Draw four compounds with molecular formula C 5 H 10 that have carbon–carbon double bonds but do not have cis–trans isomers.

Hello everyone. So in this video, we want to draw the structural members with the molecular formula of C four H eight that do not show cis or trans humanism and have a carbon carbon double bond. So it is possible to determine the structure of a molecule with the help with of the degree of saturation, which is basically the total number of pi bonds and rings within a molecule. So let's make sure we write this out. So it's engraved in our heads. Again, we're talking about the degree of unsaturated. Again, what this is is basically the total number of our pi bonds and rings within a molecule. And each degree of saturation is either because of the double bond or a ring. So the general formula to calculate the degree of saturation. So that's D O U. The equation is as follows. So we have two times our carbons plus two plus N minus H and minus X all divided by two. So let's actually explain what this equation means. So see here. So two times C the C is the number of carbons and here is going to be the number of hydrogen H is going to be the number of hydrogen and X is going to be any number of our religions. So let's actually write out what our halogen are. So our halogen is going to include flooring chlorine roaming and lastly iodine. So for the given molecular formula for C four H eight, the degree oven situation can be found using our equation right over here. So we'll go ahead and just kind of fill out this formula. Again, we're solving for the degree of saturation also that's D O U. So two times. So the number of carbons that we have in this molecule is of course four because we see we have a little number here of four. Then we're gonna go ahead and add to for our nitrogen of course zero. So we'll just add zero just to use again this template of the equation. Then we have to go ahead and subtract our hydrogen and that's just going to be eight. Since we see in a molecule here, we have a little eight. So we have eight atoms of hydrogen in our molecule. And do we have any? Hey, lights. So flooring, flooring bromine iodine, we did not. So that is just subtracting zero. Alright. So that's enough for our top part of our fraction are the numerator and we're doing all this by two. So once you put this into the calculator, we see that we get the degree of saturation being just one. So again, we said that each degree of situation is either because of the double bond or a ring. So one degree oven situation means it can be from a double bond or a ring. Since the question as for molecules with a carbon, carbon, double bond, the Apostle structures will contain only one double bond. So the only possible structural members with a molecular formula of C four H eight is let's go ahead and draw those out. We can go ahead and draw something green. So we have this upside down y with a double bond at one of those bonds were just at the bottom here. So it does contain this carbon carbon double bond without having a CIS or trans ice firm. And the other structure that we can go ahead and draw is this exact year with one of pi bonds at the end. So this right here is going to be my final answer for this problem. These are the two possible structure with a molecular formula of C four H eight that do not show any cis or trans humanism and do have a carbon, carbon double bond.

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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

1. A Review of General Chemistry

Constitutional Isomers

Organic Chemistry

1. A Review of General Chemistry

Constitutional Isomers

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Problem 5

Textbook Question

Draw four compounds with molecular formula C₅H₁₀ that have carbon–carbon double bonds but do not have cis–trans isomers.

Verified step by step guidance

Step 1: Understand the molecular formula C5H10. This formula indicates that the compound is an alkene (CnH2n) due to the presence of a double bond, as it follows the general formula for alkenes.

Step 2: Recall that cis-trans isomerism occurs in alkenes when there are two different groups attached to each carbon of the double bond. To avoid cis-trans isomerism, ensure that at least one carbon in the double bond has two identical substituents.

Step 3: Begin by drawing a straight-chain alkene (pentene) where one of the double-bonded carbons has two identical substituents. For example, 2-methyl-1-butene has a double bond at the first carbon, and the second carbon has two hydrogens, preventing cis-trans isomerism.

Step 4: Create additional structures by varying the position of the double bond or branching the chain. For instance, 2-methyl-2-butene has a double bond at the second carbon, and the second carbon has two identical substituents (methyl groups), which also prevents cis-trans isomerism.

Step 5: Continue exploring other possibilities, such as cyclic alkenes like cyclopentene, where the ring structure inherently prevents cis-trans isomerism due to restricted rotation around the double bond. Another example is 3-methyl-1-butene, where the double bond is at the first carbon, and the third carbon has two hydrogens, avoiding cis-trans isomerism.

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Key Concepts

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

Structural Isomers

Structural isomers are compounds that have the same molecular formula but differ in the connectivity of their atoms. In the case of C5H10, various structural isomers can be formed by altering the arrangement of carbon atoms and the placement of double bonds. Understanding structural isomerism is crucial for identifying different compounds that fit the same molecular formula.

Geometric Isomerism

Geometric isomerism, also known as cis-trans isomerism, occurs due to the restricted rotation around a double bond, leading to different spatial arrangements of substituents. For a compound to exhibit cis-trans isomerism, it must have two different groups attached to each carbon of the double bond. The question specifies compounds without cis-trans isomers, indicating that the double bonds must be positioned such that this type of isomerism is not possible.

Alkenes

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond (C=C) and are characterized by their unsaturation. The presence of double bonds in alkenes allows for various structural configurations, but the specific arrangement of carbon atoms can lead to the absence of geometric isomers. Recognizing the properties and types of alkenes is essential for drawing the required compounds with the formula C5H10.

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