Practice Problems
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Practice Problems
Here is a practice problem set for your enjoyment. Solutions will be posted soon, along with another problem set.
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Re: Practice Problems
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.
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Re: Practice Problems
If you are referring to the first problem set, because it places the largest substituents anti to each other.
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Re: Practice Problems
Here are the solutions for problem set 02. A new problem set will be posted soon.
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Re: Practice Problems
Chem_Mod wrote: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..
Last edited by Mizuno_Mikaela_1D on Thu Mar 16, 2017 10:17 pm, edited 1 time in total.
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Re: Practice Problems
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?
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Re: Practice Problems
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?
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Re: Practice Problems
Alex Uy 2D wrote: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.
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Re: Practice Problems
Julia Nakamura 2D wrote: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.
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Re: Practice Problems
Chem_Mod wrote: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)
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Re: Practice Problems
Ariana de Souza 4C wrote:Chem_Mod wrote: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.
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Re: Practice Problems
Here are solution for the last problem set. Good luck on the
final everyone!!!-
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Re: Practice Problems
Chem_Mod wrote: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?
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Re: Practice Problems
Madeline_Foo_3J wrote:Chem_Mod wrote: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.
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