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O-Chem Section 1 Bonding

Sigma Bonds vs. Pi Bonds:

o Q1. Which is stronger, a sigma bond or a pi bond? Why?

o The first bond formed between two elements is always a sigma bond (σ) (e.g., a single bond, or the first bond of a double or triple bond) and involves directional head-to-head overlap of two atomic orbitals. The second and third bonds are always pi () bonds and involve side-to-side overlap of two p orbitals. Because pi bonds require side-to-side overlap, the atoms must be fairly close to one another. As the radius of either atom increases, the p orbitals are spread apart, resulting in less overlap and a weaker pi bond. This is why a C=N bond is weaker than a C=O bond.

o Comparison of sigma and pi bonds:

Q2. How does the rotational limitation of pi bonds relate to our previous study of proteins?

Hybridization

Definition: Atoms, when bonded, hybridize (i.e., mix) their higher and lower energy valence electron orbitals to form “hybrid orbitals” with intermediate energy. Carbon, for example, has two electrons in the s orbital and two electrons in the p orbital. However, carbon forms four orbitals of equivalent energy when hybridized as sp3.

Determining Hybridization: Count the number of sigma bonds and add the number of pairs of unbonded electrons. This number will equal the sum of the superscripts on one of the following hybridizations: (Note: No superscript indicates a superscript of one)

o sp, sp2, sp3, sp3d, sp3d2

Q3. What is the percent “s” character of the hybrid oxygen orbital in water?

rotation strength stability reactivity

pi bonds prevent rotation weaker less stable more

reactive

sigma bonds

allow for rotation stronger more

stable less reactive

IMPORTANT NOTE Pi bonds themselves are weaker associations than are sigma bonds. However, a double bond (one sigma bond plus one pi bond) is stronger together than a single bond (one sigma bond only). Also, triple bonds are much shorter, stronger bonds than are single bonds, but are also more reactive. Reactivity has to do with the tendency of the third bond (a pi bond) to react. Bond strength is a measure of the energy needed to completely break the two atoms apart (i.e., break all three bonds).

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Shape and Bond Angle

Valence Shell Electron Pair Repulsion Theory: The theory that predicts which shape molecules will take due to the repulsion of lone pairs of electrons. It is often abbreviated as VSEPR.

Shape and Bond Angle Are Determined by: Hybridization!

1) Hybridization

sp = Linear [180˚] sp2 = Trigonal Planar or Bent [120˚] sp3 = Tetrahedral, Trigonal Pyramidal, or Bent [109.5˚] sp3d = Trigonal Bipyramidal, Seesaw, T-Shaped, or Linear [90˚/120˚ or 180˚] sp3d2 = Octahedral, Square Pyramidal, or Square Planar [90˚]

2) Lone Pairs of Electrons

The hybridization state determines the possible shapes for that molecule, as seen in the lists on the previous page. The presence or absence of unpaired electrons determines the exact shape from among those choices. For example, an sp2 hybridized atom can take on the trigonal planar shape (no lone pairs) or the bent shape (one lone pair). All of the possible shapes for each hybridization state are diagramed below. You must know them well and be able to quickly predict the shape of any molecule. Don’t just memorize them abstractly. Use groupings and similarities to help you. (Note: sp is omitted; sp hybridized atoms are always linear)

sp2 sp3 sp3d sp3d2

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Bond Length & Bond Strength:

o Q4. Rank the following according to decreasing bond length: a) triple bonds, b) double bonds, c) single bonds.

o Q5. Rank the bonds listed above according to increasing stability.

o Q6. Rank the bonds listed above according to increasing reactivity.

o Q7. Rank the bonds listed above according to increasing bond strength.

o Q8. Describe the forces responsible for the strength of a bond. What is the relationship between potential energy and bond length?

Bond Polarity:

o Electronegativity: For the MCAT, you should have a general intuition for the relative electronegativity of common atoms. Most of this can come from the periodic table trends covered in the General Chemistry 1 lesson. It is worth knowing that carbon and hydrogen have similar, but not identical, electronegativities (carbon = 2.5; hydrogen = 2.2). Fluorine has the largest electronegativity (4.0) and Francium has the smallest (0.7). Beyond these general trends, if specific electronegativity figures are required, they will always be given.

Dipole Moment: Any time charge is not evenly distributed within a bond (i.e., when the two atoms have non-identical electronegativities) that bond will have a dipole moment, . The dipole moment can be calculated using = d ; where represents charge and d represents distance between charges. (Note: Observe that is also the symbol often used to denote partial positive and partial negative charges; i.e., + and -). You may recall seeing the dipole moment represented on paper as an arrow with a plus sign on its tail. When this convention is used, the arrow points in the direction to which electrons in the bond will be pulled by the more electronegative atom.

Molecules with two or more dipole moments can still have no NET dipole when their geometric orientation causes the dipole moments to cancel each other out. We have seen many students fail to remember this concept when evaluating a molecule. This concept explains why carbon tetrachloride is non-polar even though carbon and chlorine clearly have different electronegativities.

Sample MCAT Question

1) The H-O-H bond angle in water is measured experimentally to be 104.4°. The related angle H-N-H in ammonia is expected to be:

A) larger due to the greater electron repulsion in ammonia B) larger due to the greater electron repulsion in water C) smaller due to the greater electron repulsion in water D) smaller due to the greater electron repulsion in ammonia

Solution: Both molecules are approximately tetrahedral in shape, but experience some distortion due to the repulsion of lone pairs of electrons. In water, two lone pairs push both hydrogen substituents away from the plane shared by the two lone pairs, decreasing the H-O-H bond angle. Ammonia has only one lone pair, so the same effect is observed, but to a lesser degree. Thus, the bond angle is larger in ammonia due to the greater repulsion in water than in ammonia, or answer B.

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Forming and Breaking Bonds:

o Q9. Energy is always when a bond is formed, and when a bond is broken.

o Heat of Combustion: When molecules are combusted, all of the bonds are broken and then reformed via a radical reaction. The less stable the bond, the greater will be the heat of combustion. The more stable the bond, the lower the heat of combustion.

Q10. Draw two energy coordinate diagrams demonstrating why the combustion of a less-stable molecule results in a higher heat of combustion (∆Hcombustion) than the combustion of a more-stable molecule.

Coordinate Covalent Bonds: A coordinate covalent bond is one in which both electrons shared in the bond are donated by one atom. Usually more than one of these “donor” molecules (i.e., Lewis bases) surround and bind a single “recipient” molecule (i.e., Lewis acid; usually a metal). For the MCAT, if a molecule does not have a lone pair of electrons it will NOT participate in a coordinate covalent bond. The complex formed by the metal and the molecules forming coordinate covalent bonds with that metal is called a “coordination complex.”

o Q11. Provide multiple examples of coordinate covalent bonds. Include a human-body example.

The Octet Rule:

o Atoms of low atomic number (<20) tend to gain or lose electrons to obtain exactly eight electrons in their valence shell. This is the highly stable “noble gas configuration.” This is why sodium forms the ion Na+ and not Na2+.

o EXCEPTIONS to the octet rule:

1) Hydrogen and Helium: Stable with only two electrons in their valence shells (e.g., H2)

2) Boron and Beryllium: Stable with only six electrons in their valence shells (e.g., BF3)

3) Atoms from the third period or higher can accept more than eight electrons. Common MCAT examples include: PCl5, SF6, PO4

3- and SO42-.

Valence: The number of bonds an atom “normally” makes: carbon and its family are tetravalent; nitrogen and its family are trivalent; oxygen and its family are divalent and fluorine and its family are monovalent.

Formal Charge: The difference between the # of electrons in an atom’s valence shell when it is in its ground/elemental state and the number assigned (lone-pair electrons and 1/2 of the bonding electrons) to it in a molecule.

o Formal Charge = valence – assigned

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Resonance

Important Clarifications: o Resonance structures are a “snapshot” of the different arrangements of electrons that contribute to

the “actual structure.”

o The actual structure is a weighted average (i.e., hybrid) of all of the contributors and does NOT look exactly like any of the individual resonance structures.

o Individual resonance structures contribute differently; the most stable structures contribute the most to the actual structure and the least stable structures contribute the least.

o The actual structure does NOT resonate back and forth between forms; it is a permanent weighted hybrid of the contributing structures.

Drawing Resonance Structures:

o Atoms can never be moved.

o Single bonds can never be moved.

o All structures must obey the octet rule (excluding the exceptions outlined previously).

o All structures must have the same number of total valence electrons (i.e., the number of electrons in bonds plus the number of electrons in lone pairs). This is NOT to say that individual atoms will not change their valence. In one structure they may have a formal charge and in another they may not, but in all structures the total number of valence electrons for the structure as a whole must be constant.

o The tail of an arrow showing electron flow during resonance can only start from a lone pair, a double, or a triple bond.

Sample MCAT Question

2) Rank the bonds labeled a, b and c in the molecule below from longest to shortest bond length.

A) a > c > b B) a = b > c C) c > b > a D) c = b > a

Solution: The three bonds in question are two double bonds and one single bond. A single bond will always be longer than a double bond. However, in this molecule we must note that the ring structure shown is actually one of two equally-stable resonance structures. All six of these bonds actually have an equal partial double-bond character. This makes bond a equal to bond b in length. Comparing bonds a/b to bond c, we know that c will be shorter because oxygen is smaller than carbon, allowing for a closer, tighter double bond and more pi overlap. Answer B is thus correct.

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Ranking Resonance Structures:

o If determining the relative contribution of various resonance structures to the actual structure, the individual structure that contributes the most to the actual structure is the one that:

1) Allows the most atoms to have a full octet (unless atom is a common exception to the octet rule)

2) Has the least formal charge (i.e., no charge is better than some charge and one formal charge is better than multiple formal charges)

3) Places formal charge on the atom most receptive to that charge (i.e., it is preferable to have a negative formal charge on oxygen than on carbon because oxygen is more electronegative).

o Among similar species that both experience resonance, the more stable specie will be the one with the most possible resonance structures (i.e., ClO4

- is more stable than ClO3- because

perchlorate has four resonance forms and chlorate has only three).

o Q12. Draw all possible resonance structures for a) an amide (R-CONH2) and b) the phosphate ion (PO4

3-). In each case rank each structure in terms of its contribution to the actual structure.

Aromaticity:

o Aromatic compounds are conjugated, unsaturated ring systems that exhibit greater stability than one would expect based on either resonance or conjugation alone. For example, benzene shows far greater stability than other unsaturated compounds with three double bonds and two resonance forms. Conjugated systems are always very stable, but benzene is far more stable than a conjugated straight-chain alkene.

o Hückel’s Rule: To exhibit aromaticity, a ring system must have exactly 4n + 2 pi electrons.

Organic Nomenclature

Functional Groups: You must memorize all of these! The MCAT will refer to functional groups by either their names or their structures and expect you to understand. o Q13. Draw multiple examples of each of the following functional groups:

Alkane Alkoxy Anhydride Acyl Alkene Alkyl Aryl Aromatic Alkyne Carbonyl Benzyl Aliphatic Alcohol Hydroxyl Hydrazine Sulphone Ether Acetyl Vinyl Nitro Ester Gem-/Vic- Allyl Acetal Amine Mesyl/Tosyl Nitrile Ketal Aldehyde Carboxylic Acid Epoxide Hemiacetal Ketone Amide Enamine/Imine Hemiketal

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IUPAC Nomenclature: o Prefixes: meth- ; eth- ; prop- ; but- ; pent- ;

hex- ; hept- ; oct- ; non- ; dec-

o Also know: sec-butyl; isopropyl; isobutyl; tert-butyl

o IUPAC Rules:

1) Find the longest carbon chain.

If there is a tie, the one with the most substituents is the parent chain.

2) The terminal carbon closest to a substituent is numbered #1.

If there is a tie, look at the second substituent.

3) Order the substituents alphabetically and give each one a number to match the carbon to which it is attached.

4) If more than one of the same substituent is present, use the prefixes di, tri, tetra, etc. (i.e., 2,2-dimethylbutane).

5) When to use hyphens:

Hyphens are placed before and after substituent numbers, but NOT between standard prefixes.

6) Rules for alphabetizing:

Do NOT consider the prefix when alphabetizing the substituents if: 1) it represents a number (di-, tri-, etc.) or 2) it includes a hyphen (sec-, tert-, etc.). Do alphabetize other prefixes (isopropyl, isobutyl, etc.).

Formulas, Depictions and 3-D Representations

Familiarize yourself with each of the following ways of writing, drawing, or representing a molecule. o Q14. Draw an example of each type of structural representation below:

o Lewis Dot Structure, Line-Bond Formula, Wedge-Dash Formula, Condensed Formula, Fischer Projection (horizontal lines project out of the page; vertical lines project into the page), Newman Projection, Ball & Stick Model, and Space Filling Model.

Q15. Which of the above models is the most accurate visual representation of an actual molecule?

IMPORTANT NOTE The MCAT will almost never require you to provide a correct IUPAC name for a complex organic molecule. They will, however, refer to molecules in question stems by their IUPAC names, in which case you’ll need to recognize those molecules and be able to draw them. You must also recognize a functional group from its associated suffix. For example, the “–ol” ending denotes an alcohol, the “–oic acid” ending denotes a carboxylic acid, the “–oate” ending denotes an ester, and so on.

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Isomers

Definition: Two molecules are isomers if they have the same molecular formula but are actually different compounds.

o Isomers are frequently and consistently tested on MCAT-2015.

Predicting the Number of Isomers:

o The maximum number of optically active stereoisomers for any compound = 2n, where n is the number of chiral centers.

Types of Isomers: Know all classes of isomers extremely well. They will be on your MCAT.

1) Conformational Isomers:

These are NOT true isomers! They are the same exact molecule. When a molecule twists or rotates around its bonds these are considered “conformers” NOT isomers.

Q16. Draw the six conformers of butane and rank them according to their relative stability (Hint: some of the six conformers are of equal stability).

2) Structural Isomers:

Same formula, different bond-to-bond connectivity (For example: 2-methylpentane and 3-methylpentane are both C6H14).

3) Stereoisomers:

Same formula, same bond-to-bond connectivity, but different in the 3-D arrangement of their substituents. There are two categories of stereoisomers: enantiomers and diastereomers.

Enantiomers:

Two molecules with the same molecular formula and the same bond-to-bond connectivity that are non-identical, non-superimposable mirror images. They contain at least one chiral center.

Chirality: Often referred to as “handedness.” Any atom attached to four different substituents must be chiral. Any atom with less than four substituents cannot be chiral (and therefore cannot be an enantiomer).

Absolute Configuration (R and S): How two enantiomers are distinguished.

Assign priority to all four substituents based on molecular weight, with higher molecular weight atoms receiving higher priority. Rotate the lowest priority substituent to the back (i.e., into the page). If proceeding in order from one to three requires a clockwise motion, the absolute configuration is R; If proceeding in order from one to three requires a counter-clockwise motion, the absolute configuration is S.

Relative Configuration: Two molecules have the same relative configuration if their spatial arrangement is identical, but they have one—and only one—non-identical substituent.

Rotation of Plane-Polarized Light:

Q17. Provide a conceptual definition for each of the following: observed rotation, specific rotation, polarimeter, plane-polarized light, optically active vs. optically inactive, and racemic mixture.

Enantiomers rotate plane-polarized light. R and S enantiomers rotate this light to the same degree but in opposite directions. R enantiomers can rotate this light in a clockwise or counterclockwise direction, as can S enantiomers.

If a compound rotates light clockwise it is called (+) or d (dextrorotary)

If a compound rotates light counterclockwise = (-) or l (levorotary)

Q18. Can an optically inactive compound rotate plane-polarized light?

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Remember the following regarding enantiomers:

a) They have opposite R/S configurations at every chiral carbon (mirror images).

b) They rotate plane-polarized light to the same degree, but in different directions.

c) They have all the same physical properties (boiling point, reactivity, etc.) EXCEPT for: a) how they rotate plane polarized light and b) the products they form when reacted with another chiral compound.

Diastereomers:

Two molecules with the same formula and the same bond-to-bond connectivity that are non-identical, but are NOT mirror images. There are three kinds of diastereomers you must be familiar with for the MCAT: geometric isomers, epimers and anomers.

Notice the similarities in the definitions of enantiomers and diastereomers. Compare both definitions carefully and make sure you can easily distinguish between them.

Geometric Isomers (cis/trans): cis = same side; trans = opposite sides.

The E/Z Convention: prioritize the two constituents on each carbon by molecular weight (as you do for R/S). If the two higher priority substituents are on the same side = Z. If the two higher priority substituents are on opposite sides = E

Remember the following regarding geometric isomers:

a) Cis isomers often have a dipole moment, but trans isomers usually do NOT (i.e., cis-1,2-dichloroethene vs. trans-1,2-dichloroethene)

b) Cis isomers often experience “steric hindrance,” resulting in a higher energy molecule, but trans do NOT.

Q19. Which version of 2,3-dibromo-2-butene would have the higher heat of combustion, the cis isomer or the trans isomer? Which would have the higher boiling point?

Epimers: Diastereomers that differ at only one chiral center. Many pairs of carboydrates are epimers (e.g., glucose and galactose).

Anomers: Molecules that differ only in their spatial orientation at the anomeric carbon of a ring structure. If the anomeric OH/OR group and the CH2OH group are on the same side of the ring = Beta, if they are on the opposite side = Alpha.

4) Meso Compounds:

Molecules with two or more chiral centers that contain a plane of symmetry. This symmetry cancels out their optical activity.

To recognize them, look for a center of symmetry. However, not all molecules with a plane of symmetry are meso compounds. Next, check for mirror chiral centers across the plane of symmetry (NOT between two molecules; remember that a meso compound is one molecule).

Q20. Are meso compounds diastereomers? Are they enantiomers? Why or why not?

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Organic Chemistry Reactions

The remainder of the O-Chem material from this point forward focuses on reactions. For all reactions marked with a double asterisk (**) you should be familiar with the mechanism. For the other reactions you can focus on just the products and reactants.

Electron Flow:

o Electrons flow from high concentration to low concentration during reactions. In other words, they flow from a species with a full or partial negative charge to a species with a full or partial positive charge. Electron dense species, such as bases and nucleophiles, always attack electron poor species, which are called electrophiles.

Bases: Bases abstract protons (i.e., hydrogens). Bases are electron dense and have a full or partial negative charge. The strongest base is the one that forms the strongest, most stable bond with a hydrogen. In other words, for a strong base the acid/base equilibrium favors the conjugate acid, NOT the base. Basicity is a function of thermodynamics—it quantifies the stability of the reacted base (BH) compared to the unreacted base (B:-). Basicity says nothing about how quickly the base will react. Put another way, basicity describes how much the molecule “wants” to react, but nothing about how quickly it will do so.

Nucleophiles: Nucleophiles attack carbons (or other central atoms). Nucleophiles are electron dense and have a full or partial negative charge. The best nucleophile is the one that reacts fastest with an available electrophile. Nucleophilicity is a function of kinetics—it quantifies the rate at which a molecule reacts with a standardized electrophile. Put another way, nucleophilicity describes how “easily” or “readily” a molecule will react, but nothing about how stable the new bond will be.

Electrophiles: Electrophiles are electron poor species with a full or partial positive charge. Electrophiles accept electrons from either nucleophiles or bases.

Leaving Groups: Leaving groups are atoms or molecules that leave the parent molecule during a reaction and take both electrons from the bond with them. The best leaving groups are those that are the most stable after they leave.

Q21. Provide five examples each of bases, nucleophiles, electrophiles, and leaving groups.

Sample MCAT Question

3) An unknown molecule has an absolute configuration of S and rotates plane-polarized light 15 degrees in the clockwise direction. Which of the following molecules is enantiomeric to the unknown?

A) Absolute configuration is R; rotates plane-polarized light 15°; levorotary B) Absolute configuration is R; rotates plane-polarized light 165°; levorotary C) Absolute configuration is S; rotates plane-polarized light 15°; dextrorotary D) Absolute configuration is R; rotates plane-polarized light 15°; dextrorotary

Solution: Enantiomers must have opposite R/S configurations. You CANNOT predict dextrorotary or levorotary from R and S. However, if you know one enantiomer is dextrorotary, then the other enantiomer must be levorotary. In this case, the molecule was S and rotated light clockwise (dextrorotary), so its R equivalent must be levorotary, or Answer A.

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Reaction Types: E1/E2 and SN1/SN2

o E1 (Elimination, Unimolecular)

Rate depends on the concentration of one species only and is thus first order.

Involves two steps:

1) The dissociation of the leaving group, resulting in formation of a carbocation [slow step]. 2) The abstraction of a proton with collapse of the electrons to form a double bond (which

quenches the carbocation) [fast step].

o E2 (Elimination, Bimolecular)

Rate depends on the concentration of two species and is thus second order.

Involves one step only:

1) The single-step abstraction of a proton with collapse of the electrons to form a double bond and ejection of the leaving group.

o SN1 (Nucleophilic Substitution, Unimolecular)

Rate depends on the concentration of one species only and is thus first order.

Involves two steps:

1) The dissociation of the leaving group, resulting in formation of a carbocation [slow step]. 2) Attack of the carbocation by the nucleophile [fast step].

o SN2 (Nucleophilic Substitution, Bimolecular)

Rate depends on the concentration of two species and is thus second order.

Involves one step only:

1) The single-step “back side attack” of the electrophile with simultaneous ejection of the leaving group.

o Master all of the information presented in the following summary table:

number of steps

order of reaction

carbocation formed

methyl or hydride shifts

product stereo-chemistry favored by

E1 2 first order yes yes planar

weak bases; 3° carbons only; polar protic

solvents

SN1 2 first order yes yes racemic mixture

poor nucleophiles; 3° carbons only;

polar protic solvents

E2 1 second order no no planar strong and/or

bulky bases

SN2 1 second order no no

inversion of relative

configuration

good nucleophiles;

methyl, 1° or 2° carbons

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Alkanes

Definition: Compounds made entirely of carbon-carbon or carbon-hydrogen single bonds (e.g., gasoline, tar, crude oil, butane, methane, etc.)

Nomenclature: Alkanes are named with the –ane suffix.

o Carbon Substitution:

o Q22. Describe the difference between methyl, primary, secondary, tertiary and quaternary carbons.

Physical Properties:

o Insoluble in water; very low density; non-polar; most are oils or gases

o Melting and Boiling Point Trends: Master the trends listed below; these general principles apply not only to alkanes, but to most other functional groups we will study.

1) Boiling point increases with increasing chain length and/or molecular weight. 2) Boiling point decreases with increased branching. 3) Melting point increases with increasing chain length and/or molecular weight. 4) Straight-chain alkanes have the highest melting points. Among branched alkanes,

however, increased branching increases melting point.

o Cyclic Compounds: Ring Strain: Cycloalkanes create ring strain because they force bond angles to deviate from the

optimum tetrahedral angle of 109.5°. Cyclohexane in its chair conformation has zero ring strain. Cycloalkanes with more or less than six carbons exhibit increasing ring strain as one moves away from six. Very large rings (i.e. 10+ carbons) have enough freedom to again approximate the tetrahedral angle. Bicyclic ring systems generally exhibit more ring strain than do monocyclic rings.

Axial vs. Equatorial: On the chair conformation of cyclohexane the axial substituents rise vertically from the ring and the equatorial substituents extend horizontally. Q23. Draw both chair conformations of cyclohexane. In which position, axial or equatorial,

would a large substituent be most stable? Try to draw cis-1,2-dichlorocyclohexane with both chlorines in the more stable position (i.e., axial vs. equatorial).

Combustion of an Alkane: a radical, exothermic chain reaction with oxygen; high energy of activation.

o CH4 + 2O2 CO2 + 2H2O + Heat

o Heat of combustion is a favorite MCAT topic. This is the third time we’ve mentioned it in the lessons, so make sure you’ve got it down.

Q24. N2 is one of the most stable molecules known; will it have a high or low heat of combustion? Rank cyclopropane, cyclobutane, cyclohexane and cyclooctane according to increasing heat of combustion per –CH2 group.

o The AAMC has stated that “radical reactions” will not be on MCAT-2015. We believe this to mean that you will not need to understand radical mechanisms such as initiation, propagation, and termination. Combustion, however, is central to many concepts that remain on the MCAT-2015 topic list, and it happens to occur via a radical pathway. Remember that just because something is removed from the topics list does NOT mean you will NEVER see it. It simply means specific required background knowledge won’t be necessary. We fully EXPECT radicals to remain on the MCAT in some form; especially given that they are a major source of tissue and DNA damage and living systems have developed specific antioxidant mechanisms to counteract them.

o General Awareness of Radicals: Be generally familiar with how radicals form. The heterolytic cleavage of bonds during reactions results in paired electrons on both product and reactant. A homolytic cleavage sends ONE electron to one species and ONE to another, generating radicals. Radicals also form as a result of damage to the molecule itself. For example, radiation has sufficient energy to strip electrons off of molecules, creating radicals. That is why radiation exposure is dangerous and carcinogenic.

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Synthesis of an Alkane from an Alkene:

o Alkanes can be formed by reducing an alkene with H2 in the presence of a metal catalyst. The reaction is an example of syn addition, where both new bonds are formed on the same side. In anti addition, the new bonds are formed on opposite sides.

CH2=CH2 + H2/Pd(catalyst) CH3CH3

Alkenes

MCAT-2015 does NOT specifically test alkene reactions. However, you will be tested on the general characteristics of alkenes and you will certainly see them as part of other reactions such as beta-oxidation, dehydration of an alcohol, etc. Further, double bonds are present in almost every biomolecule, so with the new MCAT-2015 emphasis on relevance to living systems, expect to see a lot of alkenes. Whenever we instruct you that a particular subject is not required knowledge for the MCAT, that statement does NOT mean that you will never encounter that subject on an MCAT exam. Rather, it means that if you were to encounter it on an exam it would most likely be discussed in a passage or question stem. All pertinent information would be given and no prior background knowledge would be required.

Knowing the following alkene properties will suffice:

Definition: An alkene is any species with a double bond.

o Alkenes are nucleophiles. The pi electrons in the double bond will attack electrophiles forming a new bond to one of the carbons and leaving a carbocation on the other. The carbocation is then quenched by a nucleophile.

o When located one carbon away from another atom, alkenes are weakly electron withdrawing.

o Alkene Stability: Alkyl substituents (R-groups) increase alkene stability. Thus, tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted

Alkynes

Definition: An alkyne is any species with a triple bond. Alkynes are almost identical in how they react to alkenes. They are also NOT tested specifically on MCAT-2015.

Benzene Yet again, benzene reactions are NOT specifically tested by MCAT-2015, and are NOT included on the “MCAT-2015 Topic List.” However, you will need to recognize benzene and know at least the following:

Definition: A benzene ring is a six-membered ring with alternating double and single bonds. (Note: Of course, the bonds don’t really alternate; they form a conjugated pi system often drawn as a circle.)

o Three Representations: --Ph, --C6H5 and sometimes --Ar

Don’t confuse this: A “phenyl” group (--Ph) is a benzene attached directly to the primary chain; in a “benzyl” group, a –CH2 is attached to the primary chain on one side and benzene on the other; an “aryl” group (--Ar) is any aromatic ring system, which would include phenyl groups, but also many others.

C6H5: Make sure you recognize what you’re looking at when you see “C6H5.” It represents a benzene, not an alkane chain. Also, there are double bonds in a C6H5 group, so it is unsaturated. We have seen many students mistake this condensed formula, C6H5 for an alkyl substituent.

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Alcohols

Definition: An alcohol is any species with an –OH functional group.

o Alcohols behave as either nucleophiles (the lone pair on oxygen acts as a Lewis base) or as Lewis acids when they are oxidized to carbonyl groups (i.e., the oxygen accepts a pair of electrons from the O-H bond as the proton is abstracted).

Nomenclature: Alcohols are named with the –ol suffix (i.e., butanol, cyclohexanol, 1,3-hexanediol)

Physical Properties:

o Melting and Boling Point: Like alkanes, boiling point goes up with increasing molecular weight and down with increased branching. Melting point also goes up with increasing polarity and H-bonding Because alcohol substituents can be considered as branching, the effect of branching on the melting points of an ROH is variable.

o Hydrogen Bonding:

Hydrogen Bond Donors/Acceptors: F, O or N bonded to a hydrogen can act as BOTH a hydrogen bond donor and a hydrogen bond acceptor. F, O, N atoms with lone pairs that are NOT attached to a hydrogen (i.e., ROR, RCOR, etc.) can act as hydrogen bond acceptors.

Hydrogen bonding is a concept of primary importance for the MCAT. It explains most of the unique properties and behavior of alcohols and amines. For example, we said earlier that the effect of branching on melting point is less predictable for alcohols. One reason for this is because the major intermolecular force present in alkanes is the Van der Waals force—which is dramatically decreased by branching because it impedes close stacking of the molecules in the solid. In alcohols, however, hydrogen bonding dominates and the impact of Van der Waals forces becomes far less significant.

In relation to hydrocarbons with similar molecular weights, alcohols have much higher boiling points and melting points due to hydrogen bonding.

Q25. Most alcohols are highly soluble in water. Explain why.

o Alcohol Acidity:

Alcohols are less acidic than water.

Alcohol acidity increases from tertiary to secondary to primary.

Q26. Provide a conceptual explanation for the two acidity trends described above. Predict the acidity of carboxylic acids compared to alcohols and water.

**Formation of an Alkyl Halide from an Alcohol

o Via SN2: CH3OH + HCl CH3OH2+ + Cl- CH3Cl + H2O

o STEPS:

1) The alcohol acts as a nucleophile, attacking the electrophilic halide hydrogen and forming the good leaving group water plus a halide ion.

2) The halide ion attacks the central carbon via SN2, kicking off water. Don’t be confused by the fact that SN2 reactions are usually considered “one step.” In this case the protonation of the alcohol is somewhat of a “preparatory” step to transform a hydroxyl group into a good leaving group. This same preparatory step will be necessary in the SN1 reaction described below.

o Via SN1: R3COH + HCl R3COH2+ + Cl- R3C+ + H2O R3CCl

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o STEPS:

1) The alcohol acts as a nucleophile, attacking the electrophilic halide hydrogen and forming the good leaving group water plus a halide ion.

2) The good leaving group water spontaneously dissociates (rate-limiting step) leaving a carbocation.

3) The chlorine ion attacks the carbocation to form the product.

Q27. Draw a mechanism for the formation of an alkyl halide from an alcohol via both SN1 and SN2.

Oxidation of an Alcohol

o In general, adding an oxidizing agent to an alcohol will cause progression through the following algorithm. It can become a bit complex because some stronger oxidizing agents take primary alcohols straight to carboxylic acids while others stop at the aldehyde. Other reagents will oxidize secondary alcohols to ketones but not primary alcohols to aldehydes, and so forth. The MCAT would not expect you to memorize each case, so focus on recognizing the oxidizing agents themselves and general characteristics of oxidation—such as the fact that secondary alcohols can only form ketones, and tertiary alcohols cannot be oxidized at all.

1˚ Alcohols Aldehydes Carboxylic Acids

2˚ Alcohols Ketones

3˚ Alcohols cannot be oxidized further

o Common oxidizing agents include: O3, Cr2O7, CrO4, KMnO4, Jones, Collins, PCC, PDC, etc.

Q28. Potassium dichromate is often used as an oxidizing agent. A solution of dichromate ions (Cr2O7

2-) is orange, but a solution of chromium ions (Cr3+) is green. If alcohol A is reacted with potassium dichromate and produces a green solution, and alcohol B is reacted with the same reagent and produces an orange solution, what can be inferred about the two alcohols, A and B?

Reduction Synthesis of an Alcohol

o Generally speaking, reducing agents can reverse the algorithm shown above.

o Reducing agents such as NaBH4, LiAlH4 and H2/pressure reduce a carbonyl to an alcohol. NaBH4 can only reduce aldehydes and ketones; LiAlH4 and H2/pressure can reduce aldehydes, ketones, carboxylic acids and esters. The difference in strength between LiAlH4 and NaBH4 is a more common theme in O-Chem and therefore the MCAT will require this specific knowledge.

o Q29. Why can the weaker reducing agents only reduce aldehydes and ketones?

**The Pinacol Rearrangement (Polyhydroxyl Alcohols)

o vic-diol + hot acid ketone or aldehyde

o The two hydroxyl groups must be in the vicinal (vic-) position and the carbons bearing those two hydroxyl groups must be tri- or tetra-substituted by –R groups. If tri-substituted, it will yield an aldehyde, and if tetra-substituted it will yield a ketone.

o Q30. Propose a mechanism for the pinacol rearrangement.

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**Protection of Alcohols

o It is often helpful during a synthetic scheme to be able to protect alcohols from oxidation or related reaction while still allowing the reaction to occur with other functional groups on the molecule. Two common examples are given below:

Protection with TMS: ROH + TMS RO-Si(CH3)3

Protection by MOM (Mothers are very protective! MOM stands for methoxymethyl ether) ROH + Base RO- + CH3OCH2Cl (MOMCl) ROCH2OCH3 (RO-MOM)

Acidification will remove either protecting group to restore the alcohol.

Reaction with SOCl2 and PBr3

o ROH + SOCl2 RCl

o ROH + PBr3 RBr

**Formation of Tosylates/Mesylates

o Tosyl-Cl + ROH Tosyl-OR + HCl

o STEPS:

1) The alcohol attacks the tosyl or mesyl halide via SN2, kicking off a halide ion.

2) A hydrogen is abstracted by the halide ion, quenching the charge on the oxygen.

Q31. Draw a mechanism for the formation of a tosylate or mesylate. Why would we want to form a tosylate or a mesylate anyway?

**Dehydration of an Alcohol: Synthesis of an Alkene

o CH3CH2OH + H2O CH3CH2OH2+ CH2=CH2

This is an equilibrium reaction. The alkene is favored by hot, concentrated acid; the alcohol is favored by cold, dilute acid.

The major product is the most substituted, most stable alkene

o STEPS:

1) The alcohol is protonated by the acid 2) The “good leaving group water” leaves, forming a carbocation 3) Methyl or hydride shifts can occur, but only if it results in a more stable carbocation. 4) A water molecule abstracts a proton and the electrons collapse to quench the carbocation and

form an alkene.

Q32. Draw a mechanism for the dehydration of an alcohol.

**Grignard Synthesis: Production of an alcohol with extension of the carbon chain

o CH3COCH3 + CH3MgBr CH3COH(CH3)2

Produces an alcohol by adding RMgX (most frequently RMgBr; called an organometallic compound) to a carbonyl.

Results in an increase in the number of carbons.

The Grignard Reaction will also work with other electrophilic double bonds such as C=N, cyano groups, S=O and N=O

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o STEPS:

1) Due to the very low electronegativity of Mg, the R group in RMgBr gains significant electron density and more or less acts as if it were a carbanion (R:-), attacking the electrophilic carbonyl carbon. This occurs in a single step, kicking the electrons in the C=O bond up onto the oxygen.

2) The negatively charged oxygen is protonated, yielding an alcohol.

Q33. Draw a mechanism for the reaction of (CH3)2CHMgBr with 2-pentanone.

Ethers

Definition: An ether is an oxygen bonded to two –R groups (R-O-R).

Nomenclature: On the MCAT simple ethers are sometimes referred to by common names wherein each –R group is named separately (i.e. methyl ethyl ether, diethylether, methyl propyl ether, etc.). IUPAC rules call for the standard naming of the longer of the two –R groups as the parent chain, and naming of the other –R group as a substituent with the suffix “-oxy” added. (i.e., 1-ethoxyheptane).

Properties:

o Very non-reactive

o Weakly polar

o With short –R groups ethers are slightly soluble in water

o Most non-polar species are soluble in ethers

o Low boiling point (no H-bonding)

All of the above characteristics make ethers excellent solvents.

o Q34. Describe to your tutor why each attribute is desirable in an organic solvent.

**Reactions of Ethers

o You are unlikely to see an ether participate in a reaction on the MCAT. It will almost always be the solvent and it therefore should NOT be participating.

o If ethers do react, it will only be after the oxygen is protonated by strong acid. The resulting unstable intermediate could then be attacked by a nucleophile.

Epoxides

Definition: Epoxides are cyclic ethers involving one oxygen and two carbons in a three-member ring.

**Substitution of an Epoxide

o Epoxides suffer from severe ring strain, making them highly reactive.

o Acid catalyzes the reaction by protonating the oxygen and making it a better leaving group.

o Q35. Substitution of an epoxide can proceed via an SN1 or an SN2 pathway. Use your knowledge of these two types of reactions to predict and draw a mechanism for both pathways.

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Making Predictions

One of the most common—and most challenging—things you will be asked to do on MCAT O-Chem questions is to make predictions. You’ll be asked to predict reactants, products, mechanisms, acidity, basicity, nucleophilicity, SN1 vs. SN2, etc. Some of the following principles should be a review of what we have already covered, while other principles are mentioned for the first time. Be sure you truly understand each principle—these are necessary skills for MCAT O-Chem.

Electronegativity:

o The greater the difference in the electronegativity of two atoms in a bond, the more polar the bond. This shows us that oxygen forms stronger hydrogen bonds than does nitrogen.

Q36. What does this tell us about the relative reactivity of the carbons in a C=O vs. a C=N bond?

o Polarity results in partial positive and partial negative charges, which are clues when trying to identify the nucleophile and electrophile in a reaction. To be a base or nucleophile, a molecule must have a full or partial negative charge. To act as an electrophile, it must have a full or partial positive charge.

Electron Donating and Withdrawing Groups:

o Often, you’ll be asked to distinguish between the basicity, nucleophilicity, etc., of two species that appear quite similar. This usually boils down to how the atoms in question are affected by electron donating or withdrawing groups.

o To determine if a group is electron donating or withdrawing:

1) Look at the first atom from the point of attachment. Compare its electronegativity to the atoms bound to it. If it is more electronegative, it will bear a partial negative charge and if it is less electronegative, it will bear a partial positive charge.

2) Atoms with full or partial positive charges withdraw from whatever they are attached to. Atoms with full or partial negative charges donate to whatever they are attached to.

3) Hydrogen is considered neither electron donating nor withdrawing.

4) Alkenes are weakly electron withdrawing (just memorize this one).

Q37. Label each of the following as electron donating or electron withdrawing: alkyl groups, nitro groups, cyano groups (a.k.a. nitriles), sulphones, amines, carboxylic acids, esters, alcohols and quaternary amines.

Q38. Use the principles of electron donation and withdrawal to explain why the alpha hydrogens on the methyl side of a methylethyl ketone are more acidic than those on the ethyl side.

Mechanisms: SN1/SN2 and E1/E2

o Realize that SN1 competes with E1 and SN2 competes with E2. SN1 does NOT compete with SN2, nor does E1 compete with E2.

o Review the chart on this subject in the O-Chem 2 lesson. The major factors that determine which mechanism will be used are: structure of the electrophile that contains the leaving group, steric hindrance, and the strength of the nucleophile or base.

o Remember that—as far as the MCAT is concerned—a 3° carbon must react via an SN1 or E1 mechanism. This is not an absolute truth, but a convention followed by the AAMC to avoid potential confusion or unfair test items as a result of the fact that 2° carbons can proceed via either mechanism. They have simplified this issue by only using tertiary carbons for SN1 and E1 mechanisms. That said, it is always possible that a passage could carefully explore the chemistry of 2° carbons and factors causing them to proceed via either mechanism. In such a case, however, no prior knowledge of a “general rule” would be required; everything required would be in the passage.

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Acidity:

o ALWAYS look at the stability of the conjugate base! This stability, or lack thereof, will often be affected by electron donating or withdrawing groups (e.g., alcohols are weaker acids than water due to the donating effect of the -R group).

o Resonance Stabilization: This explains why the alcohol group on a carboxylic acid is more acidic than other hydroxyl groups and why alpha hydrogens are acidic.

Basicity:

o Electron donating groups increase basicity, while electron withdrawing groups decrease basicity.

Basicity vs. Nucleophilicity:

o Steric hindrance favors basicity over nucleophilicity; nucleophiles must have very little hindrance. Primary nucleophiles are most common. Secondary atoms can often act as bases or nucleophiles, depending on conditions. However, tertiary atoms will only act as bases.

o Reactivity (low stability) favors basicity over nucleophilicity (e.g., NH2- vs. RO-). If an atom

has a full negative charge it will almost always act as a base (halides are a notable exception).

Electrophilicity:

o This is the easy one—but many students still make errors here. Electrophiles are always electron poor. They will always have a full or partial positive charge. Do NOT start a reaction arrow at an electrophile. Electrophiles always get attacked by electron rich species, NOT the other way around!

Predicting Reactions & Products:

o Things to remember when predicting mechanisms and products:

1) CARBOCATIONS: The mechanism will always proceed through the most stable carbocation (unless peroxide is present). Carbocation Stability = tertiary > secondary > primary

2) STERIC HINDRANCE: If more than one mechanism, intermediate, electrophile, or nucleophile is possible, the one that involves the LEAST steric hindrance will be favored.

3) COUNT YOUR CARBONS: Be sure that the product has the correct number of carbons. Pay special attention to those reactions that add to, or take away from, the length of the carbon chain (e.g., Grignard synthesis, Hofmann degradation, aldol condensation, and acetoacetic ester synthesis.

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Sample MCAT Question

4) The Grignard Synthesis is a useful step in many synthetic pathways because the attachment of the organometallic substituent to almost any carbon transforms that carbon into a good nucleophile. Given that carbons do not act as nucleophiles under standard conditions, this functionality is best explained by the:

A) high electronegativity of the metal compared to the carbon to which it is attached B) lower electronegativity of the halogen compared to the associated metal C) ability of metals to give up valence electrons easily D) polarity of the halogen-carbon bond within the organometallic reagent

Solution: The standard form of an organometallic reagent is R-MX, where M represents a metal and X represents a halogen or other highly electronegative atom. R-MgBr is the most common form. The R group, or more specifically, the first carbon in the R group (which is attached to the metal), acts as the nucleophile. Carbon is far more electronegative than the metal, allowing it to increase its electron density dramatically via induction. Were the carbon to be attached to an element that did not give up electrons easily, the typical organometallic reagent reactivity would not occur. This makes C the correct answer.