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144 Cards in this Set
- Front
- Back
- 3rd side (hint)
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The pka of a hydrogen bonded to the sp3 garbon of propene is 42, which is greater than that of any of the carbon acids listed in table 18.1 but less than the pKa of an alkane. Explain.
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The electrons left behind then a proton is removed from propene are delocalized- they are shared by three carbon atoms. In contrast, the electrons left behind when a proton is removed from an alkane are lovalized-they belong to a single (carbon) atom. Because electron delocalization allows charge to be distributed over more than one atom, it stabilizes the base. Thus, base with delocalized electrons is more stable so it has the stronger conjugate acid. Thus, propene is a stronger acid than an alkane.
Propene, however is not as acidic as the carbon acids in table 18.1, because the electrons left behind when a proton is removed from these carbon acids are delocalized onto an oxygen or a nitrogen, which are more electronegative atoms than carbon and, therefore, better able to accommodate the electrons. PIC |
18.1
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Give an example of a beta-keto nitrile
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18.2.a
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Give an example of a beta-diester
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18.2.b
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explain why a proton can be removed from the alpha-carbon of N,N-dimethylethanamide but not from the alpha-carbon of either N-methylethanamide or ethanamide.
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a proton cannot be removed from the alpha-carbon of N-methylethanamide or ethanamide because these compounds have a hydrogen bonded to the nitrogen and this hydrogen is more acidic than the one attached to the alpha-carbon. The following resonance contributers show why the hydrogen attached to the nitrogen is more acidic than the hydrogen attached to the alpha carbon. PIC
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18.3
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Explain why an N,N-distributed amide is less acidic than an ester.
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Electron delocalization of the lone pair on nitrogen or oxygen competes with electon delocalization of the electrons left behind on the alpha-carbon when it looses a proton. The lone pair on nitrogen is more delocalized than the lone pair on oxygen, because nitrogen is better able to acomidate a positive charge. Therefore the amide competes better with the carbanion for electron delocalization, to the alpha-carbon is less acidic.
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18.4
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List the compounds in each of the following groups in order of decreasing acidity:
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18.5.a
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List the compounds in each of the following groups in order of decreasing acidity:
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18.5.b
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List the compounds in each of the following groups in order of decreasing acidity:
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The ketone is the strongest acid because there is no competion for the electrons that are left behind when the alpha-hydrogen is removed. The lactam it the weakest acid because nitrogen, being less elecronegative, can better accommodate a positive charge and therefore, is better at delocalizing its lone pair onto the oxygen. Therefore, nitrogen is better at competing with the electrons left behind when an alpha-hydrogen is removed for delocalization onto the carbonyl oxygen.
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18.5.c
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only 15% of 2,4-penanedione exists as the enol tautomer in water, but 92% exist as the enol tautomer in hexane. Exflain why this is so.
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Both the keto and enol tautomers of 2,4-pentanedione can form hydrogen bonds with water. Neither the keto nor the enol tautomer can form hydrogen bonds with hexane, but the enol tautomer can still form an intramolecular hydrogen bond. This intramolecular hydrogen bonding stabilizes the enol tautomer. Thus, the enol tautomer is more stable relative to the keto tautomer in hexane than in water.
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18.6
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draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable.
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18.8.a
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draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable.
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18.8.b
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draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable.
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18.8.c
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draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable.
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18.8.d
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draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable.
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18.8.e
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draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable.
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18.8.f
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When a dilute solution of acetaldhyde is shaken in NaOD in D2O, esplain why the methyl hydrogens are exchanged with deuterium but the aldehyde hydrogen is not.
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The aldehyde hydrogen cannot be removed by a base. The aldehyde hydrogen is not acidic, because the electrons left behind if it were to be removed cannot be delocalized.
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18.9
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why do only methyl ketones undergo the haloform reaction?
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The halogen reaction requires that a group be created that is a weaker base than hydroxide ion so that hydroxide ion is not the group eliminated from the tetrahedral intermediate. For an alkyl group to be the weaker base, it must be bonded to three halogen atoms. The only alkyl group that can be bonded to three halogens is a methyl group.
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18.10
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A ketone undergoes acid-catalyzed bromination, acid-catalyzed chlorination, and acid-catalyzed deuterium exchange at the alpha-carbon, all at about the same rate. What does this tell you about the mechanism of these reactions?
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A Br-Br bond is weakeu and easier to break than a Cl-Cl bond, which in turn is weaker and easier to break than a D-O bond. Because the rates of bromination, chloronation, and deuterium exchange are about the same, you know that breaking the Br-Br, Cl-Cl, or D-O bond takes place after the rate-determining step. Therefore the rate-determining step must be removal of the proton from the alpha carbon of the ketone.
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18.11
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Synth
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18.12.a
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Synth
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18.12.b
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Synth
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18.12.c
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Synth
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18.12.d
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how could this be prepared from a carbonyl compound with no C=C double bonds?
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18.13.a
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how could this be prepared from a carbonyl compound with no C=C double bonds?
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18.13.b
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Synth from cyclohexanone
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18.14.a
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Synth from cyclohexanone
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18.14.b
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Synth from cyclohexanone
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18.14.c
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Synth from cyclohexanone
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18.14.d
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What compound is formed when a dilute solution of cyclohexanone is shaken with NaOD in D2O for several hours?
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18.15
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explain why alkylation of an alpha-carbon works best if the alkyl halide used in the reaction is a primary alkyl halide, and why alkylation does not work at all if a tertiary alkyl halide is used.
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Alkylation of an alpha carbon is an SN2 reaction. SN2 reactions work best with primary alkyl halides because there is less steric hiderance in a primary alkyl halide than in a secondary alkyl halide. SN2 reactions don't work at all with tertiary alkyl halides because they are the most sterically hindered. Therefore in the case of tertiary alkyl halides, the SN2 reaction cannot compete with the E2 elimination reaction.
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18.16
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Synth from a ketone and alkyl halide (with double bond)
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18.17.a
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Synth from a ketone and alkyl halide
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18.17.b
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Describe how the following compound could be prepared using an enamine intermediate
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18.18.a
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Describe how the following compound could be prepared using an enamine intermediate
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18.18.b
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synth
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18.19.a
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synth
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18.19.b
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what is the aldol addition product
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18.20.a
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what is the aldol addition product
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18.20.b
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what is the aldol addition product
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18.20.c
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what is the aldol addition product
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18.20.d
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indicate the aldehyde or ketone from which 2-ethyl-3-hydroxyhexanal would be formed by an aldol addition
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18.21.a
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indicate the aldehyde or ketone from which 4-hydroxy-4-methyl-2-pentanone would be formed by an aldol addition
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18.21.b
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indicate the aldehyde or ketone from which 2,4-dicyclohexyl-3-hydroxybutanal would be formed by an aldol addition
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18.21.c
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indicate the aldehyde or ketone from which 5-ethyl-5-hydroxy-4-methyl-3-heptanone would be formed by an aldol addition
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18.21.d
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an aldol addition can be catalyzed by acids as well as by bases. Propose a mechanism for the acid-catalyzed aldol addition of propanal.
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18.22
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What product is obtained from the aldol condensation of cyclohexanone?
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18.23
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Synth from starting materials with no more than three carbons
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18.24.a
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Synth from starting materials with no more than three carbons
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18.24.b
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Synth from starting materials with no more than three carbons
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18.24.c
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products of a mixed aldol addition
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18.25.a
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products of a mixed aldol addition
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18.25.b
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products of a mixed aldol addition
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18.25.c
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products of a mixed aldol addition
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18.25.d
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Describe how the following could be prepared using an aldol addition in the first step of the synthesis.
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18.26.a
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Describe how the following could be prepared using an aldol addition in the first step of the synthesis.
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18.26.b
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Describe how the following could be prepared using an aldol addition in the first step of the synthesis.
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18.26.c
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Products
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18.28.a
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Products
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18.28.b
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Which of the following esters cannot undergo a Claisen condensation?
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A,B,D cannot undergo a Claisen condensation. A cannot because a proton cannot be removed from an sp2 carbon. B and D cannot b/c they do not have an alpha carbon.
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18.29
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Products
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18.30.a
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Products
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18.30.b
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Products
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18.30.c
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Products
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18.30.d
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synth from 3-thiomethylcyclohexanone
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18.31.a
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synth from 3-thiomethylcyclohexanone
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18.31.b
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write the mechanism for the base-catalyzed formation of a cyclic beta-keto ester from a 1,7-diester.
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18.32
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If the preference for formation of a six membered ring were not so great, what other cyclic product would be formed from the intramolecular aldol addition of 2,6-heptanedione
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18.33.a
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If the preference for formation of a six membered ring were not so great, what other cyclic product would be formed from the intramolecular aldol addition of 2,8-nonanedione
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18.33.b
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Can 2,4-pentanedione undergo an intramolecular aldol addition. Why or why not?
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No, b/c an intramolecular reaction would lead to a strained four membered ring. Therefore the intermolecular reaction will be preferred.
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18.34
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product of reaction with a base
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18.36.a
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product of reaction with a base
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18.36.b
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product of reaction with a base
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18.36.c
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product of reaction with a base
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18.36.d
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synth using a robinson annulation:
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18.37.a
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synth using a robinson annulation:
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18.37.b
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synth using a robinson annulation:
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18.37.c
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synth using a robinson annulation:
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18.37.d
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What alkyl bromides should be used in the malonic ester synthesis of propanoic acid
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methyl bromide
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18.39.a
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What alkyl bromides should be used in the malonic ester synthesis of 2-methylpropanoic acid
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methyl bromide twice
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18.39.b
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What alkyl bromides should be used in the malonic ester synthesis of 3-phenylpropanoic acid
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benzyl bromide
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18.39.c
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What alkyl bromides should be used in the malonic ester synthesis of 4-methylpentanoic acid
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isobutyl bromide
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18.39.d
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which of the following compounds would be expected to decarboxylate then heated.
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A and D can be decarboxylated
B can't be decarboxylated, b/c it doesn't have a carboxyl group the electrons left behind if C were decarboxylated cannot be delocalized onto an oxygen |
18.38
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explain why the following cannot be prepared by the malonic ester synthsis
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an SN2 reaction cannot be done on bromobenzene.
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18.40.a
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explain why the following cannot be prepared by the malonic ester synthsis
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an SN2 reaction cannot be done on vinyl bromide
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18.40.b
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explain why the following cannot be prepared by the malonic ester synthsis
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an SN2 reaction cannot be done on a tertiary alkyl halide
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18.40.c
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Starting with methyl proanoate, how could you prepare 4-methyl-3-heptanone?
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Because the starting material is an ester and the target moleule has more carbons than the starting material, a Claisen condensation appears to be a good way to start this Synth. The claisen condensation forms a beta-keto ester that can be easily alkylated at the desired carbon because it is flanked by two carbonyl groups. Acid-catalyzed hydrolysis will form a 3-oxocarboxylic acid that will decarboxylate when it is heated.
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18.41
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What alkyl bromide should be used in the acetoacetic ester synthesis of the following 2-pentanone (methyl ketones)?
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ethyl bromide
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18.42.a
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What alkyl bromide should be used in the acetoacetic ester synthesis of the following 2-octanone (methyl ketones)?
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pentyl bromide
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18.42.b
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What alkyl bromide should be used in the acetoacetic ester synthesis of the following 4-phenyl-2-butanone (methyl ketones)?
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benzyl bromide
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18.42.c
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Synth
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18.43.a
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Synth
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18.43.b
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Synth
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18.43.c
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Synth
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18.43.d
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structure ethyl acetoacetate
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18.48.a
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structure alpha-methylmalonic acid
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18.48.b
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structure beta-keto ester
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18.48.c
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structure enol tautomer of cyclopentanone
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18.48.d
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structure the carboxylic acid obtained from the malonic ester synthesis when the alkyl halide is propyl bromide
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18.48.e
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Products:
diethyl heptanedioate: (1) sodium ethoxide, (2) HCl |
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18.49.a
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Products:
entaoic acid +Pbr3 +Br2, hydrolysis |
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18.49.b
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Products:
acetone + ethyl acetate:(1) sodium ethoxide, (2) HCl |
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18.49.c
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Products:
diethyl 2-ethylhecandioate: (1) sodium ethoxide, (2) isobutyl bromide; (3) HCl, H2O + heat |
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18.49.d
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Products:
acetophenone +diehtyl carbonate: (1) sodium ethoxide (2) HCl |
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18.49.e
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Products:
1,3-cyclohexanedione + alkyl bromide + sodium hydroxide |
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18.49.g
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Products:
dibenzyl ketone + methyl vinyl ketone + excess sodium hydroxide |
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18.49.h
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Products:
cyclopentanone: (1)pyrrolidine + catalytic H+ , (2) ethyl bromide, (3) HCl, H2O |
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18.49.i
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2,7-octanedione + sodium hydroxide
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18.49.k
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Products:
gamma-butryolactone + LDA in THF, followed by methyl iodide |
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18.49.j
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Products:
diethyl 1,2-benznedicarboxylate + ethyl acetate: (1) excess sodium ethocide, (2) HCl |
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18.49.l
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Which compound decarboxylates at the lowest temperature?
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The compound on the right loses CO2 at the lowest temperature because the electrons left behind when CO2 is removed can be delocalized
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18.50
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The chemical shifts of nitromethane, dinitromethane, and trinitromethane are at delta 6.10, 4.33, 7.52. Match each chemical shift with the compound. Explain how chemical shift correlates with pKa.
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The electron withdrawing nitro group will cause the signla to occur at a higher frequency ( will have a larger chemical shift). The more acidic the hydrogen, the greater the chemical shift.
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18.51
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Explain why a racemic mixture of 2-methyl-1-phenyl-1butanone is formed when (R)-2-methyl-1-phenyl-1-butanone is dissolve in an acidic or basic aqueous solution. Give an example of another ketone that would undergo acid or base catalyzed racemization.
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a) When the ketone enolizes, the asymmetric center is lost. When the enol reforms the ketone, an asymmetric center is created, it can form the R and S enationmer equally as easily, so a racemic mixture is obtained.
b) you need a ketone that has an alpha-carbon that is an asymmetric center. If you don't want racemization to have to compete with the removal of a hydrogen from a second alpha-carbon, it is best to chose a compound that can form only one enol. |
18.52
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Product
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18.53
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synth from cyclohexanone
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18.60.a
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synth from cyclohexanone
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18.60.b
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synth from cyclohexanone
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18.60.c
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synth from cyclohexanone
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18.60.d
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synth from cyclohexanone
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18.60.e
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synth from cyclohexanone
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18.60.f
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synth
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18.62.a
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synth
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18.62.b
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synth
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18.62.c
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synth
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18.62.d
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synth
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18.62.e
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synth
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18.62.f
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Bupropion hydrochloride is an antidepressant marketed under the trade name wellbutrin. Propose a synthesis of bupropion hydroochloride, starting with benzene.
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18.63
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Reagents?
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18.64
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Products?
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18.65.a
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Products?
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18.65.b
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Products?
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18.65.c
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Show how the amino acid alanine can be synthesized from propanoic acid
show how the amino acid glycine can be synthesized from phthalimide and diethyl 2-bromomalonate. |
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18.66
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.a
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.b
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.c
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.d
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.e
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.f
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.g
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.h
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Synth using an aldol condensation or explain why an aldol condensation is not possible. EDIT
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18.67.i
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Synth using only the carbon containing compound shown
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18.68.a
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Synth using only the carbon containing compound shown
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18.68.b
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Synth using only the carbon containing compound shown
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18.68.c
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Explain why the following bromoketone froms different bicyclic compounds under different reaction conditions.
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18.69
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