quiz 1a answers - university of hultin/chem2220/archive/2002/quizzes_2002.pdfآ  acid-catalyzed...

Download Quiz 1A Answers - University of hultin/chem2220/Archive/2002/quizzes_2002.pdfآ  Acid-catalyzed hydrolysis

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  • 1A ANSWER KEY

    2.222 In-class Quiz #1 Friday January 18, 2002. CIRCLE the correct response for each of the following questions.

    1. Which of the following alkenes would react most rapidly with HBr in CH2Cl2?

    Since protonation of the alkene is rate limiting, the alkene that will react fastest will be the one that forms the most stable intermediate cation. As the textbook points out, the activation energy for this step will correlate with cation stability. Note that delocalization of the positive charge onto the aromatic ring is a highly stabilizing factor.

    2. Which one of compounds A-E below would be the major product of the following reaction?

    This reaction involves a 1,2-migration of a methyl group to convert the initially formed secondary cation into the more stable tertiary cation. Note that structure A also arises from a 1,2-shift, but that it is disfavoured by the ring strain in the cyclobutane ring. Structures D and E are less favourable since they arise from secondary cations.

    Br O

    A B C D E

    H2SO4, H2O ?

    A B C D E

    OH

    OH

    OH

    OH

    OH

  • 1A ANSWER KEY

    3. Which of the following is NOT a feasible reaction?

    As we discussed in class, the preparation of a geminal dibromide from an alkyne is not really possible in water because the dibromide will react further to form a ketone under those conditions. Your textbook doesn’t actually say that the reaction can be done in water, but in Figure 10.8 it shows the participation of H3O+, which implies the presence of water. I disagree with this. All the other reactions are ones we have discussed – notice that D is an addition of singlet dichlorocarbene, generated in situ by base treatment of chloroform.

    Br2, CH2Cl2

    Br

    H

    H

    Br

    A

    B

    C

    D

    E

    HCl, H2O

    OH

    CH3C C CH3 HBr, H2O CH3C CH2 CH3

    Br Br

    NaOH (aq.), Bu4N OH CHCl3

    Cl

    Cl H

    1) Hg(OAc)2, H2O 2) NaBH4

    OH

  • 1B ANSWERS

    2.222 In-class Quiz #1 Friday January 18, 2002. CIRCLE the correct response for each of the following questions.

    1. Which of the following alkenes would react SLOWEST with HBr in CH2Cl2?

    The initial protonation step of the reaction is rate limiting, and its activation barrier is correlated with the stability of the resulting cation. The leaststable cation will be the one substituted with an electron-withdrawing group, and a ketone is a very good EWG. All the other alkenes are substituted with H and/or donor groups.

    2. Which one of compounds A-E below would be the major product of the following reaction?

    This question is basically the same as Exercise 10.7a in the textbook. Protonation of the alkene forms a secondary cation, which rearranges to a more-stable tertiary cation by a 1,2-shift of a methyl group. Note that structures D and E are impossible to obtain under these conditions, while structure B is extremely unlikely since the conjugate base of sulfuric acid is a poor nucleophile.

    O

    A B C D E

    H2SO4, H2O ? OH OSO3H

    HO

    OH

    OH

    A B C D E

  • 1B ANSWERS

    3. Only ONE of the following IS a feasible reaction. Which one is it?

    Acid-catalyzed hydrolysis of a terminal alkyne to a ketone proceeds with heating as shown here, or at room temperature in the presence of mercuric acetate. Notice that reaction B looks a bit like examples we have seen, but that there is no driving force for a 1,2-shift when the ring is NOT aromatic! Reaction D would require a two-step mechanism consistent with a triplet carbene, but these conditions form singlet dichlorocarbene.

    I2, CH2Cl2

    I

    H

    H

    I

    A

    B

    C

    D

    E

    HCl, H2O

    OH

    CH3CH2C CH H2SO4, H2O,

    heat CH3CH2C CH3

    O

    NaOH (aq.), Bu4N OH CHCl3

    1) Hg(OAc)2, H2O 2) NaBH4 HO

    Cl Cl

    HH

  • 1C ANSWERS

    2.222 In-class Quiz #1 Friday January 18, 2002. CIRCLE the correct response for each of the following questions.

    1. Which of the following alkenes would react with HBr in CH2Cl2 to give a CHIRAL product?

    In all cases (except A) the bromide ion will attack the more-substituted position and the proton will end up on the other carbon of the alkene. The resulting bromides from A, B, D and E all have a plane of symmetry passing through the Br-substituted position, but in the case of C the molecule is not symmetrical. Of course, the chiral bromides obtained from C are racemic.

    2. Which one of compounds A-E below would be the major product of the following reaction?

    Reaction will occur fastest at the most-substituted alkene group, and of course will give the Markovnikov product. Note that if the reaction were left long enough, both alkenes would react to give a diol, but I didn’t include that option in this question.

    Br

    A B C D E

    H2SO4, H2O ?

    A B C D E

    OH

    HO

    HOOH

    HO

    HBr

    Br

    H

    Br

    H

    if addition is syn

    if addition is anti

  • 1C ANSWERS

    3. Which ONE of the following DOES NOT involve electrophilic addition to an alkene?

    The reaction in D proceeds by protonation of oxygen, loss of water, and isomerization of the double bond to the internal position. Although it is disfavoured by formation of a primary cation, the product is more stable than the reactant because the alkene group is more- substituted.

    ICl, CH2Cl2A

    B

    C

    D

    E

    HCl, H2O

    CH3CH2C CH H2SO4, H2O,

    heat CH3CH2C CH3

    O

    1) Hg(OAc)2, H2O 2) NaBH4

    Cl

    I

    OH

    OH

    OH H2SO4, H2O, heat

    OH

  • 2A ANSWER KEY

    2.222 In-class Quiz #2 Friday February 1, 2002. CIRCLE the correct response for each of the following questions.

    1. What is the Major Product of the following reaction?

    This Freidel-Crafts Alkylation proceeds via the more-stable secondary cation formed from the alkene on treatment with the Lewis Acid AlCl3. Notice that this is the same Markovnikov selectivity that you would expect in an acid-catalyzed addition to the alkene – in fact this reaction actually IS an acid-catalyzed addition from the alkene’s point of view!

    2. Which set of reagents will perform the following transformation?

    A. Zn(Hg)/HCl (aq.)

    B. Na (s), NH3 (l), EtOH C. H2 (g), Pd (cat.), AcOH

    D. 1) Hg(OAc)2, H2O; 2) NaBH4

    E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)

    Selective reduction of the alkyne by dissolving metal always gives the trans alkene product. Option A is a Clemmensen Reduction, which would remove the carbonyl, while option C would hydrogenate both the carbonyl and the alkyne. The alkyne would be reduced all the way to an alkane. Options D and E would result in the formation of ketone products via intermediate enols.

    + AlCl3 ?

    A B C D E

    ClOH

    O

    ? O

    H

    H

  • 2A ANSWER KEY

    3. Which set of compounds could be used to prepare the following molecule using the conditions specified?

    This is a Diels-Alder cycloaddition. Recall that this type of reaction always occurs between a diene and a dienophile (alkene), and that the dienophile should be substituted with an electron-withdrawing group. Also, recall that the product will be a cyclohexene ring, in which the double bond is located between what were the C2 and C3 positions of the diene.

    Option A shows conditions for a Simmons-Smith cyclopropanation, but the product is NOT a cyclopropane – note that the bridging CH2 group is attached across the ring, not to two adjacent carbons from one of the alkenes. Option E describes an oxidative cleavage of the alkenes, which would form carboxylic acids, but it would cleave BOTH alkene groups, not just the exocyclic one.

    heat? CO2CH3

    A B

    C D

    E

    CO2CH3

    + Et2Zn CH2I2+

    + CO2CH3

    CO2CH3

    + CH2H2C

    H3CO2C +

    + KMnO4

    CO2CH3

    + CO2CH3

    H3CO2C

    H3CO2C

    HO2C

    HO2C

    CO2H

    CO2H HO2C

    HO2C

  • 2A ANSWER KEY

    4. What product is obtained from the following 2-step sequence of reactions?

    The N,N-dimethylamino group is a strongly activating ortho/para director. Thus, Freidel-Crafts acylation will occur primarily para. The second reaction will be influenced by both the dimethylamino and the acyl groups, but both these groups favor nitration at the same position – ortho to the amine and meta to the ketone.

    5. Which set of reaction conditions would perform the following transformation?

    A. 1) Hg(OAc)2, H2O; 2) NaBH4

    B. H2SO4, H2O

    C. OsO4

    D. 1) O3, CH2Cl2; 2) Zn, AcOH

    E. 1) B2H6, THF; 2) H2O2, NaOH (aq.) Only Hydroboration/Oxidation can add a hydroxyl group to an alkene in an anti- Markovnikov fashion. Option A and Option B would both form the tertiary alcohol, while Option C would form a DIOL. Option D is an oxidative cleavage, which would chop off the terminal CH2 group and form a ketone in its place.

    N

    AlCl3

    O

    Cl HNO3, H2SO4 ?

    A B C D E

    N

    NO2

    O

    N

    NO2

    O

    N

    NO2

    O

    N

    NO2 O

    N

    O

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