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  • 7/28/2019 Orgo Final

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    CHEM 3311-200 Fall 2006

    Final ExamProfessor R. Hoenigman

    I pledge to uphold the CU Honor Code:

    Signature______________________________________________

    Name (printed)____________________________________________________

    Last four digits of your student ID number_______________________________

    Recitation TA_____________________________________________________

    Recitation number, day, and time______________________________________

    You have 2 hours and 30 minutes to complete this exam.No model kits or calculators allowed.

    Periodic table and scratch paper are attached.

    DO NOT TURN THIS PAGE UNTIL INSTRUCTED TO DO SO.

    Recitation Sections:

    # Day Time TA211 Monday 8 am Noel251 Monday 2 pm Carolynn291 Monday 5 pm Heather 252 Tuesday 12 pm Sam293 Tuesday 5 pm Carolynn212 Wednesday 8 am Noel253 Wednesday 1 pm Tom292 Wednesday 5 pm Heather 213 Friday 8 am Heather

    Score:

    Page 1 ____________/16 Page 6 ____________/40

    Page 2 ____________/15 Page 7 ____________/40

    Page 3 ____________/20 Page 8 ____________/24

    Page 4 ____________/30 Page 9 ____________/45

    Page 5 ____________/20 Page 10 /10

    TOTAL ____________/250

    Average Score = 145

    High Score = 235

    Low Score = 27

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    CHEM 3311, Fall 2006 Final Exam, page 1 of 12

    1. (10 pts) BenzaClin is a topical antibiotic with the structure shown below.

    A. Label the configuration of each indicated chirality center in BenzaClin.(2 pts each)B. Circle all of the terms below that describe one or more structural feature of

    BenzaClin.(1 pt each)Amine Ketone Aldehyde

    Amide Ester Sulfide

    Nitrile Ether Thiol

    O

    C

    NN

    O

    OH

    OH

    n-Pr

    S

    OPO3H2

    Cl

    H

    H

    R

    RR

    2. (6 pts) Give the IUPAC name for each of the following compounds.(2 pts each, -1 for no stereochemistry, -1 for wrong numbers)

    A.

    (3Z,5R)-3,5,6,6-tetramethyl-3-heptene

    B.Br

    OH

    (3S,4R)-4-bromo-5-ethyl-2,5-dimethyl-3-heptanol

    C.

    F

    O

    para-fluoroacetophenone (or 4-fluoroacetophenone)

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    CHEM 3311, Fall 2006 Final Exam, page 2 of 12

    3. (5 pts) Vitamin C, shown below, has a pKa of 4.1. Circle the most acidichydrogen in Vitamin C.

    OO

    HO

    OH

    C

    H H

    OH

    CH2

    OH

    4. (10 pts) Explain why an ester can be both an activating, ortho/para directorand also a deactivating, meta director. Use resonance structures to support yourdiscussion. (4 pts for resonance structures, 6 pts for explanation)

    The ester functional group can be attached to a benzene ring in one of twoways. If the ester is attached at the oxygen, a lone pair of electrons can helpstabilize the arenium ion towards ortho/para attack. The resulting resonancestructure is particularly good, since all atoms have an octet of electrons.

    O O

    R

    H Electrophile

    O O

    R

    H Elec trophile

    If the ester functional group is attached to the benzene ring at the carbonylgroup, the benzene ring become less electron-rich and is deactivated. Inaddition, the adjacent carbonyl group is a meta director, since the followingresonance structure would be destabilized(due to the adjacent positive charges)by ortho/para attack.

    O

    R

    Electrophile HO

    O

    R

    Electrophile HO

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    CHEM 3311, Fall 2006 Final Exam, page 3 of 12

    5. (15 pts) The bromination of anisole produces two isomeric bromoanisoles.Using curved arrows to show the flow of electrons, give a mechanism to accountfor the formation of these two products.

    OCH3

    Br Br FeBr3 Br Br FeBr3

    para attack ortho attack

    OCH3 OCH3

    H

    Br

    Br

    H

    Br FeBr3 Br FeBr3

    OCH3 OCH3

    Br Br

    FeBr3

    HBrFeBr3

    HBr

    6. (5 pts) Circle the compounds below that are chiral. (1 pt each)

    C C C

    H3C H

    D

    H

    cis-3-methylcyclobutanol

    Cl

    Br CH3

    CH3

    BrCl

    (1R,2S)-1,2-dimethylcyclopentane

    CH3

    OHH

    H OH

    CH3

    (+)-2-chlorobutane

    OTs

    OTs

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    CHEM 3311, Fall 2006 Final Exam, page 4 of 12

    7. (15 pts) Circle the more acidic compound in the following pairs. Give thereason for your choice in the adjacent box.(2 pts circle, 3 pts explanation)

    A.or

    B.CH3COOH or CF3COOH

    C.NO2

    HO

    OH

    NO2

    or

    8. (15 pts) Circle the compound that has the lowerheat of combustion. Give thereason for your choice in the adjacent box.(2 pts circle, 3 pts explanation)

    A. Book Problem 11.50c/d

    or

    B.

    H3C

    H3C CH3

    C(CH3)3 H3C

    H3C CH3

    C(CH3)3

    or

    C.

    or

    The allylic conjugate base is

    resonance stabilized

    Inductive effect

    More stable conjugate base (more

    resonance structures)

    Aromatic cation versus anantiaromatic anion

    More substituents are in theequatorial position

    Allylic cation is resonance stablized

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    CHEM 3311, Fall 2006 Final Exam, page 5 of 12

    9. (10 pts) Catalytic hydrogenation of 1,4-dimethylcyclopentene yields a mixtureof two products. One of these products is formed in a much greater yield than theother (the observed ratio is 10:1). Draw the products of this reaction and circlethe major product. Why is there a preference for one of these products?Book Problem 6.40

    (2 pts each product, 1 pt circle, 5 pts explanation)

    Hydrogenation is regioselective to the less hindered face of the doublebond. Thus, the alkene will attach to the catalyst on the side opposite the C4methyl group and the hydrogen will add to the opposite side, favoring formationofcis-1,3-dimethylcyclopentane.

    H2

    Pt+

    10. (5 pts) Match the following alkenes with the appropriate HH2 value. Heats of

    hydrogenation: 151 kJ/mol, 122 kJ/mol, 114 kJ/mol, 111 kJ/mol, 105 kJ/molBook Problem 6.29(1 pt each)

    _122 kJ/mol_ 1-pentene

    _111 kJ/mol_ (E)-4,4-dimethyl-2-pentene

    _114 kJ/mol_ (Z)-4-methyl-2-pentene

    _151 kJ/mol_ (Z)-2,2,5,5-tetramethyl-3-hexene

    _105 kJ/mol_ 2,4-dimethyl-2-pentene

    11. (5 pts) Explain why 3-pentanol has a higher boiling point than3-chloropentane.

    Boiling points are determined by intermolecular forces. 3-pentanol has

    greater intermolecular forces than 3-chloropentane, since the alcohol canparticipate in hydrogen bonding. Hydrogen bonds require a lot of energy to break(relatively) and thus, the alcohol has a higher boiling point.

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    CHEM 3311, Fall 2006 Final Exam, page 6 of 12

    12. (80 pts) Give the major organic product(s) of the following reactions. Be sureto clearly show stereochemistry using dashes and wedges. Write NR if noreaction occurs. Clearly label any enantiomers, diastereomers, and/or mesocompounds. (Note: some problems have more than one step.)

    A. D CH3 D CH3

    + diastereomers

    HO OH HO OH

    meso meso

    D CH3

    OsO4

    HOOH

    B.

    NBS

    h

    Br

    Br Br

    + +

    enantiomers

    C. Exam 1 problemCH3Li + NH3 CH4 LiNH2+

    D.

    Cl+AlCl3

    E.

    2-methyl-3-hexyne1) O3

    2) Zn, H2O

    O

    OH

    HO

    O

    +

    F.

    AlCl3

    toluenechromic

    acid

    CH3COClOH

    O

    NR

    G.

    BH3 + NH3 H3B NH3

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    CHEM 3311, Fall 2006 Final Exam, page 7 of 12

    12. continued. [Be sure to clearly show stereochemistry using dashes andwedges. Write NR if no reaction occurs. Clearly label any enantiomers,diastereomers, and/or meso compounds; some problems have more than onestep.]

    H.

    2-methyl-3-hexyneH2SO4

    HgSO4

    OO

    +

    I. Book Problem 12.36b

    O

    Cl

    AlCl3

    O

    J.

    OH

    p-TsCl

    pyridine

    NaSH

    DMSO

    OTs SH

    K.

    Br

    D

    NaOCH3

    CH3OH

    D

    L.

    +

    NC

    NC CN

    CN

    CH3

    CN

    CN

    CN

    CN

    CH3

    CN

    CN

    CN

    CN

    +

    enantiomers

    M.

    2,3-dimethyl-1,3-butadiene DCl D

    ClCH3

    D

    ClCH3

    + +

    D

    Clenantiomers

    N. Book Problem 6.46

    3-methyl-1-buteneaq. H2SO4

    OH

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    CHEM 3311, Fall 2006 Final Exam, page 8 of 12

    13. (12 pts) Using curved arrows to show the flow of electrons, draw amechanism for the solvolysis oftert-butyl bromide in acetic acid.

    BrBr

    HO

    OO

    O

    H

    HO

    O

    O

    O

    H2O

    O

    Br

    E1

    SN1

    HO

    O

    H

    H2O

    O

    14. (12 pts) Using curved arrows to show the flow of electrons, propose amechanism for the base catalyzed enol-keto tautomerization shown below.

    OH

    HO

    O

    O

    HO

    O

    H O

    H

    O

    HO

    HO

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    CHEM 3311, Fall 2006 Final Exam, page 9 of 12

    15. (45 pts) Propose an efficient synthesis for the following transformations. Youmay use any reagents you like. Be sure to show any intermediates. (Do not draw

    a mechanism.)

    A. 1-bromo-2-chloropropane starting from propene

    Book Problem 10.24d

    NBS

    HOOH

    Br HCl

    Cl

    Br

    B. 3-bromo-5-nitrobenzoic acid starting from toluene

    Book Problem 12.27k

    chromic

    acid

    O

    OHH2SO4

    HNO3

    O

    OH

    NO2

    O

    OH

    NO2

    Br

    FeBr3

    Br2

    C. 2-butyne starting from 1-butyne

    Br Br

    2 HBr 2 NaNH2

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    CHEM 3311, Fall 2006 Final Exam, page 10 of 12

    Extra Credit: (10 pts) Draw a parody of aromatic nomenclature. For example,below is paraphrase.

    PhRaSe

    PhRaSe

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    CHEM 3311, Fall 2006 Final Exam, page 11 of 12

    Scratch Page

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    CHEM 3311, Fall 2006 Final Exam, page 12 of 12

    Scratch Page