1. We can compare a vibrating bond (in only one dimension, ignoring swaying motions) to a mass on a spring. The bond roughly follows Hooke’s Law and either atom can be compared to the mass. A stronger bond would be analogous to a spring with a larger spring constant (k), and a weaker bond would be analogous to a spring with a smaller spring constant. The molecular weight of the atom is analogous to the size of the mass on the spring. Following the principles dictated by Hooke’s Law, a stronger spring (larger k) increases oscillation frequency according

to: f = (1/2) √(k/m). A weaker bond would therefore vibrate at a lower frequency. A smaller atom is associated with a higher vibrational frequency and a larger atom is associated with a lower vibrational frequency.

2. See spectra below.

Alcohol Carboxylic Acid (Note O-H and C=O)

Amine (notice how shallow it is) Amide (slightly overlapping C-H stretch)

FIGURE 1

Nitrile

FIGURE 1 Continued

3. Alkenes and alkynes usually register around 170nm and 180nm respectively. The two spectra below demonstrate the impact of conjugation.

FIGURE 2

4. See the diagram below. The tallest peak will always be the base peak. The molecular ion peak (a.k.a., parent peak; often labeled as M+) is usually the last peak to the right. A few very short peaks may exist to the right of the parent peak but the parent peak will be taller. The height of each peak is proportional to the number of fragments with that m/z ratio. The mass of the fragments increases as you move from left to right.

FIGURE 3

5. A linear increase in the magnetic field will create a linear increase in the force exerted on the

particles according to: F = qvBsin. For a curved flight tube with a detector at the far end, in order for any particle to strike the detector it must experience exactly the right centripetal acceleration to trace the precise curvature that results in striking the detector. As the centripetal force increases this will occur for increasingly massive particles. The graph below shows the linear relationship between magnitude of the B field and mass of particles at the detector.

FIGURE 4

6. Carbon-13 NMR on a large steroid would NOT detect every single carbon because it only detects the carbon-13 isotope, which has a relative abundance of only about 1%.

7. magnetic field

8. See spectra below.

FIGURE 5

9. An acid would protonate bases in the mixture, potentially adding a formal charge and dramatically increasing solubility in the aqueous layer (i.e., protonated amines). Bases could similarly abstract protons to create a negatively charged species—with the identical impact upon solubility (i.e. deprotonated carboxylic acids).

10. Fluted filter paper increases the surface area for filtration.

11. Whether you would want to filter hot or cold would be determined by the conditions. If you are filtering out a solid impurity filtering hot could prevent your product from crystallizing on the filter paper. However, if you are close to the freezing point of the solid then filtering hot could bring impurities back into solution. In other cases you may be trying to capture product that crystallized upon cooling of your reaction mixture. In that case you would certainly want to filter cold or else you would re-dissolve your product.

12. Vacuum filtration is performed with a Hirsch or Buchner funnel. A vacuum is created inside of the flask which creates suction to pull the filtrate through the filter paper. The filter usually has holes in it which are covered by the filter paper. The primary advantage is that it is faster than gravity filtration.

13. Vacuum distillation is used because it dramatically lowers the boiling point, allowing you to work at more manageable temperatures with substances that have much higher boiling points at atmospheric pressure. This observation is supported by the fact that liquids can be said to boil at the point where vapor pressure equals atmospheric pressure. By creating the vacuum we essentially drive down the atmospheric pressure to meet the vapor pressure.

14. The sample is usually a mixture of components which is spotted onto the paper or TLC plate near its bottom edge. The top of the solvent should always be below this line where the sample is spotted so that the mixture does not simply dissolve into the solvent.

15. Given that Rf is a ratio of the distance traveled by the component over the distance traveled by the solvent. So a number close to one means that the polarity of the component is similar to that of the solvent. Because non-polar solvents are normally used, the component is therefore relatively non-polar.

16. Smears are usually caused when the solvent is unable to dissolve and carry the components as it moves up the paper via capillary action. This can be because too much of the sample was spotted onto the paper and the solvent could not dissolve the entire sample. It can also occur if there is too big of a difference in polarity between the sample and the solvent. If the polarities are too distant the solvent cannot dissolve the sample well enough to carry it. By a similar token, if they are too similar, the components may never come out of solution to form a spot and be carried all the way to the top of the paper.

17. In column chromatography, assuming a polar matrix, the most non-polar substances will elute first, followed by increasingly more polar substances.

18. The height of the peak in a gas chromatograph is relative to the abundance of that component.

19. Higher and sharper melting points indicate better purity. Impurities lower melting point. They also broaden the range across which the crystals melt.

20. An alkene involves a sigma and a pi bond between two carbons. A carbonyl involves a sigma and a pi bond between a carbon and an oxygen. Whenever you think about pi bonds one of the first things you should consider is the amount of overlap of the p orbitals. Compared to two carbons, a carbon and an oxygen can get closer to one another because oxygen has a smaller atomic radius. This allows for more pi overlap and therefore a stronger bond.

21. This scenario would occur if the carbonyl was of the form R1COR2 and the nucleophile was something such as a Grignard Reagent that added either R1 or R2 to the carbonyl carbon. Because two of the substituents would be identical, it could not be chiral. This could also happen if water or hydroxide added to the bond, creating two –OH substituents. You are now nearing the end of the course—hopefully a question like this makes a lot of sense to you—not necessarily that the answer jumps right out at you—but that this format looks familiar. The MCAT is all about questions that require you to gather pieces of information, make predictions, use logic and inference, etc.

22. There are two resonance structures for the conjugate base formed by removal of an alpha hydrogen. Remember that neither form actually exists—the actual structure is a permanent hybrid of the two. In this case, the hybrid will look slightly more like the second structure because the negative charge is on the more electronegative oxygen.

FIGURE 6

23. These structures differ in the substituents attached to the carbonyl carbon. An electron

donating group on the carbonyl carbon will decrease its partial positive charge and thereby make it less able to stabilize the conjugate base. An electron withdrawing group will make the carbonyl better at stabilizing the conjugate base and therefore the strongest electron withdrawing group would indicate the strongest acid (because it produces the most stable conjugate base). Choice a) is the only electron withdrawing group, so it will have the most acidic alpha hydrogens. The weakest donating group of the other three is the methyl group on choice d). Amines (choice c) and hydroxyl groups (choice b) are both electron donating, but amines are better donating groups because nitrogen is less electronegative than oxygen. Therefore, in increasing order of acidity: c < b < d < a.

24. Perhaps surprisingly, substituent aldehydes are given the same “-oxo” name as ketone substituents. There really should not be any confusion, however, because if the identified carbon is terminal it must be an aldehyde and cannot be a ketone, and if it is secondary it must be a ketone and cannot be an aldehyde. Remember that aldehydes and ketones can only be substituents when there is a higher priority functional group present, such as a carboxylic acid. If the aldehyde or ketone is the top priority functional group then the carbonyl carbon is always labeled as C-1. Finally, remember that all of this is intended just to give you enough information to discern the correct structure from a name. The MCAT will never require you to give an IUPAC name to a complex molecule with various functional groups of differing priority.

25. Alkanes are non-polar and therefore insoluble in water. Aldehydes and ketones can act as hydrogen-bond acceptors when dissolved in water, with water acting as the hydrogen bond donor. Therefore, aldehydes and ketones will be far more soluble than alkanes. Finally, alcohols can hydrogen bond as both a donor and acceptor with water, so they will be the most soluble. These trends assume comparable molecular weight, chain length, etc. This is important because a small alcohol such as methanol is miscible with water, but dodecanol is considered insoluble. The boiling point of alkanes will be the lowest because their only intermolecular attraction would be van der Waals forces. Aldehydes and ketones do NOT hydrogen bond with one another, but they are both polar and will therefore have much higher boiling points than alkanes. Finally, alcohols generally have the highest boiling points due to intramolecular hydrogen bonding.

26. In order for a substitution to occur there must be a leaving group. Acid derivatives all have leaving groups: Cl- in the case of acid chlorides, -OH in the case of carboxylic acids, a carboxylate ion in the case of anhydrides, etc. The stability of these groups after they leave varies widely, but in the case of an aldehyde or ketone there are no groups that would be reasonably stable and therefore no candidates to act as leaving groups. The aldehyde hydrogen will not leave as H:-, nor will an R group leave from a ketone as a carbanion R:- (excluding rare cases where substituents on the R group dramatically change its properties—as in –RX3). In both cases the leaving group would become a strong base. Recall that strong bases never make good leaving groups. Good leaving groups must be weak bases that are stable after they leave. For this reason, aldehydes and ketones only undergo addition reactions—lacking a suitable leaving group to undergo substitution.

27. The alpha hydrogen of an aldehyde is more acidic than a comparable alpha hydrogen on a ketone because the conjugate base in the case of the aldehyde is more stable. In an aldehyde, a hydrogen is attached to the carbonyl carbon, which is defined as neither a withdrawing group nor a donating group. However, in the case of a ketone an –R group is attached to that same carbonyl carbon and –R groups are weakly electron donating. This will decrease the magnitude of the partial positive charge on the carbonyl carbon in the ketone, making it less able to stabilize the negative charge of the carbanion in the conjugate base.

28. See mechanisms below:

FIGURE 6

29. The keto and enol forms are in an equilibrium with one another that strongly favors the keto form at room temperature. The keto form is more stable because the sum of its bond energies is greater than the sum of the bond energies in the enol form. The keto form has a C=O bond, a C-C bond, and a C-H bond that are replaced by a C-O bond, a C=C bond, and an O-H bond in the enol form. C-H and O-H bonds are quite close in bond energy, while a C=C has about 250 kJ/mol more bond energy than a C-C bond (not quite double). The real difference comes in the difference between a C-O bond and a C=O bond. A C=O bond has about 450 kJ/mole more bond energy! Of course, you are not expected to know these numbers—we just included them here to illustrate the point. What you should know is that carbonyl bonds are much shorter and stronger than alkene bonds. That is the most significant difference between the two forms and is the reason the keto form is favored.

30. A 1,3-dicarbonyl can undergo an intramolecular hydrogen bond when one of the carbonyls is in the keto form and the other is in the enol form. This significantly stabilizes the enol compared to a stand-alone enol. In this condition the enol is acting as the hydrogen-bond donor and the carbonyl as the hydrogen bond acceptor. The MCAT loves alpha hydrogens so much, it wouldn’t be right to mention 1,3-dicarbonyls without also pointing out that they have ultra-acidic alpha protons on the carbon between the two carbonyl carbons—double resonance stabilization!

FIGURE 7

31. See the two resonance forms below. The one on the left is clearly the more significant

contributor to the actual structure because it has no formal charges compared to a charge separation in the structure on the right.

FIGURE 8

32. In terms of the MCAT, you should think of an α-β-unsaturated carbonyl as an electrophile.

Nucleophiles either attack the beta carbon directly, or attack the carbocation that exists in the second resonance form. Nucleophiles that can’t act as leaving groups stay attached to the carbonyl carbon (hydrides and Grignard reagents). Nucleophiles that can act as leaving groups attack the carbonyl carbon but get kicked back out and eventually attack the beta position and stay (amines, enolates, nitriles etc.). It might be tempting to think of the double bond between the alpha and beta carbons as a nucleophile that will undergo electrophilic addition. However, the withdrawing effect of the carbonyl decreases the electron density of the double bond deactivating it toward electrophilic addition.

33. Theoretically, the carboxylic acid dimer (pictured below) would have no net dipole moment. This explains its solubility in non-polar solvents.

FIGURE 9

34. See mechanism below. Without a base catalyst, this mechanism can also be visualized as a 6-

member, concerted, “ring-like” intramolecular reaction that does not require protonation of the enolate ion. Help your students visualize how his might happen.

FIGURE 10

35. A chloride ion is capable of attacking the carbonyl carbon of a carboxylic acid. However, when

the electrons in the carbonyl bond are kicked up onto the oxygen, and then collapse back down, the substituent that is the best leaving group will leave, regardless of the original structure of the molecule. Chloride ion is more stable (i.e., it is a weaker base) than hydroxide ion, so the chlorine will be kicked off to reform the acid.

36. a) ROH ; b) RNH2 ; c) RCOOH ; d) H2O

37. Because the nitrogen in an amide donates its electron pair via resonance, both the C-O and the C-N bond have double-bond character. Therefore the hydbridization of the nitrogen will be closer to sp2 than to sp3.

38. For the same reasons given in the previous question—namely the donation of the lone pair on the nitrogen into the conjugated system—nitrogen has less electron density in an amide than it would in a normal amine. Therefore, it will be less basic. One could also consider the effect of induction. The carbonyl carbon has a strong partial positive charge, which will withdraw electron density from the amine through the sigma bond.

39. See the Lewis structures below. Phosphoric Acid Sulfuric Acid Nitric Acid

FIGURE 11

40. Chlorine is the best leaving group because it is a weak base that is remarkably stable while holding a formal negative charge. This ability to hold a charge is due to the size of chlorine’s electron cloud. The negative charge is spread out across a much larger distance than it would be for a smaller atom. For this reason, bromine would be an even better leaving group, and iodine better yet. The –OCOR group is the leaving group portion of an anhydride—the portion kicked off when a nucleophile attacks one of the carbonyl carbons. It is a carboxylate ion stabilized by resonance. That stabilization makes it much more stable than other leaving groups that place a negative charge on an oxygen. Hydroxide and alkoxide are best compared together. It should be noted that under acidic conditions, hydroxide gets the nod as the better leaving group because –R groups are weak donating groups. Hydrogen, by comparison is considered neither donating, nor withdrawing. Therefore, the –R group on an alkoxide donates more electron density to an oxygen that already bears a full negative charge. Under basic conditions the acidic hydroxide gets deprotonated and does not act as a leaving group, therefore only the alkoxide RO- can act as a leaving group. Finally, an amine makes a relatively awful leaving group because when it leaves it forms the strong base NH2

-. This is a stronger base than hydroxide or alkoxide and therefore is the most unstable bearing a full negative charge. This is the reason that amides are the least reactive of the acid derivatives.

41. Aromatic amines are less basic because they donate their electron pair into the ring forming a conjugated system with the ring.

42. Alcohols will form stronger hydrogen bonds because there is a greater difference in

electronegativity between oxygen and hydrogen than there is between nitrogen and hydrogen. This greater dipole will create a stronger electrostatic attraction and therefore a stronger hydrogen bond.

43. Protein; the amine acts as a nucleophile, attacking the carbonyl carbon of another amino acid to form a peptide bond.

44. A single hydrazine molecule will react at both carbonyl carbons to release a primary amine and create the closed-ring diamide product shown. See the diagram below for electron flow:

FIGURE 12

45. Tertiary amines are not good nucleophiles because they are too sterically hindered. They are

more likely to act as a base. Even if they were to attack the carbonyl carbon, this would form an unstable quaternary amine with a full positive formal charge. This would be a better leaving group than the water formed by protonation of the carbonyl and would therefore be kicked back off anyway.

FIGURE CREDITS

1. Sources: http://wwwchem.csustan.edu/tutorials/infrared.html ; http://orgchem.colorado.edu/Spectroscopy/irtutor ; http://www.lookchem.com/chemical-dictionary/en/spectrum http://www.chemistry.ccsu.edu/glagovich/teaching/316/ir/nitrile.html

2. Source: http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/Images/polyene.gif 3. Source: http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/MassSpec/masspec1.htm 4. Source: Altius image 5. Source: http://www.mhhe.com/physsci/chemistry/carey/student/olc/graphics/carey04oc/ref/ch13nmr.html 6. Source: Altius image 7. Source: http://en.wikipedia.org/wiki/Keto-enol_tautomerism 8. Source: Altius image 9. Source: Altius image 10. Source: Altius image 11. Source: Altius image 12. Source: Altius image