In the following chapters, we will frequently discuss several fundamental principles necessary to understand the reactivity of organic molecules. These “tools” are collected here.
By definition, an organic molecule is said to be saturated if it contains no π bonds and no rings; it is unsaturated if it has at least one π bond or a ring. A saturated compound with n carbon atoms has exactly 2n + 2 hydrogen atoms, while an unsaturated compound with n carbon atoms has fewer than 2n + 2 hydrogens.
The formula below is used to determine the degree of unsaturation (d) of simple organic molecules:
* x represents the number of hydrogens and any monovalent atoms (such as the halogens: F, Cl, Br, or I).
Since the number of oxygens has no effect, it is ignored.
For nitrogen-containing compounds, replace each N by 1 C and 1 H when using this formula.
One degree of unsaturation indicates the presence of one π bond or one ring; two degrees of unsaturation means there are two π bonds (two separate double bonds or one triple bond), or one π bond and one ring, or two rings, and so on. The presence of heteroatoms can also affect the degree of unsaturation in a molecule. This is best illustrated through a series of related molecules that all have one degree of unsaturation.
Butene (C4H8) has one degree of unsaturation, since d = [(2 × 4 + 2) − 8]/2 = 1, in the form of a double bond:
4-Chlorobutene (C4H7Cl) also has one degree of unsaturation, but the number of hydrogens is different. Each halogen atom (fluorine, chlorine, bromine, iodine) or other monovalent atom “replaces” one hydrogen atom, so d = [(2 × 4 + 2)−(7 + 1)]/2 = 1:
Methoxyethene (C3H6O) also has one degree of unsaturation. Since a divalent atom can take the place of a methylene group, it doesn’t affect the degree of unsaturation, and can be ignored. The calculation for this formula, then, should look like this: [(2 × 3 + 2) − 6]/2 = 1
Methyl vinyl amine (C3H7N) has one degree of unsaturation as well. Each nitrogen (or other trivalent atom) “replaces” one carbon and one hydrogen atom. Therefore, adjust the formula to be C4H8, then do the calculation. The new formula thus gives d = [(2 × 4 + 2) − 8]/2 = 1:
Example 4-1: Determine the degree of unsaturation of each of these molecules. Which, if any, are saturated?
(a) C6H8
(b) C4H6O
(c) C20H30O
(d) C3H8O
(e) C3H5Br
Solution:
(a) d = [(2 × 6 + 2) − 8]/2 = 3.
(b) Just ignore the O, and find that d = [(2 × 4 + 2) − 6]/2 = 2.
(c) Ignoring the O, we get d = [(2 × 20 + 2) − 30]/2 = 6.
(d) Ignore the O, and find that d = [(2 × 3 + 2) − 8]/2 = 0. This molecule is saturated.
(e) Since Br is a halogen, we treat it like a hydrogen, so d = [(2 × 3 + 2) − (5 + 1)]/2 = 1.
Although hybridization theory is covered in more detail in the MCAT General Chemistry Review, it’s useful here to briefly outline how to determine an atom’s hybridization.
Every pair of electrons must be housed in an electronic orbital (either an s, p, d, or f). For example, the carbon atom in methane, CH4, has four pairs of electrons surrounding it (four single covalent bonds and no lone pairs) so it must provide four orbitals to house these electrons. Orbitals always get “used” in the following order:
So, since the carbon atom in methane must provide four orbitals, we just count: 1 … 2…3 … 4:
Therefore, the hybridization of the carbon atom in methane is s + p + p + p, which is written as sp3. The sum of the exponents in the hybridization nomenclature tells us how many orbitals of this type are used. So, in methane, there are 1 + 3 = 4 hybrid orbitals. The following table gives the hybridization of the central atom for each of the orbital geometries:
Some organic reactions proceed through carbocations (carbonium ions), while others make use of carbanions.
Carbocations, or carbonium ions, are positively charged species with a full positive charge on carbon. The reactivity of these species is determined by what type of carbon bears the positive charge. On the MCAT, carbocations will always be sp2 hybridized with an empty p orbital.
Carbanions are negatively charged species with a full negative charge localized on carbon. The reactivity of these species is determined by what type of carbon bears the negative charge.
It’s essential to understand the stabilities of reaction intermediates, because generally the reactivity of a molecule is inversely related to its stability. This means the molecules that are more stable are less reactive, while higher energy species will be more reactive. This theme will resurface over and over again in organic chemistry, and is a useful rule of thumb to keep in mind when you need to predict how a reaction might proceed.
Organic intermediates are stabilized in two major ways: Inductive effects stabilize charge through σ bonds, while resonance effects stabilize charge by delocalization through π bonds.
All substituent groups surrounding a reaction intermediate can be thought of as electron-withdrawing groups or electron-donating groups. Electron-withdrawing groups pull electrons toward themselves through σ bonds. Electron-donating groups donate (push) electron density away from themselves through σ bonds. Groups more electronegative than carbon tend to withdraw, while groups less electronegative than carbon tend to donate. On the MCAT, alkyl substituents are always electron-donating groups.
Electron-donating groups tend to stabilize electron-deficient intermediates (carbocations), while electron-withdrawing groups tend to stabilize electron-rich intermediates (carbanions). The stabilization of reaction intermediates by the sharing of electrons through σ bonds is called the inductive effect.
Example 4-2: Inductive effects frequently alter the reactivity of molecules. Justify the fact that trichloroacetic acid (pKa = 0.6) is a better acid than acetic acid (pKa = 4.8).
Solution: The chlorine atoms in trichloroacetic acid are electron withdrawing. This decreases the amount of electron density elsewhere in the molecule, especially in the O–H bond. With less electron density, the O–H bond is weaker, making it more acidic than the O–H bond in acetic acid.
An alternative explanation would be to consider the stability of the conjugate bases of these acids.
The chlorine atoms in the trichloroacetate anion distribute the negative charge better, making it more stable than the acetate anion. Therefore, because trichloroacetate anion is a weaker base (is more stable) than acetate anion, trichloroacetic acid is a stronger acid than acetic acid. Acidity will be reviewed in more detail a bit later in the Toolbox.
While induction works through σ bonds, resonance stabilization occurs in conjugated π systems. A conjugated system is one containing three or more atoms that each bear a p orbital. These orbitals are aligned so they are all parallel, creating the possibility of delocalized electrons.
Electrons that are confined to one orbital, either a bonding orbital between two atoms or a lone-pair orbital, are said to be localized. When electrons are allowed to interact with orbitals on adjacent atoms, they are no longer confined to their original “space,” and so are termed delocalized. Consider the allyl cation:
The electrons in the π bond can interact with the empty p orbital on the carbon bearing the positive charge. This is illustrated by the following resonance structures:
The electron density is spread out—delocalized—over the entire 3-carbon framework in order to stabilize the carbocation. So, we might say of the allyl cation that both the electrons and the positive charge are delocalized.
As the allyl cation demonstrates, it often happens that a single Lewis structure for a molecule is not sufficient to most accurately represent the molecule’s true structure. It is important to remember that resonance structures are just multiple representations of the actual structure. The molecule does not become one resonance structure or another; it exists as a combination of all resonance structures, although all may not contribute equally. All resonance structures must be drawn to give an accurate picture of the real nature of the molecule. In the case of the allyl cation, the two structures are equivalent and will have equivalent energy. They will also contribute equally to the delocalized picture of what the molecule really looks like. This average of all resonance contributors is called the resonance hybrid.
Benzene (C6H6) is another common molecule that exhibits resonance. Looking at a Lewis representation of benzene might lead one to believe that there are two distinct types of carbon-carbon bonds: single σ bonds (this structure of benzene has three such bonds) and double bonds (of which there are also three):
Thus one might expect two distinct carbon-carbon bond lengths: one for the single bonds, and one for the double bonds. Yet experimental data clearly demonstrate that all the C—C bond lengths are identical in benzene. All the carbons of benzene are sp2 hybridized, so they each have an unhybridized p orbital. Two structures can be drawn for benzene, which differ only in the location of the π bonds. The true structure of benzene is best pictured as a resonance hybrid of these structures. Perhaps a better representation of benzene shows both resonance contributors, like this:
Notice that these resonance structures differ only in the arrangement of their π electrons, not in the locations of the atoms. All six unhybridized p orbitals are aligned parallel with one another. This alignment of adjacent unhybridized p orbitals allows for delocalization of π electrons over the entire ring. Whenever we have a delocalized π system (aligned p orbitals), resonance structures can be drawn.
Delocalization of electrons is also observed in thiophene:
Here the sulfur atom has two pairs of non-bonding electrons. Notice that these electrons are one atom away from two π bonds. One pair of these electrons is actually in an unhybridized p orbital, such that it can be delocalized into the cyclic π system. Here are the representative resonance structures:
The other pair of electrons, however, is in a hybrid orbital and cannot delocalize into the π system. Here the delocalization of sulfur’s electrons imparts aromatic stability to the molecule. The hybridization of the sulfur is therefore most correctly represented as sp2.
Let’s consider one more example:
The nitrogen in aniline has an unshared electron pair that is one atom removed from a cyclic π system. Again these electrons can be delocalized by overlap of the lone pair-containing orbital with the p orbitals of the benzene ring. This can be demonstrated by the following resonance structures:
In this case, the delocalization of the nitrogen’s electrons disrupts the aromaticity of the benzene ring and is therefore less favorable. Experimental determination of the nitrogen’s bond angles reveals that they are actually intermediate between 120° and 109°, so the hybridization of the nitrogen can best be described not as sp2 or sp3, but as something intermediate between them. The important point, however, is that the electrons are at least somewhat delocalized into the π system. Therefore, the nitrogen’s hybridization is not strictly sp3.
Example 4-3: For the following molecules, indicate the hybridization and idealized bond angles for the indicated atoms.
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
Solution: Remember to always draw the electrons on nitrogen if they are not drawn in the structure.
(a) i) sp2, 120° ii) sp2, 120° iii) sp2, 120° (The lone pair is delocalized, so it’s not counted.) iv) sp3, 109° v) sp2, 120° vi) sp2, 120°
(b) i) sp3, 109° ii) sp3, 109° iii) sp3, 109°
(c) i) sp3, 109° ii) sp2, 120° iii) sp3, 109°
(d) i) sp2, 120° ii) sp2, 120° iii) sp2, 120° iv) sp2, 120°
(e) i) sp3, 109° ii) sp3, 109° iii) sp2, 120° iv) sp, 180° v) sp, 180°
(f) i) sp3, 109° ii) sp, 180° iii) sp3, 109° iv) sp3, 109°
So why all this focus on resonance? In general, the more stable a molecule is, the less reactive it will be. Since the delocalization of charge tends to stabilize molecules, resonance has a big impact on the reactivity of molecules.
Since it’s important to recognize molecules that are stabilized by resonance, we’ll next review the three basic principles of resonance delocalization.
1. Resonance structures can never be drawn through atoms that are truly sp3 hybridized. Remember that an sp3-hybridized atom is one with a total of four σ bonds and/or lone electron pairs.
No resonance structures possible!
No resonance structures are possible with these electrons.
No resonance structures are possible with these electrons.
2. Resonance structures usually involve electrons that are adjacent to (one atom away from) a π bond or an unhybridized p orbital. Here are some examples of molecules that are resonance stabilized:
3. Resonance structures of lowest energy are the most important. Remember that the evaluation of resonance structure stability involves three main criteria:
a. Resonance contributors in which the octet rule is satisfied for all atoms are more important than ones in which it is not. This is the most important of the three criteria listed here and takes priority over items b and c below.
b. Resonance contributors that minimize separation of charge (formal charge) are better than those with a large separation of charge.
c. In structures with formal charge(s), the more important resonance contributor has negative charges on the more electronegative atom(s), and positive charge(s) on the less electronegative atom(s).
Now that we can identify valid resonance structures of any given molecule and rank those resonance structures based on their relative energies, let’s use this information to demonstrate the close relationship between stability and reactivity by examining acidity.
While acids and bases will be covered in detail in Chapter 11 of the MCAT General Chemistry Review, we will now examine how the organic chemistry principles we’ve just discussed can help explain the acidity of a compound, as well as help us understand the relative acidity of several functional groups.
Firstly, let’s review the definition of a Brønsted-Lowry acid. Simply put, it’s a molecule that can donate a proton (H+), and once the molecule has done so, it most commonly takes on a negative charge. This deprotonated structure is referred to as the conjugate base of the acid.
The strength of an acid refers to the degree to which it dissociates (or donates its proton) in solution. The more the acid dissociates, the stronger the acid is said to be. Acids that dissociate completely are said to be strong. Most organic acids, and all organic acids you’re likely to see on the MCAT, are said to be weak acids because they do NOT dissociate completely in solution.
The strength of the acid is determined by the extent to which the negative charge on the conjugate base is stabilized. This means all you need to rank the relative acidity of organic compounds on the MCAT is your background in ranking the stability of reactive intermediates.
For example, let’s compare the acidity of an alcohol like propanol and an alkane like propane. If we compare the stability of each conjugate base, we find that the alkoxide ion is a relatively stable species compared to the carbanion since the negative charge is located on the very electronegative oxygen atom rather than on a carbon atom.
Therefore, alcohols are considerably more acidic than hydrocarbons.
Let’s next compare the relative acidities of propanol and propanoic acid.
In the carboxylate ion, the electrons on the negatively charged oxygen are adjacent to a π bond and can therefore be delocalized. This leads to greater stability of the carboxylate anion and thus to higher acidity of the conjugate acid.
Note that the two resonance structures of the carboxylate ion are equivalent, and are therefore of equal energy.
In contrast, the electrons on the oxygen of the propoxide ion below have no adjacent empty p orbital or π system. Therefore, they are localized and highly reactive, making an alkoxide ion a very strong base (much like OH−) and the alcohol a weak acid.
This makes carboxylic acids, as their name suggests, much more acidic than alcohols.
Example 4-4: Rank the following acids in order of increasing acidity:
Solution: We examine the conjugate base of each acid in order to determine which one will have the more stabilized anion.
For the acetylene, there are no possible resonance structures for its conjugate base, and the negative charge is localized on carbon, an element with low electronegativity; rank 1st as the weakest acid.
In acetone, the hydrogens next to the carbonyl are acidic because there are two resonance structures for the conjugate base of a ketone. One is stable with the negative charge on oxygen, and one is higher in energy with the negative charge on carbon. Even though it has resonance, it is less acidic than cyclopentanol (see below) because some of the charge resides on the carbon; rank 2nd.
There are no possible resonance structures for this molecule, but the negative charge resides on an electronegative oxygen; rank 3rd.
Four resonance structures are possible for the phenoxide ion because the negative charge on the oxygen is adjacent to a benzene ring. However, they are not all of equivalent energy because the negative charge resides on the less electronegative C in three of the four structures. The delocalization of charge means that a phenol (−OH group attached to a benzene ring) is more acidic than an alkyl alcohol; rank 4th as the strongest acid.
Note that the phenol on the previous page would still be less acidic than a carboxylic acid since both resonance structures of a carboxylate ion have the negative charge on oxygen, rather than the less electronegative carbon in the phenoxide ion.
To summarize, here is a general ranking of the relative acidities of the most important organic functional groups you are likely to see on the MCAT:
As we’ve just learned, the acidity of carboxylic acids compared to alcohols results from the resonance stability of the carboxylate anion. In addition, Example 4-2 briefly illustrated how electron-withdrawing substituents next to the carboxylic acid group can increase the acidity of this (or any) functional group by increasing the stability of the negative charge on the anion. To expand upon this idea, inductive effects decrease with increasing distance; the closer the electron-withdrawing group is to the acidic proton (or the negative charge on the conjugate base), the greater the stabilizing effect. The following order of acidity for the isomers of fluorobutanoic acid should help clarify this point.
The magnitude of the effect is also dependent on the strength of the electron withdrawing substituent. In general, the more electronegative a substituent is, the greater its inductive effect will be. As shown below, while trifluoro-, trichloro-, and tribromoacetic acid all have substantially lower pKa values than standard acetic acid (pKa = 4.76), the trend in their acidities mirrors the electronegativity of their respective inductive group.
Example 4-5: Rank the following nine compounds in order of decreasing acidity.
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
Electron-withdrawing substituents on phenols increase their acidity. As an example, consider para-nitrophenol. The nitro group is strongly electron withdrawing and greatly stabilizes the phenoxide ion through resonance. Once the para-nitrophenol is deprotonated, it’s easy to see how the nitro group can withdraw electrons through the delocalized π system such that the negative charge on the phenoxide oxygen can be delocalized all the way to an oxygen atom of the nitro group. This electron-withdrawing resonance stabilization of the nitro group increases the acidity of para-nitrophenol as compared to a phenol that does not have electron-withdrawing substituents.
On the other hand, consider a substituted phenol that has an electron-donating group rather than an electron-withdrawing group. A good example of this is para-methoxyphenol. Here, it is easy to see how once para-methoxyphenol is deprotonated, the negative charge on the oxygen can be destabilized by the donation of a lone pair of electrons from the methoxy oxygen so a negative charge is placed on a carbon that’s adjacent to the negatively charged phenoxide oxygen. Electron-donating groups tend to destabilize a phenoxide ion and decrease the acidity of substituted phenols.
Example 4-6: For each of the following groups of three phenols, rank them in order of decreasing acidity.
(i) 
(ii) 
(i) C > B > A. Compound C is the most acidic because of the two electron-withdrawing nitro groups. They delocalize the charge of the conjugate base, making C a stronger acid. The para-nitro group in B can also delocalize the charge by resonance, though not as well as the two nitro groups in choice C. Finally, A is the least acidic, since it has no electron-withdrawing groups to stabilize the charge.
(ii) B > A > C. Since the amino group in choice C is similar to the OCH3 group discussed above due to the lone pair of electrons on the N, it is also an electron-donating group. As such, it will decrease the acidity of the phenol, making it the least acidic of the three compounds.
Most organic reactions occur between nucleophiles and electrophiles. Nucleophiles are species that have unshared pairs of electrons or π bonds and, frequently, a negative (or partial negative, δ−) charge. As the name nucleophile implies, they are “nucleus-seeking” or “nucleus-loving” molecules. Since nucleophiles are electron pair donors, they are also known as Lewis bases. Here are some common examples of nucleophiles:
Nucleophilicity is a measure of how “strong” a nucleophile is. There are general trends for relative nucleophilicities:
1. Nucleophilicity increases as negative charge increases. For example, NH2− is more nucleophilic than NH3.
2. Nucleophilicity increases going down the periodic table within a particular group. For example, F− < Cl− < Br− < I−.
3. Nucleophilicity increases going left in the periodic table across a particular period. For example, NH2− is more nucleophilic than OH−.
Trend #2 is directly related to a periodic trend introduced in general chemistry: polarizability. Polarizability is how easy it is for the electrons surrounding an atom to be distorted. As you go down any group in the periodic table, atoms become larger and generally more polarizable and more nucleophilic.
Trend #3 is related to the electronegativity of the nucleophilic atom. The more electronegative the atom is, the better it is able to support its negative charge. Therefore, the less electronegative an atom is, the higher its nucleophilicity.
You should note that Trend #2 should only be applied for atoms within a column of the periodic table, while Trend #3 should be applied for atoms across a row of the periodic table.
Example 4-7: In each of the following pairs of molecules, identify the one that is more nucleophilic.
(a) 
(b) 
(c) 
(d) 
Solution:
(a) SH−, since by Trend #2 on the previous page, S is more nucleophilic than O.
(b) OH−, because OH− carries a negative charge, while H2O does not (Trend #1, previous page).
(c) NH2−, since F is more electronegative than N.
(d) CH3−, because N is more electronegative than C.
Electrophiles are electron-deficient species. They have a full or partial positive (δ+) charge and “love electrons.” Frequently, they have an incomplete octet. Electrophilicity is a measure of how strong an electrophile is. Since electrophiles are electron pair acceptors, they are also known as Lewis acids. Here are some common examples of electrophiles:
In all organic reactions (except free-radical and pericyclic reactions), nucleophiles are attracted—and donate a pair of electrons—to electrophiles. When the electrophile accepts the electron pair (a Lewis acid/Lewis base reaction), a new covalent bond forms between the two species, which we can represent symbolically like this:
While students of organic chemistry often associate the discussion of leaving groups with substitution and elimination reactions, these reaction types are beyond the scope of the MCAT. However, a good understanding of leaving group ability will be useful for several reactions we’ll review in Chapter 6.
Generally speaking, the biggest take-home message about leaving groups is that they are more likely to dissociate from their substrate (i.e., do their “leaving”) if they are more stable in solution. Sound familiar? Our understanding of stability and reactivity is all we need to explain relative leaving group ability. For example, leaving groups that are resonance-stabilized (like tosylate, mesylate, and acetate) are some of the best ones out there. We’ll discuss these groups in more detail in Chapter 6.
Resonance Structures of the Mesylate Leaving Group
In addition, weak bases (I−, Br−, Cl−, etc.) are good leaving groups because their negative charge is stabilized due to their large size. In fact, it’s because basicity decreases down a family in the periodic table that leaving group ability increases. This periodic trend will be true for any family, though the halogens are the most common leaving groups you’ll likely come across.
Strong bases (HO−, RO−, NH2−, etc.), on the other hand, are great electron donors because they cannot stabilize their negative charge very well, making them very reactive. As a result, these groups are more likely to stay bound to their substrate rather than dissociate in solution. As you might expect, strong bases are therefore bad leaving groups.
Now just because you’re a bad leaving group one minute doesn’t mean you can’t be made better. For example, while the −OH group of an alcohol is unlikely to dissociate as OH−, treating the compound with acid protonates a lone pair of electrons on the oxygen, thereby making the −OH into −OH2+. The altered group can dissociate as a neutral water molecule, and voila! — no negative charge to stabilize. This trick will work for any of the strong bases listed above, and is the reason why many organic reactions are acid-catalyzed.
The last item in our toolbox is a feature of organic molecules that, unlike inductive and resonance effects, contributes to instability in a molecule: ring strain. Ring strain arises when bond angles between ring atoms deviate from the ideal angle predicted by the hybridization of the atoms. Let’s examine several cycloalkanes in turn.
Cyclopropane (C3H6) is very strained because the carbon-carbon bond angles approach 60° rather than the idealized 109° for sp3 hybridized carbons.
Cyclobutane (C4H8) might be expected to have 90° bond angles. However, one of the carbons is bent out of the plane, such that all of the bond angles are 88°. The distortion of the cyclobutane ring minimizes the eclipsing of carbon-hydrogen σ bonds on adjacent carbon atoms.
The deviation of the bond angles from the normal tetrahedral 109° causes cyclopropane and cyclobutane to be high energy compounds. The strain weakens the carbon-carbon bonds and increases reactivity of these cycloalkanes in comparison to other alkanes. For example, while it is essentially impossible to cleave the average alkane C—C single bond via hydrogenation, C—C bonds in these highly strained cyclic molecules are significantly more reactive. However, they are still much less reactive than C=C double (π) bonds.
Unlike cyclopropane and cyclobutane, cyclopentane has a low degree of ring strain, and cyclohexane is strain free. Both molecules have near-tetrahedral bond angles (109°) due to the conformations they adopt. Consequently, these cycloalkanes do not undergo hydrogenation reactions under normal conditions, and react similarly to straight chain alkanes.
Constitutional (or, less precisely, structural) isomers are compounds that have the same molecular formula but have their atoms connected together differently. Take pentane (C5H12), for example. n-Pentane is a fully-saturated hydrocarbon that has two additional constitutional isomers:
Example 4-8: Draw (and name) all the constitutional isomers of hexane, C6H14. (Hint: There are five of them altogether.)
Solution:
Conformational isomers are compounds that have the same molecular formula and the same atomic connectivity, but differ from one another by rotation about a σ bond. In truth, they are the exact same molecule. For saturated hydrocarbons there are two orientations of σ bonds attached to adjacent sp3 hybridized carbons on which we will concentrate. These are the staggered conformation and the eclipsed conformation. In staggered conformations a σ bond on one carbon bisects the angle formed by two σ bonds on the adjacent carbon. In an eclipsed conformation a σ bond on one carbon directly lines up with a σ bond on an adjacent carbon. Both conformations can be visualized using either the flagged bond notation, or the Newman projection, as shown with ethane (C2H6) below.
Example 4-9: For (a) and (b), represent the flagged bond notation conformation as a Newman projection. For (c) and (d), represent the Newman projection using flagged bond notation, and be sure to label which bond you are looking down when translating from the Newman projection.
(a) 
(b) 
(c) 
(d) 
(a) 
(b) 
(c) 
(d) 
Using these notations, we turn our attention to the conformational analysis of hydrocarbons as demonstrated for n-butane.
The σ bonds should actually directly line up with each other. For clarity here, they are not directly aligned.
It’s important to note, however, that there are an infinite number of conformations for a molecule that has free rotation around a C—C bond, and that all of these other conformations are energetically related to the staggered and eclipsed conformations on which we will concentrate. For example, relative to the carbon atom in the rear of a Newman projection, the front carbon atom could be rotated any number of degrees. Any change in the rotation of one carbon, relative to its adjacent neighbor, is a change in molecular conformation.
What are the relative stabilities of the staggered conformations, the eclipsed conformations, and the infinite number of conformations that are in between them? A staggered conformation is more stable than an eclipsed conformation for two reasons. First, the staggered conformation is more stable than the eclipsed conformation because of electronic repulsion. Covalent bonds repel one another simply because they are composed of (negatively charged) electrons. That being the case, the staggered conformation is more stable than the eclipsed, since in the staggered conformation, the σ bonds are as far apart as possible, while in the eclipsed conformation they are directly aligned with one another. The other major reason the staggered conformation is more stable than the eclipsed conformation is steric hindrance. It is more favorable to have atoms attached to the σ bonds in the roomier staggered conformation where they are 60° apart, rather than the eclipsed conformation where they are directly aligned with one another. There are further aspects to consider in conformational analysis. Not all staggered conformations are of equal energy. Likewise, not all eclipsed conformations are of equal energy. There are particularly stable staggered conformations and particularly unstable eclipsed conformations. The following demonstrates this by examining all staggered and eclipsed conformations for n-butane.
We begin our discussion with the most stable conformation of n-butane. This staggered conformation is referred to as the anti conformation and arises when the two largest groups attached to adjacent carbons are 180° apart. This produces the most sterically favorable, and hence the most energetically favorable (lowest energy) conformation. Now we proceed through a series of 60° rotations around the C2−C3 σ bond until we return to the initial conformation (360°). In our first rotation, we go from the anti staggered to an eclipsed conformation and observe the relative energy maximum that results from the alignment of the methyls and hydrogens. Next as we rotate another 60° we fall again into a staggered conformation that resides in a relative energy minimum. Notice that this energy minimum is not as low as the anti conformation. In this structure the methyl substituents are closer together than in the anti conformation. They are now 60° apart; this is referred to as a gauche conformation. A gauche conformation arises when the two largest groups on adjacent carbon atoms are in a staggered conformation 60° apart. In our next 60° rotation we travel to the absolute maximum on our potential energy diagram. In this eclipsed conformation, the two methyl groups are directly aligned behind one another and are therefore in the most crowded and unfavorable environment. This conformation is referred to as the syn conformation. As we continue our rotation, we fall from the absolute energy maximum and go through the corresponding staggered and eclipsed conformations encountered before.
Example 4-10: Draw a Newman projection for the most stable conformation of each of these compounds:
(a) 2,2,5,5-tetramethylhexane (about the C3—C4 bond)
(b) 2,2-dimethylpentane (about the C2—C3 bond)
(c) 1,2-ethandiol
Solution:
(a) 
(b) 
(c) In this molecule, the gauche conformation is more stable than the anti conformation, because an intramolecular hydrogen bond can be formed in the gauche but not in the anti conformation.
Remember that it’s usually the case that the anti conformation is the more stable. In general, the two largest groups on adjacent carbon atoms would like to be anti to one another since this will minimize steric interactions. However, if the two groups are not too large and can form intramolecular hydrogen bonds with one another, then the gauche conformation can be more stable.
Thus far we’ve limited our discussion of conformational isomers to molecules with unrestricted rotation around σ bonds. Let’s now consider the conformational analysis of two very common cycloalkanes, cyclopentane (C5H10) and cyclohexane (C6H12).
In cyclopentane, the pentagonal bond angle is 108° (close to normal tetrahedral of 109°), so we might expect cyclopentane to be a planar structure. If all of the carbons of cyclopentane were in a plane, however, all of the carbon-hydrogen σ bonds on adjacent carbons would eclipse each other. In order to compensate for the eclipsed C—H σ bonds, cyclopentane has one carbon out of the plane of the other carbons and so adopts a puckered conformation. This puckering allows the carbon-hydrogen σ bonds on adjacent carbons to be somewhat staggered, and thus reduces the energy of the compound. This puckered form of cyclopentane is referred to as the “envelope” form.
If cyclohexane were planar, it would have bond angles of 120°. This would produce considerable strain on sp3 hybridized carbons as the ideal bond angle should be around 109°. Instead, the most stable conformation of cyclohexane is a very puckered molecule referred to as the chair form. In the chair conformation, four of the carbons of the ring are in a plane with one carbon above the plane and one carbon below the plane. There are two chair conformations for cyclohexane, and they easily interconvert at room temperature:
As one chair conformation flips to the other chair conformation, it must pass through several other less stable conformations including some (referred to as half-chair conformations) that reside at energy maxima and one (the twist boat conformation) at a local energy minimum (but still of much higher energy than the chair conformations). The boat conformation represents a transition state between twist boat conformations. It is important to remember, however, that all of these conformations are much more unstable than the chair conformations and thus do not play an important role in cyclohexane chemistry.
Notice that there are two distinct types of hydrogens in the chair forms of cyclohexane. Six of the hydrogens lie on the equator of the ring of carbons. These hydrogens are referred to as equatorial hydrogens. The other six hydrogens lie above or below the ring of carbons, three above and three below; these are called axial hydrogens.
There is an energy barrier of about 11 kcal/mol between the two equivalent chair conformations of cyclohexane. At room temperature there is sufficient thermal energy to inter-convert the two chair conformations about 10,000 times per second. Note that when a hydrogen (or any substituent group) is axial in one chair conformation, it becomes equatorial when cyclohexane flips to the other chair conformation. The same is also true for an equatorial hydrogen that flips to an axial position when the chair forms interconvert. This property is demonstrated for deuterocyclohexane:
These factors become important when examining substituted cyclohexanes. Let’s first consider methylcyclohexane. The methyl group can occupy either an equatorial or axial position:
Is one conformation more stable than the other? Yes. It is more favorable for large groups to occupy the equatorial position rather than a crowded axial position. For a methyl group, the equatorial position is more stable by about 1.7 kcal/mol over the axial position. This is because in the axial position, the methyl group is crowded by the other two hydrogens that are also occupying axial positions on the same side of the ring. This is referred to as a 1,3-diaxial interaction. It is more favorable for methyl to be in an equatorial position where it is pointing out, away from other atoms.
Example 4-11: In each of the following pairs of substituted cyclohexanes, identify the more stable isomer:
(a) 
(b) 
(c) 
(d) 
Solution: Draw chair conformations of each isomer and compare them to see which is more stable. As a good rule of thumb, it’s best to first put the bulkier (i.e., the larger) substituent in a roomier equatorial position and decide if it’s the more stable of the two chair conformations; it usually is. (See figures below and on the following page.)
(a) 
(b) 
(c) 
(d) 
Stereoisomerism is of major importance in organic chemistry, especially when looking at biological molecules, so several questions relating to stereochemistry routinely appear on the MCAT. Stereoisomers are molecules that have the same molecular formula and connectivity but differ from one another only in the spatial arrangement of the atoms. They cannot be interconverted by rotation of σ bonds. For example, consider the following two molecules:
Both molecules have the same molecular formula, C2H5ClO, with the same atoms bonded to each other. However, if one superimposes II onto I without any rotation, the result is:
Note that while the −CH3 and −OH groups superimpose, the −Cl and −H do not. Likewise, if we rotate Molecule II so that the −OH is pointing directly up (12 o’clock) and the −CH3 is pointing at about 7 o’clock, and then attempt to superimpose II on I, the result is:
While the −Cl and the −H groups are now superimposed, the −CH3 and the −OH are not. No matter how one rotates Molecules I and II, two of the substituent groups will be superimposed, while the other two will not. Hence they are indeed different molecules: They are stereoisomers.
Any molecule that cannot be superimposed on its mirror image is said to be chiral, while a molecule that can be superimposed on its mirror image has a plane of symmetry and is said to be achiral. It’s important that you be able to identify chiral centers. For carbon, a chiral center will have four different groups bonded to it. Note that since a carbon atom has four different groups attached to it, it must be sp3 hybridized with (approximately) 109° bond angles and tetrahedral geometry. Such a carbon atom is also sometimes referred to as a stereocenter, a stereogenic center, or an asymmetric center.
Example 4-12: Identify all the chiral centers in the following molecules and determine how many possible stereoisomers each compound has by placing a star next to each chiral center. (Note: the number of possible stereoisomers equals 2n, where n is the number of chiral centers.)
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(a) This molecule has no chiral centers.
(b) This molecule has 1 chiral center and, therefore, 2 possible stereoisomers:
(c) There is 1 chiral center and, therefore, 2 possible stereoisomers:
(d) This molecule has 1 chiral center and, therefore, 2 possible stereoisomers:
(e) There are 2 chiral centers and, therefore, 4 possible stereoisomers:
(f) There are 2 chiral centers, which would seem to indicate 4 possible stereoisomers:
However, there are only 3, because the following “2” molecules are actually the same:
(g) This molecule has no chiral centers.
(h) This molecule has 9 chiral centers and, therefore, 29 = 512 possible stereoisomers:
(i) Although there are two chiral centers,
there are 3, not 4 stereoisomers, because—see (f) above—the following “two” molecules are actually the same:
Chiral centers (carbon atoms bearing four different substituents) can be assigned an absolute configuration. There is an arbitrary set of rules for assigning absolute configuration to a stereocenter (known as the Cahn-Ingold-Prelog rules), which can be illustrated using Molecule A:
1. Priority is assigned to the four different substituents on the chiral center according to increasing atomic number of the atoms directly attached to the chiral center. Going one atom out from the chiral center, bromine has the highest atomic number and is given highest priority, #1; oxygen is next and is therefore #2; carbon is #3, and the hydrogen is the lowest priority group, #4:
If isotopes are present, then priority among these are assigned on the basis of atomic weight with the higher priority being assigned to the heavier isotope (since they are all of the same atomic number). For example, the isotopes of hydrogen are 1H, 2H = D (deuterium), and 3H = T (tritium), and for the following molecule, we’d assign priorities as shown:
If two identical atoms are attached to a stereocenter, then the next atoms in both chains are examined until a difference is found. Once again this is done by atomic number. Note the following example:
2. A multiple bond is counted as two single bonds for both of the atoms involved. For example:
3. Once priorities have been assigned, the molecule is rotated so that the lowest priority group points directly away from the viewer. Then simply trace a path from the highest priority group to the lowest remaining priority group. If the path traveled is clockwise, then the absolute configuration is R (from the Latin rectus, right). Conversely, if the path traveled is counterclockwise, then the absolute configuration is S (from the Latin sinister, left).
Note: The two-dimensional representation (on the left) of the following hypothetical molecule is known as the “Fischer projection,” named after famous organic chemist Emil Fischer.
The Fischer projection is a simplification of the actual three-dimensional structure. In the Fischer projection, as shown on the right, vertical lines are assumed to go back into the page, and horizontal lines are assumed to come out of the page.
The Fischer projection will be very important in our discussion of carbohydrates and will be covered extensively in future chapters.
Example 4-13: Assign absolute configurations to the following molecules.
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
Solution:
(a) R. Either rotate the molecule so the lowest priority group is in the back,
or simply trace it as it stands and invert the configuration (since the lowest priority group is coming toward you):
(b) R. The lowest priority group is already pointing away from you and the trace is clockwise.
(c) S. Recall Fischer notation for molecules, note that the lowest priority group is pointing away from you, and the trace is counterclockwise.
(d) R. The lowest priority group is neither going into nor coming out of the plane of the page. One method is to rotate the molecule so the lowest priority group is in the back and redraw the molecule. Since the path is traveled clockwise, the configuration is R.
Here’s a trick to help in the rotation of molecules. Exchanging two groups on a chiral center necessarily changes the absolute configuration. So in this case, it is perhaps most convenient to exchange any two groups such that the lowest priority group is going into the page:
Note that this trace is going counterclockwise. Remember, however, that we exchanged two groups (the hydrogen and the deuterium), which necessarily changes the absolute configuration. Since the counterclockwise trace in the altered molecule means an S configuration, the true configuration is R.
(e) Because this molecule is not chiral, we cannot assign it an absolute configuration.
(f) S. Rotate so the lowest priority group is in back,
or exchange two groups, −H and −NH2:
Clockwise trace. But remember that two groups on the chiral center were exchanged, so the absolute configuration of the given molecule is the opposite; therefore, S.
(g) R. Rotate so the lowest priority group is in the back:
It is important to be able to identify chiral centers because, as we have seen, when there are four different groups attached to a centralized carbon, there are two distinct arrangements or configurations possible for these groups in space. Consider the following two molecules:
Molecule A has one chiral center with four different groups attached. Notice that Molecule B also has a chiral center and that the four groups attached to it are the same as those in Molecule A. Observe the mirror plane that has been drawn between Molecules A and B. Molecules A and B are mirror images of each other, but they are not superimposable; therefore, they are chiral.
These molecules are enantiomers: non-superimposable mirror images.
Enantiomers can occur when chiral centers are present. Note that two molecules that are enantiomers will always have opposite absolute configurations; for example:
What are the properties of enantiomers? That is, how do they differ from one another? Most chemical properties such as melting point, boiling point, polarity, and solubility are the same for both pure enantiomers of an enantiomeric pair. That is, the pure enantiomers shown above will have many identical physical properties.
One important property that differs between enantiomers is the manner in which they interact with plane-polarized light. A compound that rotates the plane of polarized light is said to be optically active. A compound that rotates plane-polarized light clockwise is said to be dextrorotatory (d), also denoted by (+), while a compound that rotates plane-polarized light in the counterclockwise direction is said to be levorotatory (l), also denoted by (−). The magnitude of rotation of plane-polarized light for any compound is called its specific rotation. This property is dependent on the structure of the molecule, the concentration of the sample, and the path length through which the light must travel.
A pair of enantiomers will rotate plane-polarized light with equal magnitude, but in opposite directions. For example, pure (+)-2-bromobutanoic acid has a specific rotation of +39.5°, while (−)-2-bromobutanoic acid has a specific rotation of −39.5°.
What do you think the specific rotation of an equimolar mixture of the two enantiomers above will be? Since one enantiomer will rotate plane-polarized light in one direction, while the other enantiomer will rotate light by the same magnitude in the opposite direction, the specific rotation of a 50/50 mixture of enantiomers —a racemic mixture—is 0°. Therefore, a racemic mixture of enantiomers, also known as a racemate, is not optically active.
Example 4-14: What is the specific rotation of the R enantiomer of 2-bromobutanoic acid? Of the S enantiomer?
Solution:
The magnitude of rotation cannot be predicted; it must be experimentally determined. It just so happens in this case that the R enantiomer has the (+) rotation [while the S enantiomer has the (−) rotation.] But be careful: This is only coincidental. (+) and (−) say nothing about whether the absolute configuration is R or S. There is no correlation between the sign of rotation and the absolute configuration.
In the preceding discussions on stereoisomerism we have focused on molecules that have only one chiral center. What about molecules with multiple stereocenters? Remember that the number of possible stereoisomers is 2n, where n is the number of chiral centers. If there is one chiral center, then there are two possible stereoisomers: the enantiomeric pair R and S. Two chiral centers means there are four possible stereoisomers. Consider the following molecule (3-bromobutan-2-ol), for example:
Each of the two chiral centers in 3-bromobutan-2-ol can have either R or S absolute configuration. This leads to four possible combinations of absolute configurations at the chiral centers. Both carbons could be of the S configuration or both could be of the R configuration; or, the left carbon could be R and the right carbon S, or vice versa. Here are the four possible combinations:
What’s the relationship between Molecules I and II? Each of the two chiral centers in Molecule I is of the opposite configuration of Molecule II: S, S vs. R, R. Note that they are non-superimposable mirror images:
Therefore, these molecules are enantiomers. What about Molecules III and IV? Once again, each of the two chiral centers in Molecule III is of the opposite configuration of those in Molecule IV. This makes Molecules III and IV an enantiomer pair, just as we noted for Molecules I and II on the previous page. Is there a relationship between Molecules I and III?
By mentally moving Molecule III to the left and aligning it over Molecule I, we see that the right chiral centers of both molecules are directly superimposable (S superimposes onto S). Also note that no matter what we do, we cannot get the left chiral centers of Molecules I and III to superimpose (S does not superimpose onto R).
Molecules I and III are diastereomers. Diastereomers are stereoisomers that are not enantiomers. That is, diastereomers are stereoisomers that are non-superimposable, non-mirror images. The same is true for Molecules I and IV. One of the chiral centers is of the same absolute configuration, while the other chiral center is of the opposite configuration:
The figure below summarizes all possible stereochemical relationships between isomers containing two stereocenters. Inverting at least one, but not all, of the chiral centers within a molecule will form a diastereomer of that molecule. Enantiomers can be formed by inverting every stereocenter within the molecule.
Example 4-15: For each pair of molecules below, state the relationship between them.
(a) 
(b) 
(c) 
(d) 
(e) 
Solution:
(a) The molecules are superimposable and therefore identical.
(b) The molecules are identical. (The left carbon is not a chiral center.)
(c) Diastereomers.
(d) Enantiomers.
(e) Diastereomers.
While the structures of diastereomers are similar, their physical and chemical properties can vary dramatically. They can have different melting points, boiling points, solubilities, dipole moments, specific rotations, etc. Most importantly for the MCAT, the specific rotation of diastereomers is also different, but there is no relationship between the specific rotations of diastereomers as there is for enantiomers. There is no way to predict the specific rotation of one diastereomer if you know the degree of rotation of another.
Nature has evolved intricate mechanisms for the bio-synthesis of enantiomerically pure, optically active compounds. For example, L-(+)-tartaric acid, D-(+)-fructose, and L-(+)-valine are all isolated as a single enantiomer from their respective biological sources.
Unfortunately, the laboratory syntheses of enantiomerically pure compounds is often laborious and expensive. It is generally more time- and cost-effective to synthesize chiral targets as racemic mixtures. For example, the reduction of 2-butanone (achiral) to 2-butanol (chiral) yields both enantiomers.
If only one enantiomer of 2-butanol is needed, the two alcohols must be separated. Since enantiomers have identical chemical and physical properties, separating a racemic mixture is a nontrivial process called resolution.
The traditional method for resolving a racemic mixture is through the use of an enantiomerically pure chiral probe, or resolving agent, that associates with the components of the mixture through either covalent bonds or intermolecular forces (like hydrogen bonds or salt interactions). The resulting products will be diastereomers, capable of separation due to their different physical properties.
For example, racemic (±)-2-amino-2-phenylacetic acid can be resolved with enantiomerically pure (1R,4R)-(+)-10-camphorsulfonic acid, as shown in the figure below.
Protonation of the amine by the sulfonic acid produces two diastereomeric salts with different chemical and physical properties. In this particular instance, these salts have different solubilities; the R salt precipitates as a crystalline solid while the S salt remains dissolved in the filtrate. A simple filtration process is used to separate the two, which can be released from the probe and isolated as enantiomerically pure material in a subsequent work-up step.
Epimers are a subclass of diastereomers that differ in their absolute configuration at a single chiral center (only one stereocenter is inverted). To illustrate epimeric relationships, let’s look at the Fischer projections of some sugars (see Chapter 7):
The prefix D on the name of these molecules refers to the orientation of the hydroxyl group (−OH) on the highest-numbered chiral center in a Fischer projection (C-5 in these cases). When the hydroxyl group is on the right of this carbon in the Fischer projection, the molecule is a D sugar. (When the hydroxyl group is on the left, the molecule is an L sugar.)
You must understand that D and L, like R and S, are entirely unrelated to optical activity, (+) or (−). 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.
• R or S = absolute configuration (structure)
• D or L = relative configuration (structure)
• (+) or (−) = observed optical rotation (property)
Concerning the three sugars above, we see that D-glucose and D-galactose differ in stereochemistry at only one chiral center (C-4). Thus, D-glucose and D-galactose are said to be C-4 epimers, and C-4 is called the epimeric carbon. Likewise, D-glucose and D-allose differ in structure at a single chiral center (C-3). D-Glucose and D-allose are C-3 epimers, with C-3 being the epimeric carbon.
What about D-galactose and D-allose? What is the relationship between these two molecules? We can see that these two sugars differ at two chiral centers (C-3 and C-4). At least one, but not all, of the stereocenters have been inverted. Therefore they are diastereomers, but NOT epimers. Note that all epimers are diastereomers, but not all diastereomers are epimers.
Epimers that form as a result of ring closure are known as anomers. For the MCAT, anomers will be encountered only with regard to sugar chemistry. To illustrate anomerism, consider D-glucose. Open-chain glucose exists in equilibrium with cyclic glucose, known as glucopyranose. Cyclization occurs when the C-5 hydroxyl group attacks the carbonyl (C=O) carbon, C-1. This converts a carbon with three substituents to a carbon with four different substituents. Thus, a new stereocenter is formed (C-1), and it can assume one of two possible forms: with the hydroxyl group down, it is α; with the hydroxyl group up, it is β. It is the orientation at C-1 that distinguishes the two anomers, and C-1 is known as the anomeric center (or anomeric carbon).
Let’s look at another molecule with more than one stereocenter. Consider 2,3-butanediol:
Upon inspection, we determine that there are two chiral centers and therefore four possible stereoisomers. Notice that both chiral centers have the same groups attached to them: −H, −CH3, −OH, and −CH(OH)CH3. When the same four groups are attached to two chiral centers, the molecule can have an internal plane of symmetry. Let’s examine this a little more closely. We first consider the R, R stereoisomer and the S, S stereoisomer of 2,3-butanediol:
There are two things to notice here. First, I and II are non-superimposable mirror images and therefore enantiomers. Second, in both I and II there is no internal plane of symmetry. This is demonstrated for Molecule II:
The −OH groups line up on the two chiral centers, but the −CH3 groups and −H atoms do not. The optical rotation of a 50/50 mixture of molecules I and II would measure zero because this is a racemic mixture.
Now look at the R, S stereoisomer and its mirror image:
It turns out that Molecules III and IV are directly superimposable and therefore identical. This is because there is an internal plane of symmetry within the molecule.
When there’s an internal plane of symmetry in a molecule that contains chiral centers, the compound is called a meso compound. Actually then, 2,3-butanediol has only three stereoisomers, not four. Molecules I and II are enantiomers, while III and IV are the same molecule. Molecule III (or IV) is an example of a meso compound. Meso compounds have chiral centers but are not optically active (so they are achiral) because one side of the molecule is a mirror image of the other. In a sense, the optical activity imparted by one side of the molecule is canceled by its other side.
Example 4-16: Which of the following molecules are optically active?
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(a) This molecule is optically active. It has two chiral centers, but no internal mirror plane. Therefore it is not a meso compound and will rotate plane-polarized light.
(b) This molecule is a meso compound due to its two chiral centers and internal mirror plane. It will be optically inactive. Be sure to look for rotations around σ bonds in order to find the mirror planes of some molecules.
(c) This molecule has no chiral centers, so will have no optical activity.
(d) By rotating around the C-2 to C-3 bond to put the molecule into an eclipsed conformation, you can see that there is an internal mirror plane in the molecule. Since C-2 and C-3 are also chiral centers with four different substituents, this is a meso compound, and will be optically inactive.
(e) This molecule has three chiral centers (the two bridgehead carbons are chiral), but no plane of symmetry. It is therefore chiral and optically active.
(f) There is no mirror image in this molecule even though it has two chiral centers (they have the same absolute configuration). It will therefore be optically active.
(g) This molecule does have an internal mirror plane, and its two chiral centers have opposite absolute configurations. It is therefore meso, and not optically active.
Geometric isomers are diastereomers that differ in orientation of substituents around a ring or a double bond. Cyclic hydrocarbons and double bonds (alkenes) are constrained by their geometry, meaning they do not rotate freely about all bonds. So, there’s a difference between having substituents on the same side of the ring (or double bond) and having substituents on opposite sides. For example, the following are geometric isomers of 1,2-dimethylcyclohexane:
Priority of substituent groups is assigned the same way as for absolute configuration. On C-1, the methyl group is given higher priority than the H, and the same is true on C-2. The molecule in which the two higher-priority groups are on the same side is termed cis, and the molecule in which the two higher-priority groups are on opposite sides of the ring is termed trans.
The same-side/opposite-side substituent relativity also occurs with double bonds, but in this case the stereochemistry is officially designated by (Z) or (E). The (Z)/(E) notation is a completely unambiguous way to specify the appropriate stereochemistry at the double bond. In this system, a high and low priority group are assigned at each carbon of the double bond based on atomic number, just as with absolute configuration. If the two high priority groups are on the same side, the configuration at the double bond is Z (from the German zusammen, meaning together). On the other hand, if the two high priority groups are on opposite sides of the double bond, the configuration is referred to as E (from the German entgegen, meaning opposite). Be aware, that the MCAT may also use the terms cis and trans when referring to double bonds. However, this is usually reserved for the case when there is one H attached to each carbon of the double bond, as shown below. The geometric isomers of 2-bromo-1-chloropropene and of 1,2-dibromoethene are shown below:
• Saturated compounds have the general formula CnH2n+2; unsaturated molecules contain rings or π bonds.
• As the substitution of carbocations increases, so does their stability due to the inductive effect; carbanions are more stable when they are less substituted.
• Resonance stabilization results from the ability of π electrons or charge to move and delocalize through a system of conjugated π bonds or unhybridized p orbitals.
• Brønsted-Lowry acids are proton donors, and are stronger (dissociate more) when their conjugate bases are most stable in solution.
• Acidity of carboxylic acids results from the resonance stability of the carboxylate anion.
• Electron withdrawing groups increase the acidity of carboxylic acids by stabilizing the negative charge of the carboxylate anion via the inductive effect.
• Nucleophiles are Lewis bases and are electron rich, while electrophiles are Lewis acids and are electron deficient.
• Nucleophiles are stronger when negatively charged, less electronegative, or larger in size.
• Leaving groups are more likely to leave as their stability in solution increases (uncharged and/or larger groups are usually better LGs).
• Compounds with the same molecular formula are known as isomers; structural, or constitutional isomers differ by the connectivity of atoms in the molecule.
• Conformational isomers differ by rotation around a σ bond.
• Stereoisomers have the same atom connectivity, but different spatial orientation of atoms.
• Chiral molecules have chiral centers (carbon with four different substituents), are not superimposable on their mirror image, and rotate plane-polarized light.
• Enantiomers are non-superimposable mirror images and have opposite absolute configuration at all chiral centers.
• Enantiomers rotate plane-polarized light an equal magnitude, but in opposite direction, therefore a 50:50 mixture of enantiomers, or a racemic mixture, is not optically active.
• The process of separating a mixture of enantiomers is called resolution, and entails conversion of the racemic mixture into a pair of separable diastereomers temporarily before converting back to the pure, chiral enantiomers.
• Diastereomers are stereoisomers that are not mirror images; they differ in absolute configuration for at least one, but not all carbons.
• Epimers are diastereomers that differ in absolute configuration at only one stereocenter.
• Geometric isomers are diastereomers that are cis/trans (or Z/E) pairs on a ring or double bond. When highest priority groups are on the same side of a ring or bond the molecule is cis (or Z); when they’re on opposite sides, the compound is trans (or E).
• Meso compounds are achiral molecules with chiral centers and an internal mirror plane.
1. In the molecule below, what are the hybridizations of C1, C2, C3, and C4 respectively?
A) sp, sp, sp2, sp2
B) sp, sp, sp, sp2
C) sp2, sp2, sp, sp
D) sp2, sp, sp, sp
2. Which of the following structures represents the most stable possible resonance structure for acetic acid (CH3CO2H)?
A) 
B) 
C) 
D) 
3. Rank the conformations of 2-aminoethanol by increasing stability.
A) anti < gauche < eclipsed
B) eclipsed < anti < gauche
C) gauche < anti < eclipsed
D) eclipsed < gauche < anti
4. The most stable conformation of the following substituted cyclohexane has the methyl groups in which of the following positions?
A) 2 Equatorial and 1 axial
B) All axial
C) All equatorial
D) 2 Axial and 1 equatorial
5. How many stereoisomers are possible for cortisone acetate (shown below)?
A) 32
B) 64
C) 128
D) 256
6. What is the correct IUPAC name for the following molecule?
A) (E)-3-heptenoic acid
B) (E)-4-heptenoic acid
C) (Z)-3-heptenoic acid
D) (Z)-4-heptenoic acid
7. Which of the following is the strongest nucleophile?
A) CN−
B) OH−
C) CH3OH
D) NH3
8. Which of the following answer choices lists two pairs of diastereomers?
A) I, III and II, IV
B) I, II and II, III
C) I, III and I, IV
D) II, IV and III, IV
In mammalian systems, aromatic hydrocarbons are enzymatically metabolized by cytochrome P450 into arene oxides when ingested or inhaled. Arene oxides are compounds in which one of the double bonds of an aromatic ring has been converted into an epoxide. These molecules can rearrange to form phenols, which are harmlessly excreted. As shown in Figure 1, arene oxide rearrangement requires the formation of an intermediate carbocation and subsequent hydride shift.
Figure 1 Arene oxide rearrangement
Benzo[a]pyrene, found in tobacco smoke and automobile exhaust, is one of the most troublesome natural arene oxides because it is a procarcinogen due to its bioconversion to a number of harmful molecules such as the 7,8-diol epoxide shown in Figure 2. The danger of the 7,8-diol epoxide lies in the unwillingness of the epoxide ring to rearrange and form a phenol. Instead, the diol epoxide intercalates in DNA and is covalently bound by 2′-deoxyguanosine nucleosides, leading to point mutations in the course of DNA replication.
Figure 2 Oxidation of Benzo[a]pyrene
1. Determine the absolute configurations of Carbons 7 and 8 of the diol epoxide shown in Figure 2.
A) 7R, 8R
B) 7R, 8S
C) 7S, 8S
D) 7S, 8R
2. All of the following statements about epoxides are correct EXCEPT:
A) epoxides and ethers bear the same leaving groups.
B) ring strain and torsional strain increase the free energy of an epoxide.
C) epoxide hydrolysis under basic conditions produces trans-1,2-diols.
D) epoxide oxygen atoms are capable of donating hydrogen bonds.
3. Rank the following four substances in order of increasing reactivity with ethylene oxide.
I. Hydroxide
II. Ammonia
III. Methide
IV. Water
A) III < I < II < IV
B) IV < II < I < III
C) IV < II < III < I
D) II < IV < I < III
4. Which labeled atom in the molecule shown below would be the fastest site of reaction with a thiolate (R-S−)?
A) 1
B) 2
C) 3
D) 4
1. B Both C1 and C2 make up the triple bond. They are both sp hybridized so you can eliminate choices C and D. C3 is part of an allene. The bonds that it forms with its neighbors are linear (180°), so it is also sp. You can eliminate choice A, which leaves choice B as the correct choice. C4 has a double bond and two single bonds so it is sp2.
2. B Good resonance structures must do the following: obey the octet rule, accrue the fewest charges possible, and place negative charges on electronegative atoms and positive charges on electropositive atoms (listed in priority). Further, resonance structures don’t represent oxidation or reduction of molecules, and as such the total charge of each structure must be the charge of the molecule. This eliminates choices C and D. Choices A and B are both valid resonance structures, but only choice B places a full octet on all non-H atoms.
3. B Since this is a ranking question, look for obvious extremes and eliminate answers. Choices A and C should be eliminated because the eclipsed conformation is always the least stable due to sterics and electron repulsions in aligned bonds. Choice D is the more enticing answer of the remaining two because the general rule of thumb is that the anti conformation is the most stable because the bulky groups are farthest apart, while they are 60° apart in a gauche conformation. This question is tricky, however, because in this case there is intramolecular hydrogen bonding which can occur in the gauche conformation, making it the most stable one (eliminate choice D).
4. A Choice B can be eliminated before analyzing the structure since if all three substituents could be axial, then by a ring flip, all three could also be equatorial. The more stable chair conformation puts substituents in an equatorial position since they are less sterically crowded than those in axial positions. Similarly, if choice A is true of the molecule, then by ring flip, choice D must be also. Therefore choice D should be eliminated since it has more axial substituents. Between choices A and C, only choice A fits the compound shown because the relationship between the methyl groups on Carbons 1 and 2 is trans and the relationship between the methyl groups on Carbons 1 and 4 is cis. In this conformation, the methyl groups on Carbons 1 and 2 would be found in the equatorial position and the methyl group on Carbon 4 would be axial, making A the better choice.
5. B The maximum number of stereoisomers is given by the formula 2n, where n equals the number of stereocenters. Cortisone acetate has six stereocenters and 26 = 64. Five of the six ring junctures are chiral centers (all sp3 carbons), as is the carbon with the OH substituent. Choice A would correspond to five stereocenters, choice C would require seven stereocenters, and choice D would correspond to a compound with eight stereocenters.
6. A When naming a compound, number the carbons starting at the end nearest a functional group (the carboxylic acid in this case). Based on the position of the double bond in the molecule, you can eliminate choices B and D. Since the two largest substituents on each carbon of the double bond are on opposite sides of the bond, the double bond has E stereochemistry; therefore, eliminate choice C.
7. A Nucleophiles are electron rich. While neutral compounds that have lone pairs can be nucleophilic, negatively charged nucleophiles tend to be stronger (eliminate choices C and D). The stronger nucleophile is the more reactive nucleophile; more reactive corresponds to less stable. Therefore, the nucleophile that is less able to stabilize a negative charge will be the stronger nucleophile. For choices A and B, the negative charge resides on the C and O, respectively. Since carbon is less electronegative than oxygen, it is therefore less able to stabilize a negative charge, making cyanide the best nucleophile (eliminate choice B).
8. B Stereoisomers in which all stereocenters are inverted are enantiomers, while stereoisomers with at least one, but not all, inverted chiral centers are diastereomers. Molecules I and III are an enantiomeric pair (eliminate choices A and C); Molecules II and IV are enantiomers as well (eliminate choice D). All other pairs of molecules are diastereomers.
1. D
Note that the configuration for Carbon 7 reads as R, but because the hydroxyl group occupies an into-plane position, this assignment is incorrect. One way to get around this is to exchange the hydroxyl group’s position with the hydrogen also on this carbon. Doing so will place the hydroxyl group in a correct position, which results in a clockwise trace. This reads as R but because two groups have been exchanged, the true configuration is S. Once the configuration of Carbon 7 is determined to be S, choices A and B can be eliminated.
2. D Epoxides and ethers all bear an OR leaving group so choice A can be eliminated. Choice B is incorrect because epoxides, due to their 3-point ring structure, possess ring strain and torsional strain, which imparts a relatively high free energy. Epoxide hydrolysis (basic or acidic) will produce trans-1,2-diols, also eliminating choice C. Choice D is the only answer choice consistent with the EXCEPT wording of the question because epoxides are only capable of accepting hydrogen bonds, not donating them.
3. B Each of the four substances listed will act as nucleophiles with ethylene oxide (C2H4O). The strongest nucleophile has a negative charge on the least electronegative atom, which is carbon. Choices B and D both list methide as the strongest nucleophile (eliminate choices A and C). The weakest nucleophile is going to be listed first. The choice is between water and ammonia. Since nitrogen is less electronegative than oxygen, NH3 is a stronger nucleophile than water (eliminate choice D).
4. B Thiolate is a relatively weak base, but a great nucleophile. This means it will be attracted to the electrophilic carbons of the molecule, C-2 and C-3, and effect an epoxide ring opening (eliminate choices A and D). Because the question asks about the fastest site of reaction, the least sterically hindered carbon is the best choice. Thiolate will have easier access to the primary carbon, so eliminate choice C.
5. A The nature of the substituents on each ring act to stabilize or destabilize the carbocation that forms. The methoxy substituent in choice A is an electron-donating group, which will stabilize the carbocation through resonance due to the lone electron pairs on the oxygen.
On the other hand, the nitro group in choice B will destabilize the intermediate since one resonance structure, shown below, puts two atoms with a positive charge next to each other (eliminate choice B). Choice C affords no increase or decrease in stability. Choice D will show mild stabilization due to the inductive donation offered by its methyl group (also shown below).
Overall, choice A shows the strongest carbocation stabilization since resonance effects are generally stronger than inductive ones.
