Use the hf6 31g model to obtain equilibrium

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Unformatted text preview: s a conjugated molecule? Elaborate. CH3 H C H C C CH2CH3 H H2 C H H C H Conjugated vs. Non-Conjugated Cyclic Dienes: 1,4-cyclohexadiene is known to be slightly (4 kJ/mol) more stable than its isomer, 1,3-cyclohexadiene. This is a surprising result as the former is a non-conjugated diene whereas the latter is a conjugated diene. Use the HF/6-31G* model to obtain equilibrium geometries for the conjugated dienes, 1,3-cyclohexadiene, 1,3-cycloheptadiene and 1,3-cyclooctadiene and their nonconjugated isomers, 1,3-cyclohexadiene, 1,3-cycloheptadiene and 1,3-cyclooctadiene. Assign the preferred isomer and calculate the energy difference for each pair. Do the calculations reproduce the experimental result for cyclohexadiene? Do they show like preferences for cycloheptadiene and cyclooctadiene? Aromaticity Whereas coplanar arrangements of two or more connected π bonds show modest energy stabilization, much larger effects are noted for some cyclic arrangements of π bonds. Specifically, molecules with 4n+2 π electrons in a planar or nearly planar arrangement are known to be unusually stable or aromatic, and are likely to exhibit chemistry that is quite distinct from that of unsaturated molecules (even conjugated unsaturated molecules). The archetypical example is benzene with three carbon-carbon double bonds (involving six π electrons) confined in a planar six-membered ring. One way to “measure” the aromatic stabilization of benzene is to compare the energy of adding H2 (leading to 1,3-cyclohexadiene) with the energy of adding hydrogen to 1,3-cyclohexadiene (leading to cyclohexene), or the energy of adding hydrogen to cyclohexene (leading to cyclohexane). H2 H2 H2 2 4 k J /m o l -110 kJ/mol - 1 2 0 k J /m o l 42 The three steps are similar insofar as each trades a CC π bond and an HH bond for two CH bonds. Normally, one would tend to think of such a trade as energetically favorable as σ bonds are stronger than π bonds. However, the first hydrogenation step is actually slightly endothermic, while the remaining two steps are both strongly exothermic. The difference between the first and second (or third) steps (~140 kJ/mol) reflects the additional (aromatic) stabilization resulting from arrangement of three π bonds in a planar six-membered ring. The individual hydrogenation reactions do not conserve the numbers of each kind of chemical bond (each reaction does maintain overall bond count), and would not be expected to benefit fully from cancellation of errors. However, the difference between the first and second reactions (or between the first and third reactions) which actually relates to aromatic stabilization can be expressed as a reaction that does maintain individual bond counts. Previous experience with reactions of this type suggests that its energy should be reasonably well described by quantum chemical methods that do not take electron correlation into account. + 2 1,3,5-Cyclohexatriene: The six carbon-carbon bonds in benzene are all the same, intermediate in length between “normal” single and double bonds. The hypothetical molecule 1,3,5-cyclohexatriene is identical to benzene except that the CC bonds alternate between single and double bonds. 1,3,5-cyclohexatriene is not an energy minimum and, if given the chance, will collapse to benzene. Is the energy difference between benzene and 1,3,5-cyclohexatriene roughly the same as the aromatic stabilization of benzene (~ 140 kJ/mol; see discussion above), or is it significantly smaller? To decide, use the HF/631G* model to obtain an equilibrium geometry for benzene and an energy for 1,3,5cyclohexatriene, assuming a fixed geometry with alternating single and double CC bonds set to 1.54Å and 1.32Å. Aromaticity of Thiophene: Thiophene is a likely candidate for an aromatic molecule with two double bonds and an out-of plane lone pair on sulfur making a total of six π electrons. S Compare reaction energies for the individual steps in the hydrogenation of thiophene to tetrahydrothiophene. As was the case for benzene, addition of the first equivalent of hydrogen breaks one of the double bonds and leading to loss of aromaticity. Addition of the second equivalent only breaks a double bond. Therefore, the difference in energy between the two provides a measure of the aromatic stabilization. 43 X X = S thiophene X = NH pyrrole X = O furan X H2 X H2 X = S tetrahydrothiophene X = NH tetrahydropyrrole X = O tetrahydrofuran Use the HF/6-31G* model to obtain geometries and energies for thiophene, dihydrothiophene and tetrahydrothiophene as well as hydrogen, and calculate hydrogenation energies for the two steps. Is the difference in hydrogenation energies comparable to that between the corresponding first and second hydrogenation energies for benzene (see discussion above) or is it significantly smaller or greater? Comment on the aromaticity of thiophene. Perform analogous calculations and analyses on pyrrole (X=NH) and furan (X=O). Do either or both of these molecules show aromaticity? Ela...
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This note was uploaded on 02/22/2010 for the course CHEM N/A taught by Professor Head-gordon during the Spring '09 term at University of California, Berkeley.

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