Christina White's Very Awesome Webpage

For the 10th and final blog of the semester and ever for organic chemistry, Dr. Mullins explained that we would need to check out Christina White's webpage and explain how the reaction deals with the schemes we have studied in class.  One publication that I found very interesting and quickly noticed similarities in the reaction was for the publication 19.  The reaction deals with reaction schemes listed in chapter 26 of the organic chemistry book.  The reaction scheme is shown as:

1

Check out this website for the overall reaction scheme(Fig. 1) because it would not upload:
http://www.scs.illinois.edu/white/pubs/pub19.pdf 1

In Christina White's publication is titled as "Synthesis of Complex Allyclic Esters via C-H Oxidation vs C-C Bond Formation."  This publication deals with the C-H bond oxidation method for the synthesis of allyclic esters.  This reaction scheme is a olefin metathesis which reacts with a grubbs catalyst to produce the final products.  The olefin metathesis occurs in the presence of a complex transition metal catalyst that contains a carbon-metal bond. 2 The first step in the olefin metathesis, two molecules of the starting alkene are arranged adjacent to each other. 2  Then, the double bonds are broken which forms two new double bonds using carbon atoms that were not previously bonded to each other in the starting alkenes. 2 The final two products show that the overall reaction and experiment was completed correctly.

Sources

1) http://www.scs.illinois.edu/white/pubs/pub19.pdf
2) Organic Chemistry. Smith 2nd edition

Michael Reaction

For blog assignment #9, Dr. Mullins explained that we would need to find and work out a problem that deals with anything within chapters 24 and 26.  After reading through both chapters I decided to pick a problem on something that I've had difficulty with trying to understand the concept.  So I decided to work out a problem that involves the Michael reaction from chapter 24.  A Michael reaction involves two carbonyl compounds: the enolate of one carbonyl compound and an alpha, beta-unsaturated carbonyl compound.  The Michael reaction is when the conjugate addition (1,4-addition) of a resonance-stabilized enolate to the beta carbon of an alpha, beta-unsaturated carbonyl system.  In this Michael reaction, the enolate acts as the nucleophile and attacks the beta carbon, which is the key step to this mechanism. The Michael reaction always forms a new carbon-carbon bond on the beta carbon of the Michael acceptor.  The question is below:

Draw the product formed from a Michael reaction with the given starting materials using ^-OEt, EtOH.

In step 1, the enolate is formed by a base, ^-OEt, removing an acidic proton between the two carbonyl groups.  The oxygen atom from the base attacks the hydrogen and the pair of electrons and a negative charge are given to the alpha carbon on the enolate.  This lone pair of electrons and negative charge on this carbon are eventually used as a nucleophile in step 2.  The following reaction depicts step 1 in this Michael reaction:

In step 2, the nucleophilic attack of the enolate to the beta carbon of the alpha, beta-unsaturated carbonyl compound takes place.   After this nucleophilic attack occurs, a new carbon-carbon bond is formed and a stabilized enolate is formed.  Also, after the nucleophilic attack of the enolate on the beta carbon occurs, the electron pair of the double bond and a negative charge are given to the alpha carbon on this compound.  The following reaction depicts step 2 in this reaction:

In step 3, the lone pair and negative charge of the alpha carbon on the compound protonate and attack the hydrogen atom from EtOH.  With this protonation if the enolate and a new carbon-hydrogen bond is formed.  The following step in this reaction depicts step 3:

The Michael reaction involves first the formation of the enolate then the nucleophilic attack at the beta carbon and then protonation.  With these steps being taken correctly the end product should end as such:

I would like to say that I think this was the best idea for a blog yet because we can find something in these chapters that we dont understand then research it and construct a problem over it.  The Michael reaction was a difficult reaction to understand, however after constructing a question and working it out I feel a lot more confident with this reaction.


Sources


1)  Organic Chemistry. Janice Gorzynski Smith. 2nd edition

Extra Credit Seminar

  On April 15, 2011 there was a seminar about tobacco smoking during pregnancy and the effects and results of smoking while pregnant.  The seminar was titled, "Biomarkers of exposure and relationship to genetics" and the speaker was Dr. Steven R. Myers.  Dr. Myers is the associate professor for the Department of Pharmacology and Toxicology at the University of Louisville School of Medicine.
  Dr. Myers started his seminar talking about smoking while pregnant and the things that can happen to the baby as a resultant of this situation.  The main problem with smoking while pregnant is that the baby gets less food and oxygen if a woman smokes while pregnant than if a woman doesn't smoke while pregnant.  He also explained the risks that come with smoking while being pregnant and here are a few listed:
   - Stillborn baby
   - Premature birth
   - Miscarriage
   - Sudden Infant Death Syndrome (SIDS)
   - Low birth-weight babies

  He then went into talking about tobacoo and the chemical make-up and the many types of diseases that occur from smoking.  Tobacco is composed of 4,000 chemicals and cause multiple types of cancers including lip, esophagus, pancreas, lung and kidney cancer.  Many respiratory diseases that can also occur are tuberculosis, pnuemonia, influenza, bronchitis and asthma.  Other health consequences that occur are:
   - a 60% increase in risk of developing lower respiratory illness
   - impaired lung function
   - increase of frequency of middle ear infections

  Dr. Myers then discussed the biomarker aspect to the tobacco and why someone would want to study it.  A biomarker is a molecular, biochemical, or cellular alterations that are measurable in biological media, such as human tissues, cells or fluids.  He then explained that the reasoning for someone studying biomarkers are to find the many chemicals, mutagens and carcinogens found in tobacco.  There are three criteria that must be met for a biomarker to be useful and they are:
   1. Specificity
        - direct result of chemical or contaminent
   2. Sensitivity
        - earliest detectable change
   3. Practicality
        - cheaper way to get answer

Dr. Myers was a very informative speaker and I really enjoyed the seminar and being able to speak with him after the seminar.  During the seminar he talked about a compound that is found in tobacco smoke and it is called 4-aminobiphenol.  The structure is below

Hell-Volhard-Zelinsky reaction

The Hell-Volhard-Zelinsky involves the halogenation of carboxylic acids at the alpha carbon of the carbonyl group.  The basic reaction reacts as such:

The first step in the mechanism of this reaction involves the substitution of a bromide on the hydroxyl(OH) group of the carboxylic acid.  The catalyst PBr3 causes this bromide to replace the OH group which causes the formation of the carboxylic acid bromide.  The present acyl bromide is then tautomerized to form an enol which reacts with the Br2 which creates the second placement of a bromide on the compound.  The following depicts the mechanism of this reaction:

An example of a synthesis where this Hell-Volhard-Zelinsky mechanism is used is in the formation of dimethylketene.  In this synthesis the starting reactant is isobutyric acid which then reacts with PBr2 to form the product of propanoyl bromide or a-bromoisobutyryl bromide.  This product is then reacted with zinc and some heat and results in the overall final product of dimethylketene.  The Hell-Volhard-Zelinsky mechanism is used in the first step of the reaction.  The steps in this reaction are:

The Hell-Volhard-Zelinsky reaction of a halogenation of carboxylic acids at the alpha carbon is named after three chemists.  Their names are Carl Magnus von Hell ( German chemist), Jacob Volhard (German chemist) and Nikolay Zelinsky who was a Russian chemist.  This reaction is being discussed in the Organic Chemistry 2 class at Campbellsville University.


Sources:
1.  "Hell-Volhard-Zelinsky Halogenation." Wikipedia, the Free Encyclopedia. Web. 21 Apr. 2011. http://en.wikipedia.org/wiki/Hell-Volhard-Zelinsky_halogenation.
2.  Organic Syntheses Website. Web. 21 Apr. 2011. http://orgsyn.org/orgsyn/default.asp?formgroup=basenpe_form_group.

Methyl trans-Cinnamate

Methyl trans-cinnamate is the methyl ester of cinnamic acid.  Methyl trans-cinnamate gives off the appearance of a white to light yellow fused crystalline mass with a strong, aromatic odor.  It is found naturally in a variety of plants and fruits, most specifically strawberries.  It is mostly used in the fragrant and flavor industries as an agent for perfumes and such.  The structure for methyl trans-cinnamate is:

The properties of Methyl trans-cinnamate:
   - Melting Point: 260-262°C
   - Boiling Point: 34-38°C
   - Water Solubility: Insoluble
   - Molecular Weight: 162.19 g/mol
   - Linear Formula: C6H5CH=CHCO2CH3
   - IUPAC name: Methyl (E)-3-Phenylprop-2-enoate

The carboxylic acid the ester is derived from is:

An example of one reaction where methyl trans-cinnamate is converted to another derivative:

Sources:

1. "Methyl Cinnamate." Chemical Book. Web. 9 Apr. 2011. http://www.chemicalbook.com/ChemicalProductProperty_EN_CB5206328.htm.
2.  Methyl Cinnamate." Wikipedia, the Free Encyclopedia. Web. 9 Apr. 2011. http://en.wikipedia.org/wiki/Methyl_cinnamate.
3.  "Cinnamic Acid | RM.com ®." Magick, Wicca, Paganism and Other Esoteric Knowledge | RM.com ®. Web. 9 Apr. 2011. http://www.realmagick.com/cinnamic-acid/.

Methyltriisopropoxytitanium

Methyltriisopropoxytitanium is the product of the reaction of titanium tetraisopropoxide in ether and of titanium tetrachloride.  Methyllithium in ether is then added to the cooled reactants to produce the final product, Methyltriiopropoxytitanium.  The IUPAC name for this final product is methyltitanium triisopropoxide.  The newly formed carbon-carbon bonds is the methyl group located on the titanium ion. 

The reaction scheme for this reaction is listed below for A. :

1

In a reaction involving an organolithium reagent the organometallic compound has a bond between a carbon and a lithium atom.  Since lithium is very electronegative, the charge of the overall bond is placed on the carbon atom which creates a carbanion.  For this reaction, A, the only step of the reaction involves the organolithium reagent acting on the titanium atom and forming the newly bonded Ti-C bond.  The first reactant for the reaction is an organotitanium compound. This product of Methyltriiospropoxytitanium is extremely flammable liquid and vapor and can cause severe eye and skin irritation.  The target organs for this compound are the respiratory system and eyes.  This overall reaction of Methyltriisopropoxytitanium involves an organolithium reagent which alters the overall product.  2

Source

1. "Organic Syntheses Prep." Organic Syntheses Website. Web. 03 Apr. 2011. http://orgsyn.org/orgsyn/prep.asp?prep=v81p0014.

2. MSDS Methyltitanium(IV) Triisopropoxide, 1M Solution in THF CAS 18006-13-8 MSDS Methyltris(isopropoxy)titanium(IV)." MSDS & Custom Synthesis Organic Synthesis Bio-Synthesis Suppliers. HBCChem. Web. 03 Apr. 2011. http://www.chemcas.org/drug/analytical/cas/18006-13-8.asp.

Aspartic Acid

  Aspartic acid is a non-essential and acidic amino acid.  Aspartic acid is abbreviated as Asp or D.  Aspartic acid was first recognized in 1868 when it was first isolated from legumin in plant seeds.  Aspartic acid is a non-essential amino acid for mammals which means that enough aspartic acid is synthesized by the body from oxaloacetic acid which is formed in metabolism of carbohydrates.  Aspartic acid is found in many different sources of food such as dairy, beef, poultry and sprouting seeds.  Here is a high quality representation of the structure of aspartic acid:

  Aspartic acid's molecular formula is C₄H₇NO₄ and has a molecular weight of 133.10 grams/mole. Aspartic acid is the one of only two amino acids that give a negatively charged carboxylic group on the side chain.  The functional groups present within aspartic acid are an amine and two carboxylic groups.  Aspartic acids are play a vital role as acids in enzyme active centers and also function in maintaining the solubility and ionic character of proteins.  This amino acid is one of the only two acidic amino acids.  There are three pKa values for aspartic acid which are: 1) alpha-carboxylic - 2.10;   2) alpha-amino -  9.82;  3) side chain  -  3.86.  The side chain is a carboxyl group located on the amino acid.  The isoelectric point for aspartic acid is 2.77 pH. 
 
Sources:
1.  Mayer, Michael. "Aspartic acid information page. All about aspartic acid and the role it plays in your diet." Zest for Life vitamins and supplements. 24 Jan. 2011. 24 Mar. 2011 http://www.anyvitamins.com/aspartic-acid-info.html.

2.  Kirste, Burkhard. "Aspartic Acid." Institut für Chemie und Biochemie an der FU Berlin. 23 Jan. 1998. 24 Mar. 2011 http://www.chemie.fu-berlin.de/chemistry/bio/aminoacid/asp_en.html.

3.  "Amino Acids - Aspartic Acid." The Biology Project. 24 Sept. 2003. 24 Mar. 2011 http://www.biology.arizona.edu/biochemistry/problem_sets/aa/aspartate.html.

4.  Parrill, Abby. "Amino Acid Structures." Michigan State University. 4 Feb. 1997. Department of Chemistry. 24 Mar. 2011 http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html.

Ibuprofen

  The fourth blog assignment was to find an article containing a synthesis of a common drug and to list any of the Electrophilic Aromatic Substitution reactions in the synthesis.  After thoroughly looking through many websites and journals I finally found an article that had a synthesis for a common drug and that drug is...IBUPROFEN!
  Ibuprofen is an over the counter drug which is used to relieve pain of symptoms of arthritis, fever and other occurrences for pain relief.  It was developed in the 1960s and was discovered by Steward Adams in 1961 and also patented in the same year.  Ibuprofen functions as an inhibitor for an enzyme that directs formation of several mediators of inflammation.
  There are five steps in the synthesis of ibuprofen.  The first step is the electrophilic aromatic substitution step where a Friedel-Crafts acylation occurs. The isobutyl group is a ortho and para director and a benzene activator.  However, the para isomer is formed because there is steric hindrance on the ortho isomer.  The next few steps involve the reduction of the ketone to an alcohol, then conversion of alcohol to a chloro group, then the chloro group is substituted by a cyano group.  The last step involves the hydrolysis of the cyano compound to achieve the final product of ibuprofen.  Listed below is the synthesis:

  The site were the article can be found is:

    http://www.pharmainstitute.in/archives.htm  on the May 2008 issue

Hope this is interesting and allows you to understand electrophilic aromatic substitution in a different way with an actual synthesis of a common everyday drug.

Granny's Rendition of Aromatic Compounds

   Aromaticity is a chemical property that compounds are listed as if they qualify for the four criteria that are listed such as it must be cyclic, planar, completely conjugated and have 4n + 2 pi electrons.All aromatic compounds are based off of the basic benzene ring which is a ring with six sides and six carbons and six hydrogens.  The first criterion is that the molecule must be cyclic which means that it must be shaped like a ring or in a circle to be an aromatic compound.  The second criterion is that the molecule must be planar which means that the compound needs to be flat like a piece of paper to be aromatic.  For example, if the structure of the compound is shaped as a tub with some parts higher than others then it is not planar. 
   The third criterion is that a molecule must be completely conjugated which means that these compounds must have a p orbital on every atom.  For example, if you have a 6-sided ring, you will need three double bonds with the double and single bond followed by another double bond and so on.  The last criterion is that a molecule must satisfy Huckel’s rule and contain a particular number of pi electrons.  The particular number of pi electrons must meet the 4n + 2 pi electron rule.  For instance, if you use something like pencils you can have 2, 6, 10, etc. because n equals 0, 1 and 2.  Hope this helps clear up how to distinguish aromatic compounds from nonaromatic ones.

Exam Uno Questions

After taking the first Organic Chemistry exam, I wondered why there was not a question that required us to label and count the number of ¹H NMR signals for different compounds.  After working hard and completing the Sapling homework I fully understood and knew I was able to answer these questions.  On Sapling, there were a few questions that covered this portion of Organic Chemistry.  When the exam was handed out I was confused on why there wasnt any of these choices of questions on there.  The correct way to answer these types of questions is to make sure to understand that the number of NMR signals equals the number of different types of protons in a compound.  The main principle of finding the number of signals is that protons in different environments give different NMR signals.  The same NMR signals are given by protons being equivalent to each other.  For example, a compound such as CH₃OCH₃ has only one ¹H NMR signal.  The reasoning for this is that each methyl group CH₃ is bonded to the same group which makes both methyl groups have the same signal.  However, for a compound such as CH₃CH₂Cl there are two different signals because the hydrogens are two carbons away from the chlorine for one signal and for the other signal the hydrogens are located one carbon away.  After working on the sapling problems and reading over this section in the book I was confident that I would be able to complete a question on the number of ¹H NMR signals.  However, since there was not a question on the exam containing this subject I was a little confused but this is still a great part of Organic Chemistry that needs to be understood.

Mass Spectrometry

   Hey guys, This is John Harbold and I will be discussing and analyzing one major part of Organic Chemistry 2.  This discussion will be about an area of Organic Chemistry that I have had difficulty with and how I've overcome the obstacles and become very knowledgeable in that area.  The area of Organic Chemistry that I have had multiple problems with has been identifying and understanding the plot of the mass spectrum and determining which compound is being plotted.
    An example of a compound that was difficult to correlate with the mass spectrum of the graph is 2-chloropropane.  To determine which graph correlates with the 2-chloropropane I first realized that I needed to calculate the molecular weight for each of the common isotopes for chlorine which are 35Cl and 37Cl.  When plotting a compound with chlorine in it there will be two peaks due to the two common isotopes.  Chlorine has a height ratio from the first peak(M peak) to the second peak(M+2 peak) of 3:1.  The mass of the molecular ion of 2-chloropropane with the 35Cl isotope is 78 which is the M peak while the mass of the molecular ion of 2-chloropropane with the 37Cl isotope is 80 which is the M+2 peak.  When plotting each of these in the graph the M peak will be at 78 while M+2 peak is at 80 and the M+2 peak is 1/3 the relative abundance of the M peak. 
    Reading throughout ch. 13 of the 2nd edition Smith Organic Chemistry book was very helpful in overcoming this problem and better understanding this concept.  Remember ALWAYS READ YOUR BOOK!