Chemistry 2030 Survey of Organic Chemistry Fall Semester 2015 Dr. Rainer Glaser

Transcription

1 Chemistry 2030 Survey of Organic Chemistry Fall Semester 2015 Dr. Rainer Glaser Examination #2 Reactions of Alkenes & Alkynes, Chemistry of Aromatic Compounds, and Stereochemistry Thursday, October 8, 2015, 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) The acid-catalyzed hydration of 1-butene affords one alcohol, the Markovnikov product. Draw the structures of the substrate and the product. Provide the name of the product. (6 p.) (b) Now consider the hydroboration-oxidation of 1-butene. Show the structure of the final product formed by addition of borane, BH 3, to 1-butene and subsequent workup with an alkaline solution of H 2 O 2. The alcohol formed in this hydroboration-oxidation sequence is the (Markovnikov, anti-markovnikov) product. Give the name of the alcohol formed. (4 points) (c) Provide the structures of the two products of the ozonolysis of 2-pentene (O 3 ; Zn, H + ). (4 p.) (d) Consider the acid-catalyzed hydration of 2-butyne. 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 structural formulas of phenol and styrene. (4 points) Phenol Styrene (b) Consider the structure of TNT, i.e., 2,4,6-trinitrotoluene. The methyl group is in position 1, the socalled _ipso_ position. The methyl group in position 1 and the nitro group in position 2 are (ortho, meta, para) relative to each other. The methyl group in position 1 and the nitro group in position 4 are (ortho, meta, para) relative to each other. The nitro groups in positions 2 & 4 are (ortho, meta, para) relative to each other. The nitro groups in positions 2 & 6 are (ortho, meta, para) relative to each other. (5 points) (c) The model shows the aromatic hydrocarbon naphthalene. Draw the structure of naphthalene on the right. Below, show all possible products of monochlorination of naphthalene. (5 points) Naphthalene Isomers of Chloronaphthalene. [Hint: How many different types of H occur in naphthalene?] (d) The resonance energy of benzene is about 36 kcal/mol. The 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 about 28.6 kcal/mol of cyclohexene, we would have expected that the hydrogenation of benzene is much more exothermic, i.e., about 85.8 kcal/mol. (6 points) 3

4 Question 3. Electrophilic Aromatic Substitution: S E Ar Mechanism. (15 points) (a) Nitronium Ion: Provide complete structural formulas for the formation reaction of the electrophile which is the reactive species in the nitration of benzene to form nitrobenzene by way of electrophilic aromatic substitution. (5 points) (b) Provide reagents and any catalyst needed to convert benzene into chlorobenzene and draw the complete, unabbreviated structural formula of chlorobenzene. (5 points) (c) Provide reagents and any catalyst needed to convert benzene into acetophenone by Friedel-Crafts acylation and draw the complete, unabbreviated structural formula of acetophenone. (5 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 serine, H 2 N CH(CH 2 OH) CO 2 H (O in red, N in blue). Do the two models show the same molecule or do they show different enantiomers (same, different)? Provide the R or S label for each model. (6 points) Model #1: R Model #2: S (b) The amino acid serine has the structure H 2 N CH(CH 2 OH) CO 2 H and contains one chiral C atom. 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 2 -OH: 3 C(O H H) Priority of COOH: 2 C(O O O) (c) Perspective drawing of (S)-enantiomer of serine. 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 go behind the paper plane. (4 p.) (d) Draw the complete structure of the (E)- isomer of 1-bromo-2-chloro-1-butene. (4 points) 7