BLOG 10 = LAST BLOG!!

Wow! This spring semester is almost over and this is the last blog you’ll be seeing from me, well dealing with Organic Chemistry anyway. The blog assignment for blog #10 was to search for a publication article from Christina White’s, “White Lab.” Christina White is a chemist with the Department of Chemistry at the University of Illionois Urbana-Champaign.

The article I found that involved a topic that we had discussed in Organic Chemistry II was, “A General and Highly Selective Chelate-Controlled Intermolecular Oxidative Heck Reaction.” The Heck Reaction is a chemical reaction where an unsaturated halide reacts with an alkene, a base,  and a palladium catalyst to form a substituted alkene.¹ The intermolecular Heck reaction is unique among cross-coupling reactions due to the direct formation of C-C bonds from vinylic C-H bonds of α-olefins.² An olefin, another name for alkene, is an unsaturated chemical compound containing at least one carbon-to-carbon double bond.³

Below is the general scheme for a Heck Reaction:

Below is Dr. White’s reaction scheme for this publication where the Heck Reaction occurs in the first step of the reaction.

It’s hard to believe Organic Chemistry has come to an end. I hope everyone has enjoyed reading my blog posts. Good luck to the lucky individuals enrolled in Organic I in the Fall J.

References:
1.  http://en.wikipedia.org/wiki/Heck_reaction
2. http://www.scs.illinois.edu/white/pubs/pub11.pdf
3. http://en.wikipedia.org/wiki/Olefin

Blog 9: Potential Exam Questions

This blog assignment was to search for problems that could possibly appear on our Organic Chemistry II Exam 5. I found a problem that utilizes The Michael Reaction.

EXAMPLE QUESTION:
Draw the product of The Michael Reaction:

ANSWER:

The CN attaches where the C=C was located on the pentane ring.



Further examples of The Michael Reaction are located in Chapter 24, Section 24.8 (pp. 934) of our Organic Chemistry textbook.

References:

1. http://usm.maine.edu/~newton/Chy251_253/Lectures/MichaelAdditions/MichaelFS.html 

Organic Chemistry 2- EXTRA CREDIT

On Thursday, April 21^st, 2011, Dr. Steven Myers, from the University of Louisville Department of Pharmacology and Toxicology, gave a seminar titled, “Tobacco Smoking During Pregnancy/Biomarkers of exposure and relationship of genetics.”

Dr. Myers stated that the biggest risks for smoking during pregnancy are that the baby gets less food and oxygen. Also, certain risks are increased such as, miscarriages, stillborn births, premature births, low-birth weight babies, babies who have an increased risk of dying with SIDS, babies with a high risk of respiratory illnesses, ADHD, and lower IQ levels.

The most common cancers in women who smoke are lung and urinary bladder. Cancers such as lip, oral cavity, pharynx, esophagus, pancreas, larynx, lung, trachea, bronchus, urinary bladder, and kidney are among the highest risk. Cardiovascular issues are at a higher risk, such as rheumatic, ischaematic, and pulmonary heart disease, as well as hypertension, atherosclerosis, and aortic aneurysm.

One cigarette contains around 50 different compounds. Dr. Myers and his team study the molecular epidemiology, the incorporation of biomarkers into analytical epidemiological studies. Biomarkers are any change produced by an environmental contamination. They test specificity, direct results of exposure; sensitivity, the change produced by the contamination; and practicality, the cheapest ways to get the same answer. In pregnant women, they test the amniotic fluid as the biomarker. The amniotic fluid tells everything the fetus has been exposed to. An increase of 1-hydroxypyrene levels will appear in the amniotic fluid with the more cigarettes that are smoked. After a birth, cord blood is tested because maternal blood does not mix with cord blood so it can be tested to see exactly what both the mother and infant were exposed to. 4-aminobiphenyl is the most carcinogenic compound found in cigarettes and is very hard on the bladder.

I really enjoyed this seminar. Dr. Myers did a fantastic job on explaining, at an undergraduate level, his research and the harms of smoking during pregnancy. Since I plan to specialize in obstetrics and gynecology, I especially enjoyed this seminar.

Hell-Volhard-Zelinsky Halogenation Reaction

The Hell-Volhard-Zelinsky halogenation reaction was named after three chemists, Carl Magnus von Hell, Jacob Volhard, and Nikolay Zelinsky. This reaction is unlike other halogenation reactions because it takes place in the absence of a halogen carrier, instead, it is initiated by a catalytic amount of PBr₃, after which one molar equivalent of Br₂ is added.¹

The PBr₃ replaces the carboxylic OH with a Br, making a carboxylic acid bromide. The acyl bromide can be tautomerized to an enol, which will react with the Br₂ to brominates a seond time at the α position.¹ A nucleophilic acyl substitution reaction occurs when the hydrolysis of the α-bromo acyl bromide occurs spontaneously in neutral to acid aqueous solution and yields the α-bromo carboxylic acid. If little nucleophilic solvent is present, the reaction of the α-bromo acyl bromide with the carboxylic acid yields the α-bromo carboxylic acid product and creates the acyl bromide intermediate.¹ Below is an example of the scheme of the Hell-Volhard-Zelinksy halogenation reaction.

Sources:

1. http://en.wikipedia.org/wiki/Hell-Volhard-Zelinsky_halogenation

γ-Butrolactone

γ-Butrolactone, also known under the IUPAC name Dihyrdofuran-2(3H)-one, is used as a stain remover, superglue remover, and paint stripper. It is also used in the human body because GBL is a prodrug of GHB. GBL is consumed by a human and then transfered into GHB by lactonase enzymes in the blood. GHB stands for γ-Hydroxybutyric acid. At one time, GBL was sold under the names Revivarant or Renewtrient as a human growth hormone. In high doses, it acts as a sedative and can make a human unconscious. With low doses, it has a euphoric effect. GBL is commonly called “liquid ecstasy”. If excessive doses are taken irrational behavior, severe sickness, coma, or death can occur. GBL is addicting and withdrawals can lead to delirium tremens, seizures, and can cause death in many circumstances. In layman terms, do NOT consume GBL.

γ-Butrolactone has a melting point of -43.53° C and a boiling point of 204° C. The pH of this compound is 4.5, making it acidic.  It is soluble in water, as well as CCl4, acetone, bezene, methanol, and ethyl ether. GBL can be synthesized from gamma-hydroxybutyric acid (GHB) by removal of water or by distillation from such a mixture. It may also be obtained via oxidation of tetrahydrofuran (THF).¹ One such process, which affords GBL in yields of up to 80%, utilizes bromine generated in situ from an aqueous solution of sodium bromate and potassium hydrogen sulfate.¹  Another process can proceed by using commercially-available calcium hypochlorite in the presence of activating acetic acid and an appropriate solvent such as acetonitrile.¹

Example Synthesis:



Example Reaction Where γ-Butrolactone is Converted to Another Caboxylic Acid:



Sources:

1. http://en.wikipedia.org/wiki/Gamma-Butyrolactone

Blog 6: Grignard Reagent

"SYNTHESIS OF ELECTRON-DEFICIENT SECONDARY PHOSPHINE OXIDES AND SECONDARY PHOSPHINES: BIS[3,5-BIS(TRIFLUOROMETHYL)PHENYL]PHOSPHINE OXIDE and BIS[3,5-BIS(TRIFLUOROMETHYL)PHENYL]PHOSPHINE," was the title of the article from http://orgsyn.org. The above synthesis undergoes a Grignard reaction. A Grignard reaction is an organometallic chemical reaction in which alkyl- or aryl-magnesium halides (Grignard reagents) act as nucleophiles and attack electrophilic carbon atoms that are present within polar bonds to yield a carbon-carbon bond, thus altering hybridization about the reaction center.¹ In this reaction, the Grignard reaction occurs in step 1 with i-PrMgCl / THF reacting with the substituent Br on the benzene ring to synthesize MgBr. Then, it reacted with (EtO)₂ POH/ THF in step 2 to synthesize a POH bond where the substituent MgBr was and connected to an  additional benzene ring with the same substituents, thus forming 5 new C-C bonds and 3 new C=C bonds (highlighted above). In step 3, the compound reacts with DIBAL-H and MTBE/cyclohexane to eliminate the double bonded oxygen from the P bond.

References:

1.      http://en.wikipedia.org/wiki/Grignard_reaction

Proline

Between 1899 and 1906, Hermann Emil Fischer discovered the amino acid Proline. Proline, abbreviated as Pro or P, is one of the twenty DNA-encoded amino acids. The systematic name for Proline is Pyrrolidine-2-carboxylic acid. It is not considered an essential amino acid; therefore the human body can synthesize it. Proline is the only cyclic amino acid and is biosynthetically derived from the amino acid,  L-glutamate and its immediate precursor is the imino acid (S)-1-pyrroline-5-carboxylate (P5C).¹ An imino acid is any molecule that contains both imino (>C=NH) and carboxyl (-C(=O)-OH) functional groups.² Proline is used a systemmatic catalyst in organic reactions. L-Proline is an osmoprotectant and therefore is used in many pharmaceutical, biotechnological applications.¹ Proline is thermophilic and can withstand an environment with extreme osmotic stress, thus why proline is widely used in pharmaceuticals. In studying Immunology, proline plays a significant role. Proline-rich polypeptides (PRP), such as the proline-rich polypeptide in ovine colostrum has an effect on skin permeability and immune response.³

When researching this amino acid, I found it interesting that, “proline is the only amino acid that does not form a blue/purple colour when developed by spraying with ninhydrin for uses in chromatography. Proline, instead, produces an orange/yellow colour.”¹ In our Organic I and II labortories, we use chromatography. The pK value for the amide is 11.0 and the pK value for the carboxyl group is 2.0.⁴

1. http://en.wikipedia.org/wiki/Proline

2. http://en.wikipedia.org/wiki/Osmoprotectant

3.  http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1457668/
4. http://www.biology.arizona.edu/biochemistry/problem_sets/aa/proline.html

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

http://0-www.lexisnexis.com.library.acaweb.org/lnacui2api/results/docview/docview.do?start=2&sort=RELEVANCE&format=GNBFI&risb=21_T11400955493

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.