For each NMR spectrum, propose a structure consistent with the spectrum and the additional information provided. <IMAGE> a. Elemental analysis shows the molecular formula to be C 8 H 7 OCl. The IR spectrum shows a moderate absorption at 1602 cm –1 and a strong absorption at 1690 cm –1 . b. The mass spectrum shows a double molecular ion of ratio 1:1 at m / z 184 and 186. <IMAGE>

Hey everyone. Let's do this problem and says propose the structure of an unknown compound. Using the NMR spectrum And the additional data provided here the elemental analysis confirmed the molecular formula of the compound to B C eight H 70. B. R. The IR spectrum shows two stretches at 16 and 16 05. The mass spectrum shows a double molecular ion peak with equal intensities at mass over charge ratio of 1 98 and 200. So when we're given the molecular formula and we want to determine a structure, it's helpful to calculate the index of hydrogen deficiency or the I. H. D. To determine any elements of unsaturated in. So the formula for this is two times the number of carbons plus two plus the number of nitrogen minus the number of hydrogen minus the number of halogen all over two, plugging in the values for our molecular formula, that would be two times eight plus two plus zero nitrogen ins minus seven Hydrogen is and -1 halogen All over two gives us an i. h. d. of five. So what does that tell us? Well, I know that Benzene Gives us a value of four and having one value left over. That means we could also have a ring or double bond. one. That would be a value of one. And our NMR spectrum confirms that we have a Benzene because I see some signals in the aromatic proton region. So that makes sense. So we'll just see if we have a ring or double bond and we'll use our other data to help us determine which one. So our Ir spectrum, we have the signals 16 And 1605. So we know that a carbonell group C double bond O has a signature signal around 1700. So this first value 16 92 is pretty close to that, but it's a little bit lower, which tells us we have a conjugated carbon eel group. So that means we're going to have our double bond instead of a ring and it's lower. This value is lower than an unconscious gated carbonell group. Because when we have the conjugated carbonell group, there will be a weaker C double bond O bond because of the D localization of those electrons. Okay, so that's how we know this is a conjugated carbonell. And the 1605 signal confirms that we have a benzene ring with the C double bond C. H signals there. Okay. Also, our mass spectrum Shows molecular ion peaks at 198 and 200. If you calculate the atomic mass of our molecular formula C eight H 70 B. R, You get a value of 1 99.04. So That shows that we have M - -1 and M plus one proton. Right, so all of this data will help us get closer to our structure. So we know we have a conjugated carbon Neil group. So let's draw that and we know we have a benzene ring. So if we have a benzene ring and our carbon group is conjugated. Let me draw it this way, then that means our carbon eel group must be directly off of the benzene ring. Right. And let's look at our signals in our proton NMR. So we have our aromatic protons down here around 7.5 to 8 PPm. So we have two protons, one proton, two protons. That's going to represent these protons, A. B. C. And those will be peaks A. B. C. And R. C. proton is actually more D. Shielded than be because our um substitue in group is going to be a meta directing group. So therefore the electron density will be more around protons. B. Okay, continuing with our spectrum, we have this peak at around 4.5 ppm. Which is going to represent a. D. Shielded methylene group, which tells me that they're going to be in between this carbonell group and our bromine atom. Right? And that's why the signal is a single it because these two protons have no neighboring protons. So that gives us this structure, leaving B two B. The correct answer 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

Mass Spect:Fragmentation

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

Mass Spect:Fragmentation

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Problem 38

Textbook Question

For each NMR spectrum, propose a structure consistent with the spectrum and the additional information provided.
<IMAGE>
a. Elemental analysis shows the molecular formula to be C₈H₇OCl. The IR spectrum shows a moderate absorption at 1602 cm^–1 and a strong absorption at 1690 cm^–1.
b. The mass spectrum shows a double molecular ion of ratio 1:1 at m/z 184 and 186.
<IMAGE>

Verified step by step guidance

Step 1: Analyze the molecular formula C8H7OCl provided in the problem. This indicates the presence of 8 carbons, 7 hydrogens, 1 oxygen, and 1 chlorine atom. The chlorine atom suggests the possibility of isotopic peaks in the mass spectrum due to its natural abundance (35Cl and 37Cl).

Step 2: Examine the IR spectrum data. The absorption at 1602 cm⁻¹ suggests the presence of a C=C bond, likely in an aromatic ring. The strong absorption at 1690 cm⁻¹ indicates a carbonyl group (C=O), which is consistent with the molecular formula.

Step 3: Analyze the mass spectrum data. The double molecular ion peaks at m/z 184 and 186 in a 1:1 ratio confirm the presence of chlorine, as the isotopes 35Cl and 37Cl contribute to this pattern.

Step 4: Interpret the NMR spectrum. The peaks at approximately 7-8 ppm correspond to aromatic protons, indicating the presence of an aromatic ring. The integration values (2H, 2H, 1H) suggest specific proton environments within the aromatic ring. The peak at approximately 4 ppm corresponds to 2H, likely from a CH2 group adjacent to an electronegative atom or group.

Step 5: Combine all the data. The molecular formula, IR spectrum, mass spectrum, and NMR spectrum suggest a structure with an aromatic ring, a carbonyl group, and a chlorine atom. The aromatic ring likely contains substituents that account for the observed proton environments in the NMR 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, their environment, and connectivity in a molecule. Peaks in the NMR spectrum correspond to different chemical environments of protons, and their integration values indicate the relative number of protons contributing to each peak.

Infrared (IR) Spectroscopy

IR spectroscopy is used to identify functional groups in organic compounds by measuring the absorption of infrared light, which causes molecular vibrations. Specific absorption bands correspond to particular functional groups; for example, a strong absorption around 1690 cm-1 typically indicates a carbonyl (C=O) group, while a moderate absorption at 1602 cm-1 may suggest the presence of an aromatic ring. Understanding these absorptions helps in deducing the functional groups present in the compound.

Mass Spectrometry

Mass spectrometry is an analytical technique that measures the mass-to-charge ratio of ions to identify and quantify molecules in a sample. The presence of a double molecular ion at m/z 184 and 186 suggests isotopic variants of the same compound, likely due to the presence of chlorine, which has isotopes with different masses. This information, combined with the molecular formula and other spectral data, aids in confirming the molecular structure of the compound.

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