An alkyl halide reacts with an alkoxide ion to form a compound whose 1 H NMR spectrum is shown here. Identify the alkyl halide and the alkoxide ion. ( Hint : See Section 9.15.) <IMAGE>

Hello everyone. Let's do this problem. It says the proton N M R spectrum of the compound created by the reaction of an alkyl hali and an al side ion is shown here. What are the structures of the alkyl hali and al ion? And we are given the proton N M R spectrum which has four signals from left to right. We have a double it with integration two H A multiple with integration one H A singlet with integration nine H and A double it with integration six H. And we are given four answer choices. Answer choice A gives us Trt Butyl chloride and iso butoxide anion. Answer choice B gives us neo penal chloride and iso Propox anion. Answer choice C gives us neopet oxide and isopropyl chloride. And answer choice D gives us TP pet to side anion and iso butyl chloride. So let's look at our spectrum and see what we can gather from this. So I see this large singlet that has the integration nine hydrogens and that is going to be for the three methyl groups of a trt beetle group. And that makes sense because in a turtle group that central carbon does not allow for any other neighboring protons, right? Because the central carbon is bonded to four carbon groups already. So that explains the slit with an integration of nine H. But notice that all of our answer answer choices have this Trt beel group. So that does not eliminate any answer choices. But there is another characteristic of the TRT bel group that can eliminate some answer choices. And that happens to do with this doublet a little more downfield around 3.29 P PM. And that has to do with the methylene protons, right, two hydrogens. And this doublet tells us that it's going to be next to the isopropyl group. That is illustrated by these two signals of one hydrogen and six hydrogen. Because if those two hydrogens were next to the turtle group, as seen in answer choices B and C, we would have another singlet, right. This carbon would give us a singlet because there are no neighboring protons. And this carbon in neo penal chloride would also give us a singlet because being next to that turtle group and chlorine atom, there are no neighboring protons. But as we see in answer choices A and B or sorry, A and D, that methylene group is next to the isopropyl group, which will make it a doublet specifically that methane proton, right. And this methylene group in answer choice D will also give a doublet. So we can eliminate answer choices. C and B OK. Now how do we decide between answer traces A and D, they both give us the following structure, an alcohol or sorry, an oxygen atom that has a turtle group on one side and a carbon, then an isopropyl group. But we know that when we react in oxide ion with an Alk Hala, the alco ha line should be the less hindered group. So if we look at answer choices A and D starting with answer choice. A notice that the chlorine atom is right next to that tt bel group versus an answer choice D that chlorine atom is next to a primary carbon, right? And so in answer choice A that was a tertiary carbon. So in answer choice A that alky Halide is going to be more hindered due to the trt beutel group. So we can eliminate answer choice A which leaves answer choice D to be our correct answer because we have that less hindered Alky Hali. And we will result in the structure that gives us this proton and A R spectra. That's it for this problem. 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 61

Textbook Question

An alkyl halide reacts with an alkoxide ion to form a compound whose ¹H NMR spectrum is shown here. Identify the alkyl halide and the alkoxide ion. (Hint: See Section 9.15.)
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Verified step by step guidance

Step 1: Analyze the 1H NMR spectrum provided. The spectrum shows three distinct signals: a singlet at approximately 1.2 ppm corresponding to 9 hydrogens, a doublet at approximately 1.0 ppm corresponding to 6 hydrogens, and a quartet at approximately 3.5 ppm corresponding to 2 hydrogens. Additionally, there is a small singlet at approximately 2.2 ppm corresponding to 1 hydrogen.

Step 2: Interpret the signal at 1.2 ppm (9 H, singlet). This is characteristic of a tert-butyl group ((CH3)3C), where the three methyl groups are equivalent and produce a singlet due to the lack of neighboring hydrogens.

Step 3: Interpret the signal at 1.0 ppm (6 H, doublet). This is indicative of two equivalent methyl groups adjacent to a single hydrogen atom, which causes splitting into a doublet. This suggests the presence of an ethyl group (-CH2CH3).

Step 4: Interpret the signal at 3.5 ppm (2 H, quartet). This is characteristic of a methylene group (-CH2-) adjacent to a methyl group (-CH3), which causes splitting into a quartet. This further supports the presence of an ethyl group.

Step 5: Combine the information to deduce the reactants. The tert-butyl group likely comes from the alkoxide ion (tert-butoxide, (CH3)3CO-), and the ethyl group likely comes from the alkyl halide (ethyl bromide, CH3CH2Br). The reaction between tert-butoxide and ethyl bromide forms tert-butyl ethyl ether ((CH3)3COCH2CH3), consistent with the NMR spectrum.

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Key Concepts

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

Alkyl Halides

Alkyl halides are organic compounds containing a carbon atom bonded to a halogen atom (F, Cl, Br, I). They are important in organic synthesis and can undergo nucleophilic substitution reactions. Understanding the structure and reactivity of alkyl halides is crucial for predicting the products of reactions, such as those involving alkoxide ions.

Alkoxide Ions

Alkoxide ions are negatively charged species formed by deprotonation of alcohols. They are strong nucleophiles and are commonly used in organic reactions, particularly in the formation of ethers and in nucleophilic substitutions. Recognizing the structure and reactivity of alkoxide ions is essential for determining the outcome of reactions with alkyl halides.

1H NMR Spectroscopy

1H NMR (Proton Nuclear Magnetic Resonance) 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, as indicated by chemical shifts and integration of peaks. Analyzing the NMR spectrum helps identify the alkyl halide and alkoxide ion involved in the reaction by correlating peak patterns with molecular structures.

An alkyl halide reacts with an alkoxide ion to form a compound whose 1H NMR spectrum is shown here. Identify the alkyl halide and the alkoxide ion. (Hint: See Section 9.15.)<IMAGE>

Key Concepts

Alkyl Halides

Alkoxide Ions

1H NMR Spectroscopy

An alkyl halide reacts with an alkoxide ion to form a compound whose ¹H NMR spectrum is shown here. Identify the alkyl halide and the alkoxide ion. (Hint: See Section 9.15.)
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