Which of these reactions are likely to produce both elimination and substitution products? a. 2-bromopentane + NaOCH 3 b. 3-bromo-3-methylpentane + NaOMe. (Me = methyl, CH 3 ) c. 2-bromo-3-ethylpentane + NaOH

Welcome back. Everyone can, the given reactions form both substitution and elimination products. And we're given three reactions. What we can notice is that the first reaction, the second one and the third one, all of them contain sodium compounds. Since sodium is an iron spectator, we are simply analyzing hydroxide which has a negative charge. Then we have meth oxide which has a negative charge on oxygen and metox side. Once again. Now both hydroxide and meth oxide, those are strong Niles and strong basis. So they can participate in SN two and E two reactions. Specifically, they are very likely to participate in both substitution and elimination. So we can say that our answer will essentially depend on the type of substrate and we want to draw those substrates. So what about the first one? Well, essentially the first substrate is to chloro we methyl tens, we're going to draw pens, which is the parent carbon two has a chloral group. Carbon three contains a methyl group. Now this is a secondary substrate, secondary substrates can participate in both substitution and elimination. They are not too hysterically hindered or in essence to reaction. So that's perfectly fine even though primary substrates are better because they're less sternly hindered. However, elimination is generally preferred more, right? Because it is a slightly more aster hindered substrate than a primary or a muscle substrate. However, elimination is possible and substitution is possible, right? So that means the first structure can participate in both substitution and elimination. What about the second structure? Once again, we have a pensa derivative. So we're going to draw a five member chain and now the substations are too chloral or methyl. And once again, there's nothing different. We're still considering the type of substrate and we have a secondary substrate. Now, it's less hysterically hindered at the, at the electro pilic center, meaning the substitution reaction S and two is easier compared to reaction one. And of course, elimination is also possible and more preferred, right? Because we have a strong base and a slightly more hysterically hindered substrate than let's say a primary or a methyl substrate. And now let's consider the third reaction. We have a pena derivative. The subscripts are at three chloral, 234, try muffle. So chi three and four. So what we can see is that this substrate is tertiary and we have to recall that whenever we're using a strong Nile, tertiary substrates would more likely participate in and one reactions, right. So if we have a tertiary substrate, it would need a weak Nile. If we have a strong nhi, it would not allow us to perform substitution. So we can say that for this tertiary Nile essence one is preferred. But the problem is that the Nile is strong and we need a weak Nile, right. So substitution would not take place in an essence reaction. It would need a weak nuclear file such as let's say methanol, right, and not met oxide. However, tertiary substrates wood easily undergo elimination reactions. E two right. This is really favorable. If we use a strong base, we can use a tertiary substrate for an elimination reaction. So we can say that for the third reaction S and two is not possible, it would need to participate in an sn one reaction. And therefore the third, the third pair would not participate in a substitution reaction. Meaning it is the only pair that cannot participate in substitution and elimination. So what is our answer? Well, essentially pairs one and two can form both substitution and elimination products. So that be answer B thank you 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

8. Elimination Reactions

SN1 SN2 E1 E2 Chart (Big Daddy Flowchart)

Organic Chemistry

8. Elimination Reactions

SN1 SN2 E1 E2 Chart (Big Daddy Flowchart)

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5:55 minutes

Problem 24b

Textbook Question

Which of these reactions are likely to produce both elimination and substitution products?
a. 2-bromopentane + NaOCH₃
b. 3-bromo-3-methylpentane + NaOMe. (Me = methyl, CH₃)
c. 2-bromo-3-ethylpentane + NaOH

Verified step by step guidance

Step 1: Understand the nature of the alkyl halide in each reaction. The structure of the alkyl halide determines whether the reaction will favor substitution (SN1 or SN2) or elimination (E1 or E2). For example, primary, secondary, and tertiary alkyl halides behave differently due to steric hindrance and carbocation stability.

Step 2: Analyze the nucleophile/base used in each reaction. Sodium methoxide (NaOCH3) and sodium hydroxide (NaOH) are strong bases and nucleophiles, which can promote both substitution and elimination reactions. Strong bases favor elimination (E2), while strong nucleophiles favor substitution (SN2).

Step 3: Consider the reaction conditions. For secondary alkyl halides, such as 2-bromopentane and 2-bromo-3-ethylpentane, both substitution and elimination are possible because the secondary carbon can undergo SN2 substitution or E2 elimination depending on the strength of the base/nucleophile and steric factors.

Step 4: Examine the structure of 3-bromo-3-methylpentane. This is a tertiary alkyl halide, which is highly sterically hindered. Tertiary alkyl halides typically undergo elimination (E2) rather than substitution (SN2) when reacting with strong bases like NaOMe. However, substitution (SN1) is possible under certain conditions due to carbocation stability.

Step 5: Summarize the likelihood of elimination and substitution for each reaction. For (a) 2-bromopentane + NaOCH3, both SN2 and E2 are likely. For (b) 3-bromo-3-methylpentane + NaOMe, elimination (E2) is more likely due to steric hindrance, but SN1 substitution could occur under specific conditions. For (c) 2-bromo-3-ethylpentane + NaOH, both SN2 and E2 are possible due to the secondary alkyl halide and strong base/nucleophile.

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Key Concepts

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

Nucleophilic Substitution Reactions

Nucleophilic substitution reactions involve the replacement of a leaving group in a molecule by a nucleophile. In organic chemistry, these reactions can occur via two main mechanisms: SN1 and SN2. The choice of mechanism depends on factors such as the structure of the substrate and the strength of the nucleophile. Understanding these mechanisms is crucial for predicting the products of reactions involving alkyl halides.

Elimination Reactions

Elimination reactions involve the removal of a small molecule, often HX or H2O, from a larger molecule, resulting in the formation of a double bond. The two primary types of elimination reactions are E1 and E2. E1 reactions are unimolecular and typically occur in two steps, while E2 reactions are bimolecular and occur in a single concerted step. Recognizing conditions that favor elimination over substitution is essential for predicting reaction outcomes.

Regioselectivity and Stereoselectivity

Regioselectivity refers to the preference of a chemical reaction to yield one structural isomer over others, while stereoselectivity indicates the preference for one stereoisomer over another. In reactions that can produce both substitution and elimination products, understanding these concepts helps predict which products will be favored based on the substrate structure and reaction conditions. Factors such as sterics and electronics play a significant role in determining the regio- and stereochemical outcomes.

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