Effient (prasurgel)

The drug I chose to analyze the synthesis of was Effient, which is the brand name for prasurgel. The structure of Effient is shown below:

The paper which describes the synthesis of this drug was not given on the Chemistry By Design website, which was used as a source for this information.The synthesis of Prasurgel consists of three main steps, with a subsequence which is used to synthesize one of the reactants. The first reaction is shown below:

This reaction is an esterification, which we have discussed in class. This is followed by the second reaction:

(Chemistry By Design). This is a standard nucleophilic substitution reaction. The mechanism of this step involves the removal of the hydrogen atom from the nitrogen, which results in the nitrogen atom being negatively charged and a good nucleophile. Thus, when the bond between the halogen, chlorine, in the reactant the nucleophile forms a new sigma bond replaces the halogen (Smith, 2011). The reactant which is used in this second step is not commercially available. Therefore, it is synthesized using a subsequence which is summarized below:

Finally, there is a third and final step which is summarized by the figure below:

This reaction forms a ketone with the use of pyridine, which we have also discussed in class (Chemistry By Design). 

Sources:

Effient (prasurgel). Chemistry By Design. http://chemistrybydesign.oia.arizona.edu/app.php

Smith, J. G. Organic Chemistry, 3rd ed.; Mc-Graw Hill: New York, 2011.

Artificial Amino Acid

This week's assignment challenged us to synthesize our own amino acid. My amino acid is named 2-amino-3-phenyl-butanoic acid and is pictured below:

I would synthesize this amino acid by first reacting 2-phenylpropanol (Chem-Spider) with PCC in order to form an aldehyde which would then be reacted with ammonium chloride and sodium cyanide which would replace the aldehyde with an amino group and a cyano group. This intermediate would then be reacted with hydronium in order to form the final product, 2-amino-3-phenyl-butanoic acid (Williamson). The reaction scheme is pictured below:

We were also challenged to place our artificial amino acid (2-amino-3-phenyl-butanoic acid) within a pentapeptide with four amino acids which exist naturally. The pentapeptide begins with cysteine, continues with glycine, the synthetic amino acid is in the middle, histadine is the fourth amino acid, and the final amino acid is alanine. The pentapeptide is pictured below:

 

Sources:

Chem Spider. 2-phenylpropanol.  http://www.chemspider.com/Chemical-Structure.13657.html

Smith, J. G. Organic Chemistry, 3rd Ed.; McGraw-Hill: New-York, 2008.

Aromatic Substituion Reactions: Some Biological Examples

Lately, we have been studying electrophilic aromatic substitution reactions. We have studied several examples, learned several mechanisms, and performed brominations in lab. After we have understood and mastered the basics, it is important to make connections from these concepts to the world around us. Therefore, listed below are several examples of electrophilic aromatic substitution reactions which occur biologically and are relevant to our everyday lives.

The first example is of the synthesis of Thyroxine, which occurs within the thyroid:

In these reactions, hydrogen ions in the initial reactant are electrophilically substituted by Iodine. The esterifcation of the alcohol also occurs.

A second example is the drug Prontosil, which is converted by intestinal enzymes into sulfanilamide:

In these reactions, a hydrogen atom attached to the aromatic ring was replaced by a ClSO_2 ion.

A third example of electrophilic aromatic substitution is the biosynthesis of vitamin K, an essential vitamin:

As is observed in the figure, the biosynthesis of vitamin K includes the substitution of an R group for a hydrogen atom.

In conclusion, electrophilic aromatic substitution can be expanded beyond what we have learned in class. In many biologically relevant instances, including the three examples listed, these types of reactions become relevant and important to our everyday life.

Each example and its corresponding figure can be found at the following source:
The Department of Chemistry Turk Group. Electrophilic Aromatic Substitution. http://www.drtchemistry.com/CHM_224_files/Chapter%2016.pdf

The Diels-Alder Reaction: A Natural Example

Based upon our previous knowledge of the Diels-Alder class of reactions, we can make some general statements regarding this style of synthesis:

• A conjugated diene in an s-cis conformation reacts with
• A dienophile to form
• A 6-membered ring with one double bond

These basic facts and more are summarized in the following video:

My Little s-Cis Diene (Diels- Alder Reaction Animated) 

We know that this reaction can occur in the lab, but can it happen naturally?

The answer is yes. A photosynthetic fungus, Macrophoma commelinae, synthesizes the compound macrophomate synthase in complex with pyruvate through a three step proccess, the second of the reactions in this process is a diels alder reaction. The entire reaction process is illustrated below:

It can be observed in the figure the starting molecule (first molecule in the second row of the figure), which reacts to form a six- membered ring.
The diene of this reaction is known as 2-pyrone (IUPAC: pyran-2-one) and is illustrated below:

The dienophile of this reaction is ethyl propiolate (IUPAC:  Ethyl acetylenecarboxylate) and is illustrated below:

The reaction conditions of this process also include that the reaction occurs entirely within the fungus itself, and must be catalyzed by the enzyme MPS. 

Sources (In order of appearance):

Video: My Little s-Cis Diene (Diels-Alder Reaction Animated). YouTube. http://www.youtube.com/watch?v=8SfGsNdVV0Q (accessed Feb 03, 2014)

Reaction Article: Ose, Toyoyuki; Watanabe, Kenji; Mie, Takashi; Honma, Mamoru; Watanabe, Hiromi; Yao, Min; Oikawa, Hideaki; and Tanaka, Isao. Inisght into a natural Diels-Alder reaction from the structure of macrophomate synthase. Nautre. 2003, 422, 185-189.

 

2-pyrone: 2H-Pyran-2-one. Pub Chem Compound. http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=68154#itabs-2d (accessed Feb 03, 2014)

Ethyl propriolate: ethyl propiolate. Pub Chem Compound. http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=12182#itabs-2d (accessed Feb 03, 2014)

New Year: A Public Service Announcement

Hello All!

It has been almost an entire year since my first post "Tryptophan" on the blog: "General Chemistry II." As with most years, many things have changed, including the name and purpose of this blog. A lot of chemistry has been studied/learned in the last twelve months! Upon the completion of my General Chemistry II course, I went on to take both Organic Chemistry I, and Analytical Chemistry in the fall of 2013. Therefore, the blog needed a remodel, if only to reflect the growth which has occurred in the last ~365 days. Accompanying the new design is a new title: "Organic Chemistry II." A blog has once again been assigned to me, and in the pursuit of cohesiveness and posterity, I have decided to add on to the final post of "General Chemistry II." Due to the fact that most of the posts from the former identity of this blog will likely have relatively little to do with Organic Chemistry II, this post was necessitated to get the world up to speed on the changes which a year can bring.
Hopefully this will be an exciting semester full of interesting posts and organic chemistry!

-McKenzie

Blog Post #4: A Reflection...

Now that the semester has come to a close, I am writing my fourth and final blog post as part of my General Chemistry II class. As I reflect back upon the semester, I remember feeling unmotivated in the last weeks. However, I also learned a very important lesson that even though I may lose my motivation when I receive a blow like a bad quiz or test grade, my character is defined by how I react in those moments. I tried to focus my frustration into my study time, and it payed off in the end.

Overall, the most challenging concept of General Chemistry II was trying to differentiate the chapters which contained similar concepts. For example, one chapter was on Acids/Bases and another, on a different exam, was about Buffer Solutions. Also, a difficult concept to grasp was when to use an ICE table and when not to.
My advice to future Gen. Chem. II students: study and practice problems as much as you can. Also, take advantage of any extra credit opportunities because they can make the difference in a letter grade (or, in some cases, whether you pass or fail.) Lastly, don't sweat the small stuff. Yes, quizzes are hard, but they end up not being a huge part of your grade. Just do your best! :)

Blog Post #3- Investigating a Question About Buffer Solutions

Recently, a question was posed to me by a classmate:
a) Begin by defining what a buffer is and summarize 3-4 buffer characteristics
b) Solve this problem: A buffer is created by combining 150.0mL of 0.25M HCHO2 with 75.0mL of 0.20M NaOH. Determine the pH of the buffer.

While this question is undoubtedly one I need more practice on, the difficulty level does not seem up to par with the assignment. This same type of problem was given to us in the homework, and on our recent test. Personally, I feel that it is significant due to the fact I struggled with a similar type of question on the exam. However, I doubt that the instructor feels the same way.

Here is my answer to this question: 

a) A buffer is a solution which resists a change in pH by neutralizing an added acid or base. Some buffer characteristics are: 

1. A buffer contains significant amounts of both a weak acid and its conjugate base. For example, human blood is composed of the weak acid carbonic acid (H2CO3) and its conjugate base (HCO3-). 
2. When additional base is added to a buffer solution, the weak acid reacts with the base, neutralizing it.
3. When additional acid is added to a buffer solution, the conjugate base reacts with the acid, neutralizing it. 
4. Weak acids and conjugate bases by themselves do not contain sufficient bases or acids respectively to work as neutralizing/buffering agents. 

b) In order to solve this problem, I must first determine the amount of the weak acid that is converted to a conjugate base by the addition of the NaOH. This is found by comparing stoichiometric ratios presented in the equation:

HCHO₂+NaOH→ H₂O+NaCHO₂

_{From this equation, we can then set up a table to track the changes induced by the addition of the NaOH. Please note that the role of water in this reaction does not effect the outcome of our problem and thus has been omitted:}

	HCHO2	NaOH	NaCHO2
Before Addition	0.0375mol	0mol	0mol
Addition	0mol	0.015mol	0mol
After Addition	0.0225mol	≈0mol	0.015mol

Once we have determined the amount of HCHO2 present in equilibrium, we can use the concentrations of the weak acid and the conjucate base, along with the pK_a value which will be plugged into the Henderson-Hasselbalch equation in order to find the pH of the buffer solution.The Henderson-Hasselbalch equation can be used because the "x is small approximation" is relevant in this context due to the fact that the initial concentration of weak acid is much larger than its Ka value.

 The Henderson-Hasselbalch equation is as follows:

pH=pK_a+log[base] /[acid]

The pKa was determined by using the Ka value, 1.8x10^-4 in this equation: 

pKa=-logKa

This value (3.7447), along with the amounts of the weak acid (0.0225mol) and its conjugate base(0.015mol), were plugged in to the Henderson-Hasselbach equation, leaving us with this expression: 

pH=3.7447+log(0.015/0.0225)

Solved algebraically, we are left with: 

pH=3.57

My reference for this problem was:
Tro, Nivaldo J. Chemistry: A Molecular Approach. 2011. Pages 667,714-715, 722-723