Draw the Z isomer of an alkene that has a CH 3 and an H on one sp 2 carbon and isopropyl and butyl groups on the other sp 2 carbon.

Alright. Hi, everyone. So this question is asking us to write the Z I some Roman all keen if one of the SP two carbons has a CEO on ethyl groups, while the other SP two carbon has a hydrogen and ethyl groups. So recall really quickly that the Z of an all keen is going to have the higher priority substitue ints on the same side of your all king. Mhm. So let's keep that in mind when we determine the priority on each SP two carbon, right? So recall that the priority is determined by the atomic number of the atom directly attached to the SP two carbon of your all king. So the higher the atomic number, the higher the priority. So now let's apply that to each of our substitue ints on both SP two carbons, right? According to the question, the first of these has a chlorine and an ethyl group attached. So that means that this SP two carbon is directly attached to a chlorine and a carbon. Mhm. Now Chlorine has an atomic number of 17 Where as carbon has an atomic number of six. So because chlorine the atom directly attached to our SP two carbon. It has a higher atomic number than the carbon of our ethyl group. Chlorine has a higher priority. So that's one sp two carbon that we got figured out. So let's go ahead and apply that to the next one and the other SP two carbon has a hydrogen and an ethyl group attached. So we have our SP two carbon atoms being attached directly to both hydrogen and a carbon atom. Now, carbon, let me mention before it has an atomic number of six, But hydrogen has an atomic number of one. Mhm So for this other sp two carbon atoms are ethyl group is actually going to be higher priority because it has a higher atomic number than hydrogen. So now for our Z I summer are chlorine and our group have to be on the same side. So first, let's go ahead and draw all keen, right R C double bond C and let's go ahead and add our substitue ints. Now, for the sake of this video, I'm going to go ahead and put our higher priorities on the top. But what matters really is that it's on the same side of your all king. So for this sp two carbon on the left, I'm gonna go ahead and put our chlorine and our group. So here's one carbon and now the other carbon, the other SP two hybridized carbon, it has the ethyl group as the higher priority. So I'm gonna make sure to put that on the same side as my chlorine. So here's our fo group and here is gonna be your hydrogen on the bottom. So now this is going to be our final answer, which is going to correspond to option C in the multiple choice. So just to recap, right, yours, ei swimmer has your higher priority substitue ints on the same side of your all keen and you determine your priority by looking at the atomic number of the atom directly attached to the sp two carbons of your all. Keep. So if you have a higher atomic number, you have a higher priority. So with that being said, thank you very much for watching and I hope you found this helpful.

Table of contents

Skip topic navigation

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

Cis vs Trans

Organic Chemistry

4. Alkanes and Cycloalkanes

Cis vs Trans

Previous problemNext problem

4: minutes

Problem 12

Textbook Question

Draw the Z isomer of an alkene that has a CH₃ and an H on one sp² carbon and isopropyl and butyl groups on the other sp² carbon.

Verified step by step guidance

Identify the structure of the alkene: The problem specifies that the molecule is an alkene, meaning it contains a carbon-carbon double bond (C=C). Each sp² carbon in the double bond is bonded to two substituents.

Assign the substituents to each sp² carbon: One sp² carbon is bonded to a CH₃ (methyl) group and an H (hydrogen) atom. The other sp² carbon is bonded to an isopropyl group (CH(CH₃)₂) and a butyl group (CH₂CH₂CH₂CH₃).

Understand the Z isomer: The Z (zusammen) isomer of an alkene has the higher-priority substituents on the same side of the double bond. Use the Cahn-Ingold-Prelog priority rules to assign priorities to the substituents on each sp² carbon. Higher atomic number atoms or groups with greater mass take priority.

Assign priorities to the substituents: On the first sp² carbon, CH₃ has a higher priority than H. On the second sp² carbon, the butyl group has a higher priority than the isopropyl group because the butyl group has a longer carbon chain.

Draw the Z isomer: Place the higher-priority groups (CH₃ and butyl) on the same side of the double bond, while the lower-priority groups (H and isopropyl) are on the opposite side. Ensure the double bond is clearly shown, and the spatial arrangement reflects the Z configuration.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

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

Z and E Isomerism

Z and E isomerism refers to the geometric isomerism in alkenes based on the arrangement of substituents around the double bond. In Z isomers, the highest priority substituents on each carbon of the double bond are on the same side, while in E isomers, they are on opposite sides. This distinction is crucial for understanding the spatial orientation of the groups attached to the double bond.

Priority Rules (Cahn-Ingold-Prelog)

The Cahn-Ingold-Prelog priority rules are used to determine the priority of substituents attached to the double-bonded carbons in alkenes. The priority is assigned based on atomic number; the higher the atomic number, the higher the priority. If two substituents have the same atom, the next atoms in the chain are considered until a difference is found, which is essential for correctly identifying Z and E configurations.

Alkene Structure and Hybridization

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond (C=C), characterized by sp2 hybridization of the carbon atoms involved in the double bond. This hybridization leads to a trigonal planar geometry around the sp2 carbons, influencing the physical and chemical properties of the alkene, including its isomerism. Understanding the structure helps in visualizing the spatial arrangement of substituents.

Watch next

Master How to name different types of double bonds or rings with a bite sized video explanation from Johnny

Start learning