chem 243 midterm review

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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

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 .

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 .

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