Download Final Reflection from Organic Chemistry

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.