Hey everyone. So let's take a look at this ring structure. We have here a Fluorine that's attached to our ring. So we can say that Fluorine, although it's not a great leaving group, it is a leaving group when compared to the methyl groups, which are not leaving groups at all. And we're using a strong base in the form of hydride ion. So we know that an E2 reaction is going to occur. We have an issue here though, because when it comes to this chair conformation of Cyclohexane, E2 can only happen if the leaving group, which is 4ene, is trans to a beta hydrogen. In addition to this, the Beta Hydrogen and the Leaving Group need to be in the axial positions. Right now, Fluorine is in an equatorial position. But there's a little trick we can use here. We can ring flip the whole chair to put them in the right positions. We could also do this technique where we rotate all the groups clockwise one space. And what I mean by that is I'm going to redraw the chair, And I'm going to rotate all the groups clockwise one space. And we have to make sure that we maintain the same direction. So here Fluorine is pointing down. I'm going to rotate it to this carbon here. It has to maintain the position of pointing down. When we rotate everyone one space, they go from equatorial to axial or axial to equatorial. Right now this is equatorial down. By rotating it clockwise one space it's now axial down. Remember the direction is maintained, so it's still down in both cases. It's the groups that change. It was equatorial before, but now it's axial. Let's keep going. This methyl group here, what is it? It's equatorial up. I'm going to move it here. When I do that now it becomes axial up. This methyl group, which is here, is axial down. I'm going to rotate it one space so now it's over here. It's axial down here, so now it becomes equatorial down here. So this is a way of just us putting the leaving group in the absolute position like we need it to be. Now we're gonna say where the halogen is located is the alpha carbon, connected to the alpha carbon is our beta carbons, beta and beta. If this methyl group is axial up, that means that the hydrogen, which is invisible, is going to be equatorial down. And then this metal group, which is equatorial down, that means a hydrogen here has to be axial up. Now we bring in our hydride ion. Here is our beta hydrogen. Here's our beta hydrogen. Remember the beta hydrogen has to be anti or trans to the leaving group. The leaving group is this Fluorine, and the beta hydrogen needs to be in the axial position. That means that this hydrogen here is out of the question. It's in the equatorial position so we're not gonna attack it or remove it. It has to be this H here that's going to be affected. This strong base comes in with its lone pair, grabs this H, the bond breaks, forms a double bond here, kicking out my fluorine. As a result of this, we get this as our final product. So we have a double bond here. The methyl group is still attached here. Remember when you're attached to a double bonded carbon you don't have stereochemistry so I'm not really showing it as pointing straight up or pointing towards the side. And then we have this other methyl group here, which is up. So this is the way we could depict it if we wanted to keep it in the chair conformation, or we can show it 2 dimensionally here where we have our cyclohexane ring. And let me make this the double bond here. The methyl group that's attached to it, Again, it loses its stereo chem since it's attached to a double bonded carbon. And then we can say 1, 2, 3. 1, 2, 3. Position 3, we have that methyl group, which is pointing up. Remember, 'up' means it will be wedged. It maintains its stereochemistry because it's not attached directly to one of these double bonded carbons, carbons 1 or 2. So again, you can depict your answer keeping the chair conformation, or you could depict it this way. Both are acceptable answers. Alright. So these will be our final answer for this particular question. Remember, for E2 to occur, you need your beta hydrogen, and you need your Leaving Group in the form of the Halogen to be anti to each other, and they both need to be in the axial positions. Here's our Beta Hydrogen that's axial up, here's our Halogen that is axial down. They are anti to each other, meaning they're pointing opposite directions and they're both in the axial positions. So then we can undergo E2.

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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

E2 - Cumulative Practice

Organic Chemistry

8. Elimination Reactions

E2 - Cumulative Practice

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Multiple Choice

Option A: (S)-1,3-dimethylcyclohex-1-ene

Option B: (R)-1,3-dimethylcyclohex-1-ene

Option C: 1,3-dimethylcyclohex-1-ene

Option D: (S)-1,4-dimethylcyclohex-1-ene

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Verified step by step guidance

Identify the starting material and the reagent: The starting material is a bicyclic compound with a fluorine substituent, and the reagent is a hydride ion (H⁻).

Recognize the type of reaction: The presence of a good leaving group (fluorine) and a strong base (hydride ion) suggests an elimination reaction, likely an E2 mechanism.

Determine the possible elimination products: The elimination will result in the formation of a double bond. Consider the possible positions for the double bond formation, taking into account the stability of the resulting alkenes.

Consider stereochemistry: If the elimination can lead to different stereoisomers, consider the stereochemistry of the starting material and the possible stereochemical outcomes of the reaction.

Analyze the given product options: Compare the possible products from the elimination reaction with the given options to determine which one matches the expected outcome based on the reaction mechanism and stereochemistry.