Chemistry 2030 Survey of Organic Chemistry Fall Semester 2017 Dr. Rainer Glaser

Transcription

1 Chemistry 2030 Survey of Organic Chemistry Fall Semester 2017 Dr. Rainer Glaser Examination #2 Reactions of Alkenes & Alkynes, Chemistry of Aromatic Compounds, and Stereochemistry Thursday, October 5, 2017, 8:25 9:15 am Name: Answer Key Question 1. Reactions of Alkenes & Alkynes. 20 Question 2. Aromatic Compounds Names, Bonding and Properties. 20 Question 3. Electrophilic Aromatic Substitution: S E Ar Mechanism. 15 Question 4. Electrophilic Aromatic Substitution: Directing Effects. 25 Question 5. Stereoisomers Chirality & Geometrical Isomers. 20 Total 100 ALLOWED: Periodic System of the Elements (printed, w/o handwriting on it). Molecular models (you can bring pre-made models). Simple, non-programmable calculator (not really needed). NOT ALLOWED: Books. Notes. Electronic devices of any kind (other than a simple calculator). 1

2 Question 1. Reactions of Alkenes and Alkynes. (20 points) (a) Acid-catalyzed hydration of propene affords one alcohol, the Markovnikov product. Draw the structures of the initially formed protonated propene and the product alcohol. Name the product. (6 p.) (b) Now consider the hydroboration-oxidation of propene. Show the structures of the intermediate borane compound (formed by addition of borane, BH 3, to propene) and of the final product (formed by subsequent workup with an alkaline solution of H 2 O 2 ). The alcohol formed is the (Markovnikov, anti-markovnikov) product. Name the alcohol formed. (5 points) (c) Provide the structure of the product of ozonolysis (O 3 ; Zn, H + ) of cyclohexene. (3 p.) (d) Consider the acid-catalyzed hydration of propyne. This alkyne hydration also requires a strong Lewis acid as co-catalyst; usually one uses HgSO 4 (give formula of co-catalyst). Draw the structures of the primary product and of the final product. (6 p.) Primary Product (formed by hydration) Final Product (formed by tautomerization) 2

3 Question 2. Aromatic Compounds Names, Bonding and Properties. (20 points) (a) Provide complete structures (all atoms, all bonds, all lone pairs) of aniline and biphenyl. (4 points) Aniline Biphenyl (b) Consider the structure of vanillin (4-hydroxy-3-methoxybenzaldehyde). The formyl group is in position 1, the so-called _ipso position. The methoxy group in position 3 and the hydroxyl group in position 4 are (ortho, meta, para) relative to each other. The formyl group in position 1 and the hydroxyl group in position 4 are (ortho, meta, para) relative to each other. The formyl group and the methoxy group are (ortho, meta, para) relative to each other. (4 points) (c) The space-filling model shows the polycyclic aromatic hydrocarbon pyrene. Draw one complete Lewis-Kekule structure of pyrene on the right (all atoms, all bonds, all lone pairs). No need to draw all resonance forms; just draw one resonance form. Below, show all possible products of monochlorination of pyrene. You can use abbreviated formulas for the chloropyrenes and, again, just draw one resonance form for each isomer. (6 points) Pyrene Isomers of Chloropyrene. [Hint: How many different types of H occur in pyrene?] (d) The heat of hydrogenation of cyclohexene is about 28.6 kcal/mol. The complete hydrogenation of benzene to cyclohexane (name of the C 6 H 12 compound formed) releases about 49.8 kcal/mol. Based on the heat of hydrogenation of cyclohexene, one might have expected that the hydrogenation of benzene is much more exothermic, i.e., one might have expected a value of about 85.8 kcal/mol. The difference between the expected and the measured values is the resonance energy of benzene, which is about 36 kcal/mol. (6 points) 3

4 Question 3. Electrophilic Aromatic Substitution: S E Ar Mechanism. (15 points) (a) Provide reagents and any catalyst needed to convert benzene into bromobenzene and draw the complete, unabbreviated structure of bromobenzene. (5 points) (b) Provide reagents and any catalyst needed to convert benzene into acetophenone by Friedel-Crafts acylation and draw the complete, unabbreviated structure of acetophenone. (5 points) (c) Consider the sigma-complex, which occurs as an intermediate in the Friedel-Crafts acylation of benzene to form acetophenone. Draw three resonance forms of this sigma complex that inform about the delocalization of the positive charge in the benzene ring. [Here, we are not interested in the resonance within the C=O group.] (6 points) 4

5 Question 4. Electrophilic Aromatic Substitution: Directing Effects. (25 points) (a) Consider the bromination of anisol. Write the reagent in the box on top of the reaction arrow and write the required catalyst in the box below the reaction arrow. For each one of the products, indicate whether it is a minor or a major product. (5 points) (b) Outlines are shown of resonance forms of the sigma-complexes of the bromination of anisol in the ortho (top), meta (center), and para (bottom) positions. Complete the resonance forms: Add all double bonds and all lone pairs on the anisol O atom. Bromine will always have 3 lone pairs and there is no need to draw the bromine lone pairs. (12 points) 5

6 (c) Directing effects in the bromination of anisol. Explain your answer to part (a) considering the resonance forms you wrote in part (b). Note that the resonance forms in part (b) are labeled A L for your convenience. Use these labels in your explanation. Be brief and concise. (4 points) Ortho: A and B are ordinary; C is special: Plus charge right next to an atom with a lone pair allows for an additional resonance form D. Meta: E, F and G are ordinary. Para: H and I are ordinary; K is special: Plus charge right next to an atom with a lone pair allows for an additional resonance form L. More resonance forms means that the intermediate is more stable. A reaction that goes through a more stable intermediate is preferred over a reaction that goes through a less stable intermediate. The ortho and para reactions are preferred over the meta reaction because the ortho and para sigma complexes are more stable than the meta sigma complex. (d) Activation/Deactivation of the bromination of anisol. Select the correct answers for the following four questions. (4 points) The methoxy group is (activating, deactivating) with regard to an electrophilic aromatic substitution. Compared to the rate of reaction of the bromination of benzene, ortho-bromination of anisol is (much slower, slower, faster, much faster). Compared to the rate of reaction of the bromination of benzene, meta-bromination of anisol is (much slower, slower, faster, much faster). Compared to the rate of reaction of the bromination of benzene, para-bromination of anisol is (much slower, slower, faster, much faster). 6

7 Question 5. Stereoisomers Chirality & Geometrical Isomers. (20 points) (a) Two models are shown of the amino acid threonine, H 2 N CH(CH(CH 3 ) OH) CO 2 H (O in red, N in blue). For each model, provide the R or S label for the alpha-c (the carbon to which the NH 2 and COOH groups are attached). The stereoisomers shown are (the same, different). (6 points) Model #1 of Thr S Model #2 of Thr R (b) The amino acid threonine has the structure H 2 N CH(CH(CH 3 ) OH) CO 2 H and contains a chiral alpha-c. Provide the CIP priorities of the four substituents. For the two C-substituents, apply the sequence rule and provide their lists. (6 points) Priority of -H: _4_ Priority of -NH 2 : _1_ Priority of -CH(CH 3 )-OH: _3_ C(O C H) Priority of -COOH: _2_ C(O O O) (c) Provide a perspective drawing of the (R)-enantiomer of alanine. The amino acid alanine has the formula H 2 N CH(CH 3 ) CO 2 H The perspective drawing should have two bonds in the paper plane, one bond that goes behind the paper plane, and one bond that goes in front of the paper plane. In addition, the C H bond should be the bond that goes behind the paper plane. (4 points) (d) Draw the structure of the (E)-isomer of 2-bromo-3-chloro-2-pentene. Show all atoms and all bonds. (4 points) 7