P3_Reaction Energies

Is the energy that you calculate for the transfer of

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Unformatted text preview: the proton affinity trimethylamine is 84 kJ/mol greater than ammonia, whereas in water, the two are of nearly equal. Use the B3LYP/6-31G* model to obtain equilibrium geometries for the ammonia and trimethylamine and their respective protonated forms (ammonium and trimethylammonium ions). Is the energy that you calculate for the transfer of a proton between the two in accord with their relative gasphase proton affinities? Me3N + NH4+ Me3NH+ + NH3 For practical reasons, you can’t perform the equivalent B3LYP calculations in water. You can, however, model the energy of the reaction in water by explicitly including a specific number of water molecules in your calculation. The obvious thing to do is to attach the minimum number of water molecules to account for all possible amine-water hydrogen bonds. This number is four for ammonia and ammonium ion, but only one for trimethylamine and trimethylammonium ion. Obtain equilibrium geometries for the four amine/ammonium ion complexes involved in the reaction below and calculate the relative “aqueous-phase” proton affinities of trimethylamine and ammonia. Me3N…H2O + NH4+ …(H2O)4 Me3NH+…H2O + NH3…(H2O)4 Do the results show the observed trend in proton affinities in moving from the gas phase into water? Elaborate. Redo your calculations with trimethylamine and trimethylammonium ion “attached to” four water molecules instead of only one, and calculate the energy of the reaction below. Me3N…(H2O)4 + NH4+ …(H2O)4 Me3NH+…(H2O)4 + NH3…(H2O)4 Does this appear to be a better model? Elaborate. Try to identify a serious flaw in both models. Boiling Points of Ethanol and Ethylamine: While ethanol and ethylamine have similar molecular weights and molecular structures and while both can participate in three hydrogen bonds, the two molecules have very different boiling points. Ethanol is a liquid at STP whereas ethylamine is a gas. Is this because ethanol forms stronger intermolecular hydrogen bonds than ethylamine? To tell, obtain equilibrium geometries for ethanol and ethylamine and their respective hydrogen-bonded dimers, and compare hydrogen-bond energies. Use the B3LYP/6-31G* model. ethanol + ethanol ethanol … ethanol ethylamine + ethylamine ethylamine … ethylamine Which dimer is more tightly bound? Is your result consistent with the ordering of boiling points? Elaborate. 40 Exploring Chemical Concepts with Quantum Chemical Models The availability of reliable and practical quantum chemical models offers the opportunity to critically examine a number of the qualitative models that chemists have advanced over many decades. Among the most important are conjugation, aromaticity and ring strain, the first leading to molecules that are stablized and the second to molecules that are destabilized. Conjugation Molecules incorporating adjacent double bonds that are coplanar or nearly coplanar are known to be preferred over non-coplanar arrangements or arrangements in which the double bonds are separated by one or more sp3 “spacers”. Thus, the double bonds in both trans-1,3-pentadiene and in trans2-butenal are coplanar and each is more stable than its non-conjugated isomer, 1,4-pentadiene and 3-butenal, respectively. Hydrogenation of 1,3-Butadiene: Both steps in the hydrogenation of trans-1,3butadiene to 1-butene and then to n-butane involve formation of two new CH bonds resulting from the loss of an HH bond and a π bond. Use the HF/6-31G* model to obtain equilibrium geometries and energies for all molecules involved in both steps of the hydrogenation. Which step is the more exothermic? Does this support the idea that conjugated double bonds are stronger than isolated double bonds? Elaborate. H H C C H H C C H2 H C H H CH2 CH3 H2 C H CH3 CH2 CH 2 CH3 H Hydrogenation Energy of 1,3-butadiene vs. But-1-yne-3-ene: Compare the energy of the first step in the hydrogenation of 1,3-butadiene (see previous problem) with that of hydrogenation of the double bond in but-1-yne-3-ene. Use the HF/6-31G* model to obtain equilibrium geometries for all molecules in the two reactions. H C H C C vs. H C H2 CH3C CH H Which reaction is more exothermic? What does this say about the conjugation energies of two double bonds vs. a double bond and a triple bond? 41 Hydrogenation of 1,2-butadiene: Both steps in the hydrogenation of 1,2-butadiene first to 1-butene and then to n-butane involve formation of two new CH bonds resulting from the loss of an HH bond and a π bond. Use the Hartree-Fock 6-31G* model to obtain equilibrium geometries and energies for all molecules involved in both steps of the hydrogenation. Does the double bond in 1,2-butadiene appear to be stronger or weaker than that in 1-butene? Is the ordering of bond strengths consistent with the ordering of double bond lengths in the two molecules? Based on hydrogenation energies, would you conclude that the double bond in 1,2-butadiene is stronger or weaker than that in 1,3butadiene (see first problem in this section)? Would you characterize 1,2-butadiene a...
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