chem 243 midterm review
DESCRIPTION
Review of Ch 18, 20, 21TRANSCRIPT
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18.1 Name .
O
CHCH2
CHCH
H3C
a) cis-Pent-2-enal
b) cis-Pent-3-enal
c) trans-Pent-2-enal
d) trans-Pent-3-enal
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a) cis-Pent-2-enal
b) cis-Pent-3-enal
c) trans-Pent-2-enal
d) trans-Pent-3-enal
Explanation:
The aldehyde is in position one. The double bond on carbon 3 is trans.
18.1 Name .
O
CHCH2
CHCH
H3C
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18.2 Name .CH2
C CCH2 OH
OO
H3C
a) 2-Oxobutanoic acid
b) 3-Butanonecarboxylic acid
c) 2-Oxopentanoic acid
d) 3-Oxopentanoic acid
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18.2 Name .CH2
C CCH2 OH
OO
H3C
a) 2-Oxobutanoic acid
b) 3-Butanonecarboxylic acid
c) 2-Oxopentanoic acid
d) 3-Oxopentanoic acid
Explanation:
The carbon in the carboxylic acid is position one.
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18.3 Identify the chemicalname for acetone.
a) Methanal
b) Ethanal
c) Propanone
d) Butanone
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18.3 Identify the chemicalname for acetone.
a) Methanal
b) Ethanal
c) Propanone
d) Butanone
Explanation:
Acetone is called propanone or dimethyl ketone.
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18.4
a) Propan-2-one
b) Butan-2-one
c) Pentan-2-one
d) Pentan-3-one
1. CH3CH2MgCl
2. H3O+
3. Na2Cr2O7, H2SO4
O
CHH3C
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18.4
a) Propan-2-one
b) Butan-2-one
c) Pentan-2-one
d) Pentan-3-one
Explanation:
Butan-2-ol is formed in the Grignard reaction. The secondary alcohol is oxidized to a ketone with sodium dichromate.
1. CH3CH2MgCl
2. H3O+
3. Na2Cr2O7, H2SO4
O
CHH3C
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18.51. O3
2. (CH3)2SCC
H3C
H3C CH3
H
a) 2-Methylbutane-2,3-diol
b) Propanone and ethanal
c) 3-Methylbutan-2-one
d) 2-Methylbutane-1,4-diol
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18.51. O3
2. (CH3)2SCC
H3C
H3C CH3
H
a) 2-Methylbutane-2,3-diol
b) Propanone and ethanal
c) 3-Methylbutan-2-one
d) 2-Methylbutane-1,4-diol
Explanation:
Ozonolysis, followed by a mild reduction, cleaves alkenes to give aldehydes and ketones.
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18.6 CCH3C H
1. HgSO4, H2SO4
H2O
2. H+
a) CH3C(OH)=CH2
b) (CH3)2C=O
c) CH3CH2CHO
d) CH3CH=CHOH (cis)
e) CH3CH = CHOH (trans)
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18.6 CCH3C H
1. HgSO4, H2SO4
H2O
2. H+
a) CH3C(OH)=CH2
b) (CH3)2C=O
c) CH3CH2CHO
d) CH3CH=CHOH (cis)
e) CH3CH = CHOH (trans)
Explanation:
Water adds across the triple bond in a Markovnikov orientation, followed by tautomerism to form a ketone.
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18.7 CCH3C H
1. Sia2BH
2. H2O2, -OH
a) CH3C(OH)=CH2
b) (CH3)2C=O
c) CH3CH2CHO
d) CH3CH=CHOH (cis)
e) CH3CH=CHOH (trans)
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18.7 CCH3C H
1. Sia2BH
2. H2O2, -OH
a) CH3C(OH)=CH2
b) (CH3)2C=O
c) CH3CH2CHO
d) CH3CH=CHOH (cis)
e) CH3CH=CHOH (trans)
Explanation:
Water adds across the triple bond in a syn anti-Markovnikov orientation, followed by tautomerism to form an aldehyde.
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18.8 C
O
H3C OH1. 2 CH3CH2Li2. H3O+
a) Propan-2-one
b) Butan-2-one
c) Pentan-2-one
d) Pentan-3-one
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18.8 C
O
H3C OH1. 2 CH3CH2Li2. H3O+
a) Propan-2-one
b) Butan-2-one
c) Pentan-2-one
d) Pentan-3-one
Explanation:
The ethyl group adds to the carbonyl carbon.
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18.9 CH3CH2C N
1. CH3CH2CH2MgCl
2. H3O+
a) Hexan-3-one
b) Pentan-3-one
c) 4-Ethylheptan-4-ol
d) 4-Ethylheptan-4-one
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18.9 CH3CH2C N
1. CH3CH2CH2MgCl
2. H3O+
a) Hexan-3-one
b) Pentan-3-one
c) 4-Ethylheptan-4-ol
d) 4-Ethylheptan-4-one
Explanation:
The propyl group adds to the nitrile to give the magnesium salt of the imine. Hydrolysis produces the ketone.
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18.10
O
COHCH3CH2
1. SOCl2
2. LiAlH(Ot-Bu)3
a) Propanoyl chloride
b) Propanal
c) Propane
d) Propan-1-ol
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18.10
O
COHCH3CH2
1. SOCl2
2. LiAlH(Ot-Bu)3
a) Propanoyl chloride
b) Propanal
c) Propane
d) Propan-1-ol
Explanation:
An acid chloride is formed first, followed by reduction to an aldehyde.
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18.11
O
COHCH3CH2
1. SOCl2
2. (CH3CH2)2CuLi
a) Propanoyl chloride
b) Butanone
c) Pentan-2-one
d) Pentan-3-one
e) 3-Ethylpentan-3-ol
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18.11
O
COHCH3CH2
1. SOCl2
2. (CH3CH2)2CuLi
a) Propanoyl chloride
b) Butanone
c) Pentan-2-one
d) Pentan-3-one
e) 3-Ethylpentan-3-ol
Explanation:
An acid chloride is formed first, then the ethyl group replaces the chloride.
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18.12
O
CHCH3CH2
C(CH3)2Ph3P
a) 2-Methylpent-2-ene
b) cis-3-Methylpent-2-ene
c) trans-3-Methylpent-2-ene
d) cis-3-Methylpent-3-ene
e) trans-3-Methylpent-3-ene
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18.12
O
CHCH3CH2
C(CH3)2Ph3P
a) 2-Methylpent-2-ene
b) cis-3-Methylpent-2-ene
c) trans-3-Methylpent-2-ene
d) cis-3-Methylpent-3-ene
e) trans-3-Methylpent-3-ene
Explanation:
The C(CH3)2 group replaces the oxygen on the aldehyde in the Wittig reaction.
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18.13
O
CHCH3CH2
H2O
a) Propan-1-ol
b) Propan-2-ol
c) Propanal
d) Propane-1,1-diol
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18.13
O
CHCH3CH2
H2O
a) Propan-1-ol
b) Propan-2-ol
c) Propanal
d) Propane-1,1-diol
Explanation:
A hydrate is formed from the addition of water to an aldehyde.
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18.14
O
CHCH3CH2
1. HCN
2. H3O+
a) 2-Hydroxybutanenitrile
b) 2-Oxobutanenitrile
c) 2-Hydroxybutanoic acid
d) 2-Oxobutanoic acid
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18.14
O
CHCH3CH2
1. HCN
2. H3O+
a) 2-Hydroxybutanenitrile
b) 2-Oxobutanenitrile
c) 2-Hydroxybutanoic acid
d) 2-Oxobutanoic acid
Explanation:
An intermediate 2-hydroxybutanenitrile is formed. The nitrile is hydrolyzed to the carboxylic acid.
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18.15
O
CHCH2
CH2
H3C
H2NOH
H+
a) Butanal imine
b) Butanal hydrazone
c) Butanal oxime
d) Butanal semicarbazone
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18.15
O
CHCH2
CH2
H3C
H2NOH
H+
a) Butanal imine
b) Butanal hydrazone
c) Butanal oxime
d) Butanal semicarbazone
Explanation:
The N–OH replaces the oxygen of the aldehyde to form an oxime.
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18.16
O
CCH3H3C
2 CH3CH2OH
H+
a) 2,2-Diethoxypropane
b) 2-Ethoxypropan-2-ol
c) Propane-2,2-diol
d) 2-Ethoxypropane
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18.16
O
CCH3H3C
2 CH3CH2OH
H+
a) 2,2-Diethoxypropane
b) 2-Ethoxypropan-2-ol
c) Propane-2,2-diol
d) 2-Ethoxypropane
Explanation:
Two molecules of ethanol are added to the carbonyl, with loss of water, forming the acetal.
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18.17
O
CHCH2
H3C1. Ag(NH3)2, -OH
2. H+
a) Propan-1-ol
b) Propanoic acid
c) Propane-1,1-diol
d) 1-Hydroxypropanoic acid
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18.17
O
CHCH2
H3C1. Ag(NH3)2, -OH
2. H+
a) Propan-1-ol
b) Propanoic acid
c) Propane-1,1-diol
d) 1-Hydroxypropanoic acid
Explanation:
The Tollens reagent oxidizes aldehydes to carboxylic acids.
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18.18
O
CCH3CH2
H3CNaBH4
CH3CH2OH
a) Butan-2-one
b) Butan-2-ol
c) Hexan-2-one
d) Butane
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18.18
O
CCH3CH2
H3CNaBH4
CH3CH2OH
a) Butan-2-one
b) Butan-2-ol
c) Hexan-2-one
d) Butane
Explanation:Sodium borohydride reduces aldehydes and ketones to the correspondingalcohols.
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18.19
O
CCH3CH2
H3CZn(Hg)
HCl, H2O
a) Butan-2-one
b) Butan-2-ol
c) Butane
d) 2-Chlorobutane
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18.19
O
CCH3CH2
H3CZn(Hg)
HCl, H2O
a) Butan-2-one
b) Butan-2-ol
c) Butane
d) 2-Chlorobutane
Explanation:
The Clemmensen reduction reduces the carbonyl to a methylene.
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18.20
O
CCH3CH2
H3C1. NH2NH2
2. -OH, heat
a) Butan-2-one hydrazone
b) Butan-2-one oxime
c) Butan-2-one imine
d) Butane
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18.20
O
CCH3CH2
H3C1. NH2NH2
2. -OH, heat
a) Butan-2-one hydrazone
b) Butan-2-one oxime
c) Butan-2-one imine
d) Butane
Explanation:
An intermediate hydrazone is formed, which is then reduced to an alkane. This is the Wolff-Kishner reduction.
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a) 2-Nitropropanoic acid
b) 3-Nitrobutanoic acid
c) 2-Aminopropanoic acid
d) 3-Aminobutanoic acid
e) 3-Aminopentanoic acid
20.1 Name .
O
COHCH2
CHH3C
NH2
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a) 2-Nitropropanoic acid
b) 3-Nitrobutanoic acid
c) 2-Aminopropanoic acid
d) 3-Aminobutanoic acid
e) 3-Aminopentanoic acid
Explanation:
The carbon of the carboxylic acid is at position 1.
20.1 Name .
O
COHCH2
CHH3C
NH2
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20.2 Name .H3C
CC
COH
OH
H
a) (E)-2-Butanoic acid
b) (Z)-2-Butanoic acid
c) (E)-But-2-enoic acid
d) (Z)-But-2-enoic acid
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20.2 Name .H3C
CC
COH
OH
H
a) (E)-2-Butanoic acid
b) (Z)-2-Butanoic acid
c) (E)-But-2-enoic acid
d) (Z)-But-2-enoic acid
Explanation:
The groups are trans in the but-2-enoic acid.
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20.3 Name .
O
CCH2HO
CH2C
OH
O
a) Di-butanoic acid
b) Pentanedioic acid
c) Ethanedioic acid
d) Propanedioic acid
e) Butanedioic acid
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20.3 Name .
O
CCH2HO
CH2C
OH
O
a) Di-butanoic acid
b) Pentanedioic acid
c) Ethanedioic acid
d) Propanedioic acid
e) Butanedioic acid
Explanation:
The structure has four carbons total.
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20.4 Name .
HOOC COOH
a) Benzoic acid
b) Phthalic acid
c) Isophthalic acid
d) Terephthalic acid
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20.4 Name .
HOOC COOH
a) Benzoic acid
b) Phthalic acid
c) Isophthalic acid
d) Terephthalic acid
Explanation:
Isophthalic acid has the carboxyl groups in positions 1 and 3.
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20.5 Give the hybridizationfor the carbonyl carbon.
a) sp
b) sp2
c) sp3
d) sp4
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20.5 Give the hybridizationfor the carbonyl carbon.
a) sp
b) sp2
c) sp3
d) sp4
Explanation:
The carbonyl carbon is sp2 hybridized.
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20.6 Give the bond angleat the carbonyl carbon.
a) 90°
b) 104.5°
c) 109.5°
d) 120°
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20.6 Give the bond angleat the carbonyl carbon.
a) 90°
b) 104.5°
c) 109.5°
d) 120°
Explanation:
The carbonyl carbon has a bond angle of 120°.
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20.7 Identify the compoundwith the lowest pKa.
a) Acetic acid
b) Chloroacetic acid
c) Dichloroacetic acid
d) Trichloroacetic acid
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20.7 Identify the compoundwith the lowest pKa.
a) Acetic acid
b) Chloroacetic acid
c) Dichloroacetic acid
d) Trichloroacetic acid
Explanation:
Electron-withdrawing substituents on the carbon increase acid
strength.
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20.8 CH3CH2OHNa2Cr2O7
H2SO4
a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) Ethanal
e) Propanal
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20.8 CH3CH2OHNa2Cr2O7
H2SO4
a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) Ethanal
e) Propanal
Explanation:
A primary alcohol is oxidized to a carboxylic acid with sodium
dichromate.
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20.9 CC
H
CH3CH2 CH2CH3
H
warm, conc
KMnO4
a) Hexane-3,4-diol
b) Hexane-3,4-dione
c) Ethanoic acid
d) Propanoic acid
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20.9 CC
H
CH3CH2 CH2CH3
H
warm, conc
KMnO4
a) Hexane-3,4-diol
b) Hexane-3,4-dione
c) Ethanoic acid
d) Propanoic acid
Explanation:
The cleavage of the double bond with potassium permanganate
produces 2 moles of propanoic acid.
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20.10 CCH2CH3CH3CH2Cwarm, conc
KMnO4
a) Hexane-3,4-diol
b) Hexane-3,4-dione
c) Ethanoic acid
d) Propanoic acid
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20.10 CCH2CH3CH3CH2Cwarm, conc
KMnO4
a) Hexane-3,4-diol
b) Hexane-3,4-dione
c) Ethanoic acid
d) Propanoic acid
Explanation:
The triple bond of the alkyne is oxidized to carboxylic acids with
potassium permanganate.
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20.11 CH3CH2MgCl
1. CO2
2. H+
a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) Butanoyl chloride
e) Pentanoyl chloride
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20.11 CH3CH2MgCl
1. CO2
2. H+
a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) Butanoyl chloride
e) Pentanoyl chloride
Explanation:
Grignard reagents add to carbon dioxide to form a salt of a carboxylic
acid. Acid hydrolysis forms a carboxylic acid.
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20.12 CH3CH2Cl
1. NaCN
2. H+
a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) Butanoyl chloride
e) Pentanoyl chloride
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20.12 CH3CH2Cl
1. NaCN
2. H+
a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) Butanoyl chloride
e) Pentanoyl chloride
Explanation:
Halogen is replaced using sodium cyanide. Hydrolysis of the cyanide
group gives the carboxylic acid.
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20.13 CH3CH2COOHCH3OH
H+
a) CH3CO2CH3
b) CH3CO2CH2CH3
c) CH3CH2CO2CH3
d) CH3COCH3
e) CH3COCH2CH3
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20.13 CH3CH2COOHCH3OH
H+
a) CH3CO2CH3
b) CH3CO2CH2CH3
c) CH3CH2CO2CH3
d) CH3COCH3
e) CH3COCH2CH3
Explanation:
The Fischer esterification converts carboxylic acids to esters through
an acid-catalyzed nucleophilic acyl substitution.
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20.14 CH3CH2COOHCH2N2
a) CH3CH2NH2
b) CH3CH2CONH2
c) CH3CH2COOCH3
d) CH3NH2
e) CH3COOCH3
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20.14 CH3CH2COOHCH2N2
a) CH3CH2NH2
b) CH3CH2CONH2
c) CH3CH2COOCH3
d) CH3NH2
e) CH3COOCH3
Explanation:
Diazomethane converts carboxylic acids to methyl esters.
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20.15 CH3CH2COOH1. CH3NH2
2. heat
a) CH3CH2CONH2
b) CH3CH2COOCH3
c) CH3COOCH3
d) CH3CONHCH3
e) CH3CH2CONHCH3
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20.15 CH3CH2COOH1. CH3NH2
2. heat
a) CH3CH2CONH2
b) CH3CH2COOCH3
c) CH3COOCH3
d) CH3CONHCH3
e) CH3CH2CONHCH3
Explanation:
An amide is formed from a carboxylic acid and an amine.
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20.16 CH3CH2COOH
1. LiAlH4
2. H3O+
a) CH3CH2OH
b) CH3CHO
c) CH3CH2CH2OH
d) CH3CH2CHO
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20.16 CH3CH2COOH
1. LiAlH4
2. H3O+
a) CH3CH2OH
b) CH3CHO
c) CH3CH2CH2OH
d) CH3CH2CHO
Explanation:
The carboxylic acid is reduced to the alcohol.
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20.17 CH3CH2COOH1. SOCl2
2. LiAl[OC(CH3)3]3H
a) CH3CH2CH2OH
b) CH3CH2COCl
c) CH3CH2CHO
d) CH3CH2COOC(CH3)3
e) CH3CH2COOH
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20.17 CH3CH2COOH1. SOCl2
2. LiAl[OC(CH3)3]3H
a) CH3CH2CH2OH
b) CH3CH2COCl
c) CH3CH2CHO
d) CH3CH2COOC(CH3)3
e) CH3CH2COOH
Explanation:Thionyl chloride reacts with the carboxylic acid to form an acid chloride.The acid chloride is reduced to an aldehyde with lithiumtri(t-butoxy)aluminum hydride.
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20.18 CH3CH2COOH1. 2 CH3Li
2. H2O
a) 3-Pentanone
b) 2-Pentanone
c) Propanone
d) 2-Butanone
e) Methyl propanoate
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20.18 CH3CH2COOH1. 2 CH3Li
2. H2O
a) 3-Pentanone
b) 2-Pentanone
c) Propanone
d) 2-Butanone
e) Methyl propanoate
Explanation:A carboxylic acid reacting with two equivalents of an organolithiumreagent produces a dianion. Hydrolysis gives a ketone.
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20.19 CH3CH2COOH1. SOCl2
2. CH3OH
a) CH3CH2COCl
b) CH3CH2CHO
c) CH3CO2CH3
d) CH3CH2COOCH3
e) CH3CH2CH2Cl
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20.19 CH3CH2COOH1. SOCl2
2. CH3OH
a) CH3CH2COCl
b) CH3CH2CHO
c) CH3CO2CH3
d) CH3CH2COOCH3
e) CH3CH2CH2Cl
Explanation:
An acid chloride is formed in the first step. The acid chloride reacts
with methanol to produce the methyl ester ester.
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20.20 CH3CH2COOH1. (C OCl)2
2. CH3NH2
a) CH3CH2COCl
b) CH3CH2CONHCH3
c) CH3CONHCH2CH3
d) CH3CH2COOCH3
e) CH3CH2CH2Cl
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20.20 CH3CH2COOH1. (C OCl)2
2. CH3NH2
a) CH3CH2COCl
b) CH3CH2CONHCH3
c) CH3CONHCH2CH3
d) CH3CH2COOCH3
e) CH3CH2CH2Cl
Explanation:
Reaction with oxalyl chloride forms the acid chloride. Nucleophilic acyl
substitution with the amine forms the amide.
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21.1 Name .
a) Ethyl ethanoate
b) Propyl propanoate
c) Ethyl propanoate
d) Propyl ethanoate
e) Propyl butanoate
O
COCH2CH2CH3CH3CH2
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a) Ethyl ethanoate
b) Propyl propanoate
c) Ethyl propanoate
d) Propyl ethanoate
e) Propyl butanoate
Explanation:
The longest chain is three carbons. Propyl is the alkoxy group.
21.1 Name .
O
COCH2CH2CH3CH3CH2
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21.2 Name .
a) 3-Hydroxybutanoic acid lactone
b) 4-Hydroxybutanoic acid lactone
c) 4-Hydroxypentanoic acid lactone
d) 5-Hydroxypentanoic acid lactone
O
O
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a) 3-Hydroxybutanoic acid lactone
b) 4-Hydroxybutanoic acid lactone
c) 4-Hydroxypentanoic acid lactone
d) 5-Hydroxypentanoic acid lactone
Explanation:
A lactone is a cyclic ester. The hydroxyl is on the fifth carbon.
O
O
21.2 Name .
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21.3 Name .
a) Pentanamide
b) Butanamide
c) N-Ethylethanamide
d) N-Methylethanamide
e) N-Ethylpropanamide
O
CNHCH2CH3CH3CH2
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a) Pentanamide
b) Butanamide
c) N-Ethylethanamide
d) N-Methylethanamide
e) N-Ethylpropanamide
Explanation:
The ethyl group is attached to the nitrogen. The longest chain is
three carbons.
O
CNHCH2CH3CH3CH2
21.3 Name .
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21.4 Name .
a) 3-Aminobutanoic acid lactam
b) 4-Aminobutanoic acid lactam
c) 4-Aminopentanoic acid lactam
d) 5-Aminopentanoic acid lactam
N
O
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a) 3-Aminobutanoic acid lactam
b) 4-Aminobutanoic acid lactam
c) 4-Aminopentanoic acid lactam
d) 5-Aminopentanoic acid lactam
Explanation:
A lactam is a cyclic amide. The amino group is on the fifth carbon.
N
O
21.4 Name .
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21.5 Name .
a) Pentanenitrile
b) Butanenitrile
c) Propanenitrile
d) 2-Methylbutanenitrile
e) 3-Methylbutanenitrile
N(CH3)2CHCH2C
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a) Pentanenitrile
b) Butanenitrile
c) Propanenitrile
d) 2-Methylbutanenitrile
e) 3-Methylbutanenitrile
Explanation:
The longest chain has four carbons. The methyl is on the third carbon.
21.5 Name .
N(CH3)2CHCH2C
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21.6 Name .
a) 1-Chloroethanoyl chloride
b) 2-Chloroethanoyl chloride
c) 1-Chloropropanoyl chloride
d) 2-Chloropropanoyl chloride
O
CClCH3CH
Cl
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a) 1-Chloroethanoyl chloride
b) 2-Chloroethanoyl chloride
c) 1-Chloropropanoyl chloride
d) 2-Chloropropanoyl chloride
Explanation:
The longest chain has three carbons. Chlorines are on the second
carbon and the carbonyl carbon.
O
CClCH3CH
Cl
21.6 Name .
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If R’ or R’’ is H, then the product will be 2 carboxylic acids
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