Carrot assignment for chapter 17 is an in depth list of questions that goes into the material that was learned in this chapter and these questions are often mentioned on the exam as well.
Recall that these are "carrot" assignments, as in "carrot and stick", to encourage you to read
ahead and grasp concepts. You earn bonus points equivalent to ~5.5% of the total available by
attempting all of them throughout the semester. Compose your responses however you like, but
submit response to assignment as a pdf file.
To be classified as aromatic, which is a special type of particularly stabilizing resonance,
molecules (like benzene) or cyclic parts of molecules have to meet the following two criteria
(§17.5,):
1. Contain a ring comprised of continuously overlapping p-orbitals...with side-by-side
overlap in a pi-fashion, as opposed to sigma-fashion.
2. The number of pairs of pi electrons contained in this cyclic pi-system has to be odd
(easier to remember and explain than the alternate form of this rule, which is that the total
number of included pi electrons must = 4n+2, with n = any integer).
The p-orbitals can be parts of pi-bonds, or empty p-orbitals, orbitals with non-bonding electrons
(lone pairs)...as long as there is continuous pi-overlap around the entire ring. If a non-bonding
lone-pair is arranged in space such that it’s orbital can not overlap side-by-side with the p-
orbitals in a cyclic pi-system, then those electrons are not part of the system (see question
concerning VESPR).
SO...to be able to classify compounds as aromatic, non-aromatic, or anti-aromatic is really just a
matter of being able to do two basic tasks based on fundamentals you already know (A)
recognize whether you have continuous pi-overlap around a ring...just be able to tell if you have
continuous side-by-side overlap of p-orbitals (B) Be able to count electrons that are in that pi-
system.
Address all the following in terms of monocyclic pi-systems (just one ring).
Let’s get right to one important point. Why this requirement for an odd number of pairs of
electrons?? This will become clear if you consider the MO diagrams. Go to Figure 17.3, which
shows the continuous array of six individual p-AOs of benzene all in the same phase (only
bonding/stabilizing interactions) to form the lowest energy pi MO shown at the bottom of Figure
17.4. To construct the rest of the 6 pi-MO’s (all higher energy), the pi-AO’s are combined with
progressively more nodes just as you did for polyenes in chapter 16, but since nodes have to
bisect the system, now each node has to bisect either TWO BONDS or TWO ATOMS as it cuts
through the entire pi-system. For such cyclic systems there are TWO WAYS TO ADD EACH
NODE, producing two separate MO’s that actually have the same energy level. Two MO’s with
the same energy are termed “degenerate”. Note in Figure 17.4 there are 2 degenerate MO’s with
one node (2 ways to add one node), 2 degenerate MOs with 2 nodes (two ways to add two
nodes), and then only one with 3 nodes ... 6 AO’s combine to produce 6 pi-MO’s but spread
, over only 4 energy levels. For acyclic 1,3,5-hexatriene (non-cyclic) there would be 6 pi-MO’s at
6 different energy levels.
Question 1. Define orbital π-overlap and contrast with σ-overlap. It is crucial for this chapter
that you can recognize all cases where pi-overlap is possible. Draw examples of pi-overlap
involving each of the following, and include resonance drawings:
(a) a lone pair on a heteroatom (“non-bonding” sp3, but “looks sp2” in resonance forms)
(b) lone pair on a carbanion
(c) empty p-orbital on a carbocation
Question 2. Reproduce the molecular orbital diagram for benzene (figure 17.4). Draw six
hexagons with a p-orbital at each vertex. Now add the nodes, which have to bisect the entire pi-
system. This is a big difference between, say, benzene and its acyclic analogue 1,3,5-hexatriene.
Notice why you have pairs of degenerate orbitals (same energy). Maybe a cartoonish approach,
but just come up with a “bonding” number for each MO. Add 1 for each pi-bonding interaction,
and subtract 1 for each anti-bonding interaction. Add 0 for each non-bonding interaction. Do
the degenerate orbitals have the same “bonding” number?
Yes, degenerate orbitals always have the same bonding number because they have the
same energy level and will interact with orbitals in the same way to form bonds.
Question 3. Note that positions of the separate pi-MO drawings in the MO energy diagram for
benzene form a hexagon, the shape of benzene. It turns out that you can generate approximations
of the MO energy diagrams (minus the actual MO pics) very quickly with the Frost circle
mnemonic. Read the section about Frost circles and draw Frost circles for cyclobutadiene (2 pair
pi electrons in a quadrilateral) and for benzene (3 pair pi electrons in a hexagon). Add the pi-
electrons for both to the appropriate lowest energy MOs following rules you learned in
GenChem. For this question, respond with the reason why it is CRUCIAL whether the number of
pairs of pi-electrons is odd (possible to be aromatic) or even (possible to be antiaromatic). By
the way, benzene is aromatic and particularly stable for an unsaturated hydrocarbon, while
cyclobutadiene is antiaromatic and particularly unstable for an unsaturated hydrocarbon. For
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