Last Organic Blog

Our last blog assignment for Organic Chemistry was to go to Christina White’s (U. Illinois) webpage and find a publication to read and comment on. The article I chose was titled, A Sulfoxide-Promoted, Catalytic Method for the Regioselective Synthesis of Allylic Acetates from Monosubstituted Olefins via C-H Oxidation. The article states that sulfoxide ligation to Pd(II) salts is shown to selectively promote C-H oxidation opposed to Wacker oxidation chemistry and to control the regioselectivity in the C-H oxidation products. Also that using benzoquinone as the stoichiometric oxidant and 10 mol % of Pd(OAc)2 or Pd(O2CCF3)2 as the catalyst in a DMSO/AcOH (1:1) solution was found to be compatible with a wide range of functionality, which was shown in Table 2 in the article. The table shows amides, esters, and ethers that was compatible with the solution. I thought this table was useful because it talked about amides, esters and ethers that we have already learned in pervious chapters. The article also explores the potential role of sulfoxide as a ligand in promoting the palladium(II) catalyzed C-H oxidation pathway, and bis-sulfoxide palladium(II), this was shown in Table 1 of the article. These results demonstrated for the first time that sulfoxide ligation of Pd(II) salts can promote C-H acetates in useful yields, which was also shown in Table 2. Overall, I found this article very interesting and very informative. I also found that this article talked about organopalladium, which are compounds that contain a carbon-palladium bond that are prepared in situ during the course of a reaction, from another palladium regent such as Pd(OAC)₂, which was in chapter 26. Lastly, the pictures wouldn’t let me copy them, but are easily viewed in the article.

Resources:

Chen, Mark S. A Sulfoxide-Promoted, Catalytic Method for the Regioselective Synthesis of Allylic Acetates from Monosubstituted Olefins via C-H Oxidation. JACS, 2004, 126, 1346-1347. http://www.scs.illinois.edu/white/index.php?p=publications (accessed May 15, 2004)

Organic Test Question

Answer the 4 Parts to: Aldol Reaction

Use this picture of Aldol Reaction of Cyclopentanone to answer questions 1 and 2

1.      During step 3, what type of overall reaction occurs in the Aldol reaction?

(A) Acid Catalysed dehydration

(B) Acid Catalysed Aldol addition

(C) Electrophilic addition

(D) Acid catalysed enolization

Answer: (A) Acid Catalysed dehydration

2.      Which functional group is responsible for the formation of the stable conjugated enone?

Water

Hydroxy group

Acid catalyst

Carbonyl group

Answer: (D) Carbonyl group

True or False: Explain

3.      In the presence of hydroxide, either propanal or acetaldehyde serve as a nucleophile in an aldol reaction.

Answer True

4.      Either propanal or acetaldehyde can serve as an electrophile in an aldol reaction.

Answer: True

In the presence of hydroxide, each aldehyde can enolize to form a nucleophilic enolate anion. The enolate can react with an unionized aldehyde in the first step of an aldol condensation. The unionized aldehyde is an electrophile in this reaction.

Dr. Steven R. Myer’s, is currently Associate Professor of Pharmacology and Toxicology at the University of Louisville School of Medicine. Dr. Myer’s came to Campbellsville University and spoke about his research on: Smoking During Pregnancy.

Steve Myer’s states that, “cigarette smoke contains more than 4,000 chemicals, including truly nasty things like cyanide, lead, and at least 60 cancer-causing compounds.” When woman smoke during pregnancy, the toxic brew gets into there bloodstream and that’s the baby's only source of oxygen and nutrients. A shortage of oxygen can have devastating effects on the baby's growth and development. On average, smoking during pregnancy doubles the chances that a baby will be born too early or weigh less than 5 1/2 pounds at birth. Smoking also more than doubles the risk of stillbirth.

Every cigarette a women smokes increases the risks of her pregnancy. “A few cigarettes a day are safer than a whole pack, but the difference isn't as great as people would think, states Meyer’s. A smoker's body is especially sensitive to the first doses of nicotine each day, and even just one or two cigarettes will significantly tighten blood vessels. That's why even a light habit can have an outsize effect on the baby's health. On average, a pack-a-day habit during pregnancy will shave about a half-pound from a baby's birth weight. Smoking two packs a day throughout a pregnancy could make the baby a full pound or more lighter. While some women may welcome the prospect of delivering a smaller baby, stunting a baby's growth in the womb can have negative consequences that last a lifetime. Steve Myer’s also talked about, 4-Aminobiphenyl in cigarette’s, which is an amine derivative of biphenyl. It is used to manufacture azo dyes, known as a human carcinogen 4-Aminobiphenyl is shown below in Figure 1.¹

 Figure1

Molecular Formula: C₁₂H₁₁N

Molecular Weight: 169.22 g mol⁻¹

Melting Point: 52−54 °C

Boiling Point: 302 °C

Solubility in Water: practically insoluble in water, soluble in alcohol, ether and chloroform

Steve Myer’s said, studies are currently ongoing to determine the relationships between maternal and fetal metabolism. In an extension of Meyer’s studies, they are also developing the application of amniotic fluid as a biomarker of carcinogen exposure detected in the first trimester of pregnancy and developing assays for the detection of breast milk biomarkers that can be applied to women that are breast-feeding their neonates.

Overall, I think Steve Myer’s did a wonderful job on his presentation about his research on smoking during pregnancy, not only was his presentation interesting, but it was also very informative.

References:

1.      Sigma Aldrich. 2-aminbiphenyl. http://www.sigmaaldrich.com/catalog/ProductDetail.do?D7=0&N5=SEARCH_CONCAT_PNO%7CBRAND_KEY&N4=A42409%7CSIAL&N25=0&QS=ON&F=SPEC

The Hell-Volhard-Zelinsky halogenation reaction

The Hell-Volhard-Zelinsky halogenation reaction, shown in Figure 1, halogenates carboxylic acids at the α carbon.¹

Figure 1.

A carboxylic acid is first reacted with PBr₃ and Br₂. The PBr₃ replaces the OH group of the carboxylic acid with bromine, resulting in a carboxylic acid bromide. The acid bromide can then tautomerize to an enol, which will readily react with the Br₂ to brominate the molecule a second time at the α position. The first step results in a carboxylic acid bromide which is also brominated at the α position.

In a second step, the carboxylic acid may be recovered by reacting the carboxylic acid bromide with water or a weak aqueous acid. This results in the hydrolysis of the acid bromide, which is the displacement of the bromine directly bound to the carbonyl group by an OH group. (The hydrolysis is an example of a nucleophilic acyl substitution.) The final product is the original carboxylic acid brominated at the α position. This is shown in Figure 2 and Figure 3.¹³

Figure 2

Figure 3

The mechanism for the exchange between an alkanoyl bromide and a carboxylic acid is shown in Figure 4.¹ The α-bromoalkanoyl bromide has a strongly electrophilic carbonyl carbon because of the electron-withdrawing effects of the two bromides. The carboxylic acid will act as a nucleophile that the carbonyl oxygen of the carboxylic acid has a partial negative charge.¹

Figure 4


The formation of dimethylketene is an example of Hell-Volhard-Zelinsky mechanism. The starting reactant is isobutyric acid and forms propanoyl bromide using PBr_2. Dimethylketene is also been prepared by the treatment of α-bromoisobutyryl bromide with zinc and by the pyrolysis of isobutyrylphthalimide, dimethylmalonic anhydride, or α-carbomethoxy-α, β-dimethyl-β-butyrolactone. Dimethylketene dimer has been prepared by heating isobutyryl chloride with a tertiary amine. Pyrolysis of the dimer yields dimethylketene. These steps are shown in Figure 5.⁴

Figure 5

Dimethylketene

Molecular Formula: C₄H₆O

Melting point: 97⁰C

Boiling Point: 34⁰C

Resources:

1.      http://en.wikipedia.org/wiki/Hell-Volhard-Zelinsky_halogenation

2.      http://onlinelibrary.wiley.com/doi/10.1002/9780470638859.conrr305/abstract

3.     http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/c_acid/reaktionen_organoli_carbons.vlu/Page/vsc/en/ch/12/oc/c_acid/alfa2/alfa2.vscml.html

      4. http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv4p0348

Ethyl Butyrate

Ethyl butyrate, is an ester with the chemical formula C₆H₁₂O₂. Ethyl butyrate is soluble in propylene glycol, paraffin oil, and kerosene. Ethyl butyrate IUPAC name is Ethyl butanoate.

Molecular Formula: C₆H₁₂O₂

Melting Point: -93⁰C

Boiling Point: 121⁰C

H₂O Solubility: slighty soluble

Molecular Weight: 116.16 g/mol

Density: 0.879 g/cm³

Ethyl Butyrate is commonly used as artificial flavoring in alcoholic beverages, a solvent in perfumery products, and as a plasticizer for cellulose. In addition, Ethyl butyrate is often even added to orange juice.

Ethyl butyrate is one of the most common chemicals used in flavors and fragrances. It can be used in a variety of flavors such as, orange, cherry, pineapple, mango, guava, bubblegum, peach, apricot, fig, and plum. Industrially-speaking, it’s also one of the cheapest chemicals, which only adds to its popularity.

Ethyl butyrate can be synthesized by reacting ethanol and butyric acid. This is a condensation reaction, meaning water is produced in the reaction as a byproduct. This is shown below.

The reaction of a carboxylic acid and an alcohol yields an ester and water is eliminated.

The functional group of an ester is O=C-O. Ester formation is an equilibrium process catalyzed by an acid catalyst.

An ester is made from an alcohol and a carboxylic acid. An acid and a alcohol in ethyl butyrate is shown below.

The synthesis of ethyl 2-ethylhexanoate is an example of ethyl butanoate converted to another carboxylic acid. Here ethyl butanoate and 1-bromobutane  are reacted together to form ethyl 2-ethylhexanoate



References:

- http://www.bbc.co.uk/scotland/learning/bitesize/higher/chemistry/carbon/naming_rev4.shtml

-http://www.chemsink.com/reaction/1755986/

-http://en.wikipedia.org/wiki/Ethyl_butyrate

Grignard Reaction

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. The one I found was called Di-n-Butyldivinyltin and it had a Grignard reaction involved. The reaction is shown below.
 

In the first part of the reaction, the vinyl bromide is added to magnesium and tetrahydrofuran (THF) is added and a Grignard reagent was produced called vinylmagnesium bromide. In the reaction, vinylmagnesium bromide used as a Grignard reagent to produce Di-n-Butyldivinyltin. Grignard reagents are formed more rapidly in tetrahydrofuran and are an important tool in the formation of carbon-carbon bonds. Grignard reagents are strong bases, and are not compatible with good leaving groups or proton acids. However, they add to epoxides and carbonyl groups, and can be used to make other organometallic reagents by reacting with metal-halogen bonds. A Grignard reaction first involves the preparation of an organomagnesium reagent with the reaction of an alkyl bromide with magnesium metal. Its nucleophilic character allows it to react with the electrophilic carbon in a carbonyl group and forms the carbon-carbon bond. Alkynylmagnesium halides and alkynyllithium reagents are useful nucleophiles for C-C bond synthesis.

Sources:

-http://orgsyn.org/orgsyn/default.asp?formgroup=basenpe_form_group&dataaction=db&dbname=orgsyn

-http://www.bluffton.edu/~bergerd/classes/CEM222/Handouts/organometallics.pdf

Phenylalanine

MW: 165.19

Phenylalanine (2-Amino-3-phenylpropanoic acid) has a symbol of F.⁷ Phenylalanine is a water-soluble amino acid that builds blocks for proteins in the body, which means the body needs it for health because it cannot make it. Phenylalanine is found in three forms: L-phenylalanine, which is the natural form found in proteins; D-phenylalanine, which is a mirror image of L-phenylalanine that is made in a laboratory, and DL-phenylalanine, a combination of both the forms.¹ Phenylalanine is an essential amino acid that is nonpolar because of the hydrophobic nature of the benzyl side chain.²

Amino group: (-NH₂)

Carboxylic acid group (-COOH)

Hydrogen Group (H)

The side chain in phenylalanine composed of just carbons and hydrogens atoms.








PKA Values

The equation for pKa is: pKa = -log Ka

To find the Ka, you do this:
Ka = [H+] x [X-]/[HX]

Ka is always products over reactants.³

The Pka values of phenylalanine are 2.58 and 9.24.

Isoelectric Point

For simple amino acids, in which the R group doesn't contain any titratable groups, the isoelectric point can be calculated by averaging the pK_a values for the a-carboxylic acid and a-amino groups.⁴

α -COOH pK_a is 1.8 and  α-NH₃ pK_a is 9.1

1.8 + 9.1/ 2 = 5.45

Phenylalanine isoelectric point:  5.45

IR Spectrum⁸

 

Polypeptides

  
Amino acids consists of a central carbon atom chemically bonded to one hydrogen atom (H), one carboxylic acid group (-COOH), one amino (nitrogen-containing) group (-NH₂), and one side group unique to each amino acid (R).⁹ The side group differs in shape, size, pH, and composition and gives each amino acid its identity. Peptides are amino acids linked in specific sequences to form strands of protein. One amino acid is joined to another by a peptide bond. A peptide is formed by the carboxyl group (-COOH) and bonds to an amino group (-NH₂) of another amino acid and releases water in the process. A polypeptide contains more then 10 amino acids. When two amino acids join together, the carboxyl group of one amino acid is matched with another amino acid. A condensed reaction forms a peptide bond and releases water.⁹               

This picture shows how phenylalanine and glycine are forming a peptide bond.

Citations:

1.      University of Maryland Medical Center. Phenylalanine.

http://www.umm.edu/altmed/articles/phenylalanine-000318.htm (accessed 2011)

           

2.      Phenylalanine. Encyclopedia.

http://www.absoluteastronomy.com/topics/Phenylalanine (accessed 2011)

3.      Amino Acid Structure. Phenylalanine. http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html (accessed 2007)

4.      Amino Acid. The Acid Base Chemistry of Amino Acids. http://chemed.chem.purdue.edu/genchem/topicreview/bp/1biochem/amino2.html (accessed 2006)

5.      Amino Acid Classification for R groups. Phenylalanine. http://mcb.berkeley.edu/courses/mcb102/discussion/Handouts/Amino%20acid%20Review%20JK.pdf

6.      Identifying Amino Acids in Protein NMR Spectra. Phenylalanine. http://hoffman.cm.utexas.edu/research/aa_h_nmr.pdf

7.      X-1A x-ray Absorption Spectra. Phenylalanine. http://xray1.physics.sunysb.edu/~micros/xas/xas.html (accessed Oct. 2, 2003)

 

8.      PhotochemCAD Spectra by Category. Phenylalanine. http://omlc.ogi.edu/spectra/PhotochemCAD/html/phenylalanine.html (accessed 1998)

9.      Nutrition. Polypeptide. http://books.google.com/books?id=46o0PzPI07YC&pg=PA208&lpg=PA208&dq=phenylalanine+polypeptide&source=bl&ots=GjuuhrOHPI&sig=nddjO-qClfNNQjCVLS3JnDNX-Yg&hl=en&ei=YISLTafVBdDUgAf14o3JDQ&sa=X&oi=book_result&ct=result&resnum=9&ved=0CF4Q6AEwCA#v=onepage&q=phenylalanine%20polypeptide&f=false (accessed 2004)

electrophilic aromatic substitution:)

For this Organic blog, we are supposed to look up at least one electrophilic aromatic substitution, which is also known as (EAS) step and talk about the step in detail. After looking I found a journal titled “Regioselective Electrophilic Aromatic Substitution Reactions over Reusable Zeolites” and the authors are Keith Smith and Gamal A. El-Hiti. The article discusses how electrophilic aromatic substitution reactions are very useful in many synthetic processes, for example aromatic compounds are versatile chemical feedstocks for a wide range of industrial products, such as, pharmaceutical, agrochemicals, dyestuffs and explosives. Also talked about how nitration of aromatic compounds is one of the most important and widely studied chemical reactions. Acetyl nitrate, which is a mixture of acetic anhydride and nitric acid, was the most useful possibility. The equation looked like this:

Ac₂O + HNO₃       ↔      AcONO₂ + AcOH           

After there compounds were nitrated there yields were excellent and with high rigoselectivity under mild conditions, using stoichiometric quantities of nitric acid and acetic anhydride, which was scheme 2 shown in the journal. Also Table 1, clearly shows how high the yields were. Overall, this article was very interesting and especially helpful with aromatic electrophilic substitution.

Resources/Citation:

Smith, Keith.; El-Hiti, Gamal A.; Regioselective Electrophilic Aromatic Substitution Reactions over Reusable Zeolites. EBSCOhost [Online] 2006, Vol. 10 Issue 13, 1603-1625. http://0-web.ebscohost.com.library.acaweb.org/ehost/detail?hid=21&sid=7ebb5565-9af3-4163-9ff3-aa040d029e1c%40sessionmgr13&vid=5&bdata=JnNpdGU9ZWhvc3QtbGl2ZQ%3d%3d#db=aph&AN=22631648 (accessed 2006)

A letter to grandma about aromaticity

205 Warren Place

Campbellsville, KY 42718

February 23, 2011

Mrs. Peggy Monson

112 Lane Way

Campbellsville, KY 42718

Dear Granny,

I’m going to explain the lay out of aromaticity, which is another word for fragrance. The structures of these compounds have double bonds regularly, but they don’t in fact behave like a double bond. Reagents such as bromine for example react with benzene by substitution instead of addition. Benzene and its derivatives are referred to as aromatic because of there unique odors. The aromaticity of the benzene ring can be assessed by measuring its resonance energy. A way to this is by measuring its heat of hydrogenation which is a Kekule formula. Both the heat of hydrogenation and combustion will have double bonds present, along with its bond distances being uniform, which is measured by spectroscopy. Also it will have longer wavelengths absorption bands in the ultraviolet region of the spectrum.

Huckel’s rule determines the number of π electrons that provide stability to an unsaturated planar ring. Huckel’s rule uses the formula 4n + 2. Although most polycyclic aromatic molecules follow Huckel’s rule some don’t, like pyrene, which has sixteen π electrons and is aromatic. For benzene, n = 1 and when put in the formula equals six (4n + 2 = 6), which is an aromatic sextet. The rule was calculated for single ring molecules and generally doesn’t apply to multiring systems. Also NMR can be used to measure the degree of aromaticity in a molecule by how well its ring of π electrons can maintain an induced ring current. In other words, it can be diatrophic which is were π electrons are delocalized and proteins are attached to the ring and it is shifted downfield in the nuclear magnetic resonance spectrum or NMR, where double bonds are present.

Overall, I hope I helped you understand the concept of aromaticity and I hope to see you soon.

Sincerely you’re favorite,

Alexia

Questions I was expecting to be on test!

For chapter 13 and 14 test, I expected a lot more questions from the sapling or even questions like the sapling. I studied the sapling a lot more then anything else because it basically had a question from each section in chapter 13 and 14. I know there were 5 questions, but I was expecting more then just 5 questions from the sapling. Also there were a lot of problems involving CNMR and HNMR spectrum and with that being, I was expecting there to be a problem like number 15 and 16 on the sapling, where it ask you to tell how many unique HMNR and CNMR signals exist for each compound, but instead there hardly wasn't anything about signals on the test. I also expected there to be a problem like number 17 on sapling, where it asked you to give the chemical shifts in parts per million (ppm) and it gave you the HMNR signal in Hz and the instrument in MHz. Overall, I thought the test was pretty hard and challenging and even though I was expecting a few more questions from sapling, I think I was hoping there would be to. I also think even though sapling gives you the answers and you can even hit solution to see the right answer, you still get to see how it’s done and you get a chance to look in your book for help, which you learn more by doing so.

Organic Muddiest Point

In chapter 13, my muddiest point is basically not being able to read and understand M+1 and the base peak in a mass spectrometry. I understand that the mass spectrometry is used for determining the molecular weight of a compound and identifying components of a compound. So as I explored the internet and I found the website: http://www.cbu.edu/~ddawson/212/Handouts/Mass%20Spectrometry%20Handout.pdf. This website was a handout and it helped me so much. The handout basically stated that, if the particle of interest is not positively charged, it can not be seen. This handout gave an example of the mass spectrum of toluene. In a mass spectrometry, a molecule is vaporized and ionized, usually by a bombardment with a beam of high-energy electrons; toluene has a molecular weight of 92.  Also the M+1 peak is one mass unit greater than the mass ion peak, which is 93. Finally the base peak, which is the peak with the relative intensity of 100%, is caused by the most stable cation. For toluene, the base peak is 91; the base peak is caused by the most stable cation and is also the tallest peak in the spectrum. Lastly, the handout went on to explain how to read a mass spectrum, which said to see if the M+ peak is even or odd. If it is even the molecule contains an even number of nitrogens (0 being even), like in toluene above which is 92. If the M+ peak is odd, the molecule contains an odd number of nitrogens. It also shows a comparison between the M+ and M+2 peaks. Overall, this website/handout was a great source and a lot of help because it just simply broke it down a lot better then the book did.