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) 0000000 Editor(s) 0000000 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).

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.

D. Klein, Organic Chemistry Passage References PMID/Book Title of Publication or Book 000000000 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) 00000000 Question ID 00000000 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)

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.

0000000 0000000 References #1 PMID/Book Title of Publication or Book 000000000 000000000 0000000 Verifications #1 Yes / No The question is at the Application or higher cognitive level. Yes / No The question is based on a realistic clinical scenario. Yes / No The question has at least one close distracter, and other options have educational value. Yes / No The question is appropriate to the entry level of nursing practice. Yes / No The explanation is short and concise, yet thorough. Yes / No The question has an appropriate table/flow chart/illustration.