Assignment KetoEnol Tautomerism Using Sparton determine the

Assignment: Keto/Enol Tautomerism Using Sparton, determine the energy difference between the enol and keto forms of the compounds below. 1. Based on the energy difference, calculate the equilibrium mixture of enol and keto forms 2. at room temperature using the equation below Nenol This equation was derived from Keq- expl-AG/RT) for AG in a.u. and 298K. 1 a.u(atomic units)- 2625 ki/mol Part I 1. Open Spartan chapter 11-1 2. Record the energies of acetone and propen-2 Energy (a.u)10 0.882 1au 1909093 3. Which tautomer à lower in enersy. acetone or peopen 2-o 4. What is the mole ratio of keto/enol (Nacea/Neso) isomers at room temperature? 5. If an experiment is capable of detecting concentrations as low, as 1% of the total, would you expect to observe both keto and enol forms of acetone at room temperature? Explain. Nyuhee Nonol 10

Solution

5) Nketo/Nenol=exp(-1060)(Eketo-Eenol)=exp(-1060)(-190.8827au-(-190.8693au))=exp(-1060)(-0.0134)=1.475*10^6

equilibrium mixture of keto/enol has Nketo/Nenol=1.475*10^6

Nenol/Nketo=1/(1.475*10^6 )=6.780*10^-7

So % enol=(Nenol/Ntotal)*100

So,if number of keto form=X

then number of enol=(6.780*10^-7 )*X

% enol=(6.780*10^-7 )*X/(X+(6.780*10^-7 )*X)*100=[(6.780*10^-7 )/(1+(6.780*10^-7 )]*100

=[(6.780*10^-7 )/(1 )]*100=(6.780*10^-7 )*100=6.780*10^-5%

So,it is <<1% ,cannot be detected.