First Hourly Exam
Advanced Organic Chemistry I
1) Use the list of molecular orbital coefficients and geometry for the ethylene molecule to answer the following questions:
Standard orientation:
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Center Atomic Coordinates (Angstroms)
Number Number X Y Z
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1 6 0.662958 0.000000 -0.000003
2 6 -0.662958 0.000000 0.000001
3 1 1.256544 0.924038 0.000012
4 1 1.256573 -0.924015 0.000000
5 1 -1.256544 -0.924038 0.000005
6 1 -1.256573 0.924015 -0.000006
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Molecular Orbital Coefficients
1 2 3 4 5
O O O O O
EIGENVALUES -- -11.02156 -11.02070 -0.97677 -0.74101 -0.59369
1 1 C 1S 0.70185 0.70146 -0.17854 -0.13550 0.00000
2 2S 0.01979 0.03088 0.47290 0.41374 0.00000
3 2PX 0.00212 -0.00435 -0.11653 0.20150 0.00000
4 2PY 0.00000 0.00000 0.00000 0.00000 0.39661
5 2PZ 0.00000 0.00000 0.00000 0.00000 0.00000
6 2 C 1S 0.70185 -0.70146 -0.17854 0.13550 0.00000
7 2S 0.01979 -0.03088 0.47290 -0.41374 0.00000
8 2PX -0.00212 -0.00435 0.11653 0.20150 0.00000
9 2PY 0.00000 0.00000 0.00000 0.00000 0.39661
10 2PZ 0.00000 0.00000 0.00000 0.00000 0.00000
11 3 H 1S -0.00465 -0.00479 0.11238 0.22309 0.26040
12 4 H 1S -0.00465 -0.00479 0.11238 0.22310 -0.26040
13 5 H 1S -0.00465 0.00479 0.11238 -0.22309 -0.26040
14 6 H 1S -0.00465 0.00479 0.11238 -0.22310 0.26040
6 7 8 9 10
O O O V V
EIGENVALUES -- -0.53507 -0.45286 -0.32741 0.32258 0.60419
1 1 C 1S 0.01486 0.00000 0.00000 0.00000 0.00001
2 2S -0.01991 0.00000 -0.00001 -0.00001 -0.00005
3 2PX 0.49724 0.00002 0.00000 0.00000 0.00001
4 2PY -0.00001 0.39099 -0.00001 -0.00001 0.69004
5 2PZ 0.00000 0.00001 0.63416 0.81286 0.00001
6 2 C 1S 0.01486 0.00000 0.00000 0.00000 -0.00001
7 2S -0.01991 0.00000 0.00000 0.00000 0.00005
8 2PX -0.49724 -0.00002 0.00000 0.00000 0.00001
9 2PY 0.00001 -0.39099 0.00001 -0.00001 0.69004
10 2PZ 0.00000 0.00001 0.63416 -0.81286 -0.00001
11 3 H 1S 0.21810 0.35128 0.00000 0.00001 -0.61535
12 4 H 1S 0.21813 -0.35126 0.00000 0.00000 0.61538
13 5 H 1S 0.21810 0.35128 0.00000 0.00000 0.61535
14 6 H 1S 0.21813 -0.35126 0.00000 0.00001 -0.61538
11 12 13 14
V V V V
EIGENVALUES -- 0.68046 0.68745 0.91999 1.00241
1 1 C 1S -0.17083 -0.12286 0.00000 0.10937
2 2S 1.05905 0.80858 -0.00001 -0.85435
3 2PX 0.19025 0.52040 -0.00002 1.16598
4 2PY 0.00002 0.00001 0.94001 -0.00001
5 2PZ 0.00001 0.00001 0.00000 0.00000
6 2 C 1S 0.17083 -0.12286 0.00000 -0.10937
7 2S -1.05905 0.80858 -0.00001 0.85435
8 2PX 0.19025 -0.52040 0.00002 1.16598
9 2PY 0.00002 -0.00001 -0.94001 -0.00001
10 2PZ 0.00000 -0.00001 0.00000 0.00000
11 3 H 1S -0.63148 -0.62264 -0.59884 -0.14010
12 4 H 1S -0.63145 -0.62263 0.59885 -0.14013
13 5 H 1S 0.63148 -0.62264 -0.59884 0.14010
14 6 H 1S 0.63145 -0.62263 0.59885 0.14013
a) What is the number of occupied MO's? 8
b) Which is the HOMO?, LUMO? 8,9
c)What symmetry are the HOMO and LUMO? pi,pi
d) Ethylene contains Pi(CC) and Pi(CH2) MO's. Which are the Pi(CH2) MO's? 5,7,10,13
e) In this calculation tell whether a minimal, double zeta, or ploarized
basis set was used and explain in one simple sentence. Minimal. There is only one type of 1s,2s,2px,2py,2pz and no 3d or higher
orbitals in the basis set.
f)Construct a qualitative MO diagram for the four Pi(CH2) MO's which result
from the combination of the left- and right-side Pi(CH2) and Pi(CH2)* fragments. Using signs of the coefficients of the AO's show how
this resembles the result for the Pi(CC) system in butadiene.
2) State the four Grand Approximations used to solve the SCHroedinger Equation
in Chemistry.
i)Born-Oppenheimer Approximation (where there nuclear and electronic motions
are decoupled).
ii)the Orbital Approximation (where the many particle electronic wave function
is approximated by a product of one-electron wave functions -MO's)
iii) the Linear Combination of Atomic Orbital - Molecular Orbital Approximation
(where each MO is approximationed by a weighted sum of AO's)
iv) the gaussian aproximation (where the hydrogen-like or Slater Type Orbital
functions which are based on exp(-r) functions are replaced by weighted
sums of gaussian functions, which are based on exp(-r**2) functions).
3)Circle the atom [here I've put it in bold face] in the following molecules
that is most favored for protonation and explain why (with one simple sentence).
O=CH-CH2-NH2 because amines are much more basic than carbonyl oxygens. ]This by the way
is not an amide.]
<-N=CH-N(CH3)-CH=CH->ring This N's lone pair is perpendicular to the aromatic pi system and protonation
won't disturb aromaticity while the other N l.p. is part of that pi system
and its protonation will destroy aromaticity;
<-CH=N-CH=C(NH2)-CH=CH->ring This N's lone pair is perpendicular to the aromatic pi system and protonation won't disturb aromaticity while the other N l.p. is in resonance with that pi system and protonation will break this resonance.
4)Assign the order of the surrounding groups and the R or S designation
for the following molecules.
5) Draw a conformational energy diagram for the rotation about the CC bond
in FH2CCH2OH. Clearly designate each major point with Newman projections.
6)Which has the lowest UV absorption and why?
H2C=CH2 ......... H2C=CH-NO2 ....... H2C=CHOCH3
Conjugation decreases the HOMO-LUMO gap, whether it be by electron withdrawing
or donating groups.
7)Which is the stronger acid and why?
cyclopentadiene or cycloheptadiene
Deprotonation of cycloheptadiene gives the cycloheptadienyl cation which is a 6-pi-electron aromatic system while deprotonation
of cyclopentadiene gives the cyclopentadienyl cation which is a 4-pi-electron antiaromatic system.