Final Organic Blog

Our final blog assignment for Organic Chemistry was to go to Christina White’s (U. Illinois) webpage and find a publication to read and comment on. After a little browsing, I found an article, titled Synthesis of Complex Allylic Esters via C-H Oxidation vs C-C Bond Formation, that interested me. In this publication, there are many many reactions and compounds, but there were two things that caught my eye when looking for something related to our class. (Sorry I don't have any illustartions, I couldn't copy or save the pictures. I don't know if it was the article, my computer, or me, but if you would like to see them, follow the link below.) The first was the table on page 2, that shows linear allylic oxidation. The general scheme shows a linear alkene reacted with a carboxylic acid to form an ester. The article takes the specific esters in the table and explains how to prepare each and talks about their specific properties. The other thing that I found interesting were the schemes on page 7 and 8. Both schemes show two paths to arrive to the same molecule, which is something I have always found neat about organic chemistry. The scheme on 7 has one step that uses a Grignard reagent, something we learned about not too long ago, and the scheme on 8 uses a Grubbs catalyst, something we will have learned about by the time anyone reads this. Those were the two steps of all the reactions that stood out to me, but the publication was full of reactions and catalysts that we have learned about in organic chemistry.


Signing off one last time
-The High School Chemist


Source
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 (accessed May 1, 2011).

Potential Test Question for Exam 5

This week (second to last... whoot whoot) in organic chemistry, we were assigned to come up with a potential question for our upcoming exam. My question would be... Draw the reactants that would undergo a Claisen Reaction to give the following product.

The answer would be

(For anyone who might be impressed, I had to draw these backward so they would come out straight when I took the picture)
-The High School Chemist

Dr. Steven R. Myers Seminar

Last week, I attended a seminar given by Dr. Steven Myers, from the University of Louisville School of Medicine Department of Pharmacology and Toxicology, titled "Tobacco Smoking During Pregnancy: Biomarkers of Exposure and Relationship to Genetics." I really enjoyed this seminar because I plan on going into the medical field and find subjects such as this fascinating, and because he gave the seminar on an undergraduate level, making it easy to follow. He began his seminar by telling of all the different possible side effects of smoking while pregnant: higher rates of miscarriage, still born babies, premature birth, low birthweight, SIDS, lower I.Q., asthma, and many others. I did feel that this part of the seminar was a little slow because of repetition; he had the same side effects on multiple slides. He continued the seminar with smoking related facts: smoking is the leading cause of avoidable cancers, smoking contributes to cardiovascular, cerebrovascular, respiratory, and pediatric diseases, cigarette smoke gives off 4,000 different chemicals (e.g. 4-aminobiphenyl),

and the first 6-8 weeks of pregnancy is when the baby is most susceptible to cigarette smoke. Dr. Myers then got into what biomarkers are and what their use is to determine the effects of smoking on unborn babies. He defined a biomarker as a molecular, biochemical, or cellular alteration that can be measured (e.g. biological fluids). The questions that you have to ask when choosing a biomarker are what is the relationship between the biomarker and the disease, how long does it take the biomarker to pick up the disease, can the biomarker be obtained unobtrusively, is the biomarker easy to process, and how do genetics and race affect the biomarker? Some of the most effective biomarkers are amniotic fluid, for the first trimester, and blood. I thoroughly enjoyed Dr. Myers' seminar and have no negative comments other that it started off slow (not really a big deal door a good seminar).

-The High School Chemist

Source
http://en.wikipedia.org/wiki/File:4-aminobiphenyl_structure.svg

Hell-Volhard-Zelinsky Halogenation Reaction

This week in Organic Chemistry, we were assigned to find a piece of literature that has a Hell-Volhard-Zelinsky (HVZ) halogenation and talk about that specific reaction. I found this blog to be the most difficult one yet due to not being able to access most chemistry journal articles without paying a large fee. HVZ is when a carboxylic acid is treated with a tri-halogenated P (PCl3, PBr3, ect.), and this reaction halogenates the alpha carbon of the carboxylic acid. HVZ begins by replacing the OH group of the C.A. with a halogen. Then another halogen is added to the alpha carbon of the C.A. Finally, the original halogen is replaced with an OH group, thus making the molecule a C.A. again. The picture below is the mechanism for an HVZ reaction.

The specific HVZ that I found reacts cyclohexanecarboxylic acid with PCl3 to yield
1-chlorocyclohexanecarboxylic acid. The OH group of the C.A. is replaced with Cl, followed by chlorination of the alpha carbon. Then the molecule undergoes hydrolysis to form the final product. The reactant and product can be seen in example 1 below.

-The High School Chemist


































Sources
http://www.springerlink.com/content/t2ur113r36720270/
http://en.wikipedia.org/wiki/Hell-Volhard-Zelinsky_halogenation

Isobutyl propionate

Our assignment for Organic this week was to research and report on an ester that was assigned to us. I had isobutyl propionate (for the sake of length, I will abbreviate it IP), on which I was surprised to find a lot of information. IP has a molecular formula C7H14O2 and a molecular weight of 130.185. This ester is a colorless clear liquid, has a specific gravity of 0.86500 to 0.87100 at 25.00 °C, a melting point of -71.00 to  -70.00 °C at 760.00 mm Hg, and a boiling point of 137.00 to 138.00 °C. at 760.00 mm Hg. I expected to find this general information, but I did not expect to find information such as the odor description at 100% being, and I quote, "fruity green ether sweet tutti frutti banana" and the taste description at 10.0 ppm being "sweet, fruity, banana, tutti frutti, with rummy nuances." After further reading, it made sense when I found that IP is used as a flavor and fragrance agent. Propanoic acid and 2-methylpropanol would be the carboxylic acid and alcohol pair from which this ester is derived. IP could be reacted with NH3 to form the primary amide propanamide. That was a lot of information and I hope you find it interesting. Here is a picture of IP if you are interested.

-The High School Chemist


Sources
http://www.thegoodscentscompany.com/data/rw1020241.html
http://www.caslab.com/Isobutyl_propionate_CAS_540-42-1/

Grignard reagent in synthetic step

This week in organic chemistry, we were assigned to search for a reaction that involves the use of an organolithium, organocuprate, or Grignard reagent in at least one synthetic step. I searched for a Grignard step, and found one in the synthesis of bis[3,5-bis(trifluoromethyl)phenyl]phosphine oxide. As seen in the diagram below, i-PrMgCl and THF are added to 3,5-bis(trifluoromethyl)bromobenzene. This inserts an Mg between the benzene ring and the Br (this first step of a Grignard reagent reaction). Then, (EtO)2POH and more THF are added to the middle reactant, linking two of the substituted benzene rings together through phosphorous, and the final product ([3,5-bis(trifluoromethyl)phenyl]phosphine oxide) is made. The CAS number for this compound is 15979-14-3, the molecular weight is 474.18, and the molecular formula is C16H7F12OP. I included the second part of this synthesis because DIBAL-H (a reagent that we studied along with the Grignard reagent) is used in the reduction of the phosphorous, thus removing the double bonded oxygen.

-The High School Chemist

Sources
-http://orgsyn.org/orgsyn/default.asp?formgroup=basenpe_form_group&dataaction=db&dbname=orgsyn
-http://www.guidechem.com/cas-159/15979-14-3.html

Tyrosine

This week for Organic Chemistry, we were assigned an amino acid, and given a list of items to research and report on. I was assigned Tyrosine (2-Amino-3-(4-hydroxyphenyl)propanoic acid), which is a non-essential amino acid with a polar side chain.
The abbreviated names for Tyrosine are Tyr or Y. Tyrosine is found in a polypeptide called Peptide YY, which "is a short (36-amino acid) protein released by cells in the ileum and colon in response to feeding. In humans it appears to reduce appetite." The pKa values are COOH - 2.20, NH3+ - 9.11, and side chain (phenol) 10.07. The isoelectric point (pH) is 5.66. An interesting fact about Tyrosine is the word comes from the the Greek tyri, which means cheese (it was first discovered in cheese). Tyrosine can be found in many high protein food products such as soy products, chicken, turkey, fish, peanuts, almonds, avocados, milk, cheese, yogurt, cottage cheese, lima beans, pumpkin seeds, and sesame seeds. In the medical field, research has shown that Tyrosine can increase plasma neurotransmitter levels (particularly dopamine and norepinephrine). I really enjoyed doing this blog and found the research very interesting. There was a lot of info on this amino acid and a short paper could be written on the subject.


-The High School Chemist


Sources
- http://en.wikipedia.org/wiki/Peptide_YY
- http://en.wikipedia.org/wiki/Tyrosine
- http://www.chemie.fu-berlin.de/chemistry/bio/aminoacid/tyrosin_en.html

Electrophilic Aromatic Substitution

This week in organic chemistry, we were assigned to find a journal with at least one electrophilic aromatic substitution (EAS) step and talk about the step specifically. After much searching, I was able to find an article that did not require me to purchase it. The article is titled "Selective Nitration of Aromatic Compounds with Bismuth Subnitrate and Thionyl Chloride" and is written by Hussni A. Muathen. In part of the article, Muathen talks of how "polycyclic aromatic hydrocarbons can be efficiently mononitrated in good yields. Biphenylene, in particular, afforded 2- nitrobiphenylene as the only product in high yield." The image of this reaction would not let me save it or copy it, so if you would like to view the reaction, you can go to the source below. This nitration is performed very similarly to the nitrations in the book. The only difference between the two is the compound used to make the electrophile (NO2+) in the article is SOCl4 and the one used in the book is H2SO4. These two reagents achieve the same goal. Anyway I found this to be an interesting article and I'm glad I was able to share it. =)

-The High School Chemist

Source
http://www.mdpi.com/1420-3049/8/7/593/pdf

Grandma and aromaticity

Hey Grandma,

 I know how badly you want to learn about aromaticity, so I thought I would help you out and give you the high points. There are four main criteria that determine if a compound is aromatic or not. 

The first of these is that the compound must be cyclic. This means exactly what it sounds like; the compound must continue in a ring or cycle. 

The second criteria is the compound must be planar (flat). If there are more than six carbons forming the cycle of the compound, then it must take on a shape other than a ring to remain planar. An aromatic compound can also remain planar by fusing rings, which means that two or more rings will be attached to each other. 

The third requirement is the molecule must be completely conjugated; meaning every carbon in the compound must have a double bond on it. This can become complicated when looking at 3D models because the double bonds do not stay between the same two carbons, but for 2D models, each carbon will have a double bond on it.

The final requirement is the molecule must follow what is called Huckel’s rule. This rule states that to be aromatic, a compound must have 4n+2 pi electrons, the electrons that make up the second bond of a double bond. To figure this, count the number of double bonds and multiply by 2. If the result follows the 4n+2, then you are good to go. Anyways good luck with your search for new knowledge.

Your Favorite Grandson,

Don Kelly (The High School Chemist)

Expected, yet absent test #1 question

I was not surprised to find the types of questions that were on my Organic II test #1, as they were what I was anticipating (for the most part). Upon reflecting on the questions that would have been appropriate for this test, I came across a Sapling question (question #4) that I saw fitting. This question said "Imagine you are given the mass spectra of these two compounds, but the spectra are missing the compound names. Which peaks occur in one isomer but not the other? That is, which peaks could be used to distinguish one isomer from the other." I found this question to be useful because it requires me to analyze the two compounds closely, looking for fragments that one could form, but not the other. Some fragments will look quite similar, but will be slightly different in mass, therefore producing different peaks. Under time restraints, this requires me to work diligently and look for unique patterns that each compound has, and I find this skill to be useful.

-The High School Chemist

#1 Muddiest Point

In my Organic Chemistry II class, we were asked to blog on our muddiest or most unclear point thus far in the semester. One thing that has been discussed so far that I do not understand, is the mass spectrum of cyclic alkanes. It was explained to me in class that, when talking about methylcyclopentane, the fragment with a m/z=56 is more stable and therefore more concentrated than the m/z=69 fragment. I do not understand what makes this a more stable or more favorable fragmentation. After looking into this subject, I was not able to find anything that explained it more clearly. The most informative source that I found on this subject further confused me because, from how I read it, it goes against what I was taught in lecture. The source talks about the fragmentation of cyclic alkanes, methylcyclohexane specifically, and says "Note that the loss of the methyl side chain is perhaps the most important fragmentation event, and the M - 15 ion of m/z 83 gives the most intense signal in the spectrum." To my followers, if you can clarify my issue or provide information that you think might be of help, it would be greatly appreciated. OK, no really, someone help!




-The High School Chemist