Háskóli Íslands, raunvísindasvið
Reikniefnafræði - Verkefni 2
Haustmisseri 2013
Kennari - Hannes Jónsson
Guðjón Henning
25. september 2013
A. Dehydrogenation of cyclohexane to form benzene
Q1: Write balanced chemical equations for each one of the three dehydrogenation steps in
your report. What other products are involved?
Step by step:
C6 H12 −→ C6 H10 + H2 (i)
C6 H10 −→ C6 H8 + H2 (ii)
C6 H8 −→ C6 H6 + H2 (iii)
Overall:
C6 H12 −→ C6 H6 + 3 H2
(iv)
Q2: Write the calculated ground state energy of each molecule in your report.
The calculated ground state energies are:
C6 H12
C6 H10
C6 H8
C6 H6
:E
:E
:E
:E
= −232.91690
= −231.72815
= −230.54323
= −229.41945
Hartree = −6337.99470 eV
Hartree = −6305.64715 eV
Hartree = −6273.40382 ev
Hartree = −6242.82419 eV
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Q3: Calculate the reaction energy for each of the three steps in the dehydrogenation of
cyclohexane. Draw an energy diagram showing the three steps of the reaction. Note that it
is important to add and subtract only numbers obtained at the same level of approximation.
Begin by calculating the energy of the H2 molecule in the same level of approximation (3-21G
basis)
H2 : E = −1.12296 Hartree = −30.55731 eV
The reaction energy is calculated with the following equation
Ereaction = Eproducts − Ereactants
So we calculate for each reaction, labeled i − iv in Q1
i) C6 H12 −→ C6 H10 + H2 : {−231.72815 + (−1.12296)} − {−232.91690}
= 0.0648
ii) C6 H10 −→ C6 H8 + H2 : {−230.54323 + (−1.12296)} − {−231.72815}
= 0.0630
iii) C6 H8 −→ C6 H6 + H2 : {−229.41945 + (−1.12296)} − {−230.54323}
= 0.00083
iv) C6 H12 −→ C6 H6 + 3 H2 : {−229.41945 + 3(−1.12296)} − {−232.91690}
= 0.1286
We see that the reaction energies in processes i − iii are equal to the reaction energy in
process iv , which is what one would assume.
Energy diagram showing the three steps of the reaction in the 3-21G basis.
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Q4: Make a table of the reaction energy in the three steps of the dehydrogenation and compare
the results obtained from the two basis sets.
Now the same calculations are performed, but in the 6-31G basis set. The following table
shows the comparison between basis. The energy of the H2 molecule in this basis is
H2 : E = −1.12683 Hartree = −30.66262 eV
Table 1: Comparison of dehydrogenation of cyclohexane to form benzene in the 3-21G and
6-31G basis sets
Process
Energy (3-21G) [Hartree] Energy (6-31G) [Hartree]
C6 H12 −→ C6 H10 + H2
0.0648
0.0573
C6 H10 −→ C6 H8 + H2
0.0630
0.0560
C6 H8 −→ C6 H6 + H2
0.00083
-0.0073
C6 H12 −→ C6 H6 + 3 H2
0.1286
0.1060
Q5: Draw an energy diagram based on the numbers obtained using the second basis set.
Energy diagram showing the three steps of the reaction in the 6-31G basis.
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Q6: Are the reaction steps exothermic or endothermic? What is the energy change during hydrogenation of benzene? How well do the calculated results agree with experimental
observations?
All the reaction steps are endothermic, except for the last step in the 6-31G basis, which is
exothermic.
The energy change during the hydrogenation of benzene
C6 H6 + 3 H2 −→ C6 H12
is
E = C6 H12 − (C6 H6 + 3 H2 ) = −0.1060 Hartree (in 6-31G)
= −0.1286 Hartree (in 3-21G)
Experimental value of the energy of the hydrogenation has been measured to be −0.788
Hartree. This is comparable with the values calculated, especially in the 6-31G basis.
(However, this experimental value was obtained by "questionable methods", a quick google
search).
B. Reaction of nitronium cation with benzene
Q7: What is the ground state energy of the nitronium cation? Is the molecule bent or linear?
Are the N-O bond lengths identical or dierent? What common molecule, often in the news
these days, is isoelectronic with the nitronium cation?
The ground state energy of the nitronium cation is
E = −202.46006 Hartree = −5509.22147 eV
The molecule is linear, its geometry is
−−− N−
−−− O
O−
The calculated bond lengths are measured to be
O[3] − N [1] : 1.11144 Å
O[3] − O[2] : 2.22287 Å
N [1] − O[2] : 1.11143 Å
So one can see that the two N−O bond lengths are identical, which should tell us that there are two double bonds. That is however, not the geometry obtained from the calculations.
The CO2 molecule is isoelectronic with the nitronium cation.
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Q8: In this representation, the red areas are more negatively charged than blue areas. Where
in the molecule is the positive charge mostly located? Based on your calculation, draw the
Lewis dot structure that best represents the nitronium cation.
MEP gure, representing the charge distribution in the nitronium cation. Red represents
negative chare, blue positive charge
From the gure one can see that the positive charge is mostly located around the N atom in
the molecule.
Lewis structure representing the nitronium cation
Q9: Judging from the electrostatics, how do you expect the nitronium ion to react with the
benzene molecule (which part of the nitronium ion will most likely collide with what part of
the benzene molecule?). Draw a Lewis structure of the nitronium/benzene adduct.
Lets rst show the MEP gure for the benzene molecule
MEP gure, representing the charge distribution in the benzene molecule. Red represents
negative chare, blue positive charge
From the two MEP gures, it can be seen that the positively charged area (blue) in the
nitronium molecule bonds with the negatively charged area (red) in the benzene molecule.
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Lewis structure representing the nitronium/benzene adduct
Q10: Place a picture of the optimized structure in your report. How do you describe the
bonding (get the C-C bond lengths in the ring)?
Optimized structure of the adduct
Lewis structure representing the optimized structure. The carbon atoms in the benzene ring
are numerated from 1 to 6 for C−C bond length calculations.
The calculated bond lengths are
r12 = 1.3477 Å
r23 = 1.4123 Å
r34 = 1.4123 Å
r45 = 1.3477 Å
r56 = 1.4826 Å
r61 = 1.4826 Å
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Q11: What is the binding energy of the adduct (energy of the adduct minus the energy of
reactants, i.e. isolated nitronium ion and benzene molecule).
The binding energy is
EBinding = ENitroBenzene − (EBenzene + ENitro
= −431.96871 − (−229.41945 + (−202.46006))
= −0.0892 Hartree
= −2.47257 eV
Q12: Show a picture of the electrostatic potential map of the adduct in your report.
Electrostatic potential map of the adduct
Q13: Which proton would you expect to be easiest to remove? Why?
The proton where the electron density is the lowest (highest + charge) is expected to be the
easiest to be removed. That proton is where the H atom is located in the gure below.
The proton indicated with the red arrow is expected to be the easiest to remove
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Q14: Do you have to run a calculation on the single proton to determine its energy? Is the
total energy of your product and the proton higher or lower than that of the charged adduct?
First, lets look at the energy for the adduct
E1 ≡ Eadduct = −431.96871 Hartree
Then we look at the energy of the adduct(NitroBenzene), where the Hydrogen discussed in
Q13 has been removed, plus the energy of a single Hydrogen
E2 ≡ ENitroBenzene + EH = −431.71216 + (−0.49620)
= −432.20836 Hartree
From this it can be seen that E1 =
6 E2 and that E1 > E2 . Therefore, E2 is more stable
(less energy). This implies that it is necessary to run a calculation on the single proton to
determine its energy.
Q15: Can you name another (quite well known) aromatic compound containing nitro groups
that has similar properties regarding thermodynamical stability?
TNT (Trinitrotoluene).
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