Organic Chemistry Peer Tutoring Department University of California, Irvine

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1 Organic Chemistry Peer Tutoring Department University of California, Irvine Arash Khangholi Cassandra Amezquita Jiana Machhor OCHEM 51A Professor King Final Review Packet 1. The following questions are pertaining to a legendary compound named Machhor-45. This compound has been empirically proven to improve patient intellect 45 times the average IQ. c a a. Predict molecular shape around the atoms indicated as a, b, and c a Trigonal Pyramidal (3 groups + 1 lone pair) b Trigonal Planar (3 groups) c Bent (2 groups + 2 lone pairs) b. What are the approximate bond angles around Nitrogen a (105, 107, 109.5, 120 or 180 )? c. What are the approximate bond angles around Carbon b (105, 107, 109.5, 120 or 180 )? d. How many π bonds are present within the molecule? 5 e. Which of the listed atoms (a,b, and c) has the same bond angles as H3O +? Hydronium has 3 groups and 1 lone pair thus trigonal pyramidal. The atom that matches this description thus has the same bond angles is Nitrogen A or (a) b

2 2. Transform the condensed structure C6H5COOH, into a: a. Lewis Structure b. Skeletal Structure c. 3D structure (with wedges and dashes) 3. Describe the overlap required to produce Bonds A-F in the following molecule. Then, describe the molecular shape and bond angles around the left-most atom involved in each bond. A: Sigma bond overlap between C sp2 and Cl sp3 The left-most atom involved in bond A, the carbon, has a molecular geometry of trigonal planar, because there are three groups surrounding the carbon atom. The bond angles are 120 degrees. B: Sigma bond overlap between C sp2 and C sp2 C: Pi bond overlap between Cp and C p The left-most atom involved in bonds B and C, the carbon, has a molecular geometry of trigonal planar, because there are three groups surrounding the carbon atom. The bond angles are 120 degrees. D: Sigma bond overlap between C sp and C sp E: Pi bond overlap between C p and C p F: Pi bond overlap between C p and C p The left-most atom involved in bonds D, E, and F, the carbon, has a molecular geometry of linear, because there are two groups surrounding the carbon atom. The bond angles are 180 degrees.

3 4. For the following molecule, draw out all possible resonance structures. Add formal charges and lone pairs where needed. 5. Amoxicillin is a widely-used antibiotic drug treatment for stomach ulcers, but results in adverse side effects including diarrhea and tooth discoloration. Please identify all functional groups in this molecule.

4 6. When looking for basicity or acidity we want to look at the stability of the resultant loss or gain of the proton. 4 Factors to consider when determining basicity/acidity. 1. Element (Most important and evaluated first) - Look at the atom that bears the negative charge and use the basicity/acidity trends Basicity Acidity 2. Resonance - Resonance delocalizes or spreads the electrons of the negative charge thus making a less negative negative charge and stabilizing the conjugate base. 3. Inductive - The electronegativity of neighboring atoms can contribute to electron localization or delocalization - The inductive effect has reduced impact with increasing distance from the atom bearing the charge of the conjugate base 4. Hybridization Effect - more s character, increased electronegativity and more stability with the negative charge of the conjugate base 1. a. Rank the following compounds in order of basicity from least to greatest. Most stable = least basic Least stable = most basic All molecules have the same resonance stabilization and element effect..inductive effect of bromine makes the negative charge on oxygen less negative thus stabilizing the conjugate base. The inductive effect decreases with distance so the leftmost compound has the strongest inductive effect resulting in the most delocalization. Most delocalization = most stable = least basic. The intermediate delocalization of the further bromine on the rightmost

5 compound delocalizes to a lesser degree making it less stable and more basic. The lack of an additional electronegative atom and presence of the electron donating isobutyl group within the rightmost compound results in the least stable and strongest base being the middle compound. The atom bearing the negative charge is the only difference in the listed compounds thus trend of basicity allows us to simply rank these compounds based on their location in the periodic table. Element effect doesn t allow us to differentiate between these listed compounds. Resonance through the π bonds of the aromatic ring delocalizes the electrons of the negative charge, increasing stability, and decreasing basicity. By this token the strongest base is the leftmost compound due to the lack of resonance stabilization. Inductive effect through the nitro group of the rightmost compound further delocalizes the electrons of the negatively charged oxygen; increasing stability, and decreasing basicity. The coupled inductive and resonance effect result in the rightmost compound being the weakest base of the 3 listed compounds.

6 7. Rank the following acids from most acidic (1) to least acidic (4). A: CH3CH2NH2 (4) B: CCl3CH2OH (1) C: CH3 CH2OH (2) D: C(CH3)3 CH2OH (3) Acid A is the least acidic because of the elemental affect. Acidity increases going to the right of the periodic table, making a hydrogen bonded to a nitrogen less acidic than a hydrogen bonded to an oxygen. Acid B is the most acidic because of the electronegative chlorine atoms drawing electron density away from the electronegative oxygen atom, therefore stabilizing the conjugate base and increasing the acidity of the acid. Acid D is less acidic than Acid C because Acid D has electron-donating methyl groups, which add electron density to the electronegative oxygen atom, therefore destabilizing the conjugate base and decreasing the acidity of the acid. a. Which acid would have the strongest conjugate base? The weakest acid is the acid with the strongest conjugate base. The weakest acid is Acid A. b. Which acid would have the lowest pka? The strongest acid is the acid with the lowest pka. The strongest acid is Acid B. 8. For the following Lewis-Acid base reaction, predict the products. Appropriately label the acid and base in the reactants, and the corresponding conjugate acid and base in the products. Predict whether the equilibrium would favor the products or the reactants, and then predict the ratio of products to reactants at equilibrium. The pka of the carboxylic acid derivative on the left is approximately 5. The pka of the conjugate acid on the right is approximately 10. Equilibrium favors the weaker acid (or the acid with the higher pka). Therefore, the products are favored by a factor of Therefore, the ratio of reactants to products at equilibrium is 1:10 5.

7 9. Use IUPAC nomenclature to appropriately name this alkane. The parent chain is eight carbons long (octane). If we start labelling from the left, we ll have substituents on 4, 5, and 6. If we start labelling from the right, we ll have substituents on 3, 4, and 5. Therefore, when we start labelling from the right, we understand that we have: 3-bromo, 4-ethyl, 5-chloro, and 5-methyl. Putting the pieces together, our compound should be appropriately named: 3-bromo-5-chloro-4-ethyl-5-methyloctane 10. Identify all stereocenters in this molecule. Assign the left-most stereocenter either an R- configuration or an S-configuration. Priority #1 is the Br group, priority #2 is the carboxylic acid group, priority #3 is the carbon attached to the ring, and priority #4 is the CH3 group. The lowest priority can never be in the plane of the paper, so we first switch the CH3 group and the carboxylic acid group, to make sure that the prior is on the dashes and the latter is in the plane of the paper. Then we assign configuration going counterclockwise, so deciding S-configuration. However, because we first switched the methyl group with the carboxylic acid group, we flip the configuration to finally decide on R-configuration 11. Convert the following Lewis Structure into its corresponding Newman Projection. If you can t visualize Newman Projections, follow the three simple rules that got me through 51A! 1. Assign the left-most carbon as the carbon at the front of the Newman Projection 2. Follow the stems (or the groups in the plane of the paper) to decide if the Y-shape is going up or down for the front carbon and the back carbon 3. Assign all groups on wedges to the right-hand side, and all molecules on dashes to the lefthand side

8 12. Convert the following Newman Projection into its corresponding Lewis Structure. 13. For the following pairs of molecules, determine their chirality and their relationship to one another.

9 14. Flip the following chair conformation, then convert to it s respective planar structure. Label the most stable conformer and draw the equilibrium arrows indicating favored formation. Despite having 3 groups in the equatorial position within the conformation on the right, the most stable chair has the bulkiest groups in the equatorial position. The isobutyl group is very large and bulky so the most stable conformation is with the least steric hindrance or isobutyl in the equatorial position. The other substituents (Me, OH, and Br) are much smaller and won t cause as much steric hindrance when placed in the axial position relative to an axial isobutyl group. 15 A. Calculate the specific rotation of a solution containing 35% (+) Amezquita-35 and 65% (-) Amezquita-35. Amezquita-35 has a specific rotation of %(-) Amezquita-35-35% (+) Amezquita-35 = 30% (-) Amezquita-35 specific rotation of mixture %ee = x 100% specific rotation of pure enantiomer x (specific rotation of mixture) 30% = x 100% x(specific rotation of mixture) =.3(-7.6 ) = B. What is the observed rotation of Amezquita-35 within a solution containing 2000 mg of (+) Amezquita-35 and 1500 mg of (-) Amezquita-35 in 150 ml of solvent? The vial utilized was 20 cm long mg (+) Amezquita mg(-) Amezquita-35 = 500 mg.5 g (+) Amezquita-35 C(g/ml) so.5 g / 150 ml =.003 g/ml 20 cm long 2 dm since we want L in dm not cm. We know the specific rotation of (-) Amezquita-35 to be -7.6 thus the (+) Amezquita-35 will have a specific rotation of +7.6 since enantiomers have equal in magnitude yet opposite in sign specific rotations. (observed rotation) +7.6 = =. 003 x 2 7.6(. 003x2) = observed rotation observed rotation =.0456

10 16. Label the location of the following on the energy diagram below. a. Reactants b. Products c. Transition states d. Enthalpy Change e. Rate Determining Step 17. Rank the following compounds in order of highest boiling point (1) to lowest boiling point (4), and give a one-sentence explanation for your reasoning of each rank. Compound #1 has the highest boiling point because it contains hydrogen-bonding. Compound #4 contains the lowest boiling point because it only contains VDW interactions. Compound #2 and #3 have the same number of carbons (and both have dipole-bonding attributed to the oxygen). However, branching decreases boiling point because of the decreased surface area of the molecule, allowing Compound #3 to have a lower boiling point than Compound #2.

11 18. a. Calculate the units of unsaturation for the compound using the provided chemical formula. C5H10O Units of unsaturation = Units of unsaturation = 2(5) = 1 unit of unsaturation b. Identify the functional groups of the IR spectra The band at 1716 cm -1 indicates a carbonyl C=O bond, probably belonging to a ketone. The bands at cm -1 indicate C-H alkane stretches.

12 The assignment of protons to their respective peaks requires identification of protons and their environments. The splitting of each peak allows us to determine the # of neighboring protons on adjacent atoms. A singlet or singular peak like the one for B indicates 0 neighboring protons while the integration of 3 indicates the number of protons that the peak represents. The high levels of deshielding, due to the electronegative Oxygen are indicated by how downfield the peak resides within the NMR spectrum. Through the aforementioned information we can identify the methyl on the right of the structure to it s respective peak. The same system of splitting, integration, and environmental awareness can be repeated for the rest of the peaks. Splitting can be confusing but I think about how many times the peak is split then subtract 1 in order to calculate the # of neighboring protons. So a singlet has 0 neighbors while a septet has 6 neighbors.