Concept explainers
Interpretation:Whether in the model of 1,2-dimethylcyclopentane, the molecule in the left box is same as the molecule in the right box or not needs to be determined.
Concept Introduction:
The cis and trans isomerism concept will be applied here. Any molecule will be called cis if it follows the conditions for itthat is the same groups attached either above the plane or below the plane.For trans isomer,same groups are attached where one is above the plane and other group is below the plane or vice versa.
Answer to Problem 1CTQ
The given both structures of isomers of 1,2 dimethylcyclopentane are similar to each other. This is because when the bonds of the groups interchanges with each other, the bond’s spatial character changes. As a result of which these structures can be converted to each other without breaking of bonds.
Explanation of Solution
It is given that the left box and right box contain two molecules of 1,2-dimethylcyclopentane.Thus, it is very easy to conclude that the 2 different molecules are isomers of 1,2-dimethylcyclopentane. This is because both the molecules have same number of atoms in it.
Both the molecules are similar toeach other. Thiscan be explained as, if the interchanging of bonds takes place, similar structure can be obtained. In the interchanging of structure in left box and right box, there is no breaking of bonds takes place. For example, if the methyl group attached to the bond below the planeis interchanged with other group attached to the same atom, then the position of methyl group will be converted to above the plane from below the plane. In this process, no breaking of bond takes place then similar structure will be obtained.
Thus, the structure on the left box is similar to that on the right box.
Want to see more full solutions like this?
Chapter 7 Solutions
Organic Chemistry: A Guided Inquiry
- a) Draw one isomer of C6H14. b) Draw one isomer of C6H12- c) Draw one isomer of C6H140 that exhibits hydrogen bonding. d) Draw one isomer of C6H140 that is not capable of hydrogen bonding. BONUS: Show all locations of possible hydrogen bonding for the C6H140 isomer that you drew above in part c.arrow_forwardCheck the box under each structure in the table that is an enantiomer of the molecule shown below. If none of them are, check the none of the above box under the table. OH & .... Molecule 1 HO Molecule 4 OH -- none of the above Molecule 2 HO In.. Molecule 5 HO X Molecule 3 sil OH Molecule 6 ..... OH X Ś 000 Ar 18arrow_forwardWithout counting hydrogens, determine which one of the following CANNOT be the unknownmolecule with molecular formula C7H8NOBr , and explain your reasoning.arrow_forward
- a model of each molecule shown above: Is the molecule in the left box the same moleculeas the molecule in the right box? Use your models to answer the question, and recall that...arrow_forwardGive detailed Solution with explanation needed...please explainarrow_forwardChoose the correct answer. Choose.. Choose.. A saturated hydrocarbon with molecular formula C6H14 An alkyne with molecular formula C5H10 A ketone with molecular formula C4H80 An alkene with molecular formula C5H10 An aldehyde with molecular formula CH4O An alkene with molecular formula C5H8 A ketone with molecular formula C3H8O An aldehyde with molecular formula C2H4O Choose..arrow_forward
- Between these 2 compounds, which one will be (R) and which one is (S) configuration? Please draw to explain your answer in detailed..arrow_forwardGive detailed answer with explanation needed..don't give Handwritten answerarrow_forward2) Draw a molecule that has one R stereocenter. Do not duplicate your example from question 1.arrow_forward
- I don't understand why the first one and second one are cis and trans respectively. Wouldn't the first one be trans-1,2-dimethylcyclobutane because the torsional strain wouldn't allow the carbons to be in the same conformation. Meaning one of the carbons would be up and then the next would be down and so on. Since both methyl groups are equatorial and the first and second carbons are arranged up and down, wouldn't it be trans. Same logic for the second molecule. Carbon 1, which is attached to the methyl is down, carbon 3 which is attached to the methyl should be down also because of torsional strain, and since both methyl are in axial (or equatorial?), it would be cis. Or is it based off of the way the carbons are positioned in the picture?arrow_forwardSolve correctly please. Should correctarrow_forwardPlease read question carefully. Asking the question for a second time. thank youarrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning