14. Small bicyclic
molecules normally have the bridgehead hydrogens (or other substituents)
pointing out from the center of the molecule, but for sufficiently
large rings the in-out and even the in-in structures are
possible. Calculate the energies of the out-out, in-out, and in-in isomers
of the bicyclo[5.5.5] and bicyclo[4.4.4] alkanes. What are the relative
stabilities of the various isomers?
[For a recent discussion, see Alder, R. Tetrahedron 1990, 46, 683.]
Problems 60 and 61 are designed to test the extent of van
der Waals
interactions when hydrogens are constrained to be near one another.
60. Two of the possible tetracyclic perhydropyrenes are shown below, once in planar form, once in its preferred conformation. Note that each isomer is constrained to have two boats (actually twists). Using molecular models as a guide when entering your structures, calculate the MM energies and structures of compounds 1 and 2. Then, use the "delete" button to erase the C15-C16 bond; remove all of the H's, and re-add them; and do MM calculations of the energy and geometry of the dihydro derivatives of 1 and 2. Be sure to use molecular models in order to "locate" the most reasonable conformations of these dihydro compounds; be sure to try all of the likely rotations and flexings of the two remaining six- membered rings.
61. The complicated
structure shown to the right is related to compounds 1 and 2
from the problem above, as can be seen from its conformational drawing.
In fact, it has been numbered from 1 to 16, exactly as is compound 2.
There are 24 carbons in all, but only the first 22 are numbered in the
two drawings. As in Problem No. 60, do the MM calculation of the given
structure. Then, use "delete" to erase the C15-C16,
C17-C22, and C19-C20 bonds.
The resulting hexahydro derivative is not only related to the dihydro derivatives
of 1 and 2 from the previous problem, but also to compound
3 from Problem 58. But
unlike the old compound 3 which can relieve the internal van der
Waals interactions from the CH2 group, this compound
has two such CH2 groups near one another. Use models
to try to find "reasonable" conformations which can then be subjected
to MM calculations.
64. Non-bonded interactions
in certain polycyclic aromatic compounds can cause a twisting of the rings,
resulting in a non-planar structure. Compound 1 with R = Ph (to
the right) has been synthesized; its X-ray crystal structure indicates
that the phenanthrene rings at the left and right are 65° tilted from
one another. Compound 2, R = Ph, has not been made, but calculations
suggest that here the phenanthrenes are orthogonal. These two compounds
with R = Ph give the message "too many pi-atoms" when entered
into our MMX program (PCMODEL). However, calculations can be done on 1
with R = H and with R = CH3; and on 2 with R = H and
with R = CH3. Carry out these calculations to see if the alleged
twisting occurs. Compare your calculated geometries with those reported
(experimental or calculated) for the phenyl-substituted compounds.
[Plummer, B. F.; Russell, S. J.; Reese, W. G.; Watson, W. H.; Krawiec, M. J. Org. Chem. 1991, 56, 3219; Smyth, N.; Van Engen, D.; Pascal, R. A., Jr. J. Org. Chem., 1990, 55, 1937 and references therein.]
79. The asteranes
(Problem 73) consisted of parallel identical rings, separated by a series
of one-carbon bridges. Shown below are two sets of related molecules, either
with zero-carbon or with two-carbon bridges. Do MM calculations on the
four compounds (from cyclopropane to cyclohexane) from either Set A or
Set B. The first two members of Set A are prismane (shown) and cubane.
The first member of Set B also appeared in Problem 76; all of the members
of Set B have severe non-bonded interactions between C-H bonds.
173. Benzene and its
simple derivatives are planar. Shown to the right is [5]- helicene, a polycyclic
aromatic hydrocarbon that cannot be planar because of interference from
the two H's; the name "helicene" derives from the helical structure
that this and related molecules adopt. Do MMX calculations on [5]-as well
as [6]- and [7]-helicenes (one and two extra rings, respectively, fused
onto the given structure in a spiral arrangement) as well as on the indenohelicenes
described in the articles. Compare your computed dihedral angles with those
in Table V of the second article.
[Katz, T. J.; Sudhakar, A.; Teasley, M. F.; Gilbert, A. M.;
Geiger, W. E.; Robben, M. P.; Wuensch, M.; Ward, M. D. J. Am. Chem.
Soc. 1993, 115, 3182;
Gilbert, A. M.; Latz, T. J.; Geiger, W. E.; Robben, M. P.; Rheingold, A.
L. Ibid. 1993, 115, 3199.]
212. The polycyclic
aromatic compound to the right (note the presence of only two sp3
carbons) has been synthesized. Because of repulsion between aryl hydrogens,
the molecule cannot remain planar. In fact, both a meso and a dl
form have been isolated, each of which consists of two helical arrays (see
the article for details). Do MMX calculations on both stereoisomers; compare
your structural parameters (angles between rings, distortions from planarity)
with those cited in the article. Then, do calculations on related compounds
(of your invention) that are similarly forced to be non-planar; for example,
replace the five-membered rings with smaller or larger rings or fuse additional
aromatic rings onto the basic nucleus, and so on.
[Bradshaw, J. D.; Solooki, D.; Tessier, C. A.; Youngs, W. J. J. Am. Chem. Soc. 1994, 116, 3177.]
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