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Topic: using A-values & position (axial strain enrg) equilibria/percentage of conformer  (Read 3770 times)

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fcukstrated

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I'm having difficulty with this problem because there are two substituents, and in the in-class example there was only one substituent.  

Here is the example used in class:
"Circle the structure of the conformer that will be present in greater amount."


t-butyl=4.9 kcal/mol.  Since equatorial conformation is the product in this equilibrium, we use -4.9 kcal/mol.


equilibrium constant expression
K= [equitorial]/[axial]

at room temp deltaG=-1.36 log K  The A-value is used for the deltaG term.

-4.9/-1.36=log K
K=10^3.60=4008   , K=[equitorial]/[axial]

Let X=the percentage of axial conformer
K=(100-X)/(X)   4008X=100-X
X=100/(K+1)=100/4009=.02%

percentage of equatorial conformer=100-X=99.98%
ANSWER=EQUATORIAL CONFORMER


here is the problem I'm having trouble with..

"Circle the structure of the conformer that will be present in greater amount."
A-values (in kcal/mol)
Ph=2.90
CO2H=1.40


Using product as the RIGHT side
-1.40/-1.36=log K
K=10^1.0294=10.70   , K=[equitorial]/[axial]

Let X=the percentage of axial conformer
K=(100-X)/(X)   11.70=100-X
X=100/(K+1)=100/11.70=8.54%

percentage of equatorial conformer=100-8.54=94.45%
ANSWER=AXIAL CONFORMER
however, if product is on the LEFT side
-2.90/-1.36=log K
K=10^2.132=135.629, K=[equitorial]/[axial]

Let X=the percentage of axial conformer
K=(100-X)/(X)   135.629=100-X
X=100/(K+1)=100/136.629=.7319%

percentage of equatorial conformer=100-.7319=99.26
ANSWER=EQUITORIAL CONFORMER

as you can see, I am running in circles because there can't be two answers.  help is desperately needed..!

« Last Edit: October 14, 2005, 09:11:35 PM by fcukstrated »

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