The following 1 H NMR spectra are for four compounds, each with molecular formula of C 6 H 12 O 2 . Identify the compounds. d. <IMAGE>

Welcome back everyone to another video. The proton and A R spectrum below is produced by a compound with the molecular formula C seven H 1402. What is its structure we're given for answer choices, ABC and D AC and D are ers and B is a ketone. So we're going to analyze our spectrum and identify the structure we can start from right to left. And what we notice is that at around 1.25 parts per million, there's an integration of nine hydrogens and we observe a singlet. On the other hand, at around 1.3 parts per million, we are seeing an integration of three hydrogens and there's a triplet. Finally, there is a third signal at around 4.2 parts per million. We have an integration of two hygen and there's a quartet. Now, what we're going to do is simply apply the M plus one rule. We are going to subtract one from the number of peaks and we will get the neighboring carbon atoms for our quartet, we will have three neighbors for a triplet, we will have two neighbors and for a singlet, we will have zero neighbors. First of all, what we can do is just exclude the structures that do not correspond to three signals. For example, structure A has two equivalent metal groups. Then we have our second signal at CH, then we have our third one at a different CH. And once again, there are two equivalent metal groups. So there will be four signals instead of three, meaning we exclude option A from our possible answer for structure B, we will have one signal at the metal group, another signal at the siege two group. Then there are two equivalent metal groups that will be our third signal. The fourth one would be our siege two and the fifth one would be our sieves. We are basically our metal group. So let's label them alphabetically ABC D and E. So that's a total of five instead of three, meaning structure B is incorrect. Now, structure C, we have three equivalent metal groups because their bonnets, the same carbon atom, we also have our C two, that's our second signal and the metal group represents the third signal. So it's our possibility because we have three signals in the proton and Mr spectrum. For structure D, we once again have three equivalent muscle groups, our siege too as well as a metal group. Now, what we noticed is that our structure can be C or D, right? Because both of them have we signals in H and Mr. So let's analyze our peaks we have nine Hs and glad. So basically C and D, both of them will show nine Hs and glad because we have three equivalent metal groups. Uh We would have three plus three plus three. That's nine hygiene atoms. There are no neighbors. So indeed, in each case, we have that slut, we cannot sell it yet. Which of the two is cracked, then we have three H triplet, there should be three neighbors. So if you think about structure C signal C is three H triple, because there are two neighboring hydrogens. So we can say that this will be our triplet, but we don't have any triplet for structure D when we consider C as three H and blood because the metal group is bonded to the carbon group and there are no neighbors. So you can say that this should be the correct answer. But let's verify that. Now, we also have two H quartet. So there should be three neighbors. So if we look at peak B, there are two hygen and it will be a quartet because they are neighboring three hygen from the metal group, meaning C is the correct option to this problem. And that would be our correct answer. Thank you 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 74d

Textbook Question

The following ¹H NMR spectra are for four compounds, each with molecular formula of C₆H₁₂O₂. Identify the compounds.
d. <IMAGE>

Verified step by step guidance

Analyze the molecular formula (C6H12O2) to determine the degree of unsaturation using the formula: \( \text{Degree of Unsaturation} = \frac{2C + 2 - H + N - X}{2} \), where C is the number of carbons, H is the number of hydrogens, N is the number of nitrogens, and X is the number of halogens. For this formula, \( \text{Degree of Unsaturation} = \frac{2(6) + 2 - 12}{2} = 1 \), indicating one double bond or a ring.

Examine the 1H NMR spectra for key features such as chemical shift values, splitting patterns, and integration. For example, peaks in the range of 9-10 ppm suggest an aldehyde group, while peaks around 3-4 ppm indicate protons near electronegative atoms like oxygen.

Use the integration values to determine the relative number of protons contributing to each signal. This will help identify the symmetry and structure of the molecule. For instance, a singlet integrating to three protons near 3.5 ppm could indicate a methoxy group (\( -OCH_3 \)).

Consider the splitting patterns (singlet, doublet, triplet, etc.) to deduce the number of neighboring protons. For example, a triplet and a quartet with a 3:2 integration ratio could indicate an ethyl group (\( -CH_2CH_3 \)).

Combine all the information (degree of unsaturation, chemical shifts, integration, and splitting patterns) to propose possible structures for the compounds. Ensure that each structure matches the molecular formula (C6H12O2) and the NMR data provided.

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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, such as hydrogen-1 (1H), to provide information about the number of hydrogen atoms in different environments within a molecule. The resulting spectra display peaks that correspond to these environments, allowing chemists to infer structural details.

Chemical Shifts

Chemical shifts in NMR spectra indicate the electronic environment surrounding hydrogen atoms in a molecule. Measured in parts per million (ppm), these shifts help identify functional groups and the types of hydrogen present. For example, protons attached to electronegative atoms like oxygen will appear downfield (at higher ppm values) compared to those in aliphatic environments.

Integration and Multiplicity

Integration in NMR refers to the area under the peaks, which correlates to the number of hydrogen atoms contributing to that signal. Multiplicity, determined by the splitting of peaks, provides insight into the number of neighboring hydrogen atoms (n+1 rule). Together, these features help deduce the connectivity and arrangement of atoms within the compound, crucial for identifying the specific isomers of C6H12O2.

The following 1H NMR spectra are for four compounds, each with molecular formula of C6H12O2. Identify the compounds. d. <IMAGE>

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Shifts

Integration and Multiplicity

The following ¹H NMR spectra are for four compounds, each with molecular formula of C₆H₁₂O₂. Identify the compounds.
d. <IMAGE>