Chem 243 Midterm Review [PDF]

Zitiervorschau

O

18.1 Name H3C

CH CH

a) b) c) d)

cis-Pent-2-enal cis-Pent-3-enal trans-Pent-2-enal trans-Pent-3-enal

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.

C CH2

H

O

18.1 Name H3C

CH CH

a) b) c) d)

.

C CH2

H

cis-Pent-2-enal cis-Pent-3-enal trans-Pent-2-enal trans-Pent-3-enal

Explanation: The aldehyde is in position one. The double bond on carbon 3 is trans.

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18.2 Name H3C CH2

a) b) c) d)

2-Oxobutanoic acid 3-Butanonecarboxylic acid 2-Oxopentanoic acid 3-Oxopentanoic acid

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O

O

C

C CH2

. OH

18.2 Name H3C CH2

a) b) c) d)

O

O

C

C CH2

2-Oxobutanoic acid 3-Butanonecarboxylic acid 2-Oxopentanoic acid 3-Oxopentanoic acid

Explanation: The carbon in the carboxylic acid is position one.

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. OH

18.3 Identify the chemical name for acetone.

a) b) c) d)

Methanal Ethanal Propanone Butanone

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18.3 Identify the chemical name for acetone.

a) b) c) d)

Methanal Ethanal Propanone Butanone

Explanation: Acetone is called propanone or dimethyl ketone.

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1. CH3CH2MgCl 2. H3O+

O

18.4

C H3C

a) b) c) d)

Propan-2-one Butan-2-one Pentan-2-one Pentan-3-one

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H

3. Na2Cr2O7, H2SO4

1. CH3CH2MgCl 2. H3O+

O

18.4

C H3C

a) b) c) d)

H

3. Na2Cr2O7, H2SO4

Propan-2-one Butan-2-one Pentan-2-one Pentan-3-one

Explanation: Butan-2-ol is formed in the Grignard reaction. The secondary alcohol is oxidized to a ketone with sodium dichromate. © 2013 Pearson Education, Inc.

H3C

18.5

CH3 C

2-Methylbutane-2,3-diol Propanone and ethanal 3-Methylbutan-2-one 2-Methylbutane-1,4-diol

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C 2. (CH3)2S

H3C

a) b) c) d)

1. O3

H

H3C

18.5

CH3 C

C 2. (CH3)2S

H3C

a) b) c) d)

1. O3

H

2-Methylbutane-2,3-diol Propanone and ethanal 3-Methylbutan-2-one 2-Methylbutane-1,4-diol

Explanation: Ozonolysis, followed by a mild reduction, cleaves alkenes to give aldehydes and ketones. © 2013 Pearson Education, Inc.

18.6

1. HgSO4, H2SO4 H2O H3C

C

C

H +

2. H

a) b) c) d) e)

CH3C(OH)=CH2 (CH3)2C=O CH3CH2CHO CH3CH=CHOH (cis) CH3CH = CHOH (trans)

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18.6

1. HgSO4, H2SO4 H2O H3C

C

C

H +

2. H

a) b) c) d) e)

CH3C(OH)=CH2 (CH3)2C=O CH3CH2CHO CH3CH=CHOH (cis) CH3CH = CHOH (trans)

Explanation: Water adds across the triple bond in a Markovnikov orientation, followed by tautomerism to form a ketone. © 2013 Pearson Education, Inc.

1. Sia2BH

18.7

a) b) c) d) e)

H3C

C

CH3C(OH)=CH2 (CH3)2C=O CH3CH2CHO CH3CH=CHOH (cis) CH3CH=CHOH (trans)

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C

H 2. H2O2, -OH

1. Sia2BH

18.7

a) b) c) d) e)

H3C

C

C

H 2. H2O2, -OH

CH3C(OH)=CH2 (CH3)2C=O CH3CH2CHO CH3CH=CHOH (cis) CH3CH=CHOH (trans)

Explanation: Water adds across the triple bond in a syn anti-Markovnikov orientation, followed by tautomerism to form an aldehyde. © 2013 Pearson Education, Inc.

O

18.8

a) b) c) d)

C H3C

Propan-2-one Butan-2-one Pentan-2-one Pentan-3-one

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OH

1. 2 CH3CH2Li 2. H3O+

O

18.8

a) b) c) d)

C H3C

OH

1. 2 CH3CH2Li 2. H3O+

Propan-2-one Butan-2-one Pentan-2-one Pentan-3-one

Explanation: The ethyl group adds to the carbonyl carbon.

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1. CH3CH2CH2MgCl

18.9

CH3CH2C

N +

2. H3O

a) b) c) d)

Hexan-3-one Pentan-3-one 4-Ethylheptan-4-ol 4-Ethylheptan-4-one

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1. CH3CH2CH2MgCl

18.9

CH3CH2C

N +

2. H3O

a) b) c) d)

Hexan-3-one Pentan-3-one 4-Ethylheptan-4-ol 4-Ethylheptan-4-one

Explanation: The propyl group adds to the nitrile to give the magnesium salt of the imine. Hydrolysis produces the ketone. © 2013 Pearson Education, Inc.

O

18.10

C CH3CH2

a) b) c) d)

Propanoyl chloride Propanal Propane Propan-1-ol

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1. SOCl2 OH

2. LiAlH(Ot-Bu)3

O

18.10

C CH3CH2

a) b) c) d)

1. SOCl2 OH

2. LiAlH(Ot-Bu)3

Propanoyl chloride Propanal Propane Propan-1-ol

Explanation: An acid chloride is formed first, followed by reduction to an aldehyde.

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O

18.11

C CH3CH2

a) b) c) d) e)

Propanoyl chloride Butanone Pentan-2-one Pentan-3-one 3-Ethylpentan-3-ol

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1. SOCl2 OH

2. (CH3CH2)2CuLi

O

18.11

C CH3CH2

a) b) c) d) e)

1. SOCl2 OH

2. (CH3CH2)2CuLi

Propanoyl chloride Butanone Pentan-2-one Pentan-3-one 3-Ethylpentan-3-ol

Explanation: An acid chloride is formed first, then the ethyl group replaces the chloride. © 2013 Pearson Education, Inc.

O C

18.12 CH3CH2

a) b) c) d) e)

2-Methylpent-2-ene cis-3-Methylpent-2-ene trans-3-Methylpent-2-ene cis-3-Methylpent-3-ene trans-3-Methylpent-3-ene

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Ph3P H

C(CH3)2

O C

18.12 CH3CH2

a) b) c) d) e)

Ph3P

C(CH3)2

H

2-Methylpent-2-ene cis-3-Methylpent-2-ene trans-3-Methylpent-2-ene cis-3-Methylpent-3-ene trans-3-Methylpent-3-ene

Explanation: The C(CH3)2 group replaces the oxygen on the aldehyde in the Wittig reaction. © 2013 Pearson Education, Inc.

O H2O

18.13

C CH3CH2

a) b) c) d)

Propan-1-ol Propan-2-ol Propanal Propane-1,1-diol

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H

O H2O

18.13

C CH3CH2

a) b) c) d)

H

Propan-1-ol Propan-2-ol Propanal Propane-1,1-diol

Explanation: A hydrate is formed from the addition of water to an aldehyde.

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O 1. HCN

18.14

C CH3CH2

a) b) c) d)

2-Hydroxybutanenitrile 2-Oxobutanenitrile 2-Hydroxybutanoic acid 2-Oxobutanoic acid

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H

2. H3O+

O 1. HCN

18.14

C CH3CH2

a) b) c) d)

H

2. H3O+

2-Hydroxybutanenitrile 2-Oxobutanenitrile 2-Hydroxybutanoic acid 2-Oxobutanoic acid

Explanation: An intermediate 2-hydroxybutanenitrile is formed. The nitrile is hydrolyzed to the carboxylic acid. © 2013 Pearson Education, Inc.

O CH2

18.15 H3C

a) b) c) d)

C CH2

Butanal imine Butanal hydrazone Butanal oxime Butanal semicarbazone

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H2NOH H

H

+

O CH2

18.15 H3C

a) b) c) d)

H2NOH

C CH2

H

H

+

Butanal imine Butanal hydrazone Butanal oxime Butanal semicarbazone

Explanation: The N–OH replaces the oxygen of the aldehyde to form an oxime.

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O 2 CH3CH2OH

18.16

C H3C

a) b) c) d)

2,2-Diethoxypropane 2-Ethoxypropan-2-ol Propane-2,2-diol 2-Ethoxypropane

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CH3

H+

O 2 CH3CH2OH

18.16

C H3C

a) b) c) d)

CH3

H+

2,2-Diethoxypropane 2-Ethoxypropan-2-ol Propane-2,2-diol 2-Ethoxypropane

Explanation: Two molecules of ethanol are added to the carbonyl, with loss of water, forming the acetal. © 2013 Pearson Education, Inc.

O -

18.17

1. Ag(NH3)2, OH H3C

C CH2

a) b) c) d)

Propan-1-ol Propanoic acid Propane-1,1-diol 1-Hydroxypropanoic acid

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H

2. H+

O -

18.17

1. Ag(NH3)2, OH H3C

C CH2

a) b) c) d)

H

2. H+

Propan-1-ol Propanoic acid Propane-1,1-diol 1-Hydroxypropanoic acid

Explanation: The Tollens reagent oxidizes aldehydes to carboxylic acids.

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O

18.18

H3C

C CH2

a) b) c) d)

Butan-2-one Butan-2-ol Hexan-2-one Butane

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NaBH4 CH3

CH3CH2OH

O

18.18

H3C

C CH2

a) b) c) d)

NaBH4 CH3

CH3CH2OH

Butan-2-one Butan-2-ol Hexan-2-one Butane

Explanation: Sodium borohydride reduces aldehydes and ketones to the corresponding alcohols. © 2013 Pearson Education, Inc.

O Zn(Hg)

18.19

H3C

C CH2

a) b) c) d)

Butan-2-one Butan-2-ol Butane 2-Chlorobutane

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CH3

HCl, H2O

O Zn(Hg)

18.19

H3C

C CH2

a) b) c) d)

CH3

HCl, H2O

Butan-2-one Butan-2-ol Butane 2-Chlorobutane

Explanation: The Clemmensen reduction reduces the carbonyl to a methylene.

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O 1. NH2NH2

18.20

H3C

C CH2

a) b) c) d)

Butan-2-one hydrazone Butan-2-one oxime Butan-2-one imine Butane

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CH3

-

2. OH, heat

O 1. NH2NH2

18.20

H3C

C CH2

a) b) c) d)

CH3

-

2. OH, heat

Butan-2-one hydrazone Butan-2-one oxime Butan-2-one imine Butane

Explanation: An intermediate hydrazone is formed, which is then reduced to an alkane. This is the Wolff-Kishner reduction. © 2013 Pearson Education, Inc.

20.1 Name H3C

a) b) c) d) e)

2-Nitropropanoic acid 3-Nitrobutanoic acid 2-Aminopropanoic acid 3-Aminobutanoic acid 3-Aminopentanoic acid

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NH2

O

CH

C CH2

. OH

20.1 Name H3C

a) b) c) d) e)

NH2

O

CH

C CH2

2-Nitropropanoic acid 3-Nitrobutanoic acid 2-Aminopropanoic acid 3-Aminobutanoic acid 3-Aminopentanoic acid

Explanation: The carbon of the carboxylic acid is at position 1. © 2013 Pearson Education, Inc.

. OH

H

20.2 Name

a) b) c) d)

H3C

(E)-2-Butanoic acid (Z)-2-Butanoic acid (E)-But-2-enoic acid (Z)-But-2-enoic acid

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OH

C C

C

H

O

.

H

20.2 Name

a) b) c) d)

H3C C

C

H

O

(E)-2-Butanoic acid (Z)-2-Butanoic acid (E)-But-2-enoic acid (Z)-But-2-enoic acid

Explanation: The groups are trans in the but-2-enoic acid.

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OH

C

.

O

20.3 Name

C HO

OH

CH2 CH2

C O

a) b) c) d) e)

Di-butanoic acid Pentanedioic acid Ethanedioic acid Propanedioic acid Butanedioic acid

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.

O

20.3 Name

C HO

OH

CH2 CH2

C O

a) b) c) d) e)

Di-butanoic acid Pentanedioic acid Ethanedioic acid Propanedioic acid Butanedioic acid

Explanation: The structure has four carbons total. © 2013 Pearson Education, Inc.

.

HOOC

20.4 Name

a) b) c) d)

Benzoic acid Phthalic acid Isophthalic acid Terephthalic acid

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COOH

.

HOOC

COOH

20.4 Name

a) b) c) d)

Benzoic acid Phthalic acid Isophthalic acid Terephthalic acid

Explanation: Isophthalic acid has the carboxyl groups in positions 1 and 3.

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.

20.5 Give the hybridization for the carbonyl carbon.

a) b) c) d)

sp sp2 sp3 sp4

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20.5 Give the hybridization for the carbonyl carbon.

a) b) c) d)

sp sp2 sp3 sp4

Explanation: The carbonyl carbon is sp2 hybridized.

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20.6 Give the bond angle at the carbonyl carbon.

a) b) c) d)

90° 104.5° 109.5° 120°

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20.6 Give the bond angle at the carbonyl carbon.

a) b) c) d)

90° 104.5° 109.5° 120°

Explanation: The carbonyl carbon has a bond angle of 120°.

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20.7 Identify the compound with the lowest pKa.

a) b) c) d)

Acetic acid Chloroacetic acid Dichloroacetic acid Trichloroacetic acid

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20.7 Identify the compound with the lowest pKa.

a) b) c) d)

Acetic acid Chloroacetic acid Dichloroacetic acid Trichloroacetic acid

Explanation: Electron-withdrawing substituents on the a carbon increase acid strength. © 2013 Pearson Education, Inc.

Na2Cr2O7 20.8

CH3CH2OH H2SO4

a) b) c) d) e)

Ethanoic acid Propanoic acid Butanoic acid Ethanal Propanal

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Na2Cr2O7 20.8

CH3CH2OH H2SO4

a) b) c) d) e)

Ethanoic acid Propanoic acid Butanoic acid Ethanal Propanal

Explanation: A primary alcohol is oxidized to a carboxylic acid with sodium dichromate. © 2013 Pearson Education, Inc.

CH3CH2

20.9

C H

a) b) c) d)

Hexane-3,4-diol Hexane-3,4-dione Ethanoic acid Propanoic acid

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CH2CH3 C

warm, conc KMnO4

H

CH3CH2

20.9

C H

a) b) c) d)

CH2CH3 C

warm, conc KMnO4

H

Hexane-3,4-diol Hexane-3,4-dione Ethanoic acid Propanoic acid

Explanation: The cleavage of the double bond with potassium permanganate produces 2 moles of propanoic acid. © 2013 Pearson Education, Inc.

warm, conc

20.10

CH3CH2C

CCH2CH3 KMnO4

a) b) c) d)

Hexane-3,4-diol Hexane-3,4-dione Ethanoic acid Propanoic acid

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warm, conc

20.10

CH3CH2C

CCH2CH3 KMnO4

a) b) c) d)

Hexane-3,4-diol Hexane-3,4-dione Ethanoic acid Propanoic acid

Explanation: The triple bond of the alkyne is oxidized to carboxylic acids with potassium permanganate. © 2013 Pearson Education, Inc.

1. CO2

20.11

CH3CH2MgCl

+

2. H

a) b) c) d) e)

Ethanoic acid Propanoic acid Butanoic acid Butanoyl chloride Pentanoyl chloride

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1. CO2

20.11

CH3CH2MgCl

+

2. H

a) b) c) d) e)

Ethanoic acid Propanoic acid Butanoic acid Butanoyl chloride Pentanoyl chloride

Explanation: Grignard reagents add to carbon dioxide to form a salt of a carboxylic acid. Acid hydrolysis forms a carboxylic acid. © 2013 Pearson Education, Inc.

1. NaCN

20.12

CH3CH2Cl

+

2. H

a) b) c) d) e)

Ethanoic acid Propanoic acid Butanoic acid Butanoyl chloride Pentanoyl chloride

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1. NaCN

20.12

CH3CH2Cl

+

2. H

a) b) c) d) e)

Ethanoic acid Propanoic acid Butanoic acid Butanoyl chloride Pentanoyl chloride

Explanation: Halogen is replaced using sodium cyanide. Hydrolysis of the cyanide group gives the carboxylic acid. © 2013 Pearson Education, Inc.

CH3OH 20.13

CH3CH2COOH H

a) b) c) d) e)

CH3CO2CH3 CH3CO2CH2CH3 CH3CH2CO2CH3 CH3COCH3 CH3COCH2CH3

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+

CH3OH 20.13

CH3CH2COOH H

a) b) c) d) e)

+

CH3CO2CH3 CH3CO2CH2CH3 CH3CH2CO2CH3 CH3COCH3 CH3COCH2CH3

Explanation: The Fischer esterification converts carboxylic acids to esters through an acid-catalyzed nucleophilic acyl substitution. © 2013 Pearson Education, Inc.

20.14

a) b) c) d) e)

CH3CH2COOH

CH3CH2NH2 CH3CH2CONH2 CH3CH2COOCH3 CH3NH2 CH3COOCH3

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CH2N2

20.14

a) b) c) d) e)

CH3CH2COOH

CH2N2

CH3CH2NH2 CH3CH2CONH2 CH3CH2COOCH3 CH3NH2 CH3COOCH3

Explanation: Diazomethane converts carboxylic acids to methyl esters. © 2013 Pearson Education, Inc.

20.15

1. CH3NH2 CH3CH2COOH 2. heat

a) b) c) d) e)

CH3CH2CONH2 CH3CH2COOCH3 CH3COOCH3 CH3CONHCH3 CH3CH2CONHCH3

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20.15

1. CH3NH2 CH3CH2COOH 2. heat

a) b) c) d) e)

CH3CH2CONH2 CH3CH2COOCH3 CH3COOCH3 CH3CONHCH3 CH3CH2CONHCH3

Explanation: An amide is formed from a carboxylic acid and an amine. © 2013 Pearson Education, Inc.

1. LiAlH4

20.16

a) b) c) d)

CH3CH2OH CH3CHO CH3CH2CH2OH CH3CH2CHO

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CH3CH2COOH

2. H3O+

1. LiAlH4

20.16

a) b) c) d)

CH3CH2COOH

CH3CH2OH CH3CHO CH3CH2CH2OH CH3CH2CHO

Explanation: The carboxylic acid is reduced to the alcohol.

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2. H3O+

1. SOCl2

20.17

CH3CH2COOH 2. LiAl[OC(CH3)3]3H

a) b) c) d) e)

CH3CH2CH2OH CH3CH2COCl CH3CH2CHO CH3CH2COOC(CH3)3 CH3CH2COOH

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1. SOCl2

20.17

CH3CH2COOH 2. LiAl[OC(CH3)3]3H

a) b) c) d) e)

CH3CH2CH2OH CH3CH2COCl CH3CH2CHO CH3CH2COOC(CH3)3 CH3CH2COOH

Explanation: Thionyl chloride reacts with the carboxylic acid to form an acid chloride. The acid chloride is reduced to an aldehyde with lithium tri(t-butoxy)aluminum hydride. © 2013 Pearson Education, Inc.

1. 2 CH3Li 20.18

a) b) c) d) e)

CH3CH2COOH

3-Pentanone 2-Pentanone Propanone 2-Butanone Methyl propanoate

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2. H2O

1. 2 CH3Li 20.18

a) b) c) d) e)

CH3CH2COOH

2. H2O

3-Pentanone 2-Pentanone Propanone 2-Butanone Methyl propanoate

Explanation: A carboxylic acid reacting with two equivalents of an organolithium reagent produces a dianion. Hydrolysis gives a ketone. © 2013 Pearson Education, Inc.

1. SOCl2

20.19

a) b) c) d) e)

CH3CH2COOH

CH3CH2COCl CH3CH2CHO CH3CO2CH3 CH3CH2COOCH3 CH3CH2CH2Cl

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2. CH3OH

1. SOCl2

20.19

a) b) c) d) e)

CH3CH2COOH

2. CH3OH

CH3CH2COCl CH3CH2CHO CH3CO2CH3 CH3CH2COOCH3 CH3CH2CH2Cl

Explanation: An acid chloride is formed in the first step. The acid chloride reacts with methanol to produce the methyl ester ester. © 2013 Pearson Education, Inc.

1. (C OCl)2 20.20

CH3CH2COOH 2. CH3NH2

a) b) c) d) e)

CH3CH2COCl CH3CH2CONHCH3 CH3CONHCH2CH3 CH3CH2COOCH3 CH3CH2CH2Cl

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1. (C OCl)2 20.20

CH3CH2COOH 2. CH3NH2

a) b) c) d) e)

CH3CH2COCl CH3CH2CONHCH3 CH3CONHCH2CH3 CH3CH2COOCH3 CH3CH2CH2Cl

Explanation: Reaction with oxalyl chloride forms the acid chloride. Nucleophilic acyl substitution with the amine forms the amide. © 2013 Pearson Education, Inc.

O

21.1 Name

a) b) c) d) e)

C CH3CH2

Ethyl ethanoate Propyl propanoate Ethyl propanoate Propyl ethanoate Propyl butanoate

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.

OCH2CH2CH3

O

21.1 Name

a) b) c) d) e)

C CH3CH2

OCH2CH2CH3

Ethyl ethanoate Propyl propanoate Ethyl propanoate Propyl ethanoate Propyl butanoate

Explanation: The longest chain is three carbons. Propyl is the alkoxy group.

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.

O

21.2 Name

. O

a) b) c) d)

3-Hydroxybutanoic acid lactone 4-Hydroxybutanoic acid lactone 4-Hydroxypentanoic acid lactone 5-Hydroxypentanoic acid lactone

O

21.2 Name

a) b) c) d)

O

.

3-Hydroxybutanoic acid lactone 4-Hydroxybutanoic acid lactone 4-Hydroxypentanoic acid lactone 5-Hydroxypentanoic acid lactone

Explanation: A lactone is a cyclic ester. The hydroxyl is on the fifth carbon.

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O

21.3 Name CH CH 3 2

a) b) c) d) e)

Pentanamide Butanamide N-Ethylethanamide N-Methylethanamide N-Ethylpropanamide

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C

NHCH2CH.3

O

21.3 Name CH CH 3 2

a) b) c) d) e)

C

NHCH2CH.3

Pentanamide Butanamide N-Ethylethanamide N-Methylethanamide N-Ethylpropanamide

Explanation: The ethyl group is attached to the nitrogen. The longest chain is three carbons. © 2013 Pearson Education, Inc.

O

21.4 Name

. N

a) b) c) d)

3-Aminobutanoic acid lactam 4-Aminobutanoic acid lactam 4-Aminopentanoic acid lactam 5-Aminopentanoic acid lactam

O

21.4 Name

a) b) c) d)

N

.

3-Aminobutanoic acid lactam 4-Aminobutanoic acid lactam 4-Aminopentanoic acid lactam 5-Aminopentanoic acid lactam

Explanation: A lactam is a cyclic amide. The amino group is on the fifth carbon.

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(CH3)2CHCH2C 21.5 Name

a) b) c) d) e)

Pentanenitrile Butanenitrile Propanenitrile 2-Methylbutanenitrile 3-Methylbutanenitrile

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N .

(CH3)2CHCH2C 21.5 Name

a) b) c) d) e)

N .

Pentanenitrile Butanenitrile Propanenitrile 2-Methylbutanenitrile 3-Methylbutanenitrile

Explanation: The longest chain has four carbons. The methyl is on the third carbon.

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O

21.6 Name CH CH 3 Cl a) b) c) d)

1-Chloroethanoyl chloride 2-Chloroethanoyl chloride 1-Chloropropanoyl chloride 2-Chloropropanoyl chloride

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C

. Cl

O

21.6 Name CH CH 3

C

. Cl

Cl a) b) c) d)

1-Chloroethanoyl chloride 2-Chloroethanoyl chloride 1-Chloropropanoyl chloride 2-Chloropropanoyl chloride

Explanation: The longest chain has three carbons. Chlorines are on the second carbon and the carbonyl carbon.

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If R’ or R’’ is H, then the product will be 2 carboxylic acids