Final Blog: Publication from Christina White's Webpage

For our final blog assignment, we were asked to find a publication from Christina White's webpage and comment on the reaction and information we have learned throughtout the semester.  After looking through the website I found a project where the Heck Reaction was explored. We learned about this recently in Chapter 26.  During this reaction new C-C bonds are formed. C-C bond forming reactions are vital to the synthesis of natural products and pharmaceuticals. 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 alkenes.  This report includes an intermolecular, oxidative Heck reaction catalyzed by a versatile Pd(II)/sulfoxide catalyst.  A palldium catalyst is always used in these types of reactions.  In general, a substitution reaction in which one H atom of the alkene starting material is replaced by the R' group of the vinyl or aryl halide. Several starting materials like homoallylic carbonyl compounds, bis-homoallylic carbonyl componds, alcohols, and thiols were used to synthesize a compound with a new C-C bond.  Most of the details in this publication were hard for me to understand because they were things that we had not covered.  The Heck Reaction was explored in greater detail than we had discussed in class. The material that I did understand consisted of alkenes, inductive effects, regioselectivity, substitution reactions, oxidation, and employing methods of synthesis.  All of these things we have learned about throughout both semesters of Organic Chemistry. Follow the link below to check it out!

Sources:

1. "A General and Highly Selective Chelate-Controlled Intermolecular Oxidative Heck Reaction." J.H. Delcamp; A.P. Brucks; and M.C. White JACS, 2008. http://www.scs.illinois.edu/white/pubs/pub11.pdf. (accessed May 5 2011).
2. Smith, J.G; “Organic Chemistry;” Good-Hodge, T., Nemmers, D,. Eds.; McGraw-Hill: New York, 2008. Edition 2, pp 1009.

Potential Test Question

A potential question I see that might be on our next test involves the Robinson Annulation.  The question asks: Draw the product of the Robinson Annulation from the given starting materials using -OH in H₂O.  This question appears in the problems at the end of Chapter 24.  See page 953, problem number 24.41, part D.

The starting materials are:

The product of this reaction will form a six-membered ring and three new carbon-carbon bonds.  First an enolate will form from the first starting material.  Then nucleophilic attack of the beta carbon and protonation occurs.  Next, an intramolecular aldol reaction will take place to form a beta-hydroxy ketone. Lastly, the material will be dehyrated to form the final alpha,beta-unsaturated ketone as follows.

An easy way to draw the product of this type of reaction is to place the alpha carbon of the compound that becomes the enolate next to the beta carbon of the alpha,beta-unsaturated carbonyl compound.  Then, join the appropriate carbons together so that a new six-membered ring forms. To better understand the Robinson Annulation reaction, see section 24.9 starting on page 936 of our textbook.  I think that a problem like this might about on our next exam because it incorporates topics that we have previously discussed and other reactions we have learned about in chapter 24.

Seminar Extra Credit Summary (for Lab)

Steven R. Meyers, Ph.D Assistant Professor,

Dept. of Pharmacology and Toxicology,

University of Louisville School of Medicine

Tobacco Smoking During Pregnancy – Biomarkers of Exposure & the

Relationship to Genetics

In the state of Kentucky, 30% of mothers smoke. This causes the baby to get less food and oxygen.  As a result, there are higher rates of miscarriage, still births, premature births, low birth weights, placenta previa, low IQ levels, ADHD, and infants dying of SIDS (Sudden Infant Death Syndrome).  If the child is exposed to second hand smoke during the first two years of life they may get ear infections, asthma or bronchitis, and pneumonia.

We know that tobacco smoke causes cancer, respiratory disease, cardiovascular disease, and cerebrovascular disease in adults. Children who are exposed to tobacco smoke have a 60% increase of having a lower respiratory disease.  There is also an increase risk of developing asthma, inner ear and other ear infections causing long-term hearing problems, and an increase risk of dying of SIDS.  There is no safe time for a mother to be exposed to tobacco smoke. Although there are no adverse effects during the first two weeks, it causes a difficulty attaching to the ovum.  During the third and eighth weeks embryonic development of organs and tissues can be damaged. The most common effect is cleft lip and cleft palate due to exposure to chemicals like the ones in tobacco smoke.  During the last stages of pregnancy functional defects and problems during birth can occur.

Biomarkers are needed because there are approximately 4000 chemicals in tobacco smoke. This is the leading cause of avoidable cancer death. Biomarkers are molecular, biochemical, or cellular alterations that are measurable in biological media such as human tissue, cells, or fluids.  The earlier the biomarker is detected, the more likely the disease will be stopped.  They can be found in metabolites in the body, DNA adducts, and gene mutations.  Biomarkers are selected by determining if it is related to the disease, if it appeared at a defined stage of disease, if it can be obtain non-invasively, if it can be modulated by eliminating exposure, if it provides specificity, sensitivity, and practicality, and if it is stable and easy to measure.

Amniotic fluid can be a biomarker during the first trimester because it protects the fetus. It is filled with urine from the fetus and can tell what the fetus has been exposed to.  During a study where women had an amniocentesis, 1-Hydroxypyrene and Benzo(a)pyrenes were detected and the levels increased as the amount of cigarettes smoked increased.  Hemoglobin can also be a biomarker of carcinogens and tobacco smoke exposure.

Since amino acids undergo nucleophilic substitution with electrophiles in tobacco chemicals, looking at the enzymes that are responsible for getting rid of chemicals in the body is a good place to find biomarkers.  If a person is null genotype they do not have the genes to make these enzymes and are at an increased risk of cancers.  These people often have much more benzo(a)pyrene in the body.  The enzyme N-acetyl transferases activate and deactivate aromatic and heterocyclic amines in the body. If a person does not have this enzyme, they are also at an increased risk.  Fetuses do not develop these enzymes until two years of age.

4-Aminobiphenyl is an aromatic amine that is a strong bladder carcinogen prominent in tobacco smoke.  It is one of the most carcinogenic chemicals in the smoke that we know of.  During a study, it was found that nonsmokers had a very small amount in their system, passively exposed mothers and babies had levels of this chemical, and in smokers the levels increased as the amount of cigarettes smoked increased. 

After all of these studies, it can be seen that there is a correlation of the amount a mother smokes affecting her baby.  It is also observed that the more chemical compounds found in babies, the less they weigh.  The final study showed that there was no difference in African Americans to Caucasians but the amount if Hispanics was much lower in mothers and in the babies.  The birth weights of Hispanic babies were also greater than African Americans and Caucasians.  Overall, this presentation showed that mothers should not smoke during of after pregnancy because the effects are very dangerous and harmful to the baby.

Hell-Volhard-Zelinsky Halogenation

The Hell-Volhard-Zelinsky halogenation reaction halogenates carboxlyic acids at the alpha carbon.  Treatment with bromine and a catalytic amount of phosphorus leads to the selective α-bromination of carboxylic acids. PBr₃ replaces the carboxylic OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol, which will readily react with bromine to brominate a second time at the α position. 

Now that I have introduced what the HVZ halogenation reaction is, let's examine a specific example. This reaction was used in the invention of Cycloalkane Carboxaldoxime Carbamates, which are used as pesticides. The compounds can be prepared via the alpha halogenation of cyclic carboxylic acids under slightly modified Hell-Volhard-Zelinsky conditions (Step I) followed by esterification to give the cyclic alpha halocycloalkanecarboxylic acid ester. The active halogen is easily displaced with sodium alkyl mercaptide in alcohol to yield the alkylthio ester. Hydrolysis of the ester followed by reacting the hydrolyzed product with thionyl chloride gives the reactive acid chloride which is transformed into the aldehyde. Further reaction of the aldehyde with hydroxylamine hydrochloride in base gives the corresponding oxime which in turn can be carbamoylated with suitable reagents to form the final insecticidal agent.

Sources:

http://www.google.com/patentshl=en&lr=&vid=USPAT3721711&id=9yMuAAAAEBAJ&oi=fnd&dq=hell+volhard+zelinsky+halogenation&printsec=abstract#

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

Ester: Gamma-Valerolactone

Gamma-Valerolactone

(http://toxicopoeia.com/chemistry/molecules/GVL-sm.png)

Gamma-valerolactone is one of the more common lactones. This compound is a colorless liquid and is used in the perfume and flavor industries because of it's herbal odor.¹ It is also a constituent of crude pyroligneous acid, in dye baths (coupling agent), brake fluid, cutting oils and as a solvent for adhesives, insecticides and lacquers.² Recently, it has been said that gamma-valerolactone is a promising "green" liquid fuel. This compound is derived from furfural:

 (http://www.alanwood.net/pesticides/structures/furfural.gif)


Basic Information
Melting Point: -31°C
Boiling Point: 207-208°C
Water Solubility: Miscible
Density: 1.03 g/mL
Irritant to eyes, respiratory system, and skin


I had a difficult time finding a reaction where gamma-valerolactone was converted to another carboxylic acid derivative. I did find a reaction where gamma-butyrolactone, which is very similar, was converted to alpha-tetralone. I found this on the organic synthesis website provided by Dr. Mullins. In the notes it states that replacing the butyrolactone with 120 grams of gamma-valerolactone in an otherwise identical procedure yields 150–160 grams (79–84%) of 4-methyl-1-tetralone.³ Here is the link so you can see what I'm talking about:
http://orgsyn.org/orgsyn/default.aspformgroup=basenpe_form_group&dataaction=db&dbname=orgsyn

Thanks for reading!


Sources:
1. http://en.wikipedia.org/wiki/Gamma-Valerolactone

2. http://www.erowid.org/chemicals/ghv/ghv_info2.shtml
3.http://orgsyn.org/orgsyn/default.aspformgroup=basenpe_form_group&dataaction=db&dbname=orgsyn

Grignard Reagent Synthesis

The following is a synthesis that involves a Grignard reagent during part of the reaction to yield the final product of  Tris(2-PerFluorohexylethyl)tin hydride:

The Grignard reagent is used in step B of the sythesis to yield Tris(2-perfluorohexyl ethyl)phenyl tin.

This compound is made by frist dissolving Phenyltin trichlorid in dry benzene in a 100-mL, round-bottomed flask under argon at room temperature. The solution is slowly added to a 500-mL, three-necked flask containing the Grignard reagent at room temperature over 1 hour, while stirring.. The reaction mixture is heated at reflux overnight in an oil bath at 50°C, removed from the bath, and allowed to stand at ambient temperature for 4.5 hours, with stirring. The reaction mixture is diluted with ether, vacuum filtered into an Erlenmeyer flask, and hydrolyzed with saturated ammonium chloride solution. Excess magnesium solid is also hydrolyzed with saturated ammonium chloride separately. The mixture is transferred to a separatory funnel. The water layer is removed, and the organic layer is washed three times with 3% sodium thiosulfate. Theorganic layer is dried over magnesium sulfate and filtered under vacuum. The solvent is evaporated to dryness under reduced pressure using a rotovap. The impure product is redissolved in ether and transferred to a 50-mL pear-shaped flask. The ether is removed under reduced pressure. Kugelrohr distillation is peformed to remove a dimer impurity of (C6F13CH2CH2CH2CH2C6F13) at 0.02 mm, 100-120°C for 5 hours. The residue is further purified by column filtration over silica under pressure with hexane. The solvent is evaporated under reduced pressure to leave 17.2 g of pure compound as a colorless oil.

The spectral properties of product of product 1 are as follows: 1H NMR (CDCl3) δ:1.31 [t, 6 H, J = 8.3, 2J(119Sn-H) = 53.4], 2.31 (m, 6 H), 7.41 (s, 5 H) ; 119Sn NMR (CDCl3) − 11.7 ppm; IR (thin film) cm−1: 3100, 2950, 1238, 1190, 1144, 655 ; MS (m/z) 1161 (M+ - Ph), 891 (M+ -CH2CH2C6F13).

The spectral properties of product 3 are as follows: 1H NMR (CDCl3) δ: 1.16 [t, 6 H, J = 8.1, 2J(119Sn-H) = 53.4], 2.35 (m, 6 H), 5.27 (s, 1 H) ; 119Sn NMR (CDCl3) − 84.5 (1J(119Sn-H) = 1835) ;−1: IR (thin film) cm 1842, 1197 ; MS (m/z)1161 (M+ - H), 813 (M+ - CH2CH2C6F13).

Thin layer chromatography was performed using silica plates and eluting with

hexane. Potassium permanganate was used to visualize the spots. The Rf values for products 1 and 3 were 0.38 and 0.37, respectively.

Amino Acid: Threonine

Threonine

Threonine is an amino acid represented by Thr and T. It's molecular weight is 119 grams. It is a neutral and polar molecule. This amino acid has a hydroxyl group, a methyl group, an amino group, and a carboxylic acid group. It is one of two common amino acids that contains a chiral side chain. Threonine has two pKa values: 2.63 and 9.10 for the carboxylic acid and amino group. The isoelectric point for this amino acid is 5.64.  There are three peaks that appear on the proton NMR spectrum of threonine. The first peak is a multiplet at 4.10 to 4.15 ppm. The second peak is a doublet at 3.5 ppm. The third peak is a doublet at 1.20 to 1.25 ppm.

A small polypeptide that contains threonine is N-acetylgalactosamine. N-acetylgalactosamine is needed for communication between cells. It also has a role in the disease processes of cancer, inflammation, and immunity in the human body. It is present in the colon, intestines, retinas, sweat glands, blood vessels, ducts of kidneys, hair follicles, testes, and skin. The hydroxy side chain can undergo O-linked glycosylation. O-glycosylation is an enzymatic process that attaches gyclans (polysaccharides) to proteins, lipids, or other organic molecules. This occurs in the Golgi apparatus of eukaryotic cells.  As an essential amino acid, threonine is not synthesized in humans. Therefore, we must ingest threonine in the form of threonine-containing proteins. Foods high in threonine include cottage cheese, poultry, fish, meat, lentils, and sesame seeds.

Sources:

http://en.wikipedia.org/wiki/Threonine

http://www.ncbi.nlm.nih.gov/pubmed/8254121
http://www.online-vitamins-guide.com/images/threonine.gif

Synthesis with Electrophilic Aromatic Substitution

I found a synthesis in an article called Synthesis, Electrophilic Substitution and Structure-Activity Relationship Studies of Polycyclic Aromatic Compounds Towards the Development of Anticancer Agents, which was published in the Current Medicinal Chemistry journal in 2001. The article is about polycyclic aromatic hydrocarbons (PAH) which are considered potentially carcinogenic. Substituted PAH derivatives may serve as anticancer agents, and as chemotherapeutics. This article presents a review of their use. Electrophlic substituation reactions are used during the synthesis of these new compounds. Here is a picture of one of the two of the steps which use electrophilic aromatic substitution:

This depicts a halogenation by replacing a hydrogen atom on the benzene ring with a bromine in each reactant. These compounds underwent bromination regioselectively in a single site. There was another step of this synthesis to make anitcancer agents from substituted  polycyclic aromatic hydrocarbons which involved an electrophilic aromatic substitution.

This step is a nitration of one of the benzene rings of the reactant. During the study, it was shown that benz[a]aceanthrylene (62) and indeno[1,2,3-cd]pyrene (63) produced single mononitro derivatives 64 and 65. The conformation of the nitro derivatives was important in determining the mutagenic activities. It is proposed that suitably substituted polycyclic aromatic compounds can inhibit cancer cellgrowth even though there was a concept that these are only carcinogenic in nature. The hope is to see potent antitumor drugs with novel mechanisms of action based on the chemistry described above.

Reference: Banik B, Becker F. Synthesis, electrophilic substitution and structure-activity relationship studies of polycyclic aromatic compounds towards the development of anticancer agents. Current Medicinal Chemistry [serial online]. October 2001;8(12):1513-1533. Available from: MEDLINE, Ipswich, MA. Accessed March 6, 2011.

What makes a compound aromatic?

Most people associate aromatic compounds with the word aroma. They think that all aromatic compounds smell nice instead of knowing the specific criteria that makes terms them as aromatic. To understand these characteristics properly, I will define them in terms that everyone can understand not just people with a background in chemistry.

First, the compound must be cyclic. But what does that mean?  To understand this, think of a ring.  A ring is a circle and for a compound to be aromatic it must be shaped like this.  This is the easiest way to comprehend the first rule.

Second, the molecule must be planar.  Think of a piece of cardboard with rubber balls attached to each side. All of the balls on the top side should be equivalent.  For example, the balls should all be the same color, shape, and size. The same goes for the underside. The piece of cardboard represents the molecule and the balls represent the orbital.  All of these aspects are on the same plane, which makes the molecule planar.

Third, the molecule should be completely conjugated.  What does it mean to be completely conjugate?  To understand this, think of a group of girls who are holding hands to form a circle.  This represents the molecule and each girl represents an atom.  Aromatic molecules must have a p orbital on every atom.  So think of every girl as having the same hat on, this could represent the p orbital. 

Fourth, the molecule must satisfy Huckel’s rule and contain a particular number of pi electrons.  An aromatic compound must contain 4n + 2 pi electrons.  This rule is the hardest to understand.  Pi refers to the p orbital as explained above. The electrons in pi bonds are referred to as pi electrons.  To understand this, relate the pi electrons that are needed to make a compound aromatic with the p orbital analogy in rule three.

If all four of these rules are satisfied then a compound can be termed as aromatic.  A compound can meet the first three rules but not have the appropriate number of pi electrons, therefore it is anti-aromatic.  These compounds have 4n pi electrons.  Those compounds without 4n or 4n +2 pi electrons or ones that disobey one of the first three rules are non-aromatic. After reading this, I hope you can understand better what makes a compound aromatic. Next time you smell something nice don’t assume it’s aromatic, determine whether it is based on this criteria.

Question Not Seen on Exam 1

I was expecting to see a question about determining an unknown compound from interpreting it's mass spectrum and IR spectrum on the exam. The example on page 484 of chapter 13 is a good representation of what I was expecting. The problem and solution is as follows:

Q: What information is obtained from the mass pectrum and IR spectrum of an unknown compound X? Assume X contains the elements C, H, and O.

Step 1: Use the molecular ion to determine possible molecular formulas. In this case, the molecular ion is 88. Use an exact mass to determine a molecular formula. Divide the mass by 12 (1 C atom) to determine the maximum amount of carbons. Replace 1 C by 12 H's for another possible molecular formula. Substitute 1 O for CH4 then repeat for a possible molecular formula with 2 O atoms. There will be three possible formulas: C5H12O, C4H8O2, and C3H4O3. The possible hydrocarbons are discounted because the compound X contains an O atom. If the molecular ion had an exact mass of 88.0580, the molecular formula of X is C4H8O2.

Step 2: Calculate the degrees of unsaturation. For a compound of molecular formula C4H8O2, the maximum number of H's = 2n + 2 = 2(4) + 2 = 10. Because the compound contains only 8 H's, it has 2 fewer H's than the maximum number. Since each degree of unsaturation removes 2 H's, X has one degree of unsaturation. This means that X has one ring or one pi bond.

Step 3: Determine what functional group is present from the IR spectrum. The two major absorptions in the IR spectrum above 1500 are due to sp3 hybridized C-H bonds (3000-2850) and a C-O double bond (1740). Thus, the one degree of unsaturation in X is due to the presence of the C-O double bond.

Mass spectrometry and IR spectrometry give valuable information when determining an unknown compound and that is why I expected a problem like this to be on the test. If the structure was wanted then NMR would be needed.

Chapter 13 Muddiest Point

My muddiest point from chapter 13 is interpreting infared spectrums. I understand no observable data can be retrieved from the fingerpoint region. My main misunderstanding is how carbonyl groups are located on the spectrum. I know that each absorption depends on the particular types but do not understand how rings and double bonds are identified. I found in chapter 21 that aldehydes and ketones give a strong peak at 1700 due to the C-O double bond. I also found that the sp3 hybridized C-H bond of an aldehyde shows one or two peaks at 2700-2830. These exact positions of the carbonyl absorptions provide additional information about a compound. For example, most aldehydes have a C-O double bond peak around 1730 and ketones peak around 1715. The carbonyl absorption of cyclic ketones shifts to a higher wavenumber as the size of the ring decreases and the ring strain increases. Weaker bonds absorb at a lower frequency on the spectrum. After reading this section of chapter 21, I understand more about how to interpret carbonyl groups on infared spectrums.