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Unformatted text preview: Compare silane combustion energies with those of the corresponding hydrocarbons (see
previous problem; you need to perform B3LYP/6-31G* calculations on 2-methylpropane
and 2,2-dimethylpropane). On a per gram basis, are alkanes or silanes better fuels?
2,4-Cyclohexadiene vs. Phenol: As a rule, enols (unsaturated alcohols) are less stable
than their keto isomers (aldehydes and ketones). For example, vinyl alcohol is estimated
to be 43 kJ/mol less stable than its “keto” tautomer, acetaldehyde.
29 ∆E ≈ 43 kJ/mol Use the B3LYP/6-31G* model to obtain the equilibrium geometry and energy of 2,4cyclohexadienone and its enol tautomer, phenol.
O OH Which is more stable? If this is an exception to the “rule”, provide a rationale as to why.
What temperature would be needed in order for the higher-energy structure to be present
as 10% of the equilibrium mixture?
Addition vs. Substitution: Alkenes typically undergo addition reactions whereas
aromatic compounds typically undergo substitution reactions. For example, reaction of
bromine and cyclohexane yields trans-1,2-dibromocyclohexane not 1-bromocyclohexene,
whereas bromination of benzene yields bromobenzene not trans-5,6-dibromo-1,3cyclohexadiene.
+ Br2 Br
vs. + HBr Br
+ Br2 Br
vs. + HBr Br What is the reason for the change in preferred reaction in moving from the alkene to the
arene? Use the HF/6-31G* model to obtain equilibrium geometries and energies for
reactants and the products of both addition and substitution reactions of both cyclohexene
and benzene (four reactions in total). Is your result consistent with what is actually
observed? Are all four reactions exothermic? If one or more are not, provide a rationale
as to why.
Bromonium Ions: Addition of Br2 to an alkene, for example, cyclopentene, is usually
represented as a two-step process. In the first step, the electrophile Br+ adds to the double
– bond giving rise to a “bromonium ion” intermediate. In the second step, Br reacts with
the intermediate to give a trans brominated product. Br2 Br–
Br The 13C NMR spectrum of the intermediate cyclopentenebromonium ion has been
recorded and shows resonances at 19, 32 and 115 ppm (relative to tetramethylsilane).
This is consistent with one of two possible structures. Either the bromine is bonded to
both carbons, leaving it with the (formal) positive charge, or it bonded to only a single
30 carbon, leaving the positive charge on the other carbon. In the latter case, the bromine
needs to transfer between the two carbons with a very low energy barrier.
Br+ + H
H H Br +
H H H Br Use the B3LYP/6-31G* model to obtain equilibrium structures, relative energies and 13C
NMR chemical shifts for the two alternative structures. Is the cyclic structure better
represented as a three-membered ring (like cyclopropane or oxacyclopropane) or as a
weak complex between cyclopentene and bromine cation? Examine the carbon-carbon
bond distance to tell. Which structure is lower in energy? Which better fits the observed
C NMR spectrum, the bridged structure or an equal mixture of the two open structures?
Would you expect the NMR spectrum to change were the measurement carried out at
160K? Elaborate. Are the energetic and NMR results consistent with each other?
Hydrogenation of Acetylene: Both steps in the hydrogenation of acetylene to ethane
involve formation of two new CH bonds 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 the hydrogenation reaction. Which step is the more exothermic? Rationalize
H C C H H2 H H
C H2 C H CH3CH3 H Ketene Dimer: There are six possible structures for the dimer of ketene, H2C=C=O.
C O O C CH2 O O C C C O O H2 C CH2
C CH2 O
C H2 C C
H2 C CH2
C C C H2 C O O CH2 Before you do any calculations, predict which of these is likely to be the most stable
based on what you know about the relative stabilities of CC and CO π bonds (lost in the
dimerization) and CC and CC σ bonds (gained in the dimerization). Then, obtain
equilibrium geometries for all six using the HF/6-31G* model. Which structure is
actually preferred. Assuming thermodynamic control, are two or more structures likely to
be seen (assume 5% as the limit of detection of any structure). Is the preferred structure
in line with your prediction? Is the dimerization exothermic or endothermic? Is this
consistent with changes in bond strengths and the strain of the resulting four-membered
Molecular Phosphorous: The most stable arrangement of molecular phosphorus (white
phosphorus) is tetrahedral P4. Dissociation to two molecules of P2 is estimated to be
endothermic by >200 kJ/mol, and can only be detected upon heating to 1000 K. To the
31 contrary, the stable form of molecular nitrogen is N2, and N4 has never been detected let
Use the B3LYP/6-31G* model to obtain equilibrium geometries and energies for
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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.
- Spring '09