Soluções e Diagramas de Energia

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Chapter 13 Suggested solutions for Chapter 13 97 Suggested solution The equilibrium position is determined both by enthalpy and entropy as discussed on p. 312 of the chapter. Entropy favours the larger number of molecules as they have greater randomness so heat will drive the equilibrium to the right. Enthalpy will favour the eight-membered ring as it has no ring strain and so lower temperatures favour the left-hand side. Problem 12 Draw transition states and intermediates for this reaction and fit each on an energy profile diagram. Be careful to distinguish between transition states and intermediates. 0 0 NaOH 0 0 0 Purpose of the problem Constructing an energy profile diagram for a multistep process. Suggested solution First, a mechanism is essential. This is quite a long job as there are several steps so we must patiently work our way through them. And no short cuts are allowed! 0 0 0 OH H OH OH step 1 step 2 0 step 3 0 0 0 0 0 0 0 Step 3 must be fast as it is just a proton transfer between oxygen atoms. The decomposition of the intermediate is likely to be fast as the leaving group (carboxylate ion, about 5) is a good one. The first step, the bimolecular attack of hydroxide on the anhydride, will be the step. We need only draw the structures of the transition states and we can construct our diagram. (-) OH transition (-) 0 state OH transition state (-) 0 0 intermediate 0 (-) (-) OH 0 OH transition state 0 0 0 starting materials intermediate 0 0 0 0 OH NaOH 0 0 0 products 0 extent of reaction