Organic Lecture 5: Epimers, Anomers, and Meso Compounds

These topics are skimmed over in most of our organic classes but, nevertheless, it's important to have a cursory understanding for the MCAT.

Epimers: differ in their absolute configuration at a single chiral center; they are a subclass of diastereomers (see picture).

The prefix D on the name of molecules (i.e. D-Galactose) refers to the orientation of the hydroxyl group (-OH) on the highest-numbered chiral center in a Fischer projection. When the hydroxyl group is on the right of this carbon in the Fischer projection, the molecule is a D sugar. When it's on the left, the molecule is an L sugar.

**Note: D and L are entirely unrelated to optical activity. Distinctions between D and L (or between R and S) can be made just by looking at a drawing of the molecule, but distinctions between (+) and (-) can be made only by running experiments in a polarimeter

Once again:
R or S = absolute configuration (structure)
D or L = relative configuration (structure)
(+) or (-) = observed optical rotation (property)

Epimeric carbon: place at which stereochemistry is different between two molecules

Anomers: epimers that form as a result of ring closure; for MCAT it pertains to only sugar chemistry (i.e. glucose)

Anomeric carbon in glucose: can assume two forms--with hydroxyl group down, it is alpha...with the hydroxyl group up, it is Beta.

Meso compounds: internal plane of symmetry in a molecule that contains chiral centers; are not optically active because one side of the molecule is a mirror image of the other (this means optical activity imparted by one side is canceled out by its other side)

Organic Lecture 4: Enantionmers, Diasteriomers, Optical Activity

We are going to continue with isomerism.

Enantiomers: non-superimposable mirror images. They can only occur when chiral centers are present. NOTE: two molecules that are enantiomers will always have opposite absolute configurations (one has R, the other will have S, and vice versa).

Optical Activity: a compound that rotates the plane of polarized light is said to be optically active.

**A pair of enantiomers will rotate plane-polarized light with equal magnitude, but in opposite directions

If it rotates clockwise it is said to be dextrorotarory (d), also denoted (+)

If it rotates counterclockwise it is said to be levorotatory (l), also denoted (-)

Specific rotation: the magnitude of rotation of plane-polarized light for any compound

**Racemic mixtures are not optically active

One last EXTREMELY IMPORTANT NOTE: (+) and (-) say nothing about whether the absolute configuration is R or S. Therefore, there is no correlation between the sign of rotation and the absolute configuration.

Diastereomers: stereoisomers that are not enantiomers; they are non-superimposable, non-mirror images. Take a look at the picture for a better understanding.

Organic Lecture 3: Chirality and Determining R/S Configurations

I'll do my best to put these concepts into words but keep in mind that this is generally taught with a molecule kit so students can visually see how this works.

Any molecule that cannot be superimposed on its mirror image is said to be chiral. For the MCAT, it's important that you be able to identify chiral centers. For carbon, a chiral center will have four different groups bonded to it--therefore, it must also be sp3 hybridized with 109 degree bond angles. A chiral carbon can also be referred to as a stereocenter, a stereogenic center, or an asymmetric center.

Chiral centers can be assigned an absolute configuration. This is an arbitrary set of rules known as the Cahn-Ingold-Prelog rules.

Priority is assigned to the four different substituents according to increasing atomic number of the atoms directly attached to the chiral center.

Note on isotopes: MCAT likes to test on these. For example, the isotopes of hydrogen are 1H, 2H=D (deuterium), and 3H=T (tritium). Just assigned on the basis of atomic weight if given isotopes.

If two identical atoms are attached to a stereocenter, then the next atoms in both chains are examined until a difference is found.

A multiple bond is counted as two single bonds for both of the atoms involved. For example, if you see a carbon double bonded to an oxygen, it is treated as a carbon bonded to two oxygens.

Once priorities have been assigned, rotate the molecule so that the lowest priority groups points directly away from the viewer. Then simply trace the path from the highest priority to the lowest priority. If the path is clockwise, the absolute configuration is R. If the path is counterclockwise, the absolute configuration is S.

Note: if the lowest priority is sticking out of the page (pointed toward the viewer), trace your path and then flip it. So if you traced a clockwise path, the configuration will be an S because the lowest priority is not pointed away from the viewer.

Organic Lecture 2: Structure, Bonding, and Saturation

A quick note on hybridization: it is essentially the mixing of atomic orbitals to rationalize observed chemical and structural trends. Simply, if add an S orbital to a P orbital we get a SP hybrid. To determine the hybridization for an atom in a molecule, add the number of attached atoms to the number of non-bonding electron pairs (non-delocalized). Then use the table in the picture.

Sigma Bonds: consists of two electrons that are localized between two nuclei. It is a very strong bond bond is always the first type of bond to be formed between any two atoms; a single bond must be a sigma bond.

Pi bonds: composed of two electrons that are localized to the region that lies on opposite sides of the plane formed by the two bonded nuclei and immediately adjacent atoms (not directly between like sigma bonds). Is formed by the side-to-side aligment of two unhybridized p orbitals. The electrons in a pi bond are further from the nuclei than the electrons of a sigma bond, and therefore at a higher energy level, less stable, and form a weaker bond. Note that pi bonds prevent rotation.

**In any multiple bond, there is only one sigma bond and the remainder are pi bonds.

Saturation: a molecule is said to be saturated if it contains no pi bonds and no rings; therefore, it is unsaturated if it has at least one pi bond or a ring.

To determine the degree of unsaturation use this formula: {(2n+2)-x}/2
n=number of carbons
x=number of hydrogens (plus any monovalent atoms such as halogens like F, Cl, Br, or I.

One degree of unsaturation indicates the presence of one pi bond or one ring, two degrees indicates two pi bonds (2 separate double bonds or one triple bond), or one pi bond and one ring, two rings, etc.

**Each oxygen (or other divalent atom) "replaces" one carbon and 2 hydrogen atoms
**Each nitrogen (or other trivalent atom) "replaces" one carbon and 1 hydrogen atom

Let's touch on bond dissociation energy (BDE). By definition, it is the energy required to break a bond homolytically. In homolytic bond cleavage, one electron of the bond being broken goes to each fragment of the molecule. Two radicals form in the process. This is different from heterolytic bond cleavage where both electrons of the electron pair that make up the bond go to the same atom; this forms a cation and an anion

**The higher the bond order, the shorter and strong the bond.
**When comparing the same types of bonds, the greater the S character in the component orbital, the shorter the bond (because s-orbitals are closer to the nucleus than p-orbitals).
**The longer the bond, the weaker it is

ERROR: on my picture it should say "degree of unsaturation", not "saturation" at the bottom. Sorry

Organic Lecture 1: The Effects of Hydrocarbon Branching on MP and BP

Let me preface our first organic lecture with a brief excerpt on how to succeed in studying for this portion of the test. My advice is to approach studying organic like you would approach studying a foreign language. It's a mistake to believe that you can simply memorize this material and perform well on any organic chemistry test, especially the MCAT. There are an infinite number of possible reactions out there so you first need to understand the vocabulary (mechanisms) and grammar of organic. How does one accomplish this? Work lots of problems and seek help when needed. With this said, I understand that by reading this post most of you have already taken at least one semester of the course. If you have and did well, congratulations. It's a class that has ruined more medical careers than any other. Studying for MCAT organic will luckily be a little more mundane than the preparation you put forth for the actual class. Flash cards won't be necessary and there are no synthesis problems on the MCAT--although they do find a way to test these concepts indirectly as we will soon discover. So without further adieu, let's begin our first organic lesson with melting/boiling points and how hydrocarbon branching affects each.

For starters, melting point (mp) and boiling point (bp) are indicators of how well identical molecules interact with (attract) each other. Nonpolar molecules interact principally due to the London dispersion force, one of the intermolecular forces. Such forces can be overcome to melt a nonpolar compound or to boil a nonpolar compound. The greater the attractive force between molecules, the more energy will be required to get the compound to melt or boil. Branching is the most significant factor in determining the degree to which molecules will interact. Branching tends to inhibit van der Waals forces by reducing the surface area available for intermolecular interaction. Thus, branching tends to reduce attractive forces between molecules and to lower both melting points and boiling points.

Another influencing melting point and boiling point for hydrocarbons is molecular weight. The greater the molecular weight, the more surface area there is to interact, the greater number of van der Waals interactions, and the higher the melting point and boiling point.

So, to get to the meat of the lesson, what is the effect of hydrocarbon branching on mp and bp?

Here's the short answer: Branching will decrease mp and bp. Condensing and freezing happen for alkanes because of dispersion forces caused by temporary, induced dipoles, which are the only intermolecular forces holding nonpolar molecules like alkanes (hydrocarbons) together in the liquid and solid states. Branched alkanes have smaller dispersion forces compared to straight-chain alkanes of the same MW, so they will be harder to liquify or freeze.

Here's the long story: Let's imagine that we are cooling down a gaseous alkane like hexane that is straight-chained, versus another of the same MW that is branched, like 2,3-dimethylbutane. Remember that gases don't have any intermolecular interactions, at least not if they're ideal. As we lower the temperature, the molecules stop moving as much, and they begin to have intermolecular interactions that are due to induced, temporary dipoles called London dispersion forces.

Ok, so now we need to consider what kinds of molecules will have the strongest induced dipoles. The strength of the induced dipoles is directly proportional to the surface areas of the molecules that are coming into contact. This is intuitive, because if contact can be made over a greater area, there is a greater chance that electrons will distribute unequally at some point over that surface, causing the temporary dipole and inducing dipoles in the neighboring molecule. You may know that the shape with the smallest surface area-to-volume ratio is a sphere. So molecules that are more spherical (ie, highly branched) do not have very much surface area relative to molecules that are long and extended (straight chains). That is why branched molecules have weaker dispersion forces versus straight-chains. Since they have weaker dispersion forces, the branched molecules will tend to want to stay in the gaseous phase longer, and you'll have to cool them further to force them to condense into a liquid. This means that branched compounds have a lower bp (condense at a lower temperature) versus straight chains.

As we continue to cool, the molecules continue to move closer and closer together, and their interactions continue to increase. Eventually, we reach a point where they begin to crystallize, or at least form an amorphous solid. So we need to consider how well the molecules pack together at this point. Branched compounds are like little spheres, or like porcupines. It's hard to get them to pack well, and this means that you will have to cool them to a lower temperature to freeze them (lower mp) compared with straight chains, which can stack up nicely, more like a cord of firewood. So the mp of a branched compound will be lower than that of a straight chain, assuming that they have the same MW.

Sorry for the shoddy image. You can click on it and it will get bigger if it appears too small.

BIO Lecture 8: Glycolysis (All you need to know)

Lucky for you and me, the MCAT doesn't require the we memorize all the enzyme names and structures of the intermediates in glycolysis. Heck, we don't even really need to memorize the order of the pathway save the inputs and outputs.

Glycolysis is an extremely old pathway and is the universal first step in glucose metabolism--all cells from all domains possess the enzymes of this pathway. In short, a glucose molecule is oxidized and split into two pyruvate molecules producing a net surplus of 2 ATP and 2 NADH. It also takes place in the cytoplasm.

Hexokinase catalyzes the first step in glycolysis, the phosphorylation of glucose to G6P. As a side note, anytime you see the word "kinase" think transfer of a phosphate group.

Phosphofructokinase (PFK) catalyzes the third step, the transfer of a phosphate group from ATP to F6P to form F16bP. This is an important step in the pathway because the reaction is very favorable (a.k.a. irreversible). In addition, the PFK reaction is the primary control point for glycolysis and is known in many MCAT books as the committed step. As a general rule, very favorable steps in enzymatic pathways are the ones that are usually subject to allosteric regulation--PFK is no exception and is regulated by ATP. High concentrations of ATP inhibit PFK.

BIO Lecture 7: Outline of Respiration

Here is a big picture summary of how all of the different stages of metabolism fit together. We'll start looking at the specifics in the next lecture. Keep in mind that when glucose is oxidized there is very little ATP generated directly. Instead, the oxidation of glucose is accompanied by the reduction of high-energy electron carriers, primarily the reduction of NAD+ to NADH (remember from last lecture that when something is oxidized something else must be reduced). The energy in reduced NADH is then used to pump protons out of the interior of the mitochondria and create a proton gradient--this is what finally drives the production of ATP. You should know that oxygen is the final electron acceptor of the electron transport chain, and that anaerobic respiration is insufficient to sustain human life. In addition, fermentation produces lactic acid as a byproduct in humans, and ethanol in yeast. Finally, you should know where in the cell each stage of respiration occurs.

Here are the stages that oxidize glucose to produce CO2 and ATP:

1. Glycolysis: "glucose splitting"; anaerobic (no oxygen required) and occurs in cytoplasm; glucose is partially oxidized while it is split in half into two pyruvate molecules
--2 net ATP (4 total made, but 2 needed in investment phase)
--2 NADH produced (3 ATP in ETC for eukaryotes and 5 ATP for prokaryotes)

2. Fermentation: In short, in anaerobic conditions (without oxygen), electron transport cannot function and the limited supply of NAD+ becomes entirely converted to NADH. Therefore, fermentation has evolved to regenerate NAD+ in anaerobic conditions thereby allowing glycolysis to continue in the absence of oxygen. This process also occurs in the cytoplasm.
--0 ATP; main purpose is to reoxidize the NADH produced in glycolysis (pyruvate is the electron acceptor)

3. Pyruvate Dehydrogenase complex (PDC): the pyruvate produced in glycolysis is decarboxylated to form an acetyl group which is then attached to coenzyme A (a carrier that transfers the acetyl group into the Krebs cycle). Only occurs when oxygen is available but doesn't use oxygen. Occurs in mitochondria matrix for eukaryotes and in the cytoplasm for prokaryotes. It is an aerobic process.
--0 ATP produced
--2 NADH produced (makes 5 ATP in ETC).

4. Krebs cycle (aerobic process): the acetyl group from PDC is added to oxaloacetate to form citric acid--the citric acid is then decarboxylated and isomerized to regenerate the original oxaloacetate. Only occurs when oxygen is available but doesn't use oxygen. Occurs in the mitochondria matrix of eukaryotes and in the cytoplasm for prokaryotes. To fully oxidize the equivalent of one glucose molecule, two acetyl-CoA must be metabolized by the cycle.
Per glucose molecule:
--2 GTP produced (equivalent of 2 ATP)
--6 NADH produced (making 15 ATP in ETC)
--2 FADH2 produced (making 3 ATP in ETC)

4. Electron transport/oxidative phosphorylation: NADH and FADH2 are oxidized by ETC in the inner mitochondrial membrane for eukaryotes and across the inner cell membrane for prokaryotes. In other words, oxidation of metabolic fuels such as glucose and the oxidation of acetyl carbons to CO2 via the citric acid cycle yields the reduced cofactors NADH and FADH2. These compounds are forms of energy currency because their reoxidation--ultimately by molecular oxygen--is an exergonic reaction. The free energy thereby released is harvested to synthesize ATP. It's an aerobic process as well.
--NADH oxidized back to NAD+ and FADH2 oxidized back to FAD occur along with ATP production allowing earlier stages to continue

Summary: 30 ATP (eukaryotes) and 32 ATP (prokaryotes)

I know this is a lot to comprehend right now but the key is to understand how all of the processes function together.