which corresponds to a benzene ring (3 double bonds) and a carbonyl (1 double bond), and one ring or another double bond. So benzene ring (3), carbonyl (1), and one more (maybe Cl-substituted). So the structure must have a benzene ring and a carbonyl group. Now, looking at the IR data: strong peak at 1720 (carbonyl), broad O-H at 2700-2500. So the compound has a carbonyl and an O-H group. The options that have both are options A, B, C, D. But option B (phenyl chloroformate) has Cl-CO-O-Ph, which doesn't have an O-H. So option B is out. So we're left with A, C, D. Now, looking at the 1H NMR: 11.0 ppm (s, 1H), 8.02 (d, 2H), 7.72 (d, 2H). The singlet at 11.0 could be a COOH proton if it's present. But the expected shift is higher. Alternatively, it could be an aldehyde proton, which is around 9-10 ppm. But the IR data suggests a carbonyl, which could be a COOH or a ketone/aldehyde. Let's consider option C: 2-chlorobenzoic acid. Structure: Cl-C6H4-COOH. The aromatic protons would be at positions 3,4,5,6. The substituents at 2 (Cl) and 1 (COOH). The coupling between aromatic protons would cause splitting. The COOH proton would be at position 1, adjacent to Cl at 2. The COOH proton would be a singlet if it's not adjacent to any other protons. However, in the NMR, the singlet is at 11.0, which is lower than expected for COOH. Therefore, perhaps the COOH is not present. Wait, but the IR data shows a strong peak at 1720, which is more in line with a COOH group. So this is conflicting. Alternatively, maybe the COOH is not present, and the carbonyl is a ketone. But then the IR wouldn't show a broad O-H. Wait, that can't be. Therefore, the compound must have both a carbonyl and an O-H group. Therefore, the structure must be either a carboxylic acid or a phenol with a carbonyl group. Wait, but option C is a carboxylic acid. So the aromatic protons would be influenced by the COOH group. Let me think of the possible splitting. In 2-chlorobenzoic acid, the aromatic protons would be at positions 3,4,5,6. The substituents at 2 (Cl) and 1 (COOH). The protons at 3 and 6 are ortho to Cl and meta to COOH. The protons at 4 and 5 are meta to Cl and ortho to COOH. The coupling between 3 and 4 (ortho to Cl) would be a doublet. The coupling between 4 and 5 (meta to Cl) would be a doublet. The coupling between 5 and 6 (ortho to Cl) would be a doublet. This would create a complex splitting pattern. However, the given NMR shows two doublets, each integrating to 2H. Therefore, this suggests that there are two sets of equivalent protons. So maybe the substituents are in positions that create two sets of equivalent aromatic protons. For example, if the substituents are at 3 and 5, then the aromatic protons would be at 2,4,6. But that's not matching the options. Alternatively, if the substituents are at 4 and 5, but that's not one of the options. Alternatively, perhaps the substituents are at positions 3 and 5, but that's not an option. Let me check the options again. The options are:

A. 3-Chloro-2-hydroxybenzaldehyde

B. Phenyl chloroformate

C. 2-chlorobenzoic acid

D. 4-chlorobenzoic acid

So none of them have two substituents in positions that would create two sets of equivalent aromatic protons. Therefore, perhaps the correct answer is option C, 2-chlorobenzoic acid, with the understanding that the aromatic protons split into two doublets, even though the exact splitting isn't clear. Alternatively, maybe the splitting in the NMR is due to coupling with the aldehyde proton, but that doesn't fit. Alternatively, perhaps the singlet at 11.0 is not the COOH proton. Wait, but the COOH proton would be around 12-13 ppm. If it's at 11.0, maybe it's a different group. Alternatively, perhaps the IR is not of the compound but of a derivative. But the question states "In the lab, a chemist discovers an unnamed bottle with a white solid inside. He gathers the following spectral information...". So the IR is of the unknown compound. Therefore, the COOH group must be present. This is conflicting. Let me think of another angle. The molecular formula is C7H5ClO2. Let's calculate the possible structures. Another approach is to consider the IR and NMR data together. The IR shows a strong peak at 1720 (carbonyl) and a broad O-H (phenol or carboxylic acid). The NMR shows two aromatic doublets, which suggests two sets of equivalent protons. Let's think of the structure as a para-substituted benzene ring with two substituents, one of which is a carbonyl group and the other a Cl. Wait, but the molecular formula is C7H5ClO2. Let's think of a benzene ring with a substituent R and a Cl. The R group would be a carbonyl group (like COOH or CHO) plus another substituent. Wait, but the molecular formula must add up. Let's consider R as Cl-CO-... No, that might not work. Alternatively, perhaps the structure is Cl-C6H3-CO-O-R. Wait, but I'm not sure. Let me think of the structure as Cl-C6H4-COOH. That's 2-chlorobenzoic acid (option C). The aromatic protons would be at positions 3,4,5,6. The substituents are at 2 (Cl) and 1 (COOH). The coupling would be between adjacent protons. The protons at positions 3 and 4 would be coupled, and 4 and 5, 5 and 6. However, the given NMR shows two doublets. Therefore, maybe the substituents are in positions that create two sets of equivalent protons. For example, if the substituents are at 3 and 5, then the aromatic protons would be at 2,4,6. But that's not an option. Alternatively, perhaps the substituents are at positions 4 and 6, making the aromatic protons at 2,3,5. But again, not matching the options. Alternatively, maybe the substituents are in positions that create two sets of two equivalent protons each. For example, substituents at 2 and 6 (para positions). Then the aromatic protons would be at 3,4,5. But that's three protons, not two. Hmm. Alternatively, substituents at 2 and 3 (meta), making the aromatic protons at 4,5,6. But again, three protons. Therefore, perhaps the correct answer is option C, 2-chlorobenzoic acid, even though the NMR splitting isn't perfectly matching. Alternatively, maybe the splitting is due to coupling with the aldehyde proton, but that doesn't fit. Alternatively, perhaps the 11.0 ppm is the aldehyde proton, but the IR suggests a carbonyl, not an aldehyde. Wait, but option A is 3-chloro-2-hydroxybenzaldehyde. Structure: benzaldehyde group, with Cl at 3 and OH at 2. The aldehyde proton would be around 9-10 ppm, but the given NMR shows 11.0, which is higher. However, maybe the aldehyde is conjugated with the substituents, causing a shift. But in that case, the IR would have a strong aldehyde C=O peak around 1700, which matches the IR data. The O-H from the phenol group would be broad. The aromatic protons would be at positions 4,5,6. The substituents are at 2 (OH) and 3 (Cl). The coupling between aromatic protons would create two doublets. For example, the protons at 4 and 6 would be equivalent (meta to OH and Cl), and the protons at 5 would be between them. The coupling between 4 and 5 (ortho to Cl) and 5 and 6 (meta to OH) would create two doublets. This could result in a singlet for the aldehyde proton (if it's a singlet), but the aldehyde proton is usually a singlet. However, the given NMR shows a singlet at 11.0 (1H), which could be the aldehyde proton. Then the aromatic protons would show two doublets, each integrating to 2H. That matches the given data. Therefore, option A would fit. But how does the molecular formula fit? Option A is C7H5ClO2. Let's check: benzaldehyde (C7H6O) plus Cl and OH. Wait, benzaldehyde is C7H6O, replacing one H with Cl and one H with OH would give C7H5ClO2. Yes, that matches. So the molecular formula is correct. The IR would have a strong aldehyde peak around 1720 (if it's a conjugated aldehyde), a broad O-H from the phenol. The NMR would have the aldehyde proton as a singlet at around 10.5-11.0 ppm. The aromatic protons would show two doublets. Let's see: the aldehyde proton is at 11.0. Then the aromatic protons would be at positions 4 and 6 (ortho to Cl) and 5 (meta to Cl and adjacent to OH). The coupling between 4 and 5 (ortho to Cl) would be a doublet, and 5 and 6 (meta to Cl and adjacent to OH) would be another doublet. However, the integration is 2H each. Wait, but positions 4 and 6 are each 1H, and position 5 is 1H. That would only integrate to 3H. But the NMR shows two doublets, each integrating to 2H. Therefore, this doesn't add up. Alternatively, perhaps the substituents are in positions that create two sets of equivalent protons. For example, if the substituents are at 2 (OH) and 4 (Cl), then the aromatic protons would be at 3,5,6. The protons at 3 and 6 (ortho to Cl) would be equivalent, and the proton at 5 would be between them. The coupling would create two doublets, each integrating to 1H. But the NMR shows two doublets, each integrating to 2H. Therefore, this is conflicting. Alternatively, maybe the substituents are at positions 2 (Cl) and 4 (OH), but that's not an option. Therefore, option A might not fit the NMR data. Let me go back to option C. If the COOH group is at position 1 and Cl at 2, the aromatic protons would be at 3,4,5,6. The protons at 3 and 6 (ortho to Cl) would be equivalent. The protons at 4 and 5 (meta to Cl and adjacent to COOH) would be equivalent. Therefore, the aromatic protons would split into two doublets, each integrating to 2H. That matches the NMR data. The COOH proton would be a singlet at position 1, but the expected shift is around 12-13 ppm. However, the given NMR has a singlet at 11.0 ppm. That's a discrepancy. But maybe there's another factor, like conjugation or resonance, that shifts the COOH proton downfield. Alternatively, perhaps the sample is in a solvent that affects the shift. But this is speculative. However, all the other data fits option C better. The IR shows a strong carbonyl peak, the molecular formula matches, and the NMR shows two doublets for the aromatic protons. The aldehyde proton isn't present in this case, so the singlet at 11.0 must be the COOH proton. Therefore, despite the shift discrepancy, option C seems the most plausible.
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