b. What is the product of the reaction of acetamide with HO − ? The p K a of NH 3 is 36; the p K a of H 2 O is 15.7.

All right. Hi, everyone. So first question, let's determine the product formed in a reaction between propan amide and hydroxide. Our hint is that the P K of N H three is 38.0 and our P K of H2O is 15.7. Option A is ethanol acid. Option B is ethanol chlorine. Option C is methyl ethanolate or Ethan oid. And option D says that no reaction will take place. So before I begin, let's recall that the reactivity of Carboni compounds stems from the polarity of the Carboni group itself, right? So I'm drawing on the bottom here, our propane which is a three carbon amide. So here's our mind and recall it carbon because it's less electron negative than oxygen is going to have a partial positive charge in our car bono whereas our oxygen is going to have a partial negative charge. Now, our carbo noc carbon being partially positive is right. And because it's electro phi our base or our nuclear file is going to go ahead and attack it directly. So here we're reacting propan amide with hydroxide, which is going to be our nuclear file because it has a negative charge. So because hydroxide is nucleic, it's going to attack the electrolytic carboy carbon. And so that's actually going to force the pi electrons of our carbonnel to shift upwards towards oxygen. So from here, we get an intermediate, we get an intermediate where the oxygen of our carbon now has a negative charge. And our O H group has gone ahead and attached itself to that car bono carbon. But what's interesting to note here, what's interesting to note is that in the beginning of this process, right, the carbonic carbon was S P two hybridized. But now in our next step of the reaction, this carbon is actually S P three hybridized. So because it's S P three hybridized, we refer to this intermediate as a tetrahedral intermediate, right. So because our tetrahedral intermediate is not stable, our car bono is actually going to form once again. Now, in order for our car bottle to form once again, a pair of electrons on oxygen must move downwards in between itself and the carbon that it's attached to, right, to regenerate our pi bond in our car bottle. But in order to do that, either our hydroxy or our amino substituent must be detached from the carbon. Now, in order to consider which one of these is actually going to leave, we have to consider their relative basic right? Because recall that weaker bases, weaker base, equal, better leaving groups. So in order to determine which of these is the better leaving group, we're given the P K A s of their conjugate acids, right? Ammonia and water. So water given that it has a P K of 15. is a stronger acid than ammonia, which has a P K of 38 right? So because it's the stronger acid, yeah, we know it's going to be the weaker base. So because it is the weaker base, it is the better lead group. So when the pair of electrons on oxygen moves to form a carbona, once again, it is the O H group that is going to detach itself from our tetrahedral intermediate. And so what you'll notice is that our final product at the end of this process is Promide once again. So we started off with Propan amide and we've ended up with Propan amide at the end of this process. So what does that mean? Right. That means that there is no change in this molecule at all. Therefore, we know that no reaction must have taken place. And therefore our answer here is option d that no reaction will take place. So with that being said, thank you very much for watching. And I hope you found this helpful.

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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

22. Carboxylic Acid Derivatives: NAS

Nucleophilic Acyl Substitution

Organic Chemistry

22. Carboxylic Acid Derivatives: NAS

Nucleophilic Acyl Substitution

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5:48 minutes

Problem 7b

Textbook Question

b. What is the product of the reaction of acetamide with HO⁻? The pK_a of NH₃ is 36; the pK_a of H₂O is 15.7.

Verified step by step guidance

Identify the functional group in acetamide. Acetamide contains an amide functional group, which is characterized by a carbonyl group (C=O) directly bonded to a nitrogen atom (−NH2).

Understand the role of HO⁻ (hydroxide ion) in the reaction. HO⁻ is a strong base and can deprotonate acidic protons or participate in nucleophilic attack.

Compare the pKa values provided. The pKa of NH3 is 36, and the pKa of H2O is 15.7. Since the pKa of H2O is much lower, it indicates that water is a stronger acid than NH3. This means that the hydroxide ion (conjugate base of H2O) is unlikely to abstract a proton from NH3 (conjugate acid of the amide nitrogen).

Analyze the reaction mechanism. In the presence of HO⁻, the hydroxide ion can attack the carbonyl carbon of the amide group. This is because the carbonyl carbon is electrophilic due to the partial positive charge created by the electronegativity of the oxygen atom.

Predict the product. The nucleophilic attack by HO⁻ on the carbonyl carbon will lead to the cleavage of the C−N bond, resulting in the formation of a carboxylate ion (CH3COO⁻) and ammonia (NH3) as the products of the reaction.

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Key Concepts

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

Acetamide Structure and Reactivity

Acetamide is an amide derived from acetic acid, featuring a carbonyl group (C=O) bonded to a nitrogen atom (N). Its structure influences its reactivity, particularly in nucleophilic and electrophilic reactions. Understanding how acetamide behaves in the presence of strong bases, like hydroxide ions (HO−), is crucial for predicting the outcome of the reaction.

Acid-Base Chemistry and pKa Values

The pKa value indicates the strength of an acid in solution; lower pKa values correspond to stronger acids. In this context, the pKa of NH3 (36) suggests it is a very weak acid, while H2O (pKa 15.7) is a much stronger acid. This information is essential for determining the direction of the reaction between acetamide and hydroxide ions, as it helps predict which species will donate or accept protons.

Nucleophilic Attack and Deprotonation

In the reaction of acetamide with hydroxide ions, the hydroxide acts as a strong nucleophile, attacking the carbonyl carbon of acetamide. This leads to the formation of a tetrahedral intermediate, which can subsequently undergo deprotonation. Understanding this mechanism is key to predicting the final product of the reaction, which typically results in the formation of an amine and a hydroxyl group.

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