Organic Chemistry Topics with Explanations

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This thread will function analogously to the General Chemistry Explanations and Physics FAQ threads. We will be adding new posts periodically on organic chemistry topics that many students find difficult. Students are requested to please NOT post questions here. If you would like to ask an organic chemistry question, you should post it in the Organic Chemistry Question Thread.

Here is a list of all of the posts in this thread alphabetized by topic:

Table of Contents:

  • Post 10: Aromaticity Determination
  • Post 09: Deciding Among Organic Mechanisms (Sn1, Sn2, E1, and E2)
  • Post 05: Determining R and S Absolute Configurations
  • Post 03: The Effect of Hydrocarbon Branching on MP and BP
  • Post 02: How to Succeed in Organic Chemistry
  • Post 04: Nucleophilicity versus Basicity
  • Post 06: Purification Methods: Recrystallization
  • Post 07: Spectroscopy Methods: Infrared (IR)
  • Post 08: Spectroscopy Methods: Nuclear Magnetic Resonance (NMR) Imaging-Background

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My advice to students is to approach studying organic like you'd approach studying a foreign language. Some students mistakenly believe that they can memorize their way through the course. But this is impossible, because there are an infinite number of possible reactions out there. You do have to learn the vocabulary and "grammar" (mechanisms) of organic chem, which requires some memorization. But the real test of fluency in these types of subjects is whether you can now take what you've learned and apply it to new reactions (or make up new sentences) that you've never seen before.

That kind of ability can only be achieved by working a lot of problems, just as learning to speak another language can only be done if you spend a lot of time practicing speaking it. Ideally, you should spend an hour every day studying organic if possible. Forgo re-reading the chapters in favor of working every problem in your book (yes, all of them, even the challenge ones) and really try to work them out yourself before reading your solutions guide. Ask your TA for help as needed, attend all of the problem sessions and classes, and go to your professor's office hours every week. Students that put in this kind of effort invariably do well come finals time. Plus you have the added bonus that the prof will actually know your name and can write you a letter when you go to apply for med school.

P.S. If you run into trouble, you can always check this thread for posts on various organic chemistry topics, as well as ask us questions in the Organic Chemistry Question Thread.
 
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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 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.

Long answer: 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.

Caveat: This information is probably beyond what you'd be expected to know about isomer properties for the MCAT, but I wanted to clarify this issue since a few people have asked about it. Please note that the effect of chain branching on alkane melting points is much harder to predict compared to its effect on alkane boiling points. In general, branching lowers van der Waals overlap between the two molecules, and therefore also lowers melting points. For example, n-pentane (mp = -130 C) has a higher mp than 2-methylbutane (mp = -160 C). However, some highly symmetrical branched alkane isomers will have abnormally high melting points because of their excellent packing properties. For instance, 2,2-dimethylpropane (mp = -17 C) has a much higher mp than either n-pentane or 2-methylbutane for this reason.
 
Many students are confused by the concept of nucleophilicity and how it differs from basicity. Here is a brief explanation of both concepts, along with how they each relate to the organic substitution and elimination mechanisms.

Basicity
Basicity measures how well a substance can accept a proton (Bronsted-Lowry definition) or how readily it donates a lone pair (Lewis definition). Most MCAT questions will use the Bronsted-Lowry definition of a base. One way to gauge basicity is to use the periodic table. In the periodic table, basicity decreases from left to right (i.e., CH3- > NH2- > OH- > F-), and it also decreases going down a group (i.e., F- > Cl- > Br- > I-). The left to right trend occurs because the less electronegative elements on the left side of the table are less capable of stabilizing the negative charge present on the base. The vertical trend occurs because the larger atoms at the bottom of the table are better able to spread the negative charge around compared with the smaller atoms at the top of the group; this ability to spread charge is called polarizability, and it is further explained below in the section on nucleophiles. One can also gauge basicity by pKa. Since pKa is the negative log of Ka (the dissociation constant of an acid), species with higher pKas are more basic.

Strong bases favor E2 reactions over all other mechanisms, and E1 and Sn1 will never occur in a strongly basic medium. Sn2 is favored if the base is also a strong nucleophile and the alkyl halide is unhindered (primary). I recommend that you memorize the two most common hindered strong bases, which are LDA and t-butoxide. If you see either of these bases on the MCAT, you should immediately consider the reaction as likely to go by the E2 mechanism.

Nucleophilicity
Nucleophilicity is a measure of how polarizable an atom's or ion's electron cloud is. You can think of good nucleophiles as having relatively "sloppy" electron clouds. Nucleophilicity does not consistently follow basicity trends, except that if you are comparing two nucleophiles of the same element, the one with higher pKa will be more nucleophilic. For example, both ethoxide and acetate are oxygen nucleophiles. Ethoxide has a much higher pKa than acetate, so it will be the better nucleophile. Note that you cannot do this comparison for two nucleophiles of different elements, such as acetate versus iodide. Acetate has a much higher pKa than iodide, but it is a much weaker nucleophile. Good nucleophiles tend to be large and not too highly charged, or with the charge spread diffusely. They are not required for Sn1 reactions, but it is essential to have a good nucleophile for Sn2 reactions.
 
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Assigning Priority:
Normally, configuration is assigned with the fourth priority group pointing away from us, and the other three arranged numerically in a circle, either clockwise (R) or counterclockwise (S). It is important to remember that groups are assigned priority by atomic number, and not by the group's mass. For example, one fluorine takes priority over a chain of 20 carbons.

The biggest problem students have with assigning configuration occurs when the fourth priority group is pointing in the wrong direction, such as out of the page. There are at least three methods for handling this:

Method 1: Assume the fourth group is pointing back away from you, then switch configuration at the end
This method is simple and probably the preferred method for most people. Simply assign configuration as if the fourth priority group were pointing away from you, but since it is really pointing out toward you, the configuration is the opposite of what you obtained. For example, if you obtain a clockwise rotation, that would normally be designated as (R). But since the fourth priority group is pointing out of the page, we switch assignments and come up with the (S) configuration.

Method 2: Swap any two pairs of substituents to place the fourth priority one in the back
This method is also quite simple, but you do run the risk of making a mistake with your swapping, so I advise against this in most cases. If you want to swap substituents, you must remember to swap TWO pairs of them in order to retain your original configuration; if you only swap one pair, you will have inverted the stereocenter. Once you are finished swapping substituents, you can assign priority and configuration as normal, with the fourth priority group pointing away from you.

Method 3: Use both hands, with your thumbs pointing in the direction of the fourth priority group.
This is how I learned to assign configuration as an undergrad, although if you find it confusing, there's no need for you to change your current method. If you want to use this technique, what you do is align your thumbs (of both hands) in the same direction of the fourth priority group, and then curl your fingers. If the priorities go in the direction of the fingers on your right hand, the configuration is (R). If they go in the direction of the fingers on your left hand, the configuration is (S). The benefit of this method is that you don't have to worry about swapping groups or remember to swap configurations if your low-priority group is sticking out toward you; you merely need to point your thumbs outward, and the technique still works just as easily as if the low priority group were pointing away from you.
 
Recrystallization is a method for purifying solids in organic chemistry. Ideally, there should only be a small amount of impurity present compared to the quantity of product compound. To perform a recrystallization, you want to dissolve your impure solid compound in a small amount of hot solvent, and then allow the solution to cool slowly. You should select a solvent in which your compound is very soluble at high temperatures, and less soluble at lower temperatures. Mixtures of solvents can also be used, and the right solvent or solvent mixture must often be determined through a process of trial and error. Upon addition of the proper hot solvent, both the compound and the impurities will dissolve. As the solution slowly cools, a crystal lattice of your product molecules will begin to form. The crystal spontaneously prefers to add alike molecules (i.e., other molecules of your product compound) over differently-shaped molecules (i.e., impurity molecules) because using alike molecules makes the lattice more regular. When the crystals have finished forming, you can then separate the crystals from the remaining liquid, which is called the mother liquor, via filtration.

The keys to a successful recrystallization are that you must cool the solution SLOWLY, and you must use a MINIMAL amount of solvent. If you cool it quickly, the impurities will crash out of solution along with the product, and your lattice will not selectively contain only product. This is called precipitation. If you use a lot of solvent, your compound will not form crystals even at low temperatures, because so much solvent is still present that your compound stays dissolved in solution at any temperature.

If your product compound is still not totally pure after one round of recrystallization, you can repeat the above steps and recrystallize it again. Each round of recrystallization will make your product purer. However, you should note that you will lose some of your product every time you recrystallize, because some of your product remains in the mother liquor each time. Thus, you must strike a balance between obtaining greater purity by more rounds of recrystallization and obtaining a reasonable yield. For this reason, multiple rounds of recrystallization are rarely performed unless a very high purity is needed.
 
IR spectroscopy is used to determine which functional groups are present in an organic molecule. Each type of functional group has a distinct bond "stretch" that can be detected after subjecting your sample to infrared radiation. We actually don't use IR very often in the research lab now that we have NMR, but it is often used in undergrad organic labs instead of NMR because the instruments are cheaper and the experiment is easy to perform and interpret. Note that unlike NMR, IR generally CANNOT be used to determine the structure of a compound. It only tells you whether a given functional group is or is not present in the molecule.

What makes IR spectroscopy possible is that the bonds in molecules are constantly moving. This is hard to appreciate from a picture in a book or a model kit, both of which depict the atoms and bonds as being static. But if you could get down to the size of a molecule, you would see that the atoms and bonds are constantly in motion. There are lots of different types of motions that are possible for atoms in bonds, but the motion that we are generally looking at when we zap the molecules with IR radiation is the "stretching" motion that occurs when the atoms of a bond pull away from one another. Note that IR spectroscopy is only possible when you have an asymmetrical bond stretch; if the bond is symmetrical (say for example, a molecule of oxygen, O=O), you will not see any IR absorption peaks.

In order to perform an IR, you will probably have to put a thin coating of your compound (if it's an oil) in between two KCl plates. The KCl plates have been polished to be transparent, and KCl itself does not absorb IR radiation in most of the range where organic compounds absorb. You will obtain a spectrum that has several peaks on it, and on the x-axis you will usually see a label that says "wavenumber." This is the inverse of the wavelength of IR radiation that was absorbed. Certain organic groups tend to absorb IR radiation at characteristic wavenumbers, and you can tell whether these groups are present in the molecule or not based on whether you see those identifying peaks.

For the MCAT, there are only a few crucial IR peaks that you should know. These include the alcohol peak (broad, appears around 3300 wavenumber) and the carbonyl peak (sharp, appears around 1700 wavenumber). The presence of either of those peaks will allow you to determine that your compound contains either an -OH group or a carbonyl group respectively. One other characteristic absorption that might be helpful to you is the C-O single bond (around 1050 wavenumber), but you will probaby be fine without remembering that one.
 
Proton NMR is without a doubt the single most important spectroscopic technique for organic chemists who are attempting to determine or confirm the identity of a compound. I want to emphasize here that the theory and uses of NMR are very complex, enough so that some people earn entire Ph.D. degrees in NMR (I was not one of them!) and spend their entire careers developing NMR techniques and theory. There are dozens if not hundreds of NMR experiments that can be performed besides the carbon and proton NMR experiments that are familiar to you from your undergrad organic course. If you want to learn more about the theory behind NMR and the myriad ways it can be used, you can look into taking an NMR class at your university.

Disclaimer: Note that this is a very basic and simplified introduction to NMR, and it is intended just so that you can appreciate NMR and its usefulness to organic chemists. It's ok if you don't understand this entire post; some of these concepts are beyond what you need to know for the MCAT. But I am including this post anyway for those of you who would like to take a peek inside of the black box that is NMR.

How NMR Works (In a Nutshell)
You are hopefully familiar with the fact that electrons have spins from general chemistry. Recall that these spins can be up or down, that a given electron's spin is represented by the quantum number ms, and that ms has two possible values of +1/2 and -1/2. This is only part of the spin story in atoms, however. It turns out that certain isotopes of certain elements, including hydrogen-1 (protons) and carbon-13, have nuclear spins as well. Note that carbon-12, the most common isotope of carbon, does NOT have a nuclear spin, and therefore carbon-12 CANNOT undergo NMR.

Normally, the nuclear spins of atoms are aligned in random orientations. However, when the atoms are placed into an external magnetic field, the nuclei behave like little magnets and align themselves with the external magnetic field. NMR occurs when atoms like protons and carbon-13 that have non-zero nuclear spins are placed in a magnetic field, and then pulsed with a second magnetic field at a frequency that makes them precess like little spinning tops. This is where NMR gets its name: the second magnetic field is pulsing in such a way that it resonates with the nuclear spins of the atoms. The spinning nuclei emit radio waves as they relax back to their original orientation aligned with the first magnetic field. Depending on the "environment" in which they are in, a given proton or carbon-13 atom will emit radio waves of a characteristic frequency. This is what we detect and transform into an NMR spectrum, which we can interpret to infer a compound's structure.

Performing NMRs in the Lab
You will probably never get to perform an NMR experiment yourself unless you go into chemistry research due to the expense of the instrument and the toxicity of the solvents used. But I want to make sure that you are aware of the following two things about NMR experiments: One is that the solvents used for NMR must be deuterated. Deuterium, which is an isotope of hydrogen that has one proton and one neutron, has a zero spin, and therefore does not undergo NMR. This is a good thing; if you were to dissolve your sample into a solvent that contains protons, you would have so many non-sample protons giving off radio waves that you would not be able to see the resonance frequencies from your sample molecules. All you would see on your spectrum would be the solvent. Second, the radio waves given off by the sample molecules are not readily interpretable unless we first transform them mathematically. Luckily, the NMR instrument does this for us, and you will certainly not have to perform a Fourier transformation on the MCAT!
 
It may seem to you that there is no rhyme or reason to figuring out which mechanism will predominate for a given reaction. In the organic lab, it is actually common to get a mixture of elimination and substitution products, particularly E1 and Sn1. However, there are some rules based on reaction conditions that will help you predict which mechanism should predominate in a particular case. Let's go over some of these conditions.

1) Number of R Groups on the Substrate
This is a very important factor. Keep in mind that primary substrates (like alkyl halides) will not react via Sn1 or E1, because the carbocation formed would be too unstable. Likewise, tertiary substrates will not react via Sn2, because they are too hindered. To sum:

-Primary substrates may undergo Sn2 or E2, but not Sn1 or E1
-Tertiary substrates may undergo Sn1, E1, or E2, but not Sn2.
-Secondary substrates are tricky and may react by any of the four mechanisms, depending on other factors. Keep reading.​

2) Strength of the Base/Nucleophile
This is another very important factor. Keep in mind that carbocations are very powerful Lewis acids. You may not be used to thinking of them this way, because the Bronsted-Lowry definition of an acid doesn't include carbocations. However, the same general principle applies to carbocations that applies to all strong acids: when you mix them with strong bases, you get a neutralization reaction. That is not what we are trying to accomplish in organic reactions! Thus, E1 and Sn1 reactions, both of which have carbocation intermediates, will not occur in the presence of a strong base. In contrast, E2 requires the presence of a strong base, and a strong base will promote E2, particularly if the base is also bulky. To sum:

-A species that is BOTH a good nucleophile and a strong base (ex. hydroxide, Grignard reagents, and alkoxide ions) will tend to promote E2 or Sn2, depending on other factors like substrate hindrance and base bulkiness
-A species that is a good nucleophile but NOT a strong base (ex. cyanide, sulfur nucleophiles, azide) will tend to promote Sn2 if the substrate is not too hindered.
-A species that is a strong base but is NOT nucleophilic (ex. NaH) will often just deprotonate an acidic hydrogen on the substrate.
-A species that is BOTH a poor nucleophile and a weak base (ex. water, alcohols) will tend to promote Sn1 and E1, assuming that the substrate is hindered enough.​

3) Solvent Type
This is another important factor that is often overlooked by students. Generally the solvent will be polar, and it can either be protic or aprotic, which affects the mechanism. Thus, it is essential that you learn how to correctly identify whether a solvent is protic or aprotic. (Read the Intermolecular Interactions post in the General Chemistry Explanations thread about protic versus aprotic solvents if you are not familiar with these terms.) A protic solvent stabilizes charged species. For mechanisms that form carbocations (Sn1 and E1), this is a good thing, because it lowers the potential energy of the carbocations, making them easier to form. However, for mechanisms where the base or nucleophile is charged, a polar protic solvent will stablize the nucleophile and make it less reactive. In contrast, a polar aprotic solvent will destabilize it and make it more reactive. (This is sometimes referred to as making a "naked nucleophile.") To sum:

-Polar protic solvents favor E1 and Sn1 by lowering the energy of the carbocation intermediate and making the RDS occur more readily.
-Polar aprotic solvents favor Sn2 by increasing the energy of the nucleophile and making it more reactive.
-E2 reactions are often carried out in polar aprotic solvents or in a polar protic solvent that is the conjugate acid of the strong base (ex. t-butanol for t-butoxide).​

4) Heat and Hindrance
It is relatively easy to force the reaction to go by E2 over Sn2 by manipulating the choice of base. A large, bulky base like t-butoxide or LDA (lithium diisopropylamide) is too hindered to serve as a nucleophile, and the presence of either of those bases should immediately tell you that the correct mechanism is E2. Forcing a reaction to go by E1 instead of Sn1 is very difficult, because these two mechanisms have the same RDS (formation of the carbocation). The one thing you can do to promote elimination over substitution is to heat the reaction. In organic chemistry, we use the Greek symbol delta (looks like a little triangle over the reaction arrow) to designate that the reaction is being heated. (Note that the delta symbol means "change in" for the physical sciences, and often precedes a variable: eg. delta G = delta H - T delta S.) To sum:

-A large, bulky base (t-butoxide or LDA) should immediately clue you in that the mechanism is E2.
-A heat symbol (the Greek letter delta) over the reaction arrow tells you that the reaction will probably go by elimination rather than substitution.​

5) None of the Above
Don't get so caught up in trying to decide which mechanism will predominate for a given reaction that you don't consider the possibility that no reaction will occur at all. There are two common situations that you should watch out for where there will be no reaction.

One is if you have a primary substrate and no good nucleophile or strong base is present. You can't have E1 or Sn1 in this scenario because the substrate is primary. You can't have E2 because there isn't a strong base. And you can't have Sn2 because there isn't a good nucleophile. The correct response here is that no reaction will occur.

Another situation where you must be cautious is if you are asked about how a cyclohexane or other hindered sytem will react, and you see that you have E2-only conditions. Remember that for an E2 to occur, the base, proton, both carbons forming the new double bond, and the leaving group must all be in what is called the antiperiplanar geometry. This means that all five atoms must lie in the same plane of space, and that the proton and leaving group must be 180 degrees apart in orientation. If the ring cannot orient itself in such a way that the proton and the leaving group are coaxial (both in axial positions on the ring with one up and one down), then an E2 cannot occur. Again, the correct response here is that there will be no reaction.
 
You may be asked on the MCAT to determine whether a compound is aromatic. There are four considerations that you need to think about, regardless of whether heteroatoms, anions, or cations are present. Use this list like a flowchart:

1) Is the compound cyclic? If not, it's definitely not aromatic. If it is cyclic:

2) Does every atom in the ring have a p-orbital on it (i.e., is every atom in the ring sp2- or sp-hybridized?) If not, it's definitely not aromatic. If every ring atom does have a p-orbital on it:

3) Is the compound planar (flat)? If not, it's definitely not aromatic. If it is planar (or if you're not sure):

4) Does the compound obey Huckel's Rule? Huckel's Rule states that every aromatic compound should have 4n + 2 pi electrons, where n is any integer equal to zero or greater. Most aromatic compounds that you are familiar with have an n of 1, which means they have six pi electrons. Benzene is an example of this. If the compound has 4n electrons, then it is not aromatic; in fact, it is an especially UNstable compound called an antiaromatic compound. Cyclobutadiene is an example of this where n = 1.

Heteroaromatic compounds (rings that contain atoms besides carbon) may seem tricky, but you should treat them just like you would a ring of all carbons. The heteroatom lone pairs may or may not be part of the ring. You can tell whether they are based on whether those electrons are necessary in order for the compound to obey Huckel's Rule. For example, pyridine (six-membered aromatic ring containing nitrogen) has a nitrogen whose lone pair is NOT part of the aromatic system. This is because pyridine already has three double bonds, providing it with six pi electrons. If the nitrogen lone pair were also included, that would make eight pi electrons, which would make the compound antiaromatic. In contrast, pyrrolidine (five-membered aromatic ring containing nitrogen) has a nitrogen that IS part of the aromatic system, because there are only two double bonds in the molecule. If the nitrogen lone pair were not included, then pyrrolidine would have four pi electrons, making it antiaromatic.

This analysis explains why pyridine is an excellent base, but pyrrolidine isn't. The lone pair on the pyridine nitrogen is not part of the aromatic system, so it is free to accept a proton without disrupting the molecule's aromaticity. In contrast, pyrrolidine cannot accept a proton without disrupting its aromaticity. Thus, pyrrolidine is not a very good base.

Note that aromatic compounds can also be cationic or anionic, as long as they obey the four rules I've listed above. Two common examples you may run into are cycloheptatriene cation (seven-membered ring with three double bonds and one carbocation) and cyclopentadiene anion (five-membered ring with two double bonds and one carbanion).
 
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