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Chemistry  2030      “Survey  of  Organic  Chemistry”      

Fall  Semester  2014      Dr.  Rainer  Glaser      

Examination #2

“Reactions of Alkenes & Alkynes; Arenes and Electrophilic Aromatic Substitution; Stereochemistry”

Thursday, October 9, 2014, 8:00 - 8:50 am

Name:

Answer Key

Question 1. Reactions of Alkenes & Alkynes 20 Question 2. Aromatic Compounds – Names, Bonding and Properties 15 Question 3. Electrophilic Aromatic Substitution: Monosubstitution 20 Question 4. Electrophilic Aromatic Substitution: Disubstitution 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).

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Question 1. Reactions of Alkenes and Alkynes. (20 points) (a) The acid-catalyzed hydration of 1-methylcyclohexene affords a tertiary 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-methylcyclohexene. Show the structure of the final product formed by addition of borane, BH3, to 1-methylcyclohexene and subsequent workup with an alkaline solution of H2O2. 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-hexene (O3; Zn, H+). (4 p.)

(d) Consider the acid-catalyzed hydration of 1-butyne. This alkyne hydration also requires a strong Lewis acid as co-catalyst; usually one uses __HgSO4__ (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)

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Question 2. Aromatic Compounds – Names, Bonding and Properties. (15 points) (a) Provide complete structural formulas of aniline and para-xylene. (4 points)

Aniline

Para-Xylene

(b) Consider the structure of 1,2,4-trimethylbenzene. (3 points) The methyl groups in positions 1 and 2 are _______ (ortho, meta, para) relative to each other. The methyl groups in positions 1 and 4 are _______ (ortho, meta, para) relative to each other. The methyl groups in positions 2 and 4 are _______ (ortho, meta, para) relative to each other. (c) The model shows the polycyclic aromatic hydrocarbon (PAH) phenanthrene. Draw the structure of phenanthrene on the right. Below, show all possible products of monochlorination of phen-anthrene. (8 points)

Phenanthrene

Isomers of Chlorophenanthrene. [Hint: How many different types of H occur in phenanthrene?]

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Question 3. Electrophilic Aromatic Substitution: Monosubstitution (20 points) (a) Provide complete structural formulas (all atoms, all lone pairs, all formal charges) for the formation reaction of nitronium ion, the reactive species in the nitration of aromatic compounds by electrophilic aromatic substitution. Show nitric acid, show protonated nitric acid, and show nitronium ion. Using curved arrows, show how nitronium ion is formed from protonated nitric acid. (8 points)

(b) Draw complete, unabbreviated structural formulas of four resonance forms for nitrobenzene. [Hints: There is resonance in the benzene ring and in the NO2 group. Draw only resonance forms that keep the NO2 group overall neutral.] (6 points)

(c) Consider the sigma-complex which occurs as an intermediate in the nitration of benzene to form nitrobenzene. 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 NO2 group. It is understood that two resonance forms are possible for the NO2 group in each case; just draw the NO2 group in one way. (6 points)

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Question 4. Electrophilic Aromatic Substitution: Disubstitution (25 points) (a) Consider the bromination of nitrobenzene. 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 nitrobenzene in the ortho (top), meta (center), and para (bottom) positions. Complete the resonance forms: Add all double bonds, all lone pairs, and any formal charges. If there is a particularly stable resonance form, i.e., one that contributes more strongly than the others, then write “good” next to it. If there is a particularly unstable resonance form, i.e., one that contributes much less than the others, then write “bad” next to it. (8 points)

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(c) Directing effects in the bromination of nitrobenzene. 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” – “I” for your convenience. Use these labels in your explanation. Be brief and concise. (6 points)

Ortho-bromination: A and C have normal secondary cation centers.

B is bad; cation-site has e-withdrawing NO2 group attached!

Meta-bromination: D, E and F have normal secondary cation centers.

Para-bromination: G and H have normal secondary cation centers. I is bad; cation-site has e-withdrawing NO2 group attached!

Meta-bromination involves the most stable intermediate (or better “least unstable

intermediate”); three RFs with normal secondary cation centers. (d) Activation/Deactivation of the bromination of nitrobenzene. State whether the NO2 group is “activating” or “deactivating”: ___deactivating____. Compared to the rate of reaction of the bromination of benzene,

• ortho-bromination of nitrobenzene is __________ (much slower, slower, faster, much faster), • meta-bromination of nitrobenzene is __________ (much slower, slower, faster, much faster), and • para-bromination of nitrobenzene is __________ (much slower, slower, faster, much faster).

Explain your answers considering the resonance forms you wrote in part (b). Note that the resonance forms in part (b) are labeled “A” – “I” for your convenience. Use these labels in your explanation. Be brief and concise. (6 points)

It is always bad (i.e., high energy) to have a cationic intermediate.

To have an electron-withdrawing NO2 group attached to a cationic intermediate makes things worse! Bromination in any position of nitrobenzene is slower than bromination

of benzene.

To have an electron-withdrawing NO2 group directly attached to the cation center in one of the RFs of the charged intermediate makes things really bad! Bromination in

the ortho- and para-positions of nitrobenzene is much slower than bromination of benzene.

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Question 5. Stereoisomers – Chirality & Geometrical Isomers. (20 points) (a) Two models are shown of the amino acid aspartic acid, H2N−CH(CH2COOH)−COOH (oxygen in red, nitrogen in blue).

The two models show ___________ (the same, different) enantiomers.

Model #1 is __S__ (R, S). Model #2 is __S__ (R, S).

(6 points)

Model #1

Model #2

(b) The amino acid aspartic acid has the structure H2N−CH(CH2COOH)−COOH and contains one chiral C atom. Provide the CIP priorities of the 4 substituents. For the two C-substituents, apply the sequence rule and provide their “lists”. (6 p.)

Priority of H: _4__ Priority of NH2: _1__ Priority of CH2COOH: __3_ C(C H H) Priority of COOH: _2__ C(O O O)

(c) Perspective drawing of (R)-enantiomer of aspartic acid. The perspective drawing should have two bonds in the paper plane (one vertical), 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 and the amino group should be in the paper plane. (4 points)

(d) Draw the complete structure of the (Z)-isomer of 1,2-dibromo-1-butene. (4 points)