4. Single > Double > Triple [bond length]

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1 1. Sigma bonds are significantly stronger than pi bonds. This is because sigma bonds allow for electron density to be concentrated to a much larger degree between the two nuclei. The lowest energy state for an electron electrostatically attracted to both nuclei is between those two nuclei and as close to each nucleus as possible. In a pi bond the p orbitals overlap above and below the atom, localizing the electrons above and below the plane of the bond a higher energy state compared to the head-on overlap of a sigma bond. You can also conceptualize that pi bonds are weaker simply because we know those electrons are in a higher-energy state. It is universally true that when a bond is higher in energy it will require less energy to break it. 2. The amide functional group in a protein (created when the amino group of one amino acid attacks the carboxylic acid group of another amino acid) exhibits resonance and therefore both the C-O and C-N bonds have double-bond character. This prevents rotation a key characteristic of peptide bonds. 3. When orbital hybridize, they do so in a weighted manner such that the character of the hybrid orbital is an average of its contributors. Therefore, an sp hybrid orbital has 50% s character and 50% p character. By this same logic, the oxygen in water, which is sp 3 hybridized, exhibits 25% s character and 75% p character. Mathematically, 1s orbital and 3p orbitals equal 4 orbitals (1s + 3p = 4 orbitals). 1 out of 4, ¼, or 25% is from the s orbital. 4. Single > Double > Triple [bond length] 5. When you see the term stability on the MCAT they mean thermodynamic stability which in essentially every case will be reflected by the strength of the bond. Remember that strong bonds are a lower-energy state and weaker bonds represent a higher-energy state. Therefore, triple bonds are the most thermodynamically stable, meaning it will require the most energy to break them apart. In increasing order of stability they would be ranked: single < double < triple. (Be careful not to confuse stability with reactivity. We ll discuss that next!) 6. In this case, reactivity is measured in terms of how readily a species will react in the presence of a nucleophile or electrophile. You should definitely know that alkanes are very unreactive, and that alkenes are far more reactive. However, it is perhaps surprising that alkenes are actually slightly MORE reactive than alkynes. Still, alkynes are far more reactive than alkanes. The higher reactivity of alkenes is accounted for by the fact that the pi electrons in an alkyne are slightly more stable than the pi electrons in an alkene. One potential explanation is that the sp hybridized carbons in the alkyne have more s character and pull more tightly on those pi electrons, making them more centralized and therefore lower energy than the pi electrons in an sp 2 hybridized alkene.

2 7. Single < Double < Triple 8. A chemical bond is a balance of electrostatic forces. We know that species with opposite charges (i.e., nuclei and electrons) will be attracted to each other with a force given by Coulomb s Law: F = Kqq/r 2. However, remember that there are also repulsive forces between the two nuclei and between electrons. The bond is formed at a sort of sweet spot where potential energy is lowest. Increasing bond length or decreasing it would raise the energy of the bond and therefore weaken it. 9. Released ; required. One example that often confuses students is ATP. They have the general concept that breaking the phosphate bond in ATP releases energy. That is wrong. Breaking the ATP bond does require energy, but the new bonds formed in particular the new OH bond resulting from hydrolysis, releases more energy than was required to break the phosphate bond of ATP. 10. The diagram below shows two RXN coordinate diagrams combined, one for combustion of Reactant1 and one for Reactant2. Both reactants are hydrocarbons and therefore the products of combustion are the same carbon dioxide and water. That is the key to understanding why less stable molecules release more energy: the energy of the products is always going to be the same (i.e., the energy of CO 2 + H 2 O). Therefore, the higher the energy of the reactant the greater will be the difference in energy between the reactant and the combustion products. (Note: This diagram shows the energy of the transition state being the same for both reactions, but that is arbitrarily chosen. This is a path-independent process and so the only concern is the difference in energy between the reactants and the products). FIGURE 1

3 11. Students will see two types of coordinate covalent bonds. The first one is illustrated any time a nucleophile with a lone pair abstracts a proton: NH 3 + HCl NH 4 + Cl -. The amine binds a fourth hydrogen, but the electron belonging to that hydrogen stays with the chlorine. Both electrons that were part of the amine lone pair now form the new covalent bond between nitrogen and the fourth hydrogen. Another example, and the one we ve seen most on the MCAT, is easily recognizable. It involves a metal coordinated with one or more molecules that have lone pairs. When you see something like [Fe(NH 3 ) 4 ] 2+ or [Co(NH 3 ) 6 ] 3+ these structures should stand out because they look almost like ionic compounds, but obviously the amine does not have a charge (so they cannot be ionic). This thought would lead one to wonder how the amines are attached to those metals? The answer is via coordinate covalent bonds. A common and very applicable example of coordinate covalent bonds in the human body is the binding of iron in the heme unit of hemoglobin: Examples of coordinate covalent bonds: FIGURE 2

4 12. The possible resonance structures are drawn below. For the amide, the most stable structure is the one to the far left because there is no separation of charge. Next in line is the middle structure because it places a positive formal charge on carbon, whereas the structure to the right places that charge on nitrogen. Among the phosphate structures, the least stable will be the structure with a positive formal charge on the phosphorous because this has more formal charges and is the only one with both negative and positive formal charges. The other four structures will be equal contributors. FIGURE These functional groups should be very familiar to all tutors or can be easily verified, so we will not list them here.

5 14. Examples of each representation are given below. Note that the MCAT has occasionally referred to what we list here as a line-bond structure as a Lewis Structure. When the term Lewis Dot Structure is used it always refers to the use of dots to represent each electron. CH 3 CO(CH 2 ) 3 CH 3 Lewis Dot Line-Bond Wedge-Dash Condensed Fischer Projection Newman Projection Ball & Stick Space-Filling FIGURE The most conceptually-accurate representation in terms of what you would actually see if you could take an Incredible Voyage down to the molecular level would be the space-filling model. However, remember that those electron clouds would not be nicely color-coded for you. They are only math functions that predict where the electrons might be. If you were seeing a real methane molecule at that size and scale you would most likely see nothing. The first thing you could see would be the nucleus and you would need to have magnification many, many times that to see an electron.

6 16. Drawings of the six conformers of butane and their relative energies are diagrammed below. FIGURE Observed rotation is simply the degree to which a sample rotates plane polarized light. However, that rotation is not a universal constant for a particular molecule. Rather, it varies depending on concentration, length of the tube, etc. Specific Rotation takes these factors into account by dividing observed rotation by the length of the tube and the concentration. Specific rotation could therefore be described as observed rotation per length, per concentration unit. A polarimeter is an apparatus that measures the rotation of plane-polarized light as it passes through a sample. Planepolarized light is light that exists in only a single plane. Normally, light waves are oriented in an infinite number of planes from zero to 360. A polarizer can sift out all of these leaving, for example, only vertically oriented waves. To be optically active means that a substance does rotate planepolarized light. Optically inactive compounds do not rotate plane-polarized light. A racemic mixture is a 50/50 mix of the two absolute configurations of a compound (i.e., R and S). 18. Yes. A compound with an even number of chiral centers could be said to be rotating light internally, but the net effect of say two R chiral centers plus two S chiral centers is zero net rotation. If we accept that chiral compounds really do rotate light, then it follows that they must actually be rotating light in this case. It would not be logical to assume that because a chiral center is near (even very, very near as in a molecule) another chiral center that rotates in the opposite direction, that this proximity actually stops that chiral center from rotating light. 19. The cis- version of 2,3-dibromo-2-butene will experience greater steric hindrance, making it less stable, due to the bromine molecules being on the same side of the double bond. Less stable molecules have higher heats of combustion than do stable molecules, so the cis- version will have a higher heat of combustion than the -trans version. Although not asked for in this question, it is worth examining the boiling point and melting point trends for -cis and -trans isomers. The -cis version of 2,3-dibromo-2-butene would have a dipole moment and the trans- version would not. This would increase intermolecular forces and therefore boiling point (Bp-cis > Bp-trans). One might think that this dipole moment would also increase melting point, but the importance of stacking to

7 melting point becomes more important. Trans- molecules can stack on top of one another closely. However, -cis molecules cannot fit inside of one another and therefore are limited in how closely they can stack. This has been hard to visualize for some students. Consider that the cis- molecules are all the same size. To stack inside of one another they would have to be like Russian dolls each one smaller than the next. The closer stacking of the trans- molecules means that more energy is required to disrupt the Van der Waals forces between them (Mp-cis < Mp-trans). 20. No. Diastereomers are PAIRS of compounds that have the same formula, the same bond-to-bond connectivity, are non-identical, and are NOT mirror images. A meso compound is a SINGLE molecule with a mirror plane. Therefore, a meso compound clearly cannot be a diastereomer (or an enantiomer) with itself. 21. Bases: NH 2 -- ; OH -- ; RO -- ; H: -- ; RC: -- ; R 3 N ; H 2 O ; NH 3 ; Nucleophiles: NH 3 ; RC=CR ; H 2 O ; RMgBr ; X -- ; RCO 2 -- ; CN -- ; Electrophiles: H + ; R 2 C=O ; RX ; X 2 ; HX ; R 3 C + ; Leaving Groups:- X ; -OCOR ; -H 2 O + ; -Tosyl ; -NH 3 + ; -N 2 + (diazonium). These are NOT complete lists. Listed in no particular order (It would be a good exercise to have your students attempt to rank the lists according to strength). There are several species that can be placed in multiple categories. Hydroxide, for example, will usually be seen acting as a base on the MCAT, but will also act as a nucleophile if there are not acidic protons. For the MCAT focus on the function: if a species abstracts a proton it is acting as a base and if it attacks a carbon it is acting as a nucleophile. Most students could simply benefit from seeing and being more familiar with each of the above classes so they begin to get a better intuition for each functional group. 22. Methyl: CH 4 ; Primary: RCH 3 ; Secondary: R 2 CH 2 ; Tertiary: R 3 CH ; Quaternary: R 4 C 23. Large substituents are more stable (i.e., lower energy) when in the equatorial position. cis-1,2- dichlorocyclohexane cannot exist with both chlorines in the equatorial position. For neighboring carbons, two equatorial substituents would be trans to one another. Two neighboring substituents can be cis to one another, but one must be in the axial position and one in the equatorial position. Below is drawing of the two chair conformations of cyclohexane. FIGURE 6

8 24. Stable molecules have very low heats of combustion, so N 2 will have a very low heat of combustion. The heat of combustion per CH 2 group increases with ring strain (precisely because more ring strain is less stable). From highest to lowest, heat of combustion is as follows: cyclopropane, cyclobutane, cyclooctane, cyclohexane. Cyclooctane comes before cyclohexane because 8-membered rings are slightly more strained than 6-membered rings. Rings near members or above will have ring strain equal to that of cyclohexane. 25. Alcohols can hydrogen bond with water, creating a very strong solvent-solute attraction. The only thing that will reduce alcohol solubility is the presence of long non-polar alkyl chains. 26. Both trends are explained by the fact that alkyl substituents are weak electron donating groups. When we look at the conjugate base of water there are no alkyl groups donating electron density to destabilize the oxygen with additional negative charge. With primary, secondary and tertiary alcohols there are one, two and three donating groups respectively. Therefore, tertiary alcohols are the least stable because the conjugate base in that case is destabilized to the greatest degree by induction. Carboxylic acids experience the exact opposite effect because the carbonyl is a strong electron withdrawing group. Further (and even more importantly) the conjugate base in the case of a carboxylic acid is stabilized by resonance. 27. See the diagram below. Note that the SN1 forms a planar intermediate and could result in a racemic mixture only if all three R groups are different from one another. FIGURE The dichromate ions are becoming chromium ions as a result of the oxidation reaction. Because the UNREACTED dichromate is orange, alcohol B must not have reacted with the dichromate. This would allow us to infer that alcohol B is very likely tertiary because tertiary alcohols do not oxidize. Alcohol A did react, as indicated by the green solution. Therefore we can infer that it was either a primary or a secondary alcohol.

9 29. You can think of LiAlH 4 as producing two equivalents of hydride ions and of NaBH 4 as producing only one equivalent. When a hydride ion attacks the carbon on an aldehyde or ketone the carbonyl double bond moves up onto the oxygen. Following protonation you have an alcohol and only one hydride equivalent was required. In the case of the acid derivatives one equivalent of hydride ions will produce a similar result, but the hydroxyl or alkoxy group will still be present. Because we know that the reaction proceeds all the way to the alcohol, this group must be kicked off, which can be accomplished when a SECOND hydride equivalent attacks the carbonyl carbon. Therefore, the acid derivatives require two equivalents of hydride ions and NaBH 4 can only provide one. 30. See below: FIGURE See the mechanism below. Mesylates and Tosylates are desirable because they make very good leaving groups that will react readily with almost any nucleophile. FIGURE 10

10 32. See the mechanism below. FIGURE See mechanism below. FIGURE The fact that ethers are unreactive makes them desirable as a solvent because generally we do not want the solvent to react with the reactants or products of our reaction to form byproducts. The unique polarity of ethers allows them to dissolve many molecules, also desirable for a solvent. Finally, the low boiling point makes it easier to separate the product from the solvent via evaporation.

11 35. Epoxide rings can open via either an SN2 mechanism with good nucleophiles, or an SN1-like reaction with poor nucleophiles. In SN2 reactions the good nucleophile attacks the least substituted epoxide carbon and in SN1-like reactions the nucleophile attacks the carbocation formed on the most substituted epoxide carbon. FIGURE Comparing electronegativity, we see that the partial positive charge on the carbon in a C=O bond is greater than the partial positive charge on the carbon in a C=N bond. This indicates that carbonyls are more reactive than comparable imines. 37. Alkyl groups are weakly electron donating; nitro groups are strong withdrawing groups; cyano groups are electron withdrawing; sulphones are electron withdrawing (assuming connection to the sulfur); amines are electron donating; carboxylic acids are electron withdrawing, esters are electron withdrawing (assuming connection to the carbonyl; if connected to the oxygen of the OR group, an ester is electron donating); alcohols are electron donating; quaternary amines are strongly electron withdrawing. 38. Looking at the stability of the conjugate base, we see that a carbanion on the alpha carbon of a methyl would experience electron withdrawal due to the carbonyl, but no additional inductive effects. However, if we examine the ethyl side we see that the same carbanion on that alpha carbon would experience withdrawal due to the carbonyl, but also induction by a weakly donating methyl group (i.e., the beta carbon). This would destabilize the carbanion.

12 FIGURE CREDITS 1. Altius image. 2. Source: Altius image. 5. Source: 6. Source: 7. Source: 8. Altius image. 9. Source: Altius image. 11. Source: Adapted from: Source:

75. A This is a Markovnikov addition reaction. In these reactions, the pielectrons in the alkene act as a nucleophile. The strongest electrophile will

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