Final blog of the semester

Our final blog post for the semester was to go to Christina White's (U.Illinois) page and find a publication that interests us and talk about it. After looking for some time, I came across this publication on the Synthesis of Complex Allylic Esters via C-H Oxidation vs. C-C Bond Formation. The article talks about commonly used reaction methods to produce alylic esters. These methods are Horner-Wadsworth-Emmons (HWE), Linear Allylic Oxidation (LAO) and Grubbs Cross Metathesis (I apologize for not having any reaction schemes or diagrams on this blog because I could not download the image from the site as a picture. But you can follow the link below to take a look at it). What I found interesting yet a bit difficult to understand was the actual reaction procedures that were carried out. The procedure used one common starting material and used two different routes, Olefination and Oxidation, to come to one common product. Each route had different yields of the product.

What I found more interesting was The Horner-Wadsworth-Emmons (HWE) reaction, so I just went ahead and did a little bit of research on it.
The Horner-Wadsworth-Emmons reaction (HWE) is a chemical reaction of stabilized phosphonate carbanions with aldehydes and ketones to produce E-alkenes. It is a stereoselective reaction where the stereochemistry is set by steric approach control.  

The HWE reaction employs similar mechanism to that of the Wittig reaction which we learned not too long ago. 


Sources cited:
1.Vermeulen, Nicolass A. et al. Synthesis of Complex Allylic Esters via C-H Oxidation vs C-C Bond Formation. JACS, 2010, 132, 11323. http://www.scs.illinois.edu/white/index.php?p=publications 
2. http://www.organic-chemistry.org/namedreactions/wittig-horner-reaction.shtm
3. http://en.wikipedia.org/wiki/Horner%E2%80%93Wadsworth%E2%80%93Emmons_reaction

Blog #9

Fill in the boxes with the appropriate starting materials and name the condensation reaction.

To solve this problem, the very first clue is that the reaction product results from two different types of reactions. From the molecule, we can see that it is most probably a ring forming reaction. Highlighted below are the new bonds formed as a result of the reactions.

The solution to this problem is as shown below:

**I used Paint and my Organic text book =)

Seminar Critique

The topic of discussion for the seminar was Tobacco smoking during pregnancy and Biomarkers of exposure and relationship to genetics. The speaker was Steven R. Myers, Ph. D, an associate professor of the Department of Pharmacology and Toxicology at the University of Louisville, School of Medicine. He talked about the different chemicals found in cigarette and how these chemicals can be traced to see if they affected the mother and the baby by the use of biomarkers.

A biomarker is a molecular, biochemical, or cellular alterations that are measurable in biological media, such as human tissues, cells, or fluids. For a biomarker to be useful as a monitoring tool, the following criteria has to be met which are specificity, sensitivity and practicality. Biomarkers are needed because there are 4000 chemicals that are generated during burning and smoking of tobacco. These chemicals are the leading cause of avoidable cancer deaths in the US and the world.

To monitor the effect of smoking during pregnancy, amniotic fluid was used as a biomarker during the first trimester of pregnancy and hemoglobin as a biomarker of carcinogen and tobacco exposure.

From the research they further proved that if an expectant mother smokes during her pregnancy, certain risk factors are increased. These factors include:
- miscarriages
- stillborn babies
- premature birth
- low birth-weight babies
- placenta previa
- placental abruption
- Sudden Infant Death Syndrome (SIDS)
Because of these complications many babies are born with low IQ levels by 25-30% and with learning difficulties as a child. A reported 2,000,000 ear infections and 530,000 doctor visits for asthma are reported on a yearly basis.

People who consume tobacco may develop illnesses such as:
- cancer
- cardiovascular disease
- cerebrovascular diseases
- respiratory diseases and
- pediatric diseases.

 One of the compounds found in cigarette smoke is 4-aminobiphenol which has the structure shown below.

Hell-Volhard Zelinsky Halogenation

The Hell-Volhard Zelinsky reaction is a halogenation reaction at the alpha carbon of a carboxylic acid. The reaction is initiated by a catalytic amount of PBr3, after which one molar equivalent of Br2 is added. The carboxylic OH is replaced by the Br resulting in a carboxylic acid bromide. The acyl bromide then tautomerizes to the enol so it can brominate again at the alpha carbon. The general mechanism of the reaction is shown below:

A reaction that employs the use of the Hell-Volhard Zelinsky mechanism is the synthesis of Dimethylketene. Its IUPAC name is 2-methyl-1-propen-1-one and it has a molecular formula of C4H6O. Its structure is shown below:

The physical properties of dimethylketene are:
- molecular weight: 70.09g/mol
- boiling point: 34*C
- density: 0.775g/cm^3

The reaction scheme for the synthesis of this compound from its starting material is shown below:

The starting material is 2-methylpropanoic acid which is catalyzed by the PBr3 resulting in the substitution of the carboxylic OH by a Br and an addition of Br at the alpha carbon. The second step in the procedure is a different kind of reaction that results in the final product dimethylketene. The true product of the Hell-Volhard Zelinsky reaction is the 1,2-dibromo-2-methylpropanone. This compound has a molecular weight of 230.09g/mol.

As of to the physical properties of the compound, I could not find any information.

References:
http://www.qwiki.com/q/#!/Hell-Volhard-Zelinsky_halogenation
http://www.chemspider.com/Chemical-Structure.120167.html

e-caprolactone

ε-Caprolactone (epsilon-caprolactone) is a cyclic ester, a member of the lactone family, with a seven-membered ring with the formula (CH₂)₅CO₂ and IUPAC name of 2-oxepanone. It is a colorless, miscible liquid with most organic solvents and it is produced on a very large scale as a precursor to caprolactam. 

The physical properties of caprolactone are as follows:

- molecular weight: 114.14g/mole

- melting pt.: -1*C

- boiling pt.: 253*C

- density: 1.030g/cm^3

- solubility: > 1000mg/L (20*C)

Caprolactone is often used as a precursor to caprolactam. It is also a monomer used in the manufacture of highly specialized polymers. An example of a polymer is polyglecaprone, used as suture material in surgery. Caprolactone is prepared industrially by Baeyer-Villiger oxidation of cyclohexanone with peracetic acid.

The dominant reaction of caprolactone which is the conversion to caprolactam, involves treatment with hydroxylamine followed by sulfuric acid to give the desired lactam as shown below in a multi-step synthesis.

Works sited:

https://fscimage.fishersci.com/msds/00202.htm

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

Histidine

Our next blog assignment was to talk about an amino acid that was assigned to us. I got Histidine.

Histidine is one of the essential amino acids that is needed by the human body. It is abbreviated as His or H. Its IUPAC name is 2-amino-3-(1H-imidazol-4-yl)propanoic acid. Below is the structure of the amino acid in discussion.

Histidine has a carboxylic acid, an ammonium and an imidazole ring in its structure. Of these three functional groups, the imidazole ring is the most significant. It is a common coordinating ligand in metalloproteins, part of catalytic sites in certain enzymes and it is a nucleophile. The ring has 6 pi electrons, 4 from the 2 double bonds and 2 from a nitrogen lone pair. Histidine can exist in 4 forms, depending on the pH of the solution, and it is aromatic at all pH values.

The pKa values of the functional groups of histidine are as follows:

- carboxylic acid pKa = 1.78

- imidazole pKa = 5.97

- ammonium pKa = 8.97

The isoelectric point of the amino acid is at pH = 7.47.

Histidine can be analyzed using UV-visible absorption spectrum. The peaks represented around 220 nm result from an excitation of pi electrons in the ground state molecular orbitals to a low energy state.  

In the 15N NMR spectrum, increase in pH to about 8 causes protonation of the imidazole ring to be lost, giving rise to N-1 or N-3 tautomers. The chemical shift of N-1 drops slightly to about 190ppm while N-3 drops considerably to 145ppm. N-1 tautomer is preferred because of hydrogen bonding to neighboring ammonium. At pH 9, the chemical shifts of N-1 and N-3 are 185ppm and 170ppm respectively. Below is the 1H NMR spectrum of histidine.

Two important peptides that contain histidine in their structures that are found in mammalian cells are histidylated oligolysine (HoK) and histidylated polylysine (HpK). They aid in the transfer of nucleic acid.

Sources sited:
http://en.wikipedia.org/wiki/Histidine
http://omlc.ogi.edu/spectra/PhotochemCAD/html/histidine.html
http://www.enovatia.com/services/nmr/capillary-nmr/

Paracetamol

Our next blog post assignment was to find a synthesis of an organic compound from a chemistry publication and talk about the steps involved in the synthesis and if there were any form of Electrophilic Aromatic Substitutions occurring. Finding the publication and the article was the toughest part. I looked into many sites, including acs.pubs.org, but most of them required a fee.
Anyway, I came across the synthesis of acetaminophen, commonly known as paracetamol. It is sold as an over-the-counter drug that is used as a pain reliever and a fever reducer.

The synthesis of this compound involves 3 steps with the starting material as phenol. Below is the mechanism of how the product, paracetamol, is formed.

The very first step in the process is a nitration process. Because the -OH group is an ortho and para director and an activator of the benzene ring, the products formed from the reaction is the ortho- and the para-nitrophenol. The para-nitrophenol is taken to the next step of the reaction process where the nitro group is reduced to form 4-hydroxyaniline. This is possibly achieved by H2 with Pd-C, Fe with HCl or Sn with HCl. The last step of the process involves the addition of -COCH3 group with the help of AC2O to obtain the final desired product, paracetamol.

If you want to know more about the drug and the history behind it, here is the link to it. Enjoy!
http://www.pharmainstitute.in/drug%20of%20the%20month%20july.htm

A.E. =)

Aromaticity

Organic molecules exist in many different forms. They are generally classified into two major groups called aliphatic organic molecules and aromatic organic molecules. Aliphatic compounds are those that contain C – C single bonds and/or double bonds. Aromatic compounds also contain these bonds but they are special that they are planar, cyclic molecules with all the C atoms having conjugated double bonds, meaning that the electron clouds of all the C atoms overlap those of the adjacent C atoms’. An electron cloud is a group of negatively charged electrons revolving around the atomic nucleus. These aromatic compounds obey Huckel’s Rule, stating that the compound must have 4n + 2π electrons in order to be classified as an aromatic compound.

The reaction of aromatic compounds is such that it is much more complex than aliphatic compounds. Because of their conjugated system, they require different kinds of chemicals, temperature regulations and much more care to assist the reactants to proceed to products. Some of the reactions involving aromatic compounds are highly useful as they can be used in synthesis of many useful drugs and other compounds.

after thoughts on organic 2.1 exam...

So, the first organic exam of the semester =( not happy about it. It turned out to be more time consuming and tougher than I thought it would be. But I wish I had studied a little more and looked at much more problems than I had.
The most confusing problem in the exam was Question 11 of Section 2 where we had to match the given organic compounds to the appropriate 13C NMR spectra. I had to go back multiple times and change my answers. To fully comprehend the question, I went back and read the section on 13C NMR in the textbook and one of the practice websites Dr. M suggested us.

This is a tutorial website where you get to practice 13C that I found a bit useful.
http://www.wfu.edu/~ylwong/nmr/c13/

I think that to master this whole spectroscopy business you need to just keep practicing practicing practicing and practicing.

 

}=(((((

NMR sucks!!!!!!!!!!! i don't get it aghhhh

Infrared Spectroscopy

So I was having difficulty understanding IR spectroscopy and what Dr. M talked about on NMR and IR stuff in lab today. I couldn't understand what the graphs were supposed to mean and how we were supposed to interpret it. Upon a little extra reading from various different cites on IR spectroscopy, I pretty much get it now =)

IR Spectroscopy is used for identifying the types of functional groups present in a compound using the visible spectrum. The concept of this technique is that molecules that absorb light and specific frequencies have certain structural characteristics. These absorbed frequencies directly correspond to the type of bonding and the bond strength present between these molecules of the compound. How these frequencies are determined and how eventually the compound is identified is based on the type of vibrational mode a bond displays. In common organic compounds, these vibrational modes can be categorized into 6 vibrational modes which are symmetrical stretching, asymmetrical stretching, wagging, twisting, rocking and scissoring. Depending on the type of bonding present between the elements and also on the element itself plays an important role on the frequency, the wavelength, the absorption and how the data on that certain compound is displayed on a graph.

My sources:
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/infrared.htm
http://en.wikipedia.org/wiki/Infrared_spectroscopy

One down, millions to go.... =|

=)

finally got the hang of it...