Electrophilic Aromatic Substitution Journal

In Biotech Business Week, September 3, 2007, BIOMEDICINE; Report summarizes biomedicine study findings from Osaka City University, Department of Chemistry was published. The experiment's goal is to determine the monoxygenase activity of three copper proteins and undergo ortho-hydroxylation of phenols and later electrophilic aromatic substitution to give the final product. During this experiment, the third copper protein tested undergoes electrophilic aromatic substitution. Lithium phenolates and peroxo dicopper(II) will react by electrophilic aromatic substitution to give an oxygenated product called catechols. The researchers  determined that the monoxygenase activity is similar to that of tyrosinase because they both undergo a simple enzymatic reaction that is the same mechanism as electrophilic aromatic substition. In the conclusion, the author states, "In this case as well, the ortho-hydroxylation of phenols to catechols has been demonstrated to involve the same ionic mechanism." The ionic mechanism being electrophilic aromatic substitution.

Jesslyn

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Aromaticity

Dear Granny,

            In Organic Chemistry II, we’ve been discussing aromaticity. Aromatic compounds are a ring of carbon atoms with alternating single and double bonds. They are defined as compounds that are cyclic, planar, completely conjugated, and obey Huckel’s Rule. They received the name “aromatic” because of their distinct characteristic odors. Cyclic compounds are compounds in a ring; for example: Benzene. Planar means that the compound is flat like a sheet of paper. Conjugation is the overlap of p-orbitals in the molecule. An orbital is the cloud around the molecule that stores the electrons. P-orbitals are a dumbbell shape around the element that is attached to a double bond. Huckel’s Rule is (4n + 2), which determines the number of π electrons in the compound. Each pi electron gives 2 electrons. The easiest way I remember whether it obeys Huckel’s role, without calculating, is by remembering that the number of π electrons begins with 2 and increases by 4. So if it contains 2, 6, 10, 14, 18, 22, 26, etc.  π electrons, it is aromatic. Compounds that are cyclic, planar, and conjugated, but do not obey Huckel’s Rule, their π electrons = 4n, are considered antiaromatic. If the compound disobeys at least two of the rules to be aromatic, it is considered nonaromatic. I hope you learned something new.

Love,

 Jesslyn

Test 1

While taking the exam on Friday, I was shocked to see that we had to draw out spectrums. It was something that we hadn't practiced and during one of the first lab meetings, it was said that we wouldn't have to know how to draw them, but we would need to know how to read them. When studying, I expected that I needed to know and understand the nitrogen rule, reading and interpreting mass spectrometry, infrared spectroscopy,  ¹H NMR, and ¹³C NMR graphs. I studied primarily Ch. 14 more than Ch. 13 and I should have went over Ch. 13 more before the test. For next test, I plan to study differently. I thought there would be more Sapling problems such as #4 where the m/z was determined for two isomers and#6 labeling unknowns based on their IR absorptions.

Week 1: Muddiest Point

After studying Chapter 13 Mass Spectrometry and Infrared Spectroscopy, my muddiest point is section 13.2 Alkyl Halides and the M + 2 Peak on pg. 467. I don’t understand the ratio concepts. For example, ³⁵Cl and ³⁷Cl occur at a 3:1 ratio. The text states that if it is a 3:1 ratio, there will be a Cl in the atom. How are we supposed to determine the ratio or is it generally given information? Also, it states that the bromine is a 1:1 ratio. Why is it different from the Cl, which is 3:1? Also, understanding the M + 2 Peak is very confusing. I understand that it is the mass of the compound + 2, but what is the significance? After further reading of the text, the M + 2 peak is used for alkyl halides that have a mass slightly higher than their mass on the periodic table. With further reading, I learned that with an M + 2 peak, there will be two peaks towards the end of a graph showing the presence of the alkyl halide. Based on the text, chlorine and bromine have two isotopes and carbon, hydrogen, oxygen, nitrogen, sulfur, phosphorus, fluorine, and iodine only have one major isotope. After reviewing the sample problem for determining molecular ions in a mass spectrum of a particular compound, I feel like I understand the concept better.