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182 Cards in this Set
- Front
- Back
- 3rd side (hint)
give two names:
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3-methylpentanal
Beta-methylvaleraldehyde |
17.2.a
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give two names:
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4-heptanone
dipropyl ketone |
17.2.b
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give two names:
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2-methyl-4-heptanone
isobutyl propyl ketone |
17.2.c
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give two names:
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4-phenylbutanal
gamma-phenylbutyraldehyde |
17.2.d
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give two names:
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4-ethylhexanal
gamma-ethylcaproaldehyde |
17.2.e
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give two names:
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1-hepten-3-one
butyl vinyl ketone |
17.2.f
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name:
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6-hydroxy-3-heptanone
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17.3.a
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name:
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2-oxocyclohexylmethanenitrile
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17.3.b
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name:
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3-formylpentanamide
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17.3.c
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which is more reactive 2-heptanone or 4-heptanone
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2-heptanone is more reactive because it has less steric hindrance. There is little difference in the amount of steric hindrance provided at the carbonyl carbon by a pentyl and a propyl group because they differ at a point somewhat removed from the site of nucleophilic attack. The difference in size between a methyl group and a prpyl group is sinigicant at the site of nucleophilic attack.
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15.4.a
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which is more reactive p-nitroacetophenone or p-methoxyacetophenone
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p-nitroacetophenone is more reactive because the electron-withdrawing nitro group makes the carbonyl carbon more susceptible to nucleophilic attack compared with an electron-donating methoxy group.
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15.4.b
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what products would be formed by the reaction with CH3MgBr followed by acid
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17.5.a
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what products would be formed by the reaction with CH3MgBr followed by acid
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17.5.b
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what products would be formed by the reaction with CH3MgBr followed by acid
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17.5.c
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We saw that 3-methyl-3hexanol can be synth from the reaction of 2-pentaone with ethylmagnesium bromide. What two other combinations of ketone and grinard reagent could be used to prepare the same tertiay alcohol
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17.6
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How many isomers are obtained from the reaction of 2-pentanone with ethylmagnesium bromide followed by treatment with aqueous acid?
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Two isomers are obtained because the reaction creates an asymmetric center in the product
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17.7.a
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How many isomers are obtained from the reaction of 2-pentanone with methylmagnesium bromide followed by treatment with aqueous acid?
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Only one compound is obtained because the product does not have an asymmetric center
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17.7.b
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What is the product of the reaction of an ester with excess acetylide ion followed by the addition of pyridinium ion?
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Similar to the way a grignard reagent reacts with an ester, an acetylide ion first undergoes a nucleophilic acyl substitution reaction with the ester; this is followed by a nucleophilic addition reaction.
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17.12
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What alcohols are obtained from the reduction of 2-methylpropanal with sodium borohydride
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.
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17.13.a
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What alcohols are obtained from the reduction of cyclohexanone with sodium borohydride
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17.13.b
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What alcohols are obtained from the reduction of benzaldehyde with sodium borohydride
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17.13.c
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What alcohols are obtained from the reduction of acetophenone with sodium borohydride
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17.13.d
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What amide would you treat with LiAlH4 in order to prepare benzylmethylamine
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17.14.a
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What amide would you treat with LiAlH4 in order to prepare ethylamine
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17.14.b
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What amide would you treat with LiAlH4 in order to prepare diethylamine
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17.14.c
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What amide would you treat with LiAlH4 in order to prepare triethylamine
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.
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17.14.d
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Starting with N-benzylbenzamide, how would you make dibenzylamine
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17.15.a
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Starting with N-benzylbenzamide, how would you make benzoic acid
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17.15.b
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Starting with N-benzylbenzamide, how would you make benzyl alcohol
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17.15.c
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Can a cyanohydrin be prepared by treating a ketone with sodium cyanide?
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No, an acid must be present in the reaction mixture in order to protonate the oxygen of the cyanohydrin. Otherwise the cyano gropu will be eliminated and the reactants will be reformed.
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17.16
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Explain why aldehydes and ketones react with a weak acid such as hydrogen cyanide in the presence of -C=N, but do not react with strong acids such as HCL or H2SO4 in the presence of Cl- or HSO4-.
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Strong acids like HCl and H2SO4 have very weak conjugate bases (Cl- and HSO4-), which are excellent leaving groups. When these bases add to the carbonyl group, they are readily eliminated, reforming the starting materials. Cyanide ion is a strong enough base, so it is not eliminated unless the oxygen in the product is negatively charged.
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17.17
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How can the following compounds be prepared, starting with a carbonyl compound that has one fewer carbon atoms than the desired product?
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The starting material for the synthesis of this two-carbon compound must be formaldehyde. Addition of hydrogen cyanide followed by addition of H2 to the triple bond of the cyanohydrin forms the target compound.
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17.18.a
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How can the following compounds be prepared, starting with a carbonyl compound that has one fewer carbon atoms than the desired product?
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The addition of cyanide ions adds one carbon to the reactant, so the starting material for the synthesis of this three-carbon alpha-hydroxycarboxylic acid must be ethanal. Addition of hydrogen cyanide, followed by hydrolysis of the cyanohydrin, forms the target compound
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17.18.b
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AT what pH should imine formation be arried out if the amine's protonated form has a pKa value of 10.0?
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Figure 17.2 for imine formation when the amine is hydroxylamine (pKa=6.0) shows that the maximum rate is obtained when the pH is 1.5 units lower than the pKa of the amine. Thus if the amine has a pKa of 10, imine formation should be carried out at pH=8.5.
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17.19
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A ketone can be prepared from the reaction of a nitrile with a grignard reagent. Describe the intermediate formed in this reaction, and explain how it can be converted to a ketone.
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The nitrile reacts with the grignard reagent to form an imine which could then by hydrolyzed to a ketone.
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17.21
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mechanism for the acid-catalyzed hydrolysis of an imine to a carbonyl compound and a primary amine
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17.22.a
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mechanism for the acid-catalyzed hydrolysis of an enamine to a varbonyl compound and a secondary amine
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17.22.b
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products of
cyclopentanone +ethylamine |
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17.23.a
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products of
cyclopentanone + diethylamine |
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17.23.b
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acetophenone + hexylamine
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17.23.c
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acetophenone + cyclohexylamine
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17.23.d
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Imines can exist as stereoisomer. The isomers are named by the E,Z system of nomenclatur. (The lone pair has the lowest priority.) Draw the structure of (E)-benzaldehyde semicabazone
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17.24.a
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Imines can exist as stereoisomer. The isomers are named by the E,Z system of nomenclatur. (The lone pair has the lowest priority.) Draw the structure of (Z)-propiophenone oxime
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17.24.b
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Imines can exist as stereoisomer. The isomers are named by the E,Z system of nomenclatur. (The lone pair has the lowest priority.) Draw the structure of cyclohexanone 2,4-dinitrophenylhydrazone
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17.24.c
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Semicarbazide has two NH2 groups. Explain why only one of them forms an imine.
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The lone pair electrons that are on each of the nitrogens attached to the carbonyl group are delocalized onto the carbonyl group. These nitrogens, therefore, cannot act as nucleophiles since their lone pair electrons are not available for nucleophilic attack
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17.25
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Excess ammonia must be used when a primary amine is synthesized by reductive amination. What product will be obtained if the reaction is carried out with an excess of the carbonyl compound?
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A tertiary amine will be obtained because the primary amine synthesized in the first part of the reaction will react with the excess carbonyl compound, forming an imine that will be reduced to a secondary amine. The secondary amine will then react with the carbonyl compound, forming an enamine that will be reduced to a tertiary amine.
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17.26
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The compounds commonly known as "amino acids" are actually alpha-aminocarboxylic acids. What carbonyl compound should be used to synth the following:
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17.27.a
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The compounds commonly known as "amino acids" are actually alpha-aminocarboxylic acids. What carbonyl compound should be used to synth the following:
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17.27.b
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ignore picture.
Hydration of an aldehyde can also be catalyzed by hydroxide ion. Propose a mechanism for hydroxide-ion-catalyzed hydration. |
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17.28
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When trichloroacetaldehyde is dissolved in water, almost all of it is converted to the hydrate. Chloral hydrate, the product of the reaction, is a sedative that can be lethal. A cocktail laced with it is known-in detective novels, at least - as a "Mickey Finn." Explain why an aqueous solution of trichloroacetaldehyde is almost all hydrate.
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Electron-withdrawing groups decrease the stability of the aldehyde and increase the stability of the hydrate. Therefore, the three electron-withdrawing chlorines cause trichloroacetaldehyde to have a large equilibrium constant for hydrate formation.
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17.29
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Which of the following ketones forms the most hydrate in an aqueous solution?
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Because an electron-withdrawing substituent decreases the stability of a ketone and increases the stabilty of a hydrate, the compound with the electron-withdrawing para-nitro sustituents has the largest equilibrium constant ofr addition of water.
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17.30
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Which of the following are:
hemiacetals, acetals, hemiketals, ketals, hydrates |
hemiacetals: 7
acetals; 2,3 hemiketals: 1,8 ketals:5 hydrates: 4,6 |
17.31
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Explain why an acetal ofketal can be isolated but most hydrates cannot be isolated
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when a tetrahedral intermediate collapses, the intermediate that is formed is very unstable because of the positive charge on the sp2 oxygen atom. In the case of an acetal or ketal, the only way to form a neutral species is to reform the acetal or ketal. In the case of a hydrate, a neutral species can be formed by loss of a proton.
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17.33
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what would have been the product of the preceding reaction with LiAlH4 if the keto group had not been protected?
What reagent could you use to reduce only the keto group? |
NaBH4
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17.34
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Why don't acetals react with nucleophiles?
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An acetal has a very poor leaving group (CH3O-)
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17.35
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What puoducts would be formed from the preceeding reaction if aniline's amino group were not protected
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nitric acid is an oxidizing agent and primary amines are easily oxidized. So one product would be nitrobenzene. If excess nitric acid is present, nitrobenzene can be converted to meta-dinitrobenzene
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17.36
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In a six step synth, what is the yield of the target compound if each of the reactions employed gives an 80% yield? What would the yield be if two more steps were added to the synthesis?
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17.37
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Use a protecting group to synth
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17.38.a
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Use a protecting group to synth
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17.38.b
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what carbonyl compoud, phosphonium ylide (and alkyl halide) are required for the synth of:
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edit.
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17.39.a
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what carbonyl compoud, phosphonium ylide (and alkyl halide) are required for the synth of:
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edit.
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17.39.b
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what carbonyl compoud, phosphonium ylide (and alkyl halide) are required for the synth of:
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edit.
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17.39.c
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what carbonyl compoud, phosphonium ylide (and alkyl halide) are required for the synth of:
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edit.
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17.39.d
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which enationmer is formed when a methyl grignard reagent attacks the Re face of propiophenone
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17.40.a
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which enationmer is formed when a methyl grignard reagent attacks the Re face of benzaldehyde
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17.40.b
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which enationmer is formed when a methyl grignard reagent attacks the Re face of 2-pendanone
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17.40.c
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which enationmer is formed when a methyl grignard reagent attacks the Re face of 3-hexanone
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17.40.d
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synth from bromocyclohexane
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17.41.a
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synth from bromocyclohexane
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17.41.b
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synth from bromocyclohexane
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17.41.c
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synth from bromocyclohexane
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17.41.d
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synth from bromocyclohexane
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17.41.e
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synth from bromocyclohexane
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17.41.f
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Product:
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17.42.a
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Product:
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17.42.b
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Product:
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17.42.c
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Product:
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17.42.d
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major products
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17.44.a
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major products
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17.44.b
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major products
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17.44.c
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major products
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17.44.d
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draw the structure of isobutyraldehyde
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17.45.a
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draw the structure of 4-hexenal
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17.45.b
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draw the structure of diisopentyl ketone
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17.45.c
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draw the structure of 3-methylcyclohexanone
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17.45.d
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draw the structure of 2,4-pentanedione
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17.45.e
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draw the structure of 4-bromo-3-heptanone
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17.45.f
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draw the structure of gamma-bromocaproaldehyde
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17.45.g
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draw the structure of 2-ethylcyclopendanecarbaldehyde
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17.45.h
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draw the structure of 4-methyl-5-oxohexanal
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17.45.i
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draw the structure of benzene-1,3-dicarbaldehyde
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17.45.j
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products
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17.46.a
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products
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17.46.b
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products
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17.46.c
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products
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17.46.d
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products
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17.46.e
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products
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17.46.f
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products
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17.46.g
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products
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17.46.h
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List the following compounds in order of decreasing reactivity toward nucleophillic attack
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17.47
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synth from cyclohexanone
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19.49.a
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synth from cyclohexanone
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19.49.b
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synth from cyclohexanone
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19.49.c
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synth from cyclohexanone
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19.49.d
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synth from cyclohexanone
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19.49.e
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synth from cyclohexanone
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19.49.f
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synth from cyclohexanone
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19.49.g
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synth from cyclohexanone
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19.49.h
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synth from cyclohexanone
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19.49.i
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propose a mechanism for the following reaction
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17.50
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list the following compounds in order of decreasing Keq for hydrate formation
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17.51
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fill in the boxes
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17.52.a
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fill in the boxes
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17.52.b
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products
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17.53.a
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products
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17.53.b
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products
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17.53.c
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products
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17.53.d
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products
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17.53.e
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products
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17.53.f
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products
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17.53.g
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products
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17.53.h
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products
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17.53.i
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products
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17.53.j
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thiols can be prepared from the reaction of thiourea with an alkyl halide, followed by hudroxide-ion-promoted hydrolysis. Propose a mechanism for the reaction. What thiol would be formed if the alkyl halide employed were pentyl bromide?
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17.54
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The only organic compound obtained when compound Z undergoes the following sebuence of reactions gives the HNMR spectrum shown. ID compound Z.
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The offset shows there is a signal at ~11.8 ppm, indicating an aldehyde. The HNMR spectrum is that of benzylaldehyde. Phenylmagnesium bromide, therefore must react with a compound with one carbon atom to form an alcohol that can be oxidized by the mild oxidizing agent (MnO2) to benzaldehyde. Therefore, compound Z must be formaldehyde
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17.55
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propose a mechanism
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17.56.a
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propose a mechanism
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17.56.b
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propose a mechanism
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17.56.c
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how many signals would the product of the following reaction show in the following HNMR and CNMR spectra
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three signals in both spectrum
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17.57
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fill in the boxes
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17.58
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convert N-methylbenzamide into N-methylbenzylamine
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17.59.a
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convert N-methylbenzamide into benzoic acid
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17.59.b
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convert N-methylbenzamide into methylbenzoate
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17.59.c
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convert N-methylbenzamide into benzyl alcohol
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17.59.d
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give the products and stereoicomers
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17.60.a
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give the products and stereoicomers
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17.60.b
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give the products and stereoicomers
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17.60.c
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give the products and stereoicomers
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17.60.d
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list three different sets of reagents (each set consisting of a carbonyl compound and a grignard reagent) that could be used to prepare the followinv tertary alcohol
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17.61.a
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list three different sets of reagents (each set consisting of a carbonyl compound and a grignard reagent) that could be used to prepare the followinv tertary alcohol
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17.61.b
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give the product of the reaction of 3-methyl-2-cyclohexenone with CH3MgBr followed by H3O+
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17.62.a
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give the product of the reaction of 3-methyl-2-cyclohexenone with excess NaCN, Hcl
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17.62.b
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give the product of the reaction of 3-methyl-2-cyclohexenone with H2, Pd/C
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17.62.c
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give the product of the reaction of 3-methyl-2-cyclohexenone with Hbr
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17.62.d
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give the product of the reaction of 3-methyl-2-cyclohexenone with (CH3CH2)2CuLi followed by H3O+
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17.62.e
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give the product of the reaction of 3-methyl-2-cyclohexenone with CH3CH2SH
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17.62.f
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Norlutin and Enovid are ketones that supress ovulation. Consequently, they have been used clinically as contraceptives. For which of these compounds would you expect the infared carbonyl absorbtion (C=O stretch) to be at a higher frequency? Explain.
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enovoid would have its carbonyl stretch at a higher frequency. The carbonyl group in Norlutin has some single bond character because of the conjugated double bonds. This causes the carbon-oxygen bond to be easier to stretch than the carbon-oxygen bond in Enovid, which has isolated double bonds.
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17.63
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Product:
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17.64.a
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Product:
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17.64.b
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Product:
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17.64.c
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Product:
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17.64.d
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propose a reasonable mechanism
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17.65.a
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propose a reasonable mechanism
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17.65.b
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propose a reasonable mechanism
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17.67.a
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Product:
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17.67.b
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Unlike a phosphonium ylide, which reacts with an aldehyde or ketone to form an alkene, a sulfonium ylide reacts with an aldehyde or ketone to form an epoxide. Explain why one ylide forms an alkene, whereas the other forms an epoxide.
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The difference in the two reactions is a result of the difference in the leaving ability of a sulfonium group and a phosphonium group. Because the sulfonium group is a weaker base, it is a better leaving group. Therefore, it is elimiated by the oxyanion. In contrast, the phosphonium group is too strongly basic to be eliminated, so the oxyanion forms a four-membemed ring by sharing electrons with the positively charged phosphorus.
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17.68
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Synth
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17.69.a
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Synth
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17.69.b
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Synth
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17.69.c
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Synth
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17.69.d
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propose a mechanism
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17.70.a
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propose a mechanism
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17.70.b
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In aqeuous solution, D-glucose exists in equilibrium with two six membered rring compounds. Draw the structures of these compounds. Which of the six membered rings will be present in the greater amount?
|
The OH group on C-5 of glucose reacts with the aldehyde group in an intramolecular reaction, forming a cyclic hemiacetal. Because the reaction creates a new asymetric center, two cyclic hemiacetals can form, one with the R configuration at the new asymmetric center and one with the S configuration.
The two products can be drawn in their chair conformations by putting the largest group (CH2OH) in the equatorial position and then putting the other groups in axial or equatorial positions depending on whether they are cis or trans to one another. The hemiacetal on the left has all but one of its OH groups in the more stable equatorial position. Therefore, the hemiacetal on the right is more stable. |
17.71
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The HNMR spectrum of the alkyl bromide used to make the ylide to form a compound with molecular formula C11H14 is shown below. What product is obtained from the Wittig reaction.
|
The alkyl bromide is 1-bromo-2-phenylethane. The molecular formula of the product of the Witting reaction indicates that the ketone that reacts with the phosphonium ylide has 3 carbons.
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17.72
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In the presence of an acid catalyst, acetalaldehyde forms a trimer known as paraldehyde. Because it induces sleep when it is administered to animals in large doses, paraldehyde is used as a sedative or hypnotic. Propose a mechanism for the formation of paraldehyde.
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17.73
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The addition of hydrogen cyanide to benzaldehyde forms a compound called mandelonitrile. (R)-Mandelonitrile is formed from the hydrolysis of amygdalin, a compound found in the pits of peaches and apricots. The drug was subsequently found to be ineffective. Is (R)-mandelonitrile formed by attack of cyanide ion on the Re or the Si face of benzaldehyde?
|
Converting the Fischer projection of (R)-mandelonitrile into a wedge-and-dash structure allows you to see that attack of cyanide ion occurred on the Si face (decreasing priorities on the face closest to the observer are in a counterclockwise direction).
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17.74
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What carbonyl compound and what phosphonium ylide are needed to synth the following compounds?
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17.75.a
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What carbonyl compound and what phosphonium ylide are needed to synth the following compounds?
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17.75.b
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What carbonyl compound and what phosphonium ylide are needed to synth the following compounds?
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17.75.c
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What carbonyl compound and what phosphonium ylide are needed to synth the following compounds?
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17.75.d
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Identify compounds A and B
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17.76
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Propose a reasonable mechanism
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17.77.a
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Propose a reasonable mechanism
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17.77.b
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a compound is reacted with methylmagnesium bromide followed by acidification to form the product with the following HNMR specdrum. ID the compound.
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The compound that gives the HNMR spectrum is 2-phenyl-2-butanol
therefore, the compound that reacts with methylmagnesium bromide is 1-phenyl-1-propanone. |
17.78
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Synth, using a protecting group
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17.79.a
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Synth, using a protecting group
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17.79.b
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Synth, using a protecting group
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17.79.c
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when a cyclic ketone reacts with diagomethane, the next larger cyclic ketone is formed. Provide a mechanism.
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17.80
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The pKa values of oxaloacetic acid are 2.22 and 3.98. which carboxyl group is more acidic. Which carboxyl group is more acidic? The amount of hydrate present in an aqueous solution of oxaloacetic acid depends on the pH of the solution: 95% at pH=0, 81% at pH=1.3, 35% at pH=3.1, 13% at pH=4.7, 6% at pH=6.7, and 6% at pH=12.7. Explain this pH dependence.
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The data show that the amount of hydrate decreases with increasing pH until about pH=6 ,and that increasing the pH beyond 6 has no effect on the amount of hydrate.
A hydrate is stabilized by electron-withdrawing groups. A COOH group is electron withdrawing, but a COO- group is less so. In acidic solutions, where both carboxylic acid groups are in their acidic (COOH) forms, the compound exists as essentially all hydrate. As the pH of the solution increases and the COOH groups becom COO- groups, the amount of hydrate decreases. Above pH=6, where both carboxyl groups are in their basic (COO-) forms, there is only a small amount of hydrate |
17.81
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propose a mechanism
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17.84.a
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propose a mechanism
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17.84.b
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propose a mechanism
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17.84.c
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