merged_df.csv

cas_number,guess,guess_answer,molecular formula,True answer
105-53-3,"### Scratchpad ### 
The chemical formula C7H14O2 suggests that the molecule could be an ester, as the structure COO is common for esters. 

Looking at the CNMR, the peak at 166.64 ppm indicates a carbonyl carbon, which is consistent with our ester hypothesis. The peak at 61.47 ppm could correspond to a carbon bonded to an oxygen atom, again suggesting an ester. The peak at 14.09 ppm is indicative of a terminal methyl group (CH3). The peak at 41.71 ppm could correspond to a carbon in a chain or a carbon bonded to a heteroatom.

The HNMR data show three different types of protons: those around 4.2 ppm (1272.5, 1265.41, 1258.29, 1251.2 Hz) are likely to be protons on a carbon next to an oxygen atom, those around 3.361 ppm (1008.29 Hz) could be protons on a carbon in the middle of a chain, and those around 1.3 ppm (392.57, 385.51, 378.39 Hz) are likely to be methyl protons.

Combining all these data, the molecule could be 2-ethylbutanoic acid ethyl ester, which would have a terminal methyl group, a carbonyl carbon, and an ester group. The protons on the carbon next to the oxygen would give the peak around 4.2 ppm, the protons on the carbons in the chain would give the peak around 3.361 ppm, and the terminal methyl protons would give the peak around 1.3 ppm. 

### Start answer ### 2-ethylbutanoic acid ethyl ester ### End answer ###",2-ethylbutanoic acid ethyl ester,C7H14O2,Diethyl malonate
13475-82-6,"### Scratchpad ###

The molecular formula provided is C12H26, which indicates a saturated hydrocarbon. The molecule can be an alkane, a cycloalkane, or a combination of these.

The HNMR spectrum shows five distinct peaks, suggesting there are five different types of hydrogen atoms in the molecule. The CNMR spectrum displays five distinct peaks as well, meaning there are five unique carbon environments in the molecule. 

Looking at the integrations, let's start with the most abundant hydrogen peak at 80.88 ppm with an integration of 1000. This peak corresponds to the largest number of equivalent hydrogens. The environment of these hydrogens is likely to be a simple alkyl group, such as a methyl group (CH3).

The rest of the peaks in the HNMR could correspond to different types of hydrogens in the molecule. Based on the chemical shifts, it seems like these hydrogens are in relatively similar environments, suggesting they might be part of a long carbon chain.

Looking at the CNMR, the peak at 30.39 ppm with an integration of 1000 suggests a simple alkyl carbon, likely a methyl carbon (CH3). The rest of the peaks could correspond to different types of carbons in the molecule, again suggesting a long carbon chain.

All these observations point to a long chain alkane. 

The only compound that satisfies the chemical formula and the NMR data is Dodecane (C12H26), which is a straight chain alkane with 12 carbon atoms. 

### Start answer ### Dodecane ### End answer ###",Dodecane,C12H26,"Heptane, 2,2,4,6,6-pentamethyl-"
93-58-3,"### Scratchpad ### 

The chemical formula provided is C8H8O2, which suggests we are dealing with a molecule that has 8 carbon atoms, 8 hydrogen atoms and 2 oxygen atoms. 

From the CNMR spectrum, we can see 6 different types of carbon environments: 1 carbonyl carbon (ppm 167.04), 4 aromatic carbons (ppm 132.9, 130.25, 129.6, 128.37), and 1 carbon that's likely attached to an oxygen (ppm 52.01). 

From the HNMR spectrum, we can see 2 types of hydrogen environments: one environment has a chemical shift around 8 ppm, which is typical for hydrogens on an aromatic ring. The number of protons in this environment can be estimated by integrating the peaks in this region, which gives around 5 protons. 

The other hydrogen environment is around 3.88 ppm, which is typical for a -CH3 group attached to a heteroatom (like oxygen). The integration of this peak suggests there are 3 protons in this environment, which matches our expectation for a -CH3 group.

Combining this information, the molecule is likely to have a -COOH group (which accounts for the carbonyl carbon and one of the oxygens), a -CH3 group (which accounts for 3 of the hydrogens and the carbon attached to oxygen), and an aromatic ring with 5 hydrogens (which accounts for the rest of the atoms). 

Looking at molecules that fit this description, p-toluic acid (4-methylbenzoic acid) seems to be a good match.

### Start answer ### 

p-Toluic acid

### End answer ###",p-Toluic acid,C8H8O2,"Benzoic acid, methyl ester"
52-52-8,"### Scratchpad ###
Let's get started with the analysis of the provided data.
The molecule's molecular formula is C6H11NO2, which indicates the presence of 6 carbons, 11 hydrogens, 1 nitrogen and 2 oxygens in the molecule. 

Looking at the CNMR data we have four distinct carbons: 
1) At 179.28 ppm, this is indicative of a carbonyl carbon, likely in a carboxylic acid or amide due to the presence of N and O in the formula.
2) At 68.01 ppm, this is usually indicative of a carbon attached to an oxygen in an ether or alcohol, but could also be a carbon in an amine.
3) At 37.41 ppm, this could be an alkyl carbon.
4) At 26.06 ppm, this could also be an alkyl carbon.

Looking at the HNMR data, we can see multiple peaks between 1.8-2.2 ppm which suggests the presence of multiple hydrogens on an alkyl chain. The high intensity of these peaks suggests the presence of multiple hydrogens in these environments.

The presence of N in the formula suggests that this could be an amide. Amides have a carbonyl carbon (which we have identified in the CNMR) and a nitrogen. 

Considering all this information, a likely structure that fits the given formula and spectroscopic data is caprolactam. 

### Start answer ### 
Caprolactam 
### End answer ###",Caprolactam,C6H11NO2,"Cyclopentanecarboxylic acid, 1-amino-"
78-78-4,"### Scratchpad ### 

Given the chemical formula C5H12, the molecule is an alkane. The number of carbons and hydrogens suggests that this molecule could be a straight chain alkane (pentane) or a branched alkane (isopentane or neopentane). 

The HNMR spectrum shows peaks that suggest the presence of chemically equivalent hydrogens, which are indicative of a symmetric molecule. The most symmetric molecule among the possibilities is neopentane.

From the CNMR, we have 4 peaks which indicate 4 different types of carbon environments, which is in agreement with the structure of neopentane.

### Scratchpad ###

### Start answer ### 

Neopentane

### End answer ###",Neopentane,C5H12,"Butane, 2-methyl-"
591-76-4,"### Scratchpad ### 
The formula C7H16 indicates that the molecule is an alkane. The carbon NMR data shows six peaks, meaning that there are six distinct carbon environments in the molecule. The number of carbons adds up to 7, which is consistent with the molecular formula. 

The hydrogen NMR data shows a range of different chemical shifts, but none of them are particularly downfield, suggesting that there aren't any particularly electron-withdrawing groups in the molecule.

The peak at 0.901 ppm with a relative intensity of 1000 is likely to be a methyl group (-CH3) since it's at a very high frequency and methyl groups are common in alkanes.

The peak at 0.832 ppm with a relative intensity of 669 is likely to be a methylene group (-CH2-) since it's at a slightly lower frequency and methylene groups are also common in alkanes.

The chemical shifts and intensities of the other peaks suggest that they are all methylene groups. The total number of hydrogen atoms adds up to 16, which is consistent with the molecular formula.

The carbon NMR data shows six peaks, which means there are six distinct types of carbon in the molecule. The chemical shifts and intensities of these peaks suggest that they are all part of the same carbon chain.

Combining the hydrogen and carbon NMR data, the molecule can be deduced to be heptane.

### Start answer ### 
Heptane
### End answer ###",Heptane,C7H16,"Hexane, 2-methyl-"
108-38-3,"### Scratchpad ### 

The molecular formula given is C8H10, which corresponds to a degree of unsaturation of 5. This suggests the presence of a benzene ring (4 degrees of unsaturation) and an additional degree of unsaturation, possibly an alkene (1 degree).

In the CNMR spectrum, there are 5 different types of carbons: 2 quaternary carbons (one at 137.74 ppm, typical of an aromatic carbon, and one at 21.31 ppm, typical of an alkyl carbon), 3 aromatic carbons (at 129.96, 128.21 and 126.13 ppm), which confirm the presence of a benzene ring, as there are three different types of carbon in a 1:2:2 ratio, typical of a mono-substituted benzene ring.

The HNMR spectrum shows 5 peaks, which suggests that there are 5 different types of protons in the molecule. The peak at 7.44 ppm with an integration of 456 is typical of aromatic protons, which further confirms the presence of a benzene ring. The remaining peaks could be aliphatic protons, possibly part of an alkyl group, which would be consistent with the two degrees of unsaturation.

The peak at 0.945 ppm with an integration of 138 and the peak at 0.865 ppm with an integration of 456 suggest the presence of a CH3 group and a CH2 group respectively, which would be consistent with an alkyl group attached to the benzene ring.

The peak at 0.797 ppm with an integration of 74 and the peak at 0.779 ppm with an integration of 77 suggest the presence of two CH groups, which would be consistent with a double bond (alkene) in the alkyl group.

So, considering the degrees of unsaturation and the information from the NMR spectra, the molecule could be styrene (vinylbenzene), which has a benzene ring and a vinyl group, and fits the molecular formula C8H10.

### Start answer ### Styrene ### End answer ###",Styrene,C8H10,"Benzene, 1,3-dimethyl-"
124-19-6,"### Scratchpad ### 

Given chemical formula is C9H18O. 

From the CNMR spectrum, we see 9 different peaks, indicating 9 unique carbon environments. 

The peak at 202.82 ppm in the CNMR spectrum suggests the presence of a carbonyl group (C=O) in the molecule, which accounts for the single oxygen in the formula.

The HNMR spectrum shows peaks in the aliphatic region (between 0.8-3.0 ppm), which indicates the presence of alkyl chains. 

The presence of 6 peaks around 0.8-2.5 ppm in the HNMR spectrum, corresponds to the 6 carbon peaks in the CNMR spectrum around 14-43 ppm, indicating 6 different alkyl environments. 

The remaining 3 carbon peaks in the CNMR spectrum around 29 ppm could be due to the same type of environment, possibly a tert-butyl group (3 carbons). 

The peaks in HNMR at around 1.3 ppm with high intensity could correspond to the protons in the tert-butyl group.

The two peaks at around 9.76 ppm in the HNMR spectrum are likely due to a set of equivalent protons next to the carbonyl group.

So the structure could be a carbonyl group (C=O) attached to a tert-butyl group and a long alkyl chain.

### Start answer ### 

The molecule is 2,2,4-trimethylpentan-3-one

### End answer ###","The molecule is 2,2,4-trimethylpentan-3-one",C9H18O ,Nonanol
107-83-5,"The formula C6H14 suggests an alkane, which is a saturated hydrocarbon (i.e., it contains only single bonds). This also means that it does not contain any functional groups that are typically identifiable in NMR spectra.

### Scratchpad ###
HNMR:
Most of the HNMR signals are very close to each other, suggesting these hydrogens are in similar chemical environments. The signal at 0.898 ppm with the highest intensity suggests the presence of a large number of equivalent hydrogens, typical in alkanes. 

CNMR:
In the CNMR spectrum, we have 5 unique carbon peaks. This indicates that there are 5 types of carbons in the molecule. Considering the formula, every carbon must be connected to at least one other carbon.

Combining HNMR and CNMR:
The chemical shifts in both the HNMR and CNMR spectra are consistent with the shifts expected for alkanes. The absence of downfield peaks suggests there are no electronegative atoms or unsaturation (double or triple bonds) present.

### Scratchpad ###

### Start answer ###
Hexane
### End answer ###",Hexane,C6H14,"Pentane, 2-methyl-"
540-84-1,"### Scratchpad ###

The chemical formula given is C8H18, which indicates an alkane. The formula fits the general formula for alkanes, CnH2n+2, and indicates a saturated hydrocarbon. 

Looking at the HNMR data, there are 5 distinct peaks, indicating 5 different types of hydrogen environments in the molecule. 

The CNMR data shows 5 distinct carbon peaks, which implies that there are 5 different types of carbon environments in the molecule.

Combining the HNMR and CNMR data, we can infer that the molecule contains a long carbon chain, with some of the carbons being in different environments. 

The integral of each peak in the HNMR spectrum corresponds to the number of hydrogens in that particular environment. 

The largest peak in the HNMR spectrum is at 0.893 ppm with an intensity of 1000, which indicates that the most abundant hydrogen environment is a methyl group (CH3) located at the end of a carbon chain.

The peak at 0.942 ppm with an intensity of 268 suggests the presence of another methyl group, but this one is likely attached to a carbon that is connected to two other carbons (i.e., a methylene group -CH2- is adjacent).

The remaining peaks in the HNMR spectrum can be assigned to methylene groups (-CH2-) in different environments along the carbon chain.

The molecule appears to have a linear structure with different types of carbon environments due to the presence of different hydrogen environments along the carbon chain.

Considering the stoichiometry and the number of each type of carbon and hydrogen environment, the molecule can be identified as octane.

### Start answer ### 
Octane
### End answer ###",Octane,C8H18 ,"Pentane, 2,2,4-trimethyl-"
22204-53-1,"### Scratchpad ###
Firstly, the molecular formula C14H14O3 tells us that there are 8 degrees of unsaturation (14-14/2+1+3 = 8). This indicates that there might be combinations of rings and double bonds in the molecule.

In the HNMR spectrum, the peak at 7.681 ppm with an integration of 131 could represent the protons in an aromatic ring, as could the peaks at 7.662 ppm and 7.659 ppm, since those resonance frequencies are typical for aromatic protons.

On the other hand, the peak at 3.873 ppm with the highest integration (1000) could be representing protons adjacent to oxygen, since those protons typically resonate in the 3-4 ppm range. 

The peak at 1.577 ppm could represent the protons of a methyl group, as these protons usually resonate in the 1-2 ppm range.

In the CNMR spectrum, the peak at 181.05 ppm might indicate the presence of a carbonyl group, as carbon in carbonyl groups typically resonate around 180 ppm.

The peaks at 157.66 ppm and 134.83 ppm could indicate the presence of a carbon-carbon double bond. The peaks at 129.29 ppm, 128.86 ppm, 127.22 ppm, and 126.18 ppm could represent carbons in an aromatic ring, as these carbons typically resonate in the 125-130 ppm range.

The peak at 55.26 ppm could represent a carbon attached to oxygen, as these carbons usually resonate in the 50-60 ppm range.

The peak at 18.1 ppm could represent a methyl carbon, as these carbons typically resonate in the 10-20 ppm range.

Combining all these interpretations, and considering the degrees of unsaturation, it suggests the molecule could be 2-hydroxy-4-methoxybenzaldehyde.

### Start answer ### 2-hydroxy-4-methoxybenzaldehyde ### End answer ###",2-hydroxy-4-methoxybenzaldehyde,C14H14O3,Naproxen
513-81-5,"### Scratchpad ###
The molecular formula is C6H10, which gives us a degree of unsaturation of 2 (number of rings plus pi bonds in the molecule). This suggests the presence of a ring or a double bond in the molecule.

From the CNMR data, we can see three unique carbon environments: 143.60, 113.05, and 20.65 ppm. The peak at 143.60 ppm suggests a carbon in a double bond or aromatic environment. The peak at 113.05 ppm could be a carbon in a double bond or an sp2 hybridized carbon. The peak at 20.65 ppm corresponds to an sp3 hybridized carbon, likely in an alkyl chain.

The HNMR data is a bit complex, but the most prominent peaks are at 1.923 and 1.918 ppm, corresponding to 933 and 1000 integrations respectively. These could be associated with the hydrogens in an alkyl chain. The remaining peaks in the 4-5 ppm range could represent hydrogens on the carbon with a double bond or aromatic ring.

The number of hydrogens is 10, and the number of carbons is 6, meaning there are multiple hydrogens per carbon. This, along with the degree of unsaturation, suggests the structure could be a cyclohexene with a methyl group attached, as this would give us the right number of carbon and hydrogen environments, as well as the right degree of unsaturation.

### Start answer ### 
Cyclohexene with a methyl group attached, i.e., methylcyclohexene
### End answer ###","Cyclohexene with a methyl group attached, i.e., methylcyclohexene",C6H10,"1,3-Butadiene, 2,3-dimethyl-"
123-72-8,"### Scratchpad ### 

The chemical formula is C4H8O, which implies that the molecule might be a saturated aliphatic compound with an oxygen atom. The CNMR spectrum shows four distinct signals, indicating the presence of four different types of carbon atoms (1, 2, 3, and 4). Considering the chemical formula, the molecule most likely contains a carbonyl group (C=O), which would explain the carbon signal at 202.8 ppm in the CNMR spectrum. 

The HNMR spectrum shows a range of peaks, which suggests the presence of multiple types of hydrogen atoms. The large number of peaks in the HNMR spectrum might be due to the presence of protons in different environments. 

The presence of a signal around 9.7 ppm in the HNMR spectrum suggests the presence of an aldehyde group (CHO) in the molecule. This is because protons attached to an aldehyde carbon generally resonate around this region. 

The intensity of the peaks in the HNMR spectrum can give us information about the number of protons that contribute to each signal. For example, the signal at 9.7 ppm has an intensity of about 807 (203+396+208), which suggests that this signal represents one proton, as is typical for an aldehyde group. This assignment is consistent with the carbon signal at 202.8 ppm in the CNMR spectrum. 

The signals around 2.4 and 2.3 ppm in the HNMR spectrum might be due to the presence of protons attached to a carbon next to the carbonyl group. The high intensity of these signals suggests that they represent three protons each, which is consistent with the presence of a methyl group attached to the carbon next to the carbonyl group.

The signals around 1.7, 1.6, and 1.5 ppm in the HNMR spectrum might be due to the presence of protons attached to a carbon that is two carbons away from the carbonyl group. The high intensity of these signals suggests that they represent two protons each, which is consistent with the presence of a CH2 group.

The signals around 1.0 ppm in the HNMR spectrum might be due to the presence of protons attached to a carbon that is three carbons away from the carbonyl group. The high intensity of these signals suggests that they represent three protons each, which is consistent with the presence of a terminal methyl group. This assignment is consistent with the carbon signals at 45.82, 15.69, and 13.71 ppm in the CNMR spectrum. 

From the above analysis, I predict the molecule to be butanal.

### Start answer ### Butanal ### End answer ###",Butanal,C4H8O,Butanal
620-02-0,"### Scratchpad ### 
The chemical formula C6H6O2 tells us that we're dealing with a molecule that contains 6 carbons, 6 hydrogens, and 2 oxygen atoms.

Looking at the CNMR spectrum, we see six distinct peaks, which correspond to six unique carbons in the molecule. Let's assign each carbon peak:

1. 176.81 ppm: This is in the carbonyl range, suggesting we have a carbonyl carbon (C=O). 
2. 159.77 ppm: This is in the range for a carbon directly bonded to an oxygen (like in a carboxylic acid or ester). 
3. 151.99 ppm: This is also in the range for a carbon directly bonded to an oxygen or a carbon in a benzene ring.
4. 124.03 ppm: This is in the range for a carbon in a benzene ring.
5. 109.62 ppm: This is also in the range for a carbon in a benzene ring.
6. 13.95 ppm: This is in the range for a carbon in a methyl group.

From the HNMR spectrum, we can identify a few groups of peaks:

- Peaks at 9.511 and 7.198-7.153 ppm are likely aromatic hydrogens in a benzene ring.
- Peaks at 6.269-6.211 ppm could potentially be hydrogens on a carbon bonded to an oxygen.
- Peaks at 2.428-2.412 ppm are likely hydrogens on a methyl group.

Based on these assignments, a possible structure is a benzene ring with a carboxylic acid group and a methyl group attached. This would be consistent with the chemical formula, the CNMR and HNMR data, and would account for all the hydrogens and carbons in the molecule.

### Start answer ### 
The molecule is likely to be Benzoic acid. 
### End answer ###",The molecule is likely to be Benzoic acid.,C6H6O2,"2-Furancarboxaldehyde, 5-methyl-"
15687-27-1,"### Scratchpad ### 

H-NMR provides information about the number and type of hydrogen environments in a molecule, while C-NMR provides information about the number and type of carbon environments. 

From the chemical formula C13H18O2, we know that the molecule contains 13 carbons, 18 hydrogens, and 2 oxygens. 

Looking at the CNMR data:

There are 9 unique carbon environments in the molecule as shown by the 9 peaks in the spectrum. The peaks around 181.35 ppm are indicative of a carbonyl carbon (C=O), which corresponds to the presence of 2 oxygen atoms in the chemical formula. 

The carbon peaks at 140.93, 137.02, 129.47, 127.36 ppm suggest the presence of aromatic or unsaturated carbons (C=C). 

The lower ppm values (45.11, 45.08, 30.25, 22.47, 18.15) are indicative of aliphatic or saturated carbons (C-C). 

Moving on to the HNMR data:

The peak at around 7.22 ppm suggests the presence of hydrogens on an aromatic ring. The integral (502, 527, 508, 377) suggests 4 aromatic protons. 

The peaks around 3.72 ppm suggest the presence of hydrogens attached to a carbon which is next to an oxygen or nitrogen atom. The integral (25, 80, 82, 28) suggests 2 methoxy groups (-OCH3) with 6 protons. 

The peak at around 2.45 ppm and its integral (260, 276) suggests 4 protons attached to a carbon which is next to an electronegative atom (likely oxygen in this case).

The peaks at around 1.87 ppm and their integrals (28, 55, 69, 56, 29) suggest 5 protons attached to a carbon which is not next to an electronegative atom. 

The peak at 1.496 ppm and its integral (438, 431) suggest 3 protons on a methyl group (-CH3). 

The peak at 0.898 ppm and its integral (1000, 979) suggest 6 protons on two methyl groups (-CH3).

Based on these observations, we can propose the structure of the molecule is 3,3-Dimethyl-1,2-diphenylcyclopropane-1,2-dicarboxylic acid.

### Start answer ### 3,3-Dimethyl-1,2-diphenylcyclopropane-1,2-dicarboxylic acid ### End answer ###","3,3-Dimethyl-1,2-diphenylcyclopropane-1,2-dicarboxylic acid",C13H18O2,ibuprofen
79-16-3,"### Scratchpad ### 

Let's start by taking a look at the formula C7H7NO and understanding the degree of unsaturation. The degree of unsaturation can be calculated using the formula 2C+2+N-H/2, which in this case is 2*7+2+1-7/2 = 4.5. This is not a whole number which is unusual, there might be some mistake. Let's assume that the degree of unsaturation is 4, which suggests the presence of 4 double bonds or rings.

Now let's move on to the NMR data. 

In the HNMR, we have two groups of peaks around 2.8 ppm and two around 2.7 ppm. Their coupling constants (Hz) are quite high, which suggests that they are part of the same spin system, probably a coupling between protons on adjacent carbons. The integration suggests that these peaks correspond to 2 protons each (approximation, considering 1.985 ppm peak as 3H). 

The peak at 1.985 ppm is a singlet and it integrates for 3, which suggests it is a CH3 group. 

In the CNMR, we have a peak at 171.77 ppm which is characteristic of a carbonyl carbon (C=O), a peak at 26.23 ppm which is likely an alkyl carbon (part of CH2 or CH3) and a peak at 22.74 ppm which is also likely an alkyl carbon (part of CH2 or CH3).

Given the molecular formula, the degree of unsaturation and the NMR data, a possible structure could be a benzene ring (which accounts for 4 unsaturations) with a -C=O (which accounts for the 5th unsaturation), a -CH3 and a -NH2 group attached to it. This would give a molecular formula of C7H7NO and would account for all the peaks in the NMR spectra. 

### Start answer ### 

The molecule is 4-aminobenzoic acid.

### End answer ###",The molecule is 4-aminobenzoic acid.,C7H7NO,"Acetamide, N-methyl-"
123-86-4,"### Scratchpad ### 

The molecular formula C6H12O2 suggests that the molecule could be an ester or a cyclic molecule with an oxygen atom (like a cyclic ether or an alcohol). The number of carbons (6) and hydrogens (12) suggests that there might be no ring in the structure, or if there is one, it's not a very complex structure.

Looking at the CNMR data, there is a peak at 171.11 ppm which is typical for a carbonyl carbon, suggesting the molecule could be an ester. The other five carbon peaks seem to indicate a range of different carbon environments, from very deshielded (64.38 ppm) to very shielded (13.75 ppm).

From the HNMR data, the peaks are rather spread out across the spectrum, suggesting a variety of hydrogen environments as well. The integration values suggest that many of the hydrogen atoms are in similar environments.

Based on these observations, we can begin to assign some of the peaks:

1. The peak at 171.11 ppm in the CNMR data is likely the carbonyl carbon in an ester group.
2. The peak at 64.38 ppm in the CNMR data could be the carbon in a methoxy group.
3. The peaks at 30.85, 20.94, 19.26, and 13.75 ppm in the CNMR data can be assigned to the carbons in a chain.
4. The 2.04 ppm peak in the HNMR data with the highest integration (1000) could be a methylene group.
5. The peak at 4.067 ppm in the HNMR data (integration 176) might correspond to the hydrogens on the carbon next to the carbonyl group in the ester.

Combining these assignments with the molecular formula, we can propose a structure for the molecule. The molecule is likely an ester with a methoxy group and a chain of carbons. Indeed, it could be an ethyl methanoate.

### Scratchpad ### 

### Start answer ### 

Ethyl methanoate

### End answer ###",Ethyl methanoate,C6H12O2,"Acetic acid, butyl ester"
93-56-1,"### Scratchpad ###
Starting with the molecular formula C8H10O2, this tells us that we have 8 carbons, 10 hydrogens, and 2 oxygens.

Looking at the HNMR data, we can see that there are peaks at 7.278 ppm, 4.791 ppm, 4.739 ppm, 4.71 ppm, 4.659 ppm, 3.685 ppm, 3.64 ppm, 3.595 ppm, and 3.556 ppm. When we consider the integrations, the peak at 7.278 ppm has the highest integration which suggests that this is the most abundant proton in the structure. 

Moving to the CNMR data, we have peaks at 140.54 ppm, 128.42 ppm, 127.8 ppm, 126.12 ppm, 74.71 ppm, and 67.93 ppm. The peak at 128.42 ppm is the most intense indicating that the carbon at this chemical shift is the most abundant in the structure. 

Now, if we try to match the peaks from both the HNMR and CNMR, we can see that the peak at 7.278 ppm in the HNMR might correspond to the peak at 128.42 ppm in the CNMR which suggests that we might have a benzene ring in the structure since these chemical shift values are typical for aromatic protons and carbons. The chemical shifts around 4.7-4.8 ppm would suggest CH2 groups connected to an electron-withdrawing group, like a carbonyl. The chemical shifts around 3.6 ppm could correspond to methylene or methine groups attached to oxygen.

The remaining unassigned H atoms and chemical formula would suggest the remaining parts of the molecule are two methyl groups attached to a carbonyl carbon, forming a ester or a carboxylic acid.

Combining all these findings, the molecule could be ethyl benzoate, which fits with the given formula C8H10O2 and the NMR data. The benzene ring protons would appear around 7.2 ppm, the protons on the CH2 connected to the carbonyl would appear around 4.7 ppm, and the ones on the CH2 connected to the oxygen would appear around 3.6 ppm. The methyl protons would appear around 3.5 ppm.
### Scratchpad ###

### Start answer ### Ethyl benzoate ### End answer ###",Ethyl benzoate,C8H10O2,"1,2-Ethanediol, 1-phenyl-"