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Topic: 2-methylpropane newman projection, torsional strain  (Read 17602 times)

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lissydoll206

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2-methylpropane newman projection, torsional strain
« on: July 07, 2005, 02:27:54 AM »
Hi.

  I am working on a problem, and first, I am supposed to draw all of the Newman projection diagrams for the staggered conformational isomers of 2-methylpropane.  There is only one way that it can be drawn though, right? Because the methyl group can only go on position 2, or it would become a butane.  I will attach my diagrams that I drew.  I also needed to draw the eclipsed conformation of 2-methylpropane, and I have attached that as well.

  Finally, I am given the informationthat the torsional strain resulting from the eclipsed C-H bonds is 4.2 kJ/mol, and that for eclipsed C-H and C-CH3 bonds are 6.3 kJ/mol.  Given this information, how do you calculate the torsional strain energy for the eclipsed conformation?

  Do you add the energies together?  For example, since there are 2 eclipsed C-H and C-CH3 bonds and only one eclipsed C-H bonds,  would you add 4.2 + 6.3 + 6.3?  This would give a total torsional strain energy of 16.8 kJ/mol.

  I am searching through my textbook, but it doesn't seem to give any insight as to how to calculate the total energy.  I would appreciate if someone could tell me if I have done this problem right, and if I haven't, to give me some insight to get me going in the right direction.

  Thank you in advance! I greatly appreciate any help I can get  :)

-Melissa

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Re:2-methylpropane newman projection, torsional strain
« Reply #1 on: July 07, 2005, 12:02:42 PM »
Well, technically there are three "different" Newman projections for staggered 2-methylpropane, they just all have the same energy, right?  To be complete, you should probably draw all three.  The same goes for the eclipsed forms.

As for calculating the energy, you do just add them up.  I can't verify that your numbers are correct because I learned them in kcals not kJs.  Looks to be in the right range though.

Another common exercise, which may or may not be useful to you, is to draw an energy diagram as a function of the angle of rotation.  Start with any one conformation and then rotate around and look at how the energy changes depending on this torsional angle (from 0 to 360 degrees).  It should look like a sine wave when your done.
« Last Edit: July 07, 2005, 12:05:26 PM by movies »

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