Practice Problems

[Chem_Mod]

Here is a practice problem set for your enjoyment. Solutions will be posted soon, along with another problem set.

Problem_Set_01_Conformations.pdf

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[Chem_Mod]

Here are solutions. Another problem set will be posted shortly.

Problem_Set_01_Conformations_Solutions_01.pdf

(24.89 KiB) Downloaded 307 times

[Chem_Mod]

Here is the next problem set. Enjoy!

Problem_Set_02_Conformations.pdf

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[Morgan Zapasnik 1N]

For the second molecule (when drawing Newman projection), why does the one with the lowest energy have hydrogens on the same side? As opposed to putting one hydrogen on the left and one on the right.

[Chem_Mod]

If you are referring to the first problem set, because it places the largest substituents anti to each other.

[Chem_Mod]

Here are the solutions for problem set 02. A new problem set will be posted soon.

Problem_Set_02_Conformations_Solutions.pdf

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[Chem_Mod]

Here is the next problem set. Enjoy!

Problem_Set_03_Conformations.pdf

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[Mizuno_Mikaela_1D]

Here are solutions. Another problem set will be posted shortly.

Problem_Set_01_Conformations_Solutions_01.pdf
Problem_Set_01_Conformations_Solutions_02.jpeg

Have the solutions for drawing the chair conformations of problem set 1 conformations been posted?

I can't figure out how to delete this post but just realized it's in the post I replied to whoops..

[Alex Uy 2D]

Does it matter if the second molecule in the first problem set has the two methyl groups on the same side and the hydrogens on the same side as long as the two highest priority groups are anti to each other?

[Julia Nakamura 2D]

For the last question on Problem Set 1, how do you know whether to draw the 3 substituents in the axial or equatorial positions? So why are 2/3 of them equatorial in one conformation and 2/3 of them are axial in another conformation?

[Chem_Mod]

Does it matter if the second molecule in the first problem set has the two methyl groups on the same side and the hydrogens on the same side as long as the two highest priority groups are anti to each other?

It does matter because the stereochemistry of those carbons is set.

[Chem_Mod]

For the last question on Problem Set 1, how do you know whether to draw the 3 substituents in the axial or equatorial positions? So why are 2/3 of them equatorial in one conformation and 2/3 of them are axial in another conformation?

This information is given by the wedge and dash notation. Wedge means the bond is coming out of the plane of the paper, and dash means the bond is going into the plane of the paper. The important thing is that you can recognize that both methyls are on the opposite side relative to isopropyl.

[Ariana de Souza 4C]

If you are referring to the first problem set, because it places the largest substituents anti to each other.

But can't you have the largest groups still anti each other but then on the left side of the newman projection have CH3 and H and then on teh right side have CH3 and H so that one side doesn't have both methyl groups in close proximity to each other?

And then in that same vein, wouldn't the least stable version of that molecule have both the methyl groups eclipsing each other and not a CH3 and a H eclipsed? Since that seems to be what is happening in that third newman projection.
(sorry this picture is so obnoxiously big)

[Chem_Mod]

If you are referring to the first problem set, because it places the largest substituents anti to each other.

But can't you have the largest groups still anti each other but then on the left side of the newman projection have CH3 and H and then on teh right side have CH3 and H so that one side doesn't have both methyl groups in close proximity to each other?

And then in that same vein, wouldn't the least stable version of that molecule have both the methyl groups eclipsing each other and not a CH3 and a H eclipsed? Since that seems to be what is happening in that third newman projection.
(sorry this picture is so obnoxiously big)

What you are proposing is a different molecule. Think of the molecule in 3D. The arrangement of substituents is set. I tried to draw a picture of a dog to illustrate what I mean. Think of the dogs legs, tail, and head as the substituents. You cannot change the order of them, you can only rotate (the dog's torso). Given those constraints, the given solution is the lowest energy conformer. Does that clear things up at all?

Just a note, the molecule juxtaposed on the dog is not the same one from the problem set. It is just to illustrate the 3D nature of a molecule.

[Chem_Mod]

Here are solution for the last problem set. Good luck on the

Problem_Set_03_Conformations_Solutions.pdf

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final everyone!!!

[Madeline_Foo_3J]

Here are solutions. Another problem set will be posted shortly.

Problem_Set_01_Conformations_Solutions_01.pdf
Problem_Set_01_Conformations_Solutions_02.jpeg

for the third figure I understand why CH(CH3)2 needs to be anti to have the lowest energy, but if the two H atoms are on the same side rather than on opposite sides, would that still be considered correct?

[Chem_Mod]

Here are solutions. Another problem set will be posted shortly.

Problem_Set_01_Conformations_Solutions_01.pdf
Problem_Set_01_Conformations_Solutions_02.jpeg

for the third figure I understand why CH(CH3)2 needs to be anti to have the lowest energy, but if the two H atoms are on the same side rather than on opposite sides, would that still be considered correct?

Not for this example.