Draw all the isomers with molecular formula C 6 H 12 that contain a cyclobutane ring.

Hello, everyone. Today, we know the following problem, draw all the isomers of a cyclo pentane ring with a molecular formula C seven H 14. So a cyclo pentane ring is a cyclic structure of five carbons. And then we have to just add two more carbons. So we can first draw an ethel group on one of the carbons. And so we see here that we have seven carbons. And if we were to count all the implicit hydrogens, we would see that we have 14 hydrogens. We have two from the circled carbons of our ring like it was eight, we then add the hydrogen that is on this carbon that holds the output group that gives us nine. And then we have two from this methylene group on the ethyl that gives us 11. And then if we count the remaining three on the methyl group of the ethel group. And so that gives us 14, we then can draw another cyclo penang ring but have our metals on the same carbon. Once again, we have our two carbons or two hydrogens that are each of these methylene carbons of our ring that gives us eight and then we have three and three that's six. So eight plus six is 14. So we have another isomer, we then can draw a structure that has our metal groups on different carbons. So we can have one of the metals on a wedge and one on a dash. This just says that one of these groups is going into the page while the other is going out of the page. And so once again, we have our methylene carbons of our ring that gives us six plus the two hydrogens that are holding these metal groups that gives us eight. And then we have the three hydrogens on these methyl groups giving us six plus eight or 14. And so we can continue this with having our metal groups be adjacent to one another once again, one on a dash and one on a wedge. And it's the same principle where we have seven carbons and 14 hydrogens. Furthermore, we can draw a cycle pin ring but have the two methyl groups both on a wedge. And this works if they were both on a dash. So we do not have to draw the product where they're both on a dash. And so this also works as having seven carbons and 14 hydrogens. And so last, but not least we can draw our metals next to each other, but both on wedges and of course, they can both be on dashes. But we don't have to draw that as it is the same product. And once again, we have seven carbons and 14 hydrogens. And so with that, we have constructed all the isomers for this problem. Overall, I hope this helped until next time.

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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

4. Alkanes and Cycloalkanes

Naming Cycloalkanes

Organic Chemistry

4. Alkanes and Cycloalkanes

Naming Cycloalkanes

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4:02 minutes

Problem 90a

Textbook Question

Draw all the isomers with molecular formula C₆H₁₂ that contain a cyclobutane ring.

Verified step by step guidance

Step 1: Understand the problem. The molecular formula C6H12 indicates that the compound is unsaturated or contains a ring structure. Since the problem specifies a cyclobutane ring, the structure must include a four-membered ring with substituents or additional structural features to satisfy the molecular formula.

Step 2: Begin by drawing the base structure of cyclobutane. Cyclobutane is a four-membered ring with single bonds between the carbon atoms. This accounts for 4 carbons and 8 hydrogens (C4H8). To reach C6H12, you need to add 2 more carbons and 4 more hydrogens as substituents.

Step 3: Add substituents to the cyclobutane ring. Consider all possible ways to add two carbon atoms as alkyl groups (e.g., methyl or ethyl groups) to the ring. Ensure that the total number of hydrogens remains consistent with the molecular formula C6H12.

Step 4: Check for structural isomers. Rearrange the positions of the substituents on the cyclobutane ring to create different isomers. For example, you can place both substituents on the same carbon (geminal), on adjacent carbons (vicinal), or on opposite carbons (trans). Ensure that each structure is unique and satisfies the molecular formula.

Step 5: Verify stereoisomers. Cyclobutane can exhibit cis-trans isomerism depending on the spatial arrangement of substituents. Draw both cis and trans configurations for each structural isomer to account for stereoisomers. Confirm that all isomers are valid and distinct.

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Key Concepts

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

Isomerism

Isomerism refers to the phenomenon where compounds have the same molecular formula but differ in the arrangement of atoms. In organic chemistry, isomers can be classified into structural isomers, which differ in connectivity, and stereoisomers, which differ in spatial arrangement. Understanding isomerism is crucial for identifying and drawing all possible structures for a given molecular formula.

Cyclobutane Structure

Cyclobutane is a cyclic alkane with four carbon atoms arranged in a ring, resulting in a molecular formula of C4H8. When considering isomers of C6H12 that include a cyclobutane ring, it is essential to recognize how the cyclobutane structure influences the connectivity of the remaining carbon atoms and hydrogens, leading to various possible isomers.

Degree of Unsaturation

The degree of unsaturation indicates the number of rings and/or multiple bonds in a molecule. For the molecular formula C6H12, the degree of unsaturation can be calculated to determine how many rings or double bonds are present. In this case, the presence of a cyclobutane ring contributes to the degree of unsaturation, guiding the identification of possible isomers.

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