Christina White's Awesome Webpage

On the White Lab webpage, multiple publicatins are posted for educatio and research purposes. I could not upload the picture for the reaction I chose to put on my blog but it can be  found at the following website:
http://www.scs.illinois.edu/white/index.php?p=publications
It is under publication #20. This reaction is an example of mixed hydroxylase/desaturase activity with a small molecule for unactivated, aliphatic C-H bonds outside the realm of enzymatic catalysis. The question is what causes the switch in Fe(PDP) from pure hydroxylase to mixed hydroxylase/desaturase activity reported in the presence of remote carboxylic acid moieties on the substrate. One hypothesis made by the White group is that it is the orientation of the iron oxidant [LnFe(OH)] relative to a key carbon centered radical intermediate on the substrate. In fact, in nature's oxidation enzymes, this radical intermediate is the gatekeeper from which all kinds of fascinating reactivities can be obtained such as hydroxylations, chlorinations, desaturations, rearrangements. Carboxylic acids, known to bind to Fe(PDP) and direct the site of substrate oxidation, may play a role in altering oxidant/substrate radical orientation. Because Fe(PDP) is stereoretentive (it doesn't scramble stereocenters during oxidation) the validity of this radical intermediate was up for debate. Prior to this work, many thought that this iron catalyst operated via direct, concerted C-H insertion (like the organic oxidant TFDO). If a concerted rather than a radical mechanism was operating, this, and other biomimetic reaction pathways, would not be accessible with Fe(PDP). To address this issue, a novel taxane-based radical trap that rearranges to nortaxane was used to provide the first direct evidence for a substrate radical intermediate. Interestingly, numerous biologically active nortaxanes have been isolated; however, there biogenic origins are controversial. This result suggests that radical-induced rearrangement of the taxane skeleton to nortaxane may be happening in nature during the oxidase phase of Taxol biosynthesis. Oxidation of Taxol in nature is performed by iron hydroxylation enzymes.

Potential Exam Question

How does the Claisen reaction differ from the aldol reaction?

The Claisen condensation differs from the aldol reaction in several important ways.

(i) The aldol reaction may be catalyzed by acid or base, but most Claisen condensations require base.

(ii) In contrast to the catalytic base used for aldol reactions, a full equivalent of base (or more) must be used for the Claisen condensation. The extra base is needed because beta-ketoesters having acidic hydrogens at the alpha-carbon are stronger acids (by about 5 powers of ten) than the alcohol co-product. Consequently, the alkoxide base released after carbon-carbon bond formation (upper right structure in the mechanism diagram) immediately removes an alpha proton from the beta-ketoester product. As noted above, formation of this doubly-stabilized enolate anion provides a thermodynamic driving force for the condensation.

(iii) The aldol reaction may be catalyzed by hydroxide ion, but the Claisen condensation requires that alkoxide bases be used, in order to avoid ester hydrolysis. The specific alkoxide base used should match the alcohol component of the ester to avoid ester exchange reactions. Very strong bases such as LDA may also be used in this reaction.

(iv) The stabilized enolate product must be neutralized by aqueous acid in order to obtain the beta-ketoester product.

References:

1.       Smith, J.; Organic Chemistry; Second Edition; McGraw Hill Co.; Pages 929-930

Seminar Summary-Extra Credit for Lab



Figure 1: 4-aminobiphenyl structure. 4-aminobiphenyl is found in
tobacco smoke and hair dye, this compound is a known bladder carcinogen.

Dr. Steven Myers visited Campbellsville University on April 15, 2011, and presented a seminar concerning tobacco smoking during pregnancy. The risks of a pregnant woman smoking are miscarriage, stillborn, premature, low birth weight, Sudden Infant Death Syndrome, and the baby is deprived food and oxygen. Secondhand smoke can cause children less than five years of age to have asthma, bronchitis, pneumonia, ear infections, and croup. Tobacco is composed of over 4,000 chemicals that are known to cause cancer and diseases. The diseases range from cardiovascular to cerebrovascular to pediatric. Tobacco smoking during pregnancy can also cause placental abruption which is when the placenta separates from the wall of the uterus and the baby is denied all oxygen.

During the first two weeks of pregnancy, smoking side effects are not noticed. During weeks 3-8 of pregnancy, all of the fetal organs are developing and the effects of smoking on the baby are harsh. The baby receives less nutrition, and may be premature (reason for premature births is unknown but can be linked to smoking).  Babies do not have the protection to rid chemicals from smoking until two years after life.

A biomarker is molecular, biochemical or cellular alterations that are measurable in biological media, such as human tissue cells or fluid. During the research of effects of smoking during pregnancy, Dr. Steven Myers and his team used amniotic fluid as a biomarker. Amniotic fluid is fetal urine and can be processed to evaluate what chemicals the baby is being exposed to.

Synthetic Application of HVZ Reaction

Figure 1(above): Synthetic Application of the Hell-Volhard-Zelinsky reaction.¹



Figure2: Reaction Mechanism of the HVZ reaction.¹



The preparation of a-bromo thioesters from carboxylic acids based on the Hell-Volhard-Zelinsky reaction was developed by H.J. Liu.¹ The procedure for the reaction in Figure 1 can be found at the website referenced below. The mechanism found in Figure 2 shows the movement of electrons to form the a-halo acyl halide. In this mechanism the a-halo acyl halide undergoes one further step, hydrolysis. However, in Figure 1, the a-bromo thioester is produced before the hydrolysis step.

The Hell-Volhard-Zelinsky halogenation reaction halogenates carboxylic acids at the a carbon. 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 the Br₂ to brominate a second time at the α position.²

More information about the Hell-Volhard-Zelinsky reaction can be found at the following website:

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

References:

1.       Kurti, L; Czako, B; Strategic Applications of Named Reactions in Organic Synthesis; Elsevier’s Inc.; [online-Google Books]; Page 200-201;  http://books.google.com/books?id=mjpJmiZ9OZ8C&pg=PA200&lpg=PA200&dq=hell+volhard+zelinsky+reaction+procedure&source=bl&ots=8lgjm8WLP0&sig=ZgZWbtmXu3xww_OaUUE9bgfCuzY&hl=en&ei=yU6uTc3qJoiU0QH87cG6Cw&sa=X&oi=book_result&ct=result&resnum=3&ved=0CCAQ6AEwAg#v=onepage&q=hell%20volhard%20zelinsky%20reaction%20procedure&f=false; Accessed April 19, 2011

2.       Hell-Volhard-Zelinsky halogenation; [online] http://en.wikipedia.org/wiki/Hell-Volhard-Zelinsky_halogenation; Accessed April 19, 2011

Methyl Anthranilate

Figure 1: Methyl Anthranilate structure.¹

Methyl anthranilate, also known as MA, methyl 2-aminobenzoate, or carbomethoxyaniline, is an ester of anthranilic acid. The chemical formula for this compound is C₈H₉NO₂. Methyl anthranilate is a clear to pale yellow liquid with a melting point of 24°C and a boiling point of 256°. It shows a light blue fluorescence. It is very slightly soluble in water, and soluble in ethanol and propylene glycol.¹ Methyl anthranilate acts as a bird repellant. It is food-grade and can be used to protect corn, sunflowers, rice, fruit, and even golf courses. It is also used for the flavoring of grape Kool-Aid. Methyl anthranilate naturally occurs in Concord grapes, lemon, strawberry, wisteria, and oranges. It is used for flavoring of soft drinks, candy, gum, and drugs.²

An ester is the product of a reaction between an alcohol and an acid, usually an organic acid.³ Methyl anthranilate is derived from the reaction of methanol and anthranilic acid. Sulfuric acid is used as a catalyst during this reaction. Amides are carboxylic acids derivatives, and can be formed by the conversion of the carboxylic acid into an ammonium salt, which then produces an amide upon heating. Anthranilic acid is a carboxylic acid, which forms the amide anthranilamide, when placed into the previous conditions.⁴

References:

1.      Methyl Anthranilate; CAS No. 134-20-3; ScienceLab.com; Houston TX; 2008;  http://www.sciencelab.com/msds.php?msdsId=9924665; Accessed April 10, 2011

2.      Umeda, K; Sullivan, L; Evaluation of Methyl Anthranilate for Use as a Bird Repellant in Selected Crops; 2000; http://ag.arizona.edu/pubs/crops/az1252/az1252-1a.pdf; Accessed April 10, 2011

3.      Fromm, J; Introduction to Esters; 1998; http://www.3rd1000.com/chem301/chem301v.htm; Accessed April 10, 2011

4.      Making Amides; Jim Clark; 2004; http://www.chemguide.co.uk/organicprops/amides/preparation.html; Accessed April 10, 2011

Synthesis of 6,9,12,15,18-PENTAMETHYL-1,6,9,12,15,18-HEXAHYDRO(C60-Ih)[5,6]FULLERENE

Figure 1: The synthesis of 6,9,12,15,18-PENTAMETHYL-1,6,9,12,15,18-HEXAHYDRO(C₆₀-I_h)[5,6]FULLERENE.¹

A Grignard reagent has a formula RMgX, where X is a halogen, and R is an alkyl or aryl (based on a benzene ring) group.² Grignard reagents are made by adding the halogenoalkane to small bits of Magnesium in a flask containing ethoxyethane or commonly known as ether. Grignard reagents react with water to produce alkanes. When preparing Grignard reagents everything must be dry.³

The synthesis found in Figure 1 caught my eye when searching for a synthesis reaction containing a Grignard reagent in one step. The Grignard reagent in this synthesis is CH₃MgBr. Since carbon is considerably more electronegative than magnesium, the metal-carbon bond in this compound has a significant amount of ionic character. Grignard reagents are best thought of as hybrids of ionic and covalent Lewis structures.⁴ The new bonds formed in this reaction are carbon-mendelevium bonds, as seen in Figure 1.

References:

1.    Matsuo, Yutaka; Organic Syntheses; 2006; [online] http://orgsyn.org/orgsyn/default.asp?formgroup=basenpe_form_group&dataaction=db&dbname=orgsyn

2.    Smith, J.; Organic Chemistry; 2nd ed.; McGraw-Hill; 2008; p. 739

3.    Clark, Jim; Grignard Reagents; 2003; [online] http://www.chemguide.co.uk/organicprops/haloalkanes/grignard.html

4.    Grignard Reagents; [online] http://chemed.chem.purdue.edu/genchem/topicreview/bp/2organic/grignard.html

Glutamine

Figure 1: The Glutamine Amino Acid¹



Glutamine has been studied extensively over the past 10-15 years and has been shown to be useful in treatment of serious illness, injury, trauma, burns, and wound healing for postoperative patients.¹ Glutamine is also marketed as a supplement used for muscle growth in weightlifting and bodybuilding.² Glutamine is the most abundant naturally occurring, non-essential amino acid in the human body and one of the few amino acids that directly cross the blood-brain barrier. In the body, it is found circulating in the blood as well as stored in the skeletal muscles.¹

Dietary sources of Glutamine include beef, chicken, fish, eggs, milk, wheat, dairy products, cabbage, beets, beans, spinach, vegetable juices, and fermented foods (tofu). ² Glutamine is synthesized by the enzyme glutamine synthetase from glutamate and ammonia. The most relevant glutamine-producing tissue is the muscle mass, accounting for about 90% of all glutamine synthesized.³

The alpha-carboxylic acid functional group of glutamine has a pK_a value of 2.13 and the alpha-amino group has a pK_a value of 9.13. Glutamine also contains a side chain, H₂NC=OCH₂CH₂.⁴ Glutamine, symbolized as Gln or Q, has a molecular formula of C₅H₁₀N₂O₃. The molecular weight is 146.15 g/mol, and the isoelectric point is at 5.65 pH.⁵

Glucagon is a polypeptide containing Glutamine within the compound. The body building formula sold to increase muscle size is usually made up of Glutamine peptides. The ideal product to have for body building would contain Glutamine peptides and glutamine in the free form. Glutamine peptides are more stable and better assimilated by the body.²

Figure 2: Glucagon (plypeptide containing Glutamine)⁶

References:

1.      Lefers, Mark; Glutamine; 2004; [Web] http://groups.molbiosci.northwestern.edu/holmgren/Glossary/Definitions/Def-G/Glutamine.html; Accessed March 23, 2011

2.      BodyBuilding.com; Glutamine; [Web] http://www.bodybuilding.com/store/glutamine.html; Accessed March 23, 2011

3.      Glutamine; 2011; [Web] http://en.wikipedia.org/wiki/Glutamine; Accessed March 23, 2011

4.      Parrill, Abby; Amino Acid Structures; 1997 [Web] http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html; Accessed March 23, 2011

5.      Burkhard, Kirste; Glutamine; 1998; [Web] http://www.chemie.fu-berlin.de/chemistry/bio/aminoacid/glutamin_en.html; Accessed March 23, 2010

6.      Glucagon; 2008; [Web] http://www.lookchem.com/cas-107/107444-51-9.html; Accessed March 23, 2011

Peer-Reviewed Journal with EAS reaction

Resol and novolac are two types of phenolic resins; they are widely used in industry because of their chemical resistance, electrical insulation, and dimensional  stability. Phenolic resin-bonded textile felts can be considered to be fiber-reinforced plastic with high levels of fibers; these fibers are derived from textile scraps recycled from the textile industry. Due to the increase of phenol cost, researchers have been working to partially substitute this monomer by natural polymers that present similiar structure close to that of phenolic resin.

The best possible substitutes of phenol based on cost and availabilty are lignosulfonates, which are obtained from the sulphite process. The ammonium lignosulfonate is the best canidate to substitute phenol, because final properties of phenolic resins are better. Lignosulfonates are not very reactive with phenol and formaldehyde, so it is usual to modify their structure by methyloletion and phenolation. Although phenolation is more expensive.

The reaction mechnism associated with the formulation of lignin-novolac is explained in the following text. The alpha-carbon of the lateral chain of the phenyl-propane units is occupied by the sulfonate group. Lignosulfonates may be reacting by itself or with the phenol adding to the Beta-carbon of the lateral chain. The first step consists of the condensation between lignin fragments and the phenol present with formaldehyde. In the second step, vacuum distillation is employed to adjust the content of free phenol in accordance with the required specifications for textile felt applications. The addition of a curing agent (HMTA) with methylene groups is also necessary for the crosslinking. (1)

On page 3/14, of the following link is Figure 2: Mechanism of lignin-novolac resins synthesis substituted with lignosulfonates. Since this peer-reviewed journal is protected, I could not copy the reaction scheme onto my blog. My apologies.

http://www.ncsu.edu/bioresources/BioRes_02/BioRes_02_2_270_283_Perez_RAOE_Novolac_Resin_Lignosulfonate.pdf

References:
1. Characterization of a Novolac; Jaun Manuel Perez; Bioresources; http://www.ncsu.edu/bioresources/BioRes_02/BioRes_02_2_270_283_Perez_RAOE_Novolac_Resin_Lignosulfonate.pdf; Web; (2007)

Aromaticity

For a compound to be considered aromatic, it must satisfy the following four rules:

1.       The molecule must be cyclic.

2.       The molecule must be planar.

3.       The molecule must be completely conjugated.

4.       The molecule must satisfy Huckel’s rule.

Determining if a compound is cyclic is rather simple. If the compound is a ring and all bonds are connected to form this ring, then the compound satisfies the first rule.

 A planar molecule has all atoms in one plane; basically the molecule is “flat”.

Conjugation means the compound must have alternating double bonds. Benzene is considered completely conjugated because each pi orbital (double bond) overlaps and is separated by a sigma bond (single bond). However, 1,3-cyclohexadiene is not conjugated because there are three sigma bonds (single bonds) with no pi orbital (double bond).

Benzene                                                                                   

1,3-cyclohexadiene

Huckel’s rule applies to the number of pi electrons in the compound.  The equation used to find the acceptable number of pi electrons is 4n+2, (n=1, 2, 3, etc.). The acceptable numbers of pi electrons are 2, 6, 10, 14, 18, etc. The trend is adding 4 to each number. A pi electron is usually shown in the compound by a dot around the molecule with the electrons, or by a double or triple bond. Each double bond has two pi electrons. For example, the benzene molecule above has 3 double bonds, which means it has 6 pi electrons.

Let’s look at a few examples of compounds and find out if they are considered aromatic.

1.

Let’s begin with rule number one; the molecule must be cyclic. This molecule is considered cyclic, it is a ring and all bonds are connected to form that ring.
Rule 2: The molecule must be planar. This compound is not flat due to the carbons that are sp³ hybridized. The two hydrogen atoms connected to the carbon are in two different planes.
Rule 3: The molecule is not completely conjugated, if there were one more double bond, the molecule would satisfy this rule.
Rule 4: The molecule has 3 double bonds which means it has 6 pi electrons.
This molecule is NOT considered aromatic because it does not satisfy rule 2 or 3.

2.  

This molecule is commonly known as Pyrrole. Let’s begin with rule 1: the molecule must be cyclic. This molecule is considered cyclic; all bonds are connected to form a ring.
Rule 2: The molecule must be planar. All carbon atoms are sp2 hybridized and the hydrogen on the nitrogen is in the same plane as all the other atoms. This molecule is considered planar.
Rule 3: The molecule must be conjugated. This molecule has alternating double bonds, each pi orbital overlaps, so this molecule is considered to be conjugated.
Rule 4: Huckel’s Rule; there are 2 double bonds which equals 4 pi electrons, but there are also two pi electrons on nitrogen. Therefore there are six pi electrons, satisfying Huckel’s rule.
This molecule IS considered to be aromatic.

Exam 1

I expected to see a question like the following example on the exam:
A 1H NMR signal is observed at 157.0-Hz (downfeild of TMS) on a 135-MHz instrument.
a.) What is the observed chemical shift in ppm? 1.16 ppm
The chemical shift is found by dividing the downfield signal (Hz) by the spectrometer frequency (MHz). (157.0-Hz/135-MHz=1.16 ppm)
b.)What is the chemical shift in ppm, if the sample is analyzed with a 300.0 MHz instrument? 1.16 ppm
Chemical shift is a ratio, the value in hertz increases proportionally to the value in megahertz. The shift observed at 135-MHz will be 1.16 ppm at any other operating frequency.
c.)Where will the proton signal appear, in hertz, if the sample is analyzed with a 300.0-MHz instrument? 349 Hz
The downfield signal (Hz) is the chemical shift (ppm) multiplied by the spectrometer frequency (MHz). (1.16 ppm*300-MHz=300-Hz)

This type of problem was assigned for the Sapling Homework, and we discussed this topic in class. Dr. Mullins even spent time with an example, although there was no question even similar to this on the exam.

I felt mislead after taking the exam. In the Monday-Wednesday lab section, we did an example where we had to draw an IR spectrum for a compound provided by Dr. Mullins. I specifically remember asking Dr. Mullins aloud, if we were going to be required to draw an IR or MS spectrum on the test and his answer to my question was NO. There were multiple problems on the exam where we had to draw the spectrum for a compound. I am not implying whatsoever that we should know specifics about our exam prior to the test date. I am however saying that when ANY teacher says that something will not be on the exam, they should hold to their word. Because I know I am not alone when I say, I don't even bother to study material that is not going to be on the exam.

Muddiest Point to Date

I have read Chapter 13, and I am confident that I understand the material from this chapter. I have yet to read chapter 14 over NMR spectra thus, NMR spectra is my muddiest point to date. I plan to read the chapter and lecture on Friday will be over this subject. Hopefully this will help me to better understand Nuclear Magnetic Resonance. The homework assigned in lab found on WebSpectra should give me some practice on reading an NMR spectrum. The necessary resources to understand Nuclear Magnetic Resonance are chapter 14, WebSpectra, and lecture on Friday. I plan to take complete advantage of the previous resources to understand Nuclear Magnetic Resonance.