7. In the second issue
of Angew. Chem., Int. Ed. Engl. in 1987, the synthesis of this compound
is reported. How strained is it? What about the corresponding compound
with a central four-membered ring? What about a central eight-membered
ring? Compare these to the permethylated cyclohexane, cyclobutane,
and cyclooctane.
13. Propellanes are
tricyclo[x.y.z.0] molecules. How strained are the [2.2.2.0] and related
[2.y.z.0] compounds?
[For some recent calculations on propellanes having at least one four-membered ring, see Wiberg, K. B.; Caringi, J. J.; Matturo, M. G. J. Am. Chem. Soc. 1990, 112, 5854.]
26. Calculate the
strain energy of pyramidalized alkenes, such as those shown to the right.
[Borden, W. T. Chem. Rev. 1989, 89, 1095; Ermer, O.; Bell, P.; Mason, S. A. Angew. Chem., Int. Ed. Engl. 1989, 28, 1239; Irngartinger, H.; Deuter, J.; Charumilind, P.; Paquette, L. A. J. Am. Chem. Soc. 1989, 111, 9236; see also Luef, W.; Keese, R. Top. Stereochem. 1991, 20, 231.]
73.
The asteranes (p. 93) are also beautifully symmetrical molecules.
Shown here are triasterane and hexaasterane. Your task is to do MM calculations
on the energies and geometries of tri-, tetra-, penta-, and hexaasterane.
Discuss the variations in strain energy as the ring size changes; explain
clearly why hexaasterane is not the least strained of the series.
[Ahlquist, B.; Almenningen, A.; Benterud, B.; Traetteberg, M.; Bakken, P.; Lüttke, W. Chem. Ber. 1992, 125, 1217.]
74. Israelane
and Helvetane (p. 87) [Why are these names used?] are (CH)24
molecules. Do MM calculations on both of them. What you will discover is
that although these structures are aesthetically pleasing, they are disasters
chemically. The latter structure has not only the strain of the four-membered
rings but also has four H's from the "front" 12-membered ring
and four more from the "rear" ring pointing toward one another;
nevertheless, it minimizes to a structure not too different from that shown.
Israelane, on the other hand, has six H's on the "front"
and six more on the "rear" ring, all pointing to the center;
neither RMM nor JEB has managed to get a minimized structure whose geometry
is close to that shown. Try the MM procedure on israelane, and see what
you get. Then, try various "tricks" to hold the ring system in
this geometry; for example, connect the interior carbons to an added single
carbon or to the carbons of added three-membered or six-membered rings;
be creative!
75.
The name triblattane (p. 269) is derived from the German word for
"leaf" and is used to describe the distorted bicyclo[2.2.2]octane
nucleus which results when various short chains are introduced. The exercise,
here, is to calculate the energy and geometry of the parent compound; pay
special attention to the dihedral angle C1- C2-C3-C4.
Then do calculations on three molecules with just one bridge; let m = 2,
1, and 0. Then do calculations on three molecules with two bridges; let
m = p = 2, 1, and 0. Finally, do calculations on three molecules with three
bridges; let m = p = n = 2, 1, and 0; the last compound is the beautifully
symmetrical "cubane." Discuss the strain energy and the variation
in the dihedral angle as the number and length of the bridges are changed.
76. Problem 75
was concerned with the increasing strain of a bicyclic octane as bridges
of ever-shorter length were incorporated into it. The present problem starts
with an analogous molecule, but with two cyclopropane rings instead
of the two bridgehead carbons. Models (and MM) suggest that unlike
bicyclo[2.2.2]octane itself, there is already significant twist in the
parent structure, the one with no bridges; models also suggest the cyclopropane
rings are not quite orthogonal and also that introduction of the bridges
does not lead to a large increase in strain. Do MM calculations on the
energy, geometry, and dihedral angles for the parent compound; for the
singly-bridged compounds with m = 1 or 0; for the doubly- bridged compounds
with m = p = 1 or 0; and for the triply-bridged compounds with m = p =
n = 1 or 0. This last compound, a wonderfully symmetrical structure, is
identical to the first bis-peristylane of the next problem.
77.
Problem 72 asked for an analysis of the
[n]-peristylanes from n = 3 through 5. One can also imagine a set of bis-[n]-
peristylanes for which structures are shown for n = 3 (two perspectives)
and n = 6. Do MM calculations on the complete set from n = 3 through n
= 6; discuss the large variation in strain energy (and the principal reasons
for it) along this series; as in Problem 73, note that
the n = 6 compound is not the least strained. Also note that n = 3 is identical
to the last compound calculated in Problem 76 and that n = 5 is identical
to dodecahedrane from Problem 72.
[Lee, C.-H.; Liang, S.; Haumann, T.; Boese, R.; de Meijere, A. Angew. Chem., Int. Ed. Engl. 1993, 32, 559.]
78. There are five possible "Platonic solids," polyhedra in which all faces are identical rings. Three of these are the tetrahedron, cube, and dodecahedron, all of which have been "built" in the form of saturated organic molecules: tetrahedrane (C4H4), cubane (C8H8, Problem 75), and dodecahedrane (C20H20, Problems 72 and 77). The other two cannot be "built" as organic structures, either because they would require five bonds to carbon or because they would require inverted sp3-hybridized carbons. There are, however, many other molecules which are nearly spherical, which have symmetry almost as high as the Platonic solids, but which contain two (or more) different-sized rings (rather than all rings the same). Any one of the following five sections will constitute a reasonable MM problem set for solution.
A. Do MM calculations
on compounds 1 and 2 which consist of three- and six- membered
rings. The latter, despite its lovely shape, is a very strained
molecule (make a model!); perhaps the best way to do its MM structure is
first to enter compound 3; minimize it; replace one of the five
CH groups by a C3H3 group; minimize; and continue
the process so that eventually all five CH groups of 3 are replaced
by the three-membered rings needed for 2. Report the MM structures
and energies for 1, for 3, for each of the cyclopropane-substituted
materials, and finally for 2. You'll note that introduction of the
first cyclopropane raises the energy somewhat; introducing the second,
third, and fourth raises it a little; and introducing the final one leads
to a large increase.
B. Do MM calculations
on this set of compounds which consists of molecules containing various
combinations of three-, four-, five-, and six-membered rings. Note that
the first two structures have, respectively, two and one inverted sp3-hybridized
carbons; note, also, that the third compound is geometrically constrained
to have planar four-membered rings. As a result, the first set of three
structures is much more strained than the second set of three.
C. A compound much
in the news lately is "buckminsterfullerene" or "buckyball,"
a form of carbon having the formula C60 and consisting of fused
five- and six- membered rings in a nearly spherical array. Its saturated
version, C60H60 ("soccerane"), consisting
of 12 regular pentagons and 20 regular hexagons, has never been made. Because
that structure has too many atoms for an MMX calculation, you should do
MM calculations on two smaller molecules (shown) which contain 12 pentagons
but only three or two hexagons; the second of these is also found in Problem
77.
D. Do MM calculations
on this set of structures which contain four-membered rings in combination
with rings of varying sizes. The third compound is also found in Part B;
the fourth compound is also calculated (although not shown) in Problem
77.
E. Replacement of
any one of the CH groups of cubane by a C3H3 unit
generates a compound called "lampane." From lampane, replacement
of one or another of the three different CH groups by a C3H3
unit gives three different C12H12 compounds. Compute
the MM energy and structure of these five compounds.
80. One can imagine
the triple intramolecular [2 + 2] cycloaddition shown here, giving a saturated
molecule which is drawn with deliberately ridiculous-looking long bonds.
Do MM calculations on this cyclized compound, C21H18,
and on the analogous compounds which one would get from the quadruple and
pentuple cyclization of related compounds with a central four- or five-membered
ring; these will have formulas C28H24 and C35H30,
respectively.
101. Hexaprismane
(1) is a highly strained molecule, but various derivatives such
as seco-[6]-prismane (2), homo-[6]-prismane (3), and various
bis-homo-[6]-prismanes (such as 1,3-bishomo compound 4) have been
made. Do MM calculations on the geometry and energy of these structures
and structures related to them. Alternatively, one could do calculations
on the heptaprismane and its various seco-, homo-, and bishomo-derivatives.
[Mehta, G.; Padma, S. Tetrahedron 1991, 47, 7783, 7807; Mehta, G.; Krishna Reddy, S. H.; Padma, S. Ibid. 1991, 47, 7821]
140. Tritwistatriene, a beautifully symmetrical molecule, at 350 ·C gives a mixture of four isomeric products, all (CH)14. Compute the heats of formation and strain energies of the starting material and of its four reaction products; compare your results with those obtained by MM3 calculations reported in the reference. [Mechanisms for the formation of these products are described in the article.]

[Otten, T.; Müller-Bötticher, H.; Hunkler, D.; Fritz, H.; Prinzbach, H. Tetrahedron Lett. 1992, 33, 4153.]
151. Shown to the
right are naphthalene and perylene (a "double" naphthalene).
Compute the structures and energies of these two molecules and of the derivatives
in which two-carbon bridges, either as -CH2-CH2-
or -CH=CH-, are connected to naphthalene at C1 and C8
or at C1-C8 and C4-C5; similarly
for positions C1-C12 and C6-C7
of perylene. Compare the increase in strain with numbers reported in the
reference.
[Mitchell, R. H.; Chaudhary, M.; Williams, R. V.; Fyles, R.; Gibson, J.; Ashwood-Smith, M. J.; Fry, A. J. Can. J. Chem. 1992, 70, 1015.]
190.
In principle, removal of one, two, and then three moles of H2
from the trimethyltriphenylene derivative shown here can lead to the interesting
molecule called sumanene (from the Hindi word for sunflower). This system
with fused five- and six-membered rings resembles a structural fragment
of buckministerfullerene (C60). Do MMX calculations on the four
molecules; comment on their shape (planar vs. bowl- shaped); compare the
increase in strain energy for each dehydrogenation step with the values
in the literature reference.
[Mehta, G.; Shah, S. R.; Ravikumar, K. J. Chem. Soc., Chem. Commun. 1993, 1006.]
211. The beautifully
bowl- shaped molecule "Pinakene", C28H14,
is a fragment of C70 (one of the fullerenes). A possible synthetic
route (reminiscent of that in Problem No. 190) involves
removal of one mole of H2 from the tetramethyl-dibenzonaphthacene
(shown) followed by removal of three more. The strain energy increases
with each removal of a hydrogen molecule and creation of a five-membered
ring. Do MMX calculations on the tetramethyl compound, on pinakene, and
on the three intermediate structures. Compare your answers with those in
the cited reference. Compare the shape of pinakene calculated by MMX with
that in the article.
[Jemmis, E. D.; Sastry, G. N.; Mehta, G. J. Chem. Soc., Perkin Trans. 2 1994, 437.]
248. The "triangulanes"
are an interesting class of molecules. They are made up of repeating units
of spiropentane (C5H8, to the right). Do MMX calculations
on cyclopropane, spiropentane (also called [2]-triangulane), and the various
[3]-, [4]-, and [5]-triangulanes shown. Compare your determination of the
"incremental strain energy" with the calculated and experimental
values in the cited article. Although not mentioned by the authors, there
are two diastereomers (one meso, one chiral) possible for [5]-triangulene;
do calculations on both of them.
[Beckhaus, H.-D.; Rüchardt, C.; Kozhushkov, S. I.; Belov, V. N.; Verevkin, S. P.; de Meijere, A. J. Am. Chem. Soc. 1995, 117, 11854.]
265. A recent article has the intriguing title "Flexible backbone segments with a marked conformational preference." Consider, first, the case of 2,4-dimethylpentane. It exists nearly 100% as an equal mixture of conformations A and B, interconverting by rotation about the C2-C3 and C4-C3 bonds; the other staggered conformations (C, D, and E) all suffer from at least one severe CH3/CH3 interaction. Calculate the energies and structures of conformations A, C, D, and E. Because A and B are of equal energy, this molecule is "bi- conformational" (i.e., it populates two equally energetic conformations). So is dicyclohexylmethane (1) whose energy and structure you should calculate. [Use molecular models; use your framework from A to set up the best conformation of 1; be sure to choose the chairs that have the bridging CH2 group equatorial on both rings.] On the other hand, chiral bis-acetal 2 and, especially, methyl-substituted 3 are "mono-conformational" in that one of the conformations resembling A or B is much more stable than the other. Do MMX calculations on the two conformations for 2 and 3; determine energies and structures; compare your results with those reported.

[Hoffmann, R. W.; Kahrs, B. C. Tetrahedron Lett. 1996, 37, 4479.]
266. A recent article
is concerned with the preparation and X-Ray crystal structure of the stereoisomeric
1,2,3- tricyclohexylcyclohexanes (R = C6H11). Do
MMX calculations on the three isomers. Compare your results for heats of
formation and dihedral angles with the experimental and computed values
in the cited article.
[Columbus, I.; Hoffman, R. E.; Biali, S. E. J. Am. Chem. Soc. 1996, 118, 6890.]
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