The following NMR spectra correspond to compound of formula (A) C 9 H 10 O 2 . Propose structure, and show how it is consistent with the observed absorptions. <IMAGE>

Hey everyone, Let's do this problem. It says an unknown compound with a molecular formula C eight H 802 shows the following NMR spectra propose the structure consistent with the observed absorption. So, before I even look at my NMR spectrum, when we're given the molecular formula and trying to determine the structure of a compound, I like to calculate the index of hydrogen deficiency to determine if there are any characteristics of unsaturated in. So we follow the formula two times the number of carbons plus two minus the number of hydrogen all over. To plugging in our values for carbon and hydrogen, we get two times eight plus two minus eight all over two, Which gives us an i. h. d. of five. Alright, well, what the heck does that mean? We know that a benzene ring Can account for four of the five values and the other value can represent either a double bond or a ring. Okay, so that's some good information. We can use our proton NMR spectrum now to tell us more about this. So speaking of benzene ring, I notice here in the spectrum at around 7.3 PPm. We have five hydrogen. And that is actually in the range for benzene protons. Okay, so if we have five benzene protons, let me draw a benzene ring. That means 12345 hydrogen. There is something else on our ring. Right. One substitution group on our ring. Alright, well, what else do we see in our spectrum? I see at around 11 p p. M. In the range for a carb oxalic acid, C double bond 00. H. There's a peak there. So our substitue in could be a carb oxalic acid. But we also have one more carbon in our molecular formula. So I can't just attach this carb oxalic acid here. We need to have another carbon. Alright well, by the looks of it, we're going to have one more carbon that has two hydrogen before our car back silic acid because we have this single it at around 3.8 PPm. That has an integration of two hydrogen. So that is around where we would find an al caine hydrogen. Right, hydrogen is on an al cane carbon. But it's a little more downfield because of the d shielding from the carb oxalic acid and the benzene ring. So adding that carbon with the two hydrogen as we get this structure and removing all the carbons and the hydrogen. This is our final structure and that makes B two B. The correct answer for this problem. That's it for this one. I hope this was helpful and thanks for watching

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

15. Analytical Techniques:IR, NMR, Mass Spect

NMR Practice

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

NMR Practice

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3:50 minutes

Problem 40a

Textbook Question

The following NMR spectra correspond to compound of formula (A) C₉H₁₀O₂. Propose structure, and show how it is consistent with the observed absorptions.
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Verified step by step guidance

Step 1: Analyze the molecular formula C9H10O2. This indicates the presence of 9 carbons, 10 hydrogens, and 2 oxygens. The degree of unsaturation can be calculated using the formula: (2C + 2 - H)/2. For C9H10O2, the degree of unsaturation is (2(9) + 2 - 10)/2 = 5. This suggests the presence of rings and/or double bonds.

Step 2: Examine the NMR spectrum. The peak at approximately 12 ppm corresponds to a highly deshielded proton, likely from a carboxylic acid (-COOH) group. This is consistent with the presence of one hydrogen in this region.

Step 3: The multiplet at approximately 7-8 ppm corresponds to aromatic protons. The integration of 5H suggests a monosubstituted benzene ring, as the benzene ring typically has 5 protons when one substituent is attached.

Step 4: The singlet at approximately 4 ppm corresponds to 2 hydrogens. This is likely from a methylene group (-CH2-) attached to an electronegative group, such as oxygen. This is consistent with the presence of an ester or alcohol functional group.

Step 5: Combine the observations. The molecular formula, degree of unsaturation, and NMR data suggest the structure is a benzene ring with a carboxylic acid group (-COOH) and a methoxy group (-OCH3) attached. Verify this structure by ensuring the integration and chemical shifts match the observed spectrum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Nuclear Magnetic Resonance (NMR) Spectroscopy

NMR spectroscopy is a powerful analytical technique used to determine the structure of organic compounds. It relies on the magnetic properties of certain nuclei, primarily hydrogen (1H) and carbon (13C), to provide information about the number of hydrogen atoms in different environments within a molecule. The chemical shifts observed in the spectrum indicate the electronic environment surrounding the nuclei, allowing chemists to infer structural details.

Integration of NMR Peaks

In NMR spectroscopy, the area under each peak corresponds to the number of hydrogen atoms contributing to that signal, known as integration. For example, a peak labeled '5 H' indicates that five hydrogen atoms are in a similar environment, while '2 H' and '1 H' indicate two and one hydrogen atoms, respectively. This information is crucial for deducing the molecular structure and understanding how different hydrogen environments relate to the overall compound.

Chemical Shifts and Splitting Patterns

Chemical shifts in an NMR spectrum are measured in parts per million (PPM) and provide insight into the electronic environment of hydrogen atoms. Peaks appearing at lower PPM values typically indicate protons attached to carbons that are more electron-rich, while higher PPM values suggest protons on less electron-rich carbons. Additionally, splitting patterns (multiplicity) reveal how many neighboring hydrogen atoms are present, which helps in determining the connectivity and arrangement of atoms in the molecule.

The following NMR spectra correspond to compound of formula (A) C9H10O2. Propose structure, and show how it is consistent with the observed absorptions.<IMAGE>

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Integration of NMR Peaks

Chemical Shifts and Splitting Patterns

The following NMR spectra correspond to compound of formula (A) C₉H₁₀O₂. Propose structure, and show how it is consistent with the observed absorptions.
<IMAGE>