The 1 H NMR spectra of three isomers with molecular formula C 4 H 9 Br are shown here. Which isomer produces which spectrum? a. <IMAGE>

Hello everyone. Let's do this problem. It asks which of the three isomers with the molecular formula C six H 11 C L has the following proton N M R spectrum. We are given our spectrum and our three isomer answer choices. Answer choice A is the structure of E two chloro three methyl pent two E structure B is three chloro two methyl pent two E and structure C is E two C hex two E. So how can we use our spectrum to match our structure? Well, there are several characteristics we can gather from our proton N M R spectrum and we can compare those to the characteristics of the spectrum that would be produced by each of these isomers and see if they match. So first in this order, I like to look at the number of signals that would be produced by each structure. That's an easy way to eliminate some answer choices. Then we can look at the integration since that was given to us and that tells us how many protons are represented by that signal. Then we can look at the multiplicity. So is our signal, a singlet doublet triplet and that's based on the number of neighboring protons, right, we use the expression N plus one where N is the number of neighboring protons to the proton in question. And lastly, we look at the chemical shift which is determined by the relative location of that proton. So is it next to any electron dense groups like a chlorine atom or a double bond? Because those groups will withdraw the electron density from that proton making it more de shielded and have a higher chemical shift. All right. So let's look at the characteristics of the potential spectrum produced by each of these isomers starting with the number of signals. So answer choice A will produce how many signals? One 234 does that match our spectrum? No, our spectrum only has three signals. So just from that we can eliminate answer choice A all right answer choice B will produce 123 signals, right? Both of these methyl groups are equivalent, they are coming off of the same carbon. So that is a potential answer choice. What about answer choice C 12345. So answer choice C is also not correct. And just by process of elimination, we already know that answer. Choice B is the correct answer for this problem. And if you want to stick around, I will go over the other characteristics. So looking at signal A, what is the integration? That is 123 protons and there are two neighboring protons. So that will be a triplet. OK. That tells us that this most upfield signal is proton signal. A signal B has two hydrogens, three neighboring protons. So that will be a quartet and that matches the signal at around two P PM, which leaves signal C and that in fact has six protons, no neighboring protons. So that is a singlet and that matches the singlet at about 1.8 P PM. All right. And let's look at the chemical shift. So proton B is closer to the double bond and chlorine atom than proton signal A. So that makes sense why proton B is more downfield and proton C, those are closer to the double bond, right? Proton A is just sort of out here not close to either of those electron dense groups, but proton C is less downfield than proton B because of how many protons there are right, there are only two protons on carbon B. So those hydrogens are definitely going to feel more naked. So that is the justification of the chemical shift. So once again, answer choice B is the correct answer. Choice for this problem. That's it for this one. I hope this 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
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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
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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 54a

Textbook Question

The ¹H NMR spectra of three isomers with molecular formula C₄H₉Br are shown here. Which isomer produces which spectrum?
a. <IMAGE>

Verified step by step guidance

Analyze the molecular formula C4H9Br, which indicates the presence of four carbons, nine hydrogens, and one bromine atom. This suggests the compound is an alkyl bromide.

Examine the 1H NMR spectrum provided. The spectrum shows three distinct signals corresponding to three different types of hydrogen environments.

Interpret the signal at approximately 1 ppm labeled as '3 H'. This is likely a methyl group (-CH3) attached to a carbon that is not directly bonded to an electronegative atom, as the chemical shift is low.

Interpret the signal at approximately 2 ppm labeled as '2 H'. This is likely a methylene group (-CH2-) adjacent to the bromine atom, as the chemical shift is slightly deshielded due to the electronegativity of bromine.

Interpret the signal at approximately 0.9 ppm labeled as '6 H'. This suggests two equivalent methyl groups (-CH3) in a symmetrical environment, likely part of an isopropyl group structure. Combine this information to deduce the isomer that matches the spectrum.

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Key Concepts

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

1H NMR Spectroscopy

Proton Nuclear Magnetic Resonance (1H NMR) spectroscopy is a powerful analytical technique used to determine the structure of organic compounds. It provides information about the number of hydrogen atoms in different environments within a molecule, indicated by peaks in the spectrum. The position of these peaks (measured in parts per million, PPM) reveals the electronic environment of the protons, while the area under each peak corresponds to the number of protons contributing to that signal.

Chemical Shifts

Chemical shifts in 1H NMR spectra are indicative of the electronic environment surrounding hydrogen atoms in a molecule. Different functional groups and molecular structures cause protons to resonate at different frequencies, resulting in distinct chemical shifts. For example, protons adjacent to electronegative atoms or in alkyl chains will appear at different PPM values, allowing chemists to infer structural information about the isomers being analyzed.

Integration and Multiplicity

Integration in 1H NMR refers to the area under the peaks, which quantifies the number of protons contributing to each signal. Multiplicity, on the other hand, describes the splitting of NMR signals due to neighboring protons (n+1 rule), providing insight into the number of adjacent hydrogen atoms. Together, integration and multiplicity help in deducing the connectivity and arrangement of atoms in the isomers, which is crucial for identifying which spectrum corresponds to each isomer of C4H9Br.

The 1H NMR spectra of three isomers with molecular formula C4H9Br are shown here. Which isomer produces which spectrum?a. <IMAGE>

Key Concepts

1H NMR Spectroscopy

Chemical Shifts

Integration and Multiplicity

The ¹H NMR spectra of three isomers with molecular formula C₄H₉Br are shown here. Which isomer produces which spectrum?
a. <IMAGE>