1. OCHM - Molecular Structure
Question Information
Q-Bank
MCAT
Sim
Non-Sim
Subject
Organic Chemistry
Foundation
OCHM - Molecular Structure
Validity
5 years
Author(s)
Reyes, V. M.
Reviewer(s)
Editor(s)
Passage Media
Media ID(s)
Passage
In everyday conversation, being “aromatic” means giving off a somewhat pleasant
smell or “aroma”, such as herbs, spices and scented candles. But in organic chemistry,
being aromatic, or “aromaticity,” has nothing to do with olfaction. Rather it has every-
thing to do with how electrons distribute themselves across molecules or certain parts
of molecules, which in turn influences how the molecule reacts towards certain other
molecules.
The hallmark of an aromatic organic compound is the presence of pi electrons in a ring system that allows free delocalization of the former within the latter. Pi electrons, on the other hand, is a necessary consequence of having a conjugated
system of double bonds of carbon, which in essence is a series of sp2-hybridized carbon atoms. Consider the benzene molecule, the epitome of an aromatic molecule (see Figure 1 on next page).

2. Figure 1 The benzene molecule is a hexagonal
ring of six carbon atoms each with a
bound hydrogen atom. Considering
each carbon atom, we see it is bonded
to two other carbon atoms, and one
hydrogen, a total of three covalent
bonds. Carbon has six electrons, the
first two of which is in the first shell (or
1st energy level), 1s2, and do not take part
in chemical reactions. Those in the 2nd
shell, the next four e-s, are the valence
shell electrons – the ones chemically
active. The four orbitals available in the
2nd shell are 2s, 2px, 2py and 2pz, each of
which can accommodate two electrons.
The four electrons in the 2nd shell (the
valence e-s of carbon) occupy the four
2nd shell orbitals mentioned above, one
electron per orbital. Since each C atom
in benzene is bonded to three other
atoms, the 2s, 2px and 2py orbitals hybri-
dize into three sp2 hybrid orbitals, leaving
the 2pz orbital unhybridized. The three sp2
orbitals are now identical and lie 120ᵒ from
Figure 1
each other on a plane. Two of them are shared with the two neighboring C atoms,
and the third with the H atom. These three bonds are the three solid black lines shown emanating from each C atom in the top molecule in Figure 1; the unhybridized 2pz orbitals are shown as “dumbells” normal to the hexagonal plane of C atoms, and each of these six dumbell orbitals has one e-. What are they to do?
Since the unhybridized 2pz orbitals are next to each other and are of equal energy, they
delocalize around the ring: the upper half of each fuse together, as do the lower halves.
The result are two “donut” clouds of delocalized electrons above and below the ring, giving rise to the phenomenon of aromaticity.
We can now state Huckel’s rule for aromaticity: If a cyclic organic molecule with adjoining C atoms in sp2 hybridization state is aromatic, then it must have 4n+2
delocalized pi electrons, where n = 0, 1, 2, 3, … Benzene has six delocalized pi electrons, and 4n+2 = 6, thus n=1.

3. D. Klein, Organic Chemistry
Passage References
PMID/Book
Title of Publication or Book
D. Klein, Organic Chemistry
(N/A)
en.wikipedia.com, youtube.com
(N/A)
Author’s own lecture notes.
Question Attributes #1
Topic Blueprint
Hückel’s Rule for Aromaticity
Competency
MCAT: BS-2: Application of Concepts & Principles
Objective
To understand how to apply Huckel’s Rule in predicting
aromaticity of organic compounds.
Media ID(s)
Question ID
Question Stem #1
Using Huckel’s Rule, predict which of the following molecules is/are not aromatic.
Answer Choices #1 A)
III only B)
I and II C)
II and III D)
III and IV
Correct: D)

4. I II IV
III
Explanation #1
The correct answer is D. Molecules III and IV both have eight pi electrons, hence
4n + 2 = 8 and n = 3/2, which is non-integral. Neither is aromatic.
(Choice A) This statement is false. Molecule IV also has eight pi electrons, hence for the same reason as above, 4n + 2 = 8 and n = 3/2, which is non-integral. It is not aromatic.
(Choice B) This statement is false. Molecules I and II are both aromatic: the first has
10 pi electrons, hence 4n + 2 = 10, and n = 2; the second has 14 pi electrons, hence
4n + 2 = 14, and n = 3.
(Choice C) This statement is false. Molecule III is not aromatic but II is (see above
reasons).
Educational objective: To understand how to apply Huckel’s Rule in predicting
aromaticity of organic compounds.

5. 0000000 0000000 References #1 PMID/Book Title of Publication or Book
Verifications #1
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The question is at the Application or higher cognitive level.
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The question is based on a realistic clinical scenario.
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The question has at least one close distracter, and other options have
educational value.
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The question is appropriate to the entry level of nursing practice.
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The explanation is short and concise, yet thorough.
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The question has an appropriate table/flow chart/illustration.