Final EU Paper and Reflection

Kathleen Tait Chemistry 512 Dr. Jacobs Final EU Paper and Reflection Enduring Understanding 4. The bonding within a molecule determines its shape and...
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Kathleen Tait Chemistry 512 Dr. Jacobs

Final EU Paper and Reflection Enduring Understanding 4. The bonding within a molecule determines its shape and polarity and therefore its interactions and reactions with other molecules. Intermolecular interactions are central to the structure and function of biochemical systems, and the extent and rate of biochemical reactions govern all cellular functions. Both interactions and reactions can be understood by analyzing energetic stability of molecules and bonds. Post Course Evidence Amino acids are the building blocks (monomers) for a protein. The arrangement of the amino acids on the chain, gives the protein shape which then determines the properties of the protein and what function it will perform. All amino acids have a carbon atom which both the amino group and the carboxyl group are attached. This forms the backbone of the amino acid. This carbon atom also forms the side chain to which other molecules are bonded to make up the twenty different amino acid groups. When the carboxyl group of an amino acid reacts to the amine group of another amino acid, a peptide bond is formed. This bond is a special type of covalent bond (Nester et. al, 2007) and the arrangement of the amino acid groups in a string or linear form are known as the primary structure of the protein. The OH of the carboxyl group on one amino acid reacts with the hydrogen of the amine group on another amino acid to form a chain. Within the dehydration reaction water is given off. The reason why the OH group from one amino acid takes the H from the amine group of another amino acid group is because it is an electrophile (electron deficient) and the H from the NH2 group is a

nucleuophile (electron enriched) (Jacobs, 2007). As a result of this dehydration reaction, a type of covalent bond known as the peptide bond is formed and it is a very strong bond. Once the linear string of amino acids is complete, the secondary structure of the protein begins when hydrogen bonds begin to form between the NHs and the COs located on the backbone of the protein. Once these bonds begin to form, the protein begins to take shape in the form of an alpha helix or a beta sheet (Jacobs, 2007). The tertiary structure of the protein involves the folding of the side chains of each amino acid. Each side chain will bond with another side chain if there is an attraction (Coulombs Law). The shape will depend on whether or not the side chains of the amino acids are polar or non-polar. If the side chain is polar, than it will fold on the outside surface to be close to water. If it is not polar than it will fold into the interior to be away from water. There are a number of different bonds that can form between the side chains. These bonds include London forces, dipoles, H bonding and ionic bonding. The last structure is that of the quaternary structure which involves the bonding of the subunit side-chains in those polypeptide chains that have more than one protein (Nester, 2007). Last year in biology with Dr. Ingrid Waldron, we learned about mitosis and cytokinesis, which involves the replication of DNA and the splitting of a parent cell into two daughter cells. Every cell must have the genetic information it needs in order to function. It is important that the genetic information contained in the chromosomes of the parent cell be replicated exactly. A daughter cell must have the same genetic information as the parent. We used socks in our lab to help us to understand the phases of mitosis and how the cell is able to replicate itself. For all this to occur, the first step in

the process is the replication of the DNA, which became clear to me both in the lab, and reviewing the worksheet entitled The Big Picture (Jacobs, 2007). Deoxyribonucleic acid or DNA is responsible for the daily workings of the cells as well as reproduction. Nucleotides make up DNA. Every nucleotide consists of a deoxyribose or a ribose, one or more phosphate groups and one of five nitrogenous bases. The backbone of a DNA molecule is made up of an alternating sugar group and phosphate group. The phosphate group contains negative charges on the oxygen, thus causing the oxygen to repel from one another, which stretches and weakens the bond between the phosphate and the oxygen. This results in a weak bond, which will be easy to break. A phosphate group can form a phosphoester bond (covalent bond) with either a ribose or a deoxyribose (sugar). This chain of phosphate-sugar forms the backbone of the DNA. The third component of the DNA chain is a nitrogenous base. The four nitrogenous bases are adenine, guanine, cytosine, and thymine. These four bases are divided into two groups. Purines are double ring bases (guanine, adenine) while pyrimidines are single ring bases (thymine, cytosine). Each base will fit only with one other nitrogenous base through hydrogen bonding. The pairs are adenine-thymine (A-T) and cytosine-guanine (C-G). The A-T group will form two hydrogen bonds between them, while the cytosine-guanine will form three hydrogen bonds between them. The hydrogen bonding of the bases forms a strand (chain) of DNA in a double helix. This strand is anti-parallel. This means that one strand is opposite in direction from the other. One strand will run from 5 prime to 3 prime while the other strand runs from 3 prime to 5 prime.

The replication of the DNA begins with the strand being separated by the helicase. Quickly following the helicase is the primer RNA primase, which will allow the DNA polymerase to attach to the strand and extend the DNA strand. The DNA polymerase always moves in the direction of 5 prime to 3 prime so that one strand will have the polymerase synthesize in one direction while the other strand will have the polymerase synthesizing in the opposite direction. Once the polymerase is complete the RNase H moves in to remove the RNA primer left by the primase. Finally, The DNA ligase moves in to work on the discontinuous strand to make the backbone complete by linking the short DNA chain. Reflective Statement Before our class began, we were all asked to complete a pre-assignment so that Dr. Jacobs could access our strengths and weaknesses. I thought that I had done fairly well so I was a little surprised when I received an email from Dr. Jacobs asking me to come to her office for a little tutorial on Lewis structures. I went reluctantly and I discovered that I really did have a misunderstanding as to how Lewis structures are formed. In addition to looking at the actual problems, I found that I also did not quite make the connection concerning bonds, the type of bonds and how they are formed. Although I understood the octet rule (shared electrons totaling 8 for each element), I couldn’t figure out which element should bond to which other element. I knew that I could find the number of valence electrons for each element by using the periodic table of elements, but I did not quite understand resonance and formal charges. The tutoring session was a great help to me, but the first homework assignment and the POGIL activity gave me the impetus I needed to sit down and really think about

how to solve the problem. I discovered that there are steps, which need to be followed before completing a Lewis structure. The first step is to find the number of valence electrons each element has within the molecule. The second step is deciding which is the central atom in the structure. I discovered that the atom with the least electronegativity is the central atom. From there I can arrange the atoms around the central atom and follow the octet rule per atom except for Sulfur and Phosphate which follow the extended octet rule (POGIL activity. At this point, I have to balance the total number of valence electrons with following the octet rule. I also need to make sure that if there are any formal charges that they are placed on the correct atom. Therefore, the more electronegative the atom, the more comfortable it will be with a negative charge, while the less electronegative atoms will be more comfortable with the positive charge. It is important to make sure that the charges are put onto the correct atom. Formal charges will determine the best Lewis structure. That homework assignment allowed me the opportunity to sit and think about how atoms act with each other and that there are certain guidelines that even atoms follow to ensure that they are stable and comfortable. I was able to successfully complete the homework and I got a check (3 points). Once I had the one molecule Lewis structure under my belt, it was time for me to tackle the intermolecular forces and bonding. I have to say that I had a great deal of trouble with covalent bonding and the two types of polar and non-polar. It was difficult for me to understand and to identify what bonds are polar and non-polar as well as what molecules are polar or non-polar. This frustrated me more than the Lewis structures because it involved visualizing the shape of the molecule. That is something with which I had great difficulty. The intermolecular forces, polarity of the molecule determines its

physical properties. Knowing these forces and the polarity of the molecules can help to predict how it will behave. I knew nothing about bonds let alone polarity or non-polarity so one day in class I had an A-HA moment. Chrissy was being very patient with our inability to understand polarity when she showed us a chart that would help us to determine the molecules polarity. The chart appears below: Name of molecule

Name of molecule

Polar Bonds Non-polar Lone Pairs Symmetrical Polar Molecule

The chart acts like a checklist for identifying the type of molecule. A polar bond means that the electrons are not shared equally there is more of a pull, which may result in an asymmetrical shape, which would make it a polar molecule, especially if it has lone pairs. The shape of the molecule has a lot to do with the polarity of the atom. If the molecule is symmetrical, it usually signifies an equal pull from all directions on the central atom and is therefore considered non-polar. However, if the molecule is asymmetrical that means that there is more of a pull in one direction than another, this is usually due to lone pairs and is therefore considered to be polar. I also need to look at the electronegativity of the atom. If the electronegativity is higher than .5, than the bond is polar. Once I understood this, it became easier to understand about IMF’s. All molecules have weak bonds called London forces. The next type of bond is stronger than the London forces and is called

dipole-dipole interaction. This bond is only found in polar molecules and is caused by the continuous movement of the electrons in the electron cloud. The last type of bond is the strongest and is known as hydrogen bonding. All of these forces contribute to how the molecule will behave. Strong bonds require a lot of energy to break so those types of molecules like water will have a high boiling point, those with weak bonds like CF4 will have a low boiling point because it contains weak bonds and therefore are easier to break. The Intermolecular Forces Review Sheets and the Evaporation Lab really helped me to understand and apply what I learned to predict how molecules will act in certain situations. This class gave me not only an understanding of bonds and how they are formed; it also helped to give myself a real boost in self-confidence. I always felt stupid when it came to chemistry but I feel as if I have a great foundation on which to build my learning. I also feel strong in helping students to understand the process of bonding and polarity of molecules. I went from a shaky, unconfident teacher to one who believes in herself in finding and learning the workings of the chemistry world.