UCI DEPARTMENT OF ORGANIC CHEMISTRY PEER TUTORING REVIEW SESSION FEEDBACK EVALUATION

Transcription

1 UCI DEPARTMENT OF ORGANIC CHEMISTRY PEER TUTORING REVIEW SESSION FEEDBACK EVALUATION Quarter: Fall Date: 12/07/18 Class: Rychnovsky Final Review Tutors Names: Ying Chow, Hannah Nguyen, Joshua Torosyan COMMENTS/ SUGGESTIONS (VERY IMPORTANT!) Ying Chow: Hannah Nguyen: Joshua Torosyan: What worked best? What could be improved? What would you like to see next time? This review was interactive and engaging. Comments The presentation volume was acceptable. Comments Strongly Disagree Disagree Neither Agree or Disagree Agree Strongly Agree The presentation was visually clear and logically organized. Comments The review improved/reinforced your understanding of the material. Comments The quality of the review packet was excellent. Comments Please fill out this evaluation, even if you plan to leave early. Thank you very much.

2 This page is intentionally left blank on purpose. You can use this page as scratch work if you want.

3 Organic Chemistry Peer Tutoring Department University of California, Irvine Ying Chow Hannah Nguyen Joshua Torosyan ( torosyaj@uci.edu) Chem 51A Professor Rychnovsky [M + H] + formulas and masses C6H9N2O: C7H9O2: C7H13N2: C9H17: Final Review a. Given the m/z ratio of an unknown protonated compound, what is the molecular formula of the unknown compound? C7H9O2-H = C7H8O2 Subtract a H because these formulas are the protonated compound (M + H) b. Calculate the number of unsaturation units in this compound. (2C H)/2 (2(7) + 2-8)/2 = 4 unsaturation units c. Using the following IR spectrum and H NMR spectrum, draw a possible structure to this unknown compound. 2. a. What does H NMR measure? What does C NMR measure? H NMR measures the energy difference between the nuclear spin states of hydrogen (a proton).

4 C NMR measures the energy difference between the nuclear spin states of carbon. b. How many H NMR signals will this compound have? 6 c. How many C NMR signals will this compound have? 5 (not 4 like I accidentaly said in the review session) there are 5 different carbons and the carbons on the alkene would not be counted as one signal, but rather 2 because they are bonded to different types of carbons (aldehyde vs alkane) d. Which of these H NMR signals will be the most downfield? Why? The H of the C-H in the aldehyde will be the most downfield because it is the most deshielded. The electrons are all going towards the O (dipole) and the resonance structure contributes to the electrons going towards the O. e. Which of these H NMR signals will be the most upfield? Why? The H of the C-H in the alkyl group in between the double bonded C and the C-OH will be the most upfield because it is the most shielded. There are no electron withdrawing groups nearby. f. Which of the bonds in compound would have the highest frequency in an IR spectrum? Why? The OH bond would have the highest frequency in an IR spectum ( ). This is because it is bonded to hydrogen (lighter atoms absorb at a higher frequency). The OH bond is also the strongest single bond. The C=O bond or C=C bond is also strong, but the single bonds themselves are weaker than the OH bond. 3. Cocaine (Methyl (1R,2R,3S,5S)-3-(benzoyloxy)-8-methyl-8-azabicyclooctane-2-carboxylate) acts by inhibiting the reuptake of serotonin, norepinephrine, and dopamine. There is a marked stereoselectivity in the psychomotor stimulant effects of cocaine; this stereoselectivity does not carry over to the toxic effects of cocaine. The specific rotation of a sample of Cocaine is shown below. a. Label each stereocenter as R or S.

5 b. How many stereoisomers would this compound have? 4 2 = 16 stereoisomers c. Label each functional group. d. A sample of Cocaine with a concentration of (3.5 g cocaine /50 ml ethanol) using a 3 dm cell. Calculate the observed rotation of the sample. specific rotation = -16 concentration of solution (in g/100 ml) = 3.5 g/50 ml= 7 g/100 ml = 7 path length (in dm) = 3-16 = observed rotation/[(7)*3] observed rotation = e. Given that the specific rotation of D-Cocaine is -38 what is the % ee Cocaine in the sample? % ee = [α] mixture/ [α] pure * 100 % ee = -16 /-38 * 100 % ee = = 42% f. How many grams of each enantiomer of Cocaine are present in the sample? 3.5 g of Cocaine in the sample % ee calculated was 42% ee = 42% (-) enantiomer 100% - 42% =58% racemic mixture 58% racemic mixture = 29% (+) enantiomer + 29% (-) enantiomer 42% + 29% = 71% (-) enantiomer 29% (+) enantiomer 0.71 * 3.5 = g (-) enantiomer 0.29 * 3.5 = g (+) enantiomer g. If this Cocaine sample was mixed with 125 ml of orange juice, what would the concentration of the sample be? 3.5 g/50 ml is the original concentration 50 ml ml = 175 ml = new volume Amount of Cocaine is not changed so still 3.5 g 3.5 g/175 ml = 0.02 g/ml = 2.00 g/100 ml h. Using the concentration from part g for part a, calculate the measured rotation of the sample.

6 specific rotation = -16 concentration of solution (in g/100 ml) = 2 path length (in dm) = 3-16 = observed rotation/[(2)*3] observed rotation = Rank the following in order of increasing boiling points (1 lowest - 3 highest). a. b.

7 ^ this one was written differently in the packet! In the packet they are identical and in the key they are enantiomers!

8 There are no stereogenic centers because going left on the ring gets to the same atom as going right around the ring. There are only two isomers A and B. A is identical to D. B is identical to C. This is Dr. Rychnovsky s explanation of this problem! This one is a little confusing so I would possibly ignore it and just make sure you understand the concepts well and use worksheets, practice exams, and the book for practice.

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11 9. For the given molecule, draw all resonance structures, include formal charge, and determine which one is the most favorable. 10. a.) Convert the following molecule into three variations of the staggered Newman Projection at C3 and C4. Determine the most stable conformation. b.) Convert the following chair molecule into a Newman Projection. Assume that there are only C-H bonds present.

12 11. Convert the following Newman Projection into both chair conformations. Determine which one is favored at equilibrium. The molecule on the right is favored because the ethyl group is in the equatorial position away from other molecules. 12. Determine the type of acid-base reaction being conducted. Show arrow pushing and determine the final product. Label formal charges, if any. Specify the nucleophile and electrophile. Formaldehyde is the electrophile due to the partial positive charge on carbon. NH3 is the nucleophile. This is a Lewis Acid Base reaction. Charges cancel out in the products side, maintaining neutrality. 13. Determine the type of acid-base reaction being conducted. Show arrow pushing and determine the final product. Label formal charges, if any. Specify the nucleophile and electrophile, along with the acid and base. Determine which side equilibrium will favor.

13 The acid is the carboxylic acid. The base is OH. Pka of the acid on the left is greater than the acid on the right (water) therefore equilibrium will favor the weaker products, which is why there is a larger arrow on the right. The nucleophile is OH. The electrophile is the partially positive H attached to the O on the carboxylic acid. 14. Using the following H NMR spectra, match the compounds to the correct H NMR spectrum. b a d c a. b.

14 c. d.