Predict the major products formed when the following amines undergo exhaustive methylation, treatment with Ag 2 O, and heating. (a) hexan-2-amine (b) 2-methylpiperidine (c) N -ethylpiperidine

Welcome back everyone. The next problem says, draw the structure of the major product expected from the exhaustive methylation and treatment with A G 20 followed by heating of pentane, two amine and amine. And we have four different answer choices that we'll come back to after we figure out what we would expect. So don't forget that when we have an amine, this is not a good leaving group. So, exhaustive methylation is used to convert an amine to a quaternary ammonium salt, which is then a good leaving group resulting in a what we call a Hoffman elimination. So typically for the first step, the methylation step, the reagent we'd use is methyl iodide. So ch three I in excess, obviously, so that we make sure we add three methyl groups onto the amine. Then we see after we've methylated, we're going to treat with silver oxide and that's going to have the result of uh changing our counter ion from the iodide ion that will result from that methylation process to a hydroxide ion. And then the heat will cause an elimination reaction. And we need to note something very important about this, which is that while in general, an E two reaction will lead to the um more substituted alkene being the major product because it's more stable. So the Zits rule, in the case of a Hoffman elimination will get the less substituted alkene as the major product. And that's due to the fact that in the transition state, because of all those methyl groups, there's too much hysteric hindrance to make that zeit of rule elimination favorable. So something really just important to note about Hoffman elimination, less substitute alkene will be the major product, not because it's more stable, but because of ster hindrance, making the transition state more unfavorable if you were to generate the more substitute alkene. So let's think about this reaction. So we have our contain two amine. So a five carbon chain with the amine on carbon two. And now we treat it with that excess methyl iodide. And that's going to methylate that uh amine group and give us a quaternary salt. So we have then I minus and we've now got our chain and then we have N plus with three methyl groups attached. Now we treat it with that silver oxide and water, our by-product will be silver iodide and we'll now have the hydroxide salt. So still N plus and then three methyl groups, but now have oh minus. And we'll want that for participating in our elimination reaction because we need to get that hydrogen off the beta hydrogen off with or excuse me, the hydrogen from our carbon in the double bond off um with a strong base. Let me just scroll down here to make a little more room and we'll add heat and have our elimination reaction. The nitrogen, the electrons will go to the nitrogen that positively charted nitrogen. Now that it's a leaving group, we have our hydrogen. And now again, we need to ponder, we need the less substituted alkene. So are we going to eliminate the hydrogen on carbon one or the hydrogen on carbon three? Well, eliminating the hydrogen on carbon one will lead to the less substituted do. So, our hydroxide group here will remove our proton and the ch electrons will go down to form the double bond as our leaving group leaves. It's an E two reaction. So this is a concerted reaction. It's leaving in the anti coplanar orientation and in generating as our major product, we'll have the double bond from carbon one to carbon two. So our major product here will be pent one in. So we'll scroll back up just a tiny bit and compare this to our answer choices. So when we look at our four possibilities choice A here is our correct molecule pent one E with the double bond going from carbon one to carbon two. So looking at answer choice B this would be the minor product since this has the double bond going from carbon two to carbon three. So P two E so this will be a minor product. So Choice B is incorrect. Again, that's a little bit tricky because we're used to thinking of the more substitute alkene as the major product, but not in the case of Hoffman elimination. And then choice C has a double bond between carbon three and a substituent carbon. So it's added on another carbon from somewhere. So that's not going to be the case. And choice D has a double bond between carbon two and a substituent carbon. So as if we added on a methyl group, so C and D definitely not correct. So once again, the structure of the major product expected from the exhaustive methylation, methylation and treatment with silver oxide followed by heating which is a Hoffman elimination is choice A P one E see you in the next video.

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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
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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

26. Amines

Hofmann Elimination

Organic Chemistry

26. Amines

Hofmann Elimination

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6:29 minutes

Problem 20a,b,c

Textbook Question

Predict the major products formed when the following amines undergo exhaustive methylation, treatment with Ag₂O, and heating.
(a) hexan-2-amine
(b) 2-methylpiperidine
(c) N-ethylpiperidine

Verified step by step guidance

Step 1: Understand the process of exhaustive methylation. Exhaustive methylation involves treating the amine with excess methyl iodide (CH3I) to convert it into a quaternary ammonium salt. For each amine, identify the nitrogen atom and determine how many methyl groups can be added to it until it becomes a quaternary ammonium ion.

Step 2: Apply the Hofmann elimination reaction. After exhaustive methylation, the quaternary ammonium salt is treated with silver oxide (Ag2O) and water, which generates hydroxide ions (OH⁻). Upon heating, the hydroxide ions induce a β-elimination reaction (Hofmann elimination), leading to the formation of an alkene. Identify the β-hydrogens (hydrogens on the carbon adjacent to the carbon bonded to the nitrogen) and determine the possible alkenes that can form.

Step 3: Predict the major product based on the Hofmann rule. The Hofmann elimination typically favors the formation of the less substituted (less stable) alkene due to steric hindrance in the transition state. For each compound, analyze the possible alkenes and identify the major product based on this rule.

Step 4: For (a) hexan-2-amine, determine the structure of the quaternary ammonium salt after exhaustive methylation. Then, identify the β-hydrogens and predict the major alkene product formed after Hofmann elimination.

Step 5: Repeat the process for (b) 2-methylpiperidine and (c) N-ethylpiperidine. For each compound, determine the quaternary ammonium salt, identify the β-hydrogens, and predict the major alkene product formed after Hofmann elimination, keeping in mind the Hofmann rule.

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Key Concepts

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

Exhaustive Methylation

Exhaustive methylation is a chemical process where amines are treated with methylating agents, such as methyl iodide, to convert them into quaternary ammonium salts. This reaction typically involves the complete substitution of hydrogen atoms on the nitrogen atom of the amine with methyl groups, leading to the formation of a fully methylated product. Understanding this concept is crucial for predicting the products of the amines listed in the question.

Ag2O Treatment

Silver oxide (Ag2O) is often used in organic chemistry to facilitate the oxidation of amines to their corresponding imines or to promote the elimination of quaternary ammonium salts. When treated with Ag2O, the quaternary ammonium salts formed from exhaustive methylation can undergo dealkylation or rearrangement, leading to the formation of new products. This step is essential for understanding the transformation of the initial methylated amines.

Heating in Organic Reactions

Heating is a common technique in organic chemistry that can drive reactions to completion, promote rearrangements, or facilitate elimination processes. In the context of the question, heating the products formed after exhaustive methylation and Ag2O treatment can lead to further reactions, such as the formation of alkenes or other stable products. Recognizing the role of heat in these reactions is vital for predicting the final products of the amines.

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