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!