The following NMR spectra correspond to compound of formula (C) C 6 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 the molecular formula C four H shows the following NMR spectrum propose the structure consistent with the observed absorption. So, before I even look at my proton NMR spectrum, I'm going to use the molecular formula to calculate the index of hydrogen deficiency because that will tell me any characteristics of on saturation, such as double bonds rings, etcetera. And that's helpful for determining the structure. So, the formula for this is two times the number of carbons plus two minus the number of hydrogen, all divided by two, plugging in our values, we get two times four plus two minus six. All over two, Which gives us an i. h. d. of two. So, what does that tell us? That means we have either two double bonds, two rings or one of each. Okay, so let's use our spectrum now to narrow down what our structure can look like. Well, this peak down here at around 11 PPm, jumps out to me because that is characteristic of a car box silic acid. All right, so, that narrows down our structure a little bit. What else do we have? Well, we have two peaks kind of close to each other around six ppm 6.5 ppm. And those are characteristic of an al keen alkaline hydrogen. And we have a doublet at around two PPm, which is characteristic of ch three protons and I know it's going to be ch three protons because it says there are three hydrogen. Okay, so thinking about my I. H. D. I know we have one double bond from our carb oxalic acid. So we're not gonna have any rings, two rings at least. And the peak from the alkaline hydrogen confirm that we have two double bonds and no rings. Alright, so, let's try and draw our structure. We know there are four carbons. 12341 of my end carbons has to be for the carb oxalic acid. Right? That's always on the end of the structure. All right. And we know we have a methyl group on the other end. So that leaves are all keen to be right in between Carbons two and 3, and they each have one proton. Right? So it all makes sense. So let me remove the carbons and the hydrogen is to draw the official structure, the fancy structure. So, this is the final structure for our product. Which makes see the correct choice for this problem. That's it for this one. I hope this video is 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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Problem 40c

Textbook Question

The following NMR spectra correspond to compound of formula (C) C₆H₁₀O₂. Propose structure, and show how it is consistent with the observed absorptions.
<IMAGE>

Verified step by step guidance

Analyze the molecular formula (C6H10O2) to determine the degree of unsaturation. Use the formula: Degree of Unsaturation = (2C + 2 - H + N - X)/2, where C = number of carbons, H = number of hydrogens, N = number of nitrogens, and X = number of halogens. For this compound, calculate the degree of unsaturation to identify the presence of rings or double bonds.

Examine the NMR spectra for key features. Look for chemical shifts, splitting patterns, and integration values. For example, chemical shifts in the range of 9-10 ppm suggest aldehydes, 2-3 ppm suggest protons near electronegative atoms, and 0-2 ppm suggest alkyl protons. Integration values will help determine the number of protons contributing to each signal.

Identify functional groups based on the NMR data and the molecular formula. For example, if you observe a peak around 170-180 ppm in the 13C NMR spectrum, it may indicate the presence of a carbonyl group (C=O). Combine this information with the degree of unsaturation to narrow down possible structures.

Propose a structure that fits the molecular formula and is consistent with the NMR data. For example, if the degree of unsaturation is 2 and the NMR data suggests a carbonyl group and an ester functional group, consider structures like esters or cyclic ketones.

Verify the proposed structure by matching all observed NMR signals to the protons and carbons in the structure. Ensure that the chemical shifts, splitting patterns, and integration values align with the proposed structure. If discrepancies arise, revisit the analysis and adjust the structure accordingly.

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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, their environment, and connectivity in a molecule. The resulting spectra display peaks that correspond to different chemical environments, allowing chemists to deduce structural features.

Chemical Shift

Chemical shift refers to the position of a peak in an NMR spectrum, which indicates the electronic environment surrounding a nucleus. It is measured in parts per million (ppm) and varies based on factors such as electronegativity and hybridization of nearby atoms. Understanding chemical shifts is crucial for interpreting NMR data, as they help identify functional groups and the overall structure of the compound.

Integration and Multiplicity

Integration in NMR refers to the area under a peak, which correlates to the number of protons contributing to that signal. Multiplicity describes the splitting pattern of a peak, indicating how many neighboring protons are present (n+1 rule). Together, integration and multiplicity provide insights into the number of hydrogen atoms in different environments and their connectivity, essential for constructing the molecular structure.

The following NMR spectra correspond to compound of formula (C) C6H10O2. Propose structure, and show how it is consistent with the observed absorptions.<IMAGE>

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Shift

Integration and Multiplicity

The following NMR spectra correspond to compound of formula (C) C₆H₁₀O₂. Propose structure, and show how it is consistent with the observed absorptions.
<IMAGE>