Amino Acid

Serine 

Chemical Features:

• Non-aromatic hydroxyl
• Hydrophilic due to the hydrogen bonding capacity of the hydroxyl group
• Molecular formula: C3H7NO3
• Molecular Weight: 105 g
• Appearance: white crystals or powder
• Soluble in water
• Melting point: 246 degrees Celsius

pKa Values:

• 2.21 (carboxyl)
• 9.15 (amino)

Isoelectric Point: (pH at which this amino acid carries no net electrical charge)

• 5.68

Functional Group:

• Hydroxyl

NMR Info:

• Three main peaks: Peak at 4.65, Peak at 3.75, Peak at 3.65
• Pictures can be found at: https://sharepoint.cisat.jmu.edu/isat/klevicca/Web/Amino/Serine_2001/9.htm

IR Info:

A better picture of the IR spectroscopy can be found at: http://webbook.nist.gov/cgi/cbook.cgi?Spec=C56451&Index=0&Type=IR&Large=on

Reaction Info:

• The biosynthesis of serine starts with the oxidation of 3-phosphoglycerate to 3-phosphohydroxypyruvate and NADH Reductive amination of this ketone followed by hydrolysis gives serine. Serine hydroxymethyltransferase catalyzes the reversible, simultaneous conversion of L-serine to glycine. 
• Racemic serine can be prepared from methly acrylate via several steps. It is also naturally produced when UV light illuminates simple ices such as a combination of water, methanol, hydrogen cyanide, and ammonia, suggesting that it may be easily produced in cold regions of space. 

Small Polypeptide:

References:

http://www.biology.arizona.edu/biochemistry/problem_sets/aa/serine.html

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

http://www.chemie.fu-berlin.de/chemistry/bio/aminoacid/serin_en.html

https://sharepoint.cisat.jmu.edu/isat/klevicca/Web/Amino/Serine_2001/9.htm

http://webbook.nist.gov/cgi/cbook.cgi?Spec=C56451&Index=0&Type=IR&Large=on

Electrophilic Aromatic Substitution from a Peer-Reviewed Journal

Aromatic compounds have multiple double bonds so the compounds cannot undergo addition reactions. Therefore, they react by electrophilic aromatic substitution reactions. The aromaticity of the ring system is preserved. I found a peer-reviewed journal through Ebsco entitled, Reaction of N,N-Dimethylaniline with N-Cyanoazoles according to Electrophilic Aromatic Substitution Pattern, that discusses an electrophilic aromatic substitution reaction. This article was found in the Russian Journal of Organic Chemistry.


N-Cyanoazoles are highly reactive organic compounds. These compounds are important intermediate products that are involved in the synthesis of numerous derivatives containing an azole ring. N-cyanoazoles are capable of acting as electrophilic reagents in elctrophlic aromatic substitution reactions because of the high polarity of the cyano group.

This journal reported on the synthesis of some imines by electrophilic aromatic substituion of hydrogen in N,N-dimethylaniline by N-cyanoimidazole,  N-cyanobenzimidazole,  N-cyano-1,2,3-benzotriazole, and N-cyano-2-methylimidazoles. These N-cyanoazoles, Ia–Id, were prepared according to certain standard procedures.

They demonstrated that the previously listed N-cyanoazoles reacted with N,N-dimethylaniline in the presence  of anhydrous aluminum bromide to give para-substituted imines in good yields (65-74%). Nitrobenzene was found to be appropriate solvent for carrying out these reaction. reactions. The high polarity and dissolving
power of nitrobenzene ensure the homogeneity of the reaction mixture. However, nitrobenzene is not involved in the reaction because of the presence of a strong electron-withdrawing nitro group in its molecule.


Here is the picture of the substitution:

Reference:
Chunaev, A. O., Stepanov, E. A., & Purygin, P. P. (2010). Reaction of N, N-dimethylaniline with N-cyanoazoles according to electrophilic aromatic substitution pattern. Russian Journal of Organic Chemistry, 46(3), 459-460. doi:10.1134/S1070428010030309

Aromaticity

(Speaking to my 15 year old brother)

Benzene is the easiest compound to use when trying to explain aromaticity. Benzene is a six carbon ring with alternating double bonds (3 π bonds).

http://t3.gstatic.com/images?q=tbn:ANd9GcRRetZejfNY9WM7hYn6fyz182OUkKJkkfDg_xul20bIXsJ_j6tq

 This is a picture of Benzene. Each point of the ring represents a carbon atom. The three extra lines you see represent double bonds, also called π bonds. A chemical bond is simply and attraction between two atoms. A double bond simply means there are two bonds between the atoms and the electrons are shared.

Aromaticity describes the special stability that Benzene has because of the structure. Since Benzene is considered Aromatic is follows Huckel’s rule. This criteria is as follows:

The molecule must cyclic which means there is a ring of atoms. It is easy to tell that the atoms are placed in a ring by drawing a circle around the molecule.

The molecule must be planar which means that all atoms in the molecule must lie in the same plane.

3.      The molecule has to be completely conjugated which means there has to a p orbital on very atom. A p orbital explains the wave-like behavior of the electrons. Below is a picture of the Benzene molecule drawn with a p orbital on each atom.

http://t0.gstatic.com/images?q=tbn:ANd9GcTG76fwSOJElip2ZUQnjt5sfALo5XjhLfkVtHsQnUTwUsl_DnkFyA

 And finally, the molecule must contain 4n+2 π electrons. This confirms that the molecule is stable because all the bonding orbitals are filled with paired electrons. You can count the number of π electrons a molecule has and set this equation equal to that number. If n is equal to 0 or any other positive number the rule has been met.  A π electron resides in a π bond. 3 π bonds = 6 π electrons because there are 2 π electrons in each double bond. Here is an example of how to confirm this rule applies using Benzene.

4n + 2 = 6

4n = 4

n = 1

Since n is a positive integer, the rule is confirmed. 

Sample Question Not Seen On Test

I anticipated a problem on determining the number or protons giving rise to an NMR signal. The example listed in chapter 14 on page 505 is a good example of this type of problem. The problem and solution are listed below.

Question:
A compound of molecular formula C9H10O2 gives the following integrated 1H NMR spectrum. How many protons give rise to each signal?

(The spectrum is shown in the book on page 505 at the bottom of the page.)

Answer:
Step 1: Determine the number of integration units per proton by dividing the total number of integration units by the total number of protons.

• Total number of integration units: 54+23+33=110 units
• Total number of protons = 10
• Divide: 110 units/10 protons = 11 units per proton

Step 2: Determine the number of protons giving rise to each signal.

• To determine the number of H atoms giving rise to each signal, divide each integration value by the answer of Step 1 and round to the nearest whole number. 

                      Solution: Signal A - 54/11 = 4.9 = 5 H
                                    Signal B - 23/11 = 2.1 = 2 H
                                    Signal C - 33/11 = 3H

The area under an NMR signal is proportional to the number or absorbing protons. The height of each step is proportional to the area under the peak, which is in turn proportional to the number of absorbing protons. Knowing the molecular formula of a compound and integration values from its 1H NMR spectrum gives the actual number of protons responsible for a particular signal.

M+2 Peak

My muddiest point to date in Organic Chemistry deals with M+2 Peak. I understand what it is but I am a little unsure of how it works. I turned to the internet to better understand this concept.
            The M+2 peak in mass spectrum is found if the chlorine or bromine is present. The two molecular ion peaks (M+ and M+2) each contain one chlorine atom. The chlorine can either be ³⁵Cl and ³⁷Cl. ³⁵Cl isotope has a formula mass of 78 and ³⁷Cl has a formula mass of 80. A ratio of 3:1 describes the peak heights. This tells us that chlorine contains 3 times as much of the ³⁵Cl isotope as the ³⁷Cl one. Therefore, there will be 3 times more molecules that contain the lighter isotope than the heavier one.
            Compounds that contain bromine have a different ratio. They have a 1:1 ratio. Bromine has two different isotopes as well, ⁷⁹Br and ⁸¹Br. If there are two lines in a molecular region with a gap of 2 m/z units and are of almost equal heights, a bromine atom is present in the molecule.
            After reading the information on this website I feel that I better understand the M+2 peak. I have put the link at the bottom of the blog if you would like to go check the website out for yourself!


http://www.chemguide.co.uk/analysis/masspec/mplus2.html