chem 243 midterm review

© 2013 Pearson Education, Inc.

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


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


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


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.


18.3 Identify the chemicalname for acetone.

a) Methanal

b) Ethanal

c) Propanone

d) Butanone


18.3 Identify the chemicalname for acetone.

a) Methanal

b) Ethanal

c) Propanone

d) Butanone

Explanation:

Acetone is called propanone or dimethyl ketone.


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


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


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


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.


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)


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.


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)


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.


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


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.


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


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.


18.10

O

COHCH3CH2

1. SOCl2

2. LiAlH(Ot-Bu)3

a) Propanoyl chloride

b) Propanal

c) Propane

d) Propan-1-ol


18.10

O

COHCH3CH2

1. SOCl2

2. LiAlH(Ot-Bu)3


b) Propanal

c) Propane

d) Propan-1-ol

Explanation:

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


18.11

O

COHCH3CH2

1. SOCl2

2. (CH3CH2)2CuLi


b) Butanone

c) Pentan-2-one

d) Pentan-3-one

e) 3-Ethylpentan-3-ol


18.11

O

COHCH3CH2

1. SOCl2

2. (CH3CH2)2CuLi


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.


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


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.


18.13

O

CHCH3CH2

H2O

a) Propan-1-ol

b) Propan-2-ol

c) Propanal

d) Propane-1,1-diol


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.


18.14

O

CHCH3CH2

1. HCN

2. H3O+

a) 2-Hydroxybutanenitrile

b) 2-Oxobutanenitrile

c) 2-Hydroxybutanoic acid

d) 2-Oxobutanoic acid


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.


18.15

O

CHCH2

CH2

H3C

H2NOH

H+

a) Butanal imine

b) Butanal hydrazone

c) Butanal oxime

d) Butanal semicarbazone


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.


18.16

O

CCH3H3C

2 CH3CH2OH

H+

a) 2,2-Diethoxypropane

b) 2-Ethoxypropan-2-ol

c) Propane-2,2-diol

d) 2-Ethoxypropane


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.


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


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.


18.18

O

CCH3CH2

H3CNaBH4

CH3CH2OH

a) Butan-2-one

b) Butan-2-ol

c) Hexan-2-one

d) Butane


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.


18.19

O

CCH3CH2

H3CZn(Hg)

HCl, H2O

a) Butan-2-one

b) Butan-2-ol

c) Butane

d) 2-Chlorobutane


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.


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


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.


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


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


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


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.


20.3 Name .

O

CCH2HO

CH2C

OH

O

a) Di-butanoic acid

b) Pentanedioic acid

c) Ethanedioic acid

d) Propanedioic acid

e) Butanedioic acid


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.


20.4 Name .

HOOC COOH

a) Benzoic acid

b) Phthalic acid

c) Isophthalic acid

d) Terephthalic acid


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.


20.5 Give the hybridizationfor the carbonyl carbon.

a) sp

b) sp2

c) sp3

d) sp4


20.5 Give the hybridizationfor the carbonyl carbon.

a) sp

b) sp2

c) sp3

d) sp4

Explanation:

The carbonyl carbon is sp2 hybridized.


20.6 Give the bond angleat the carbonyl carbon.

a) 90°

b) 104.5°

c) 109.5°

d) 120°


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°.


20.7 Identify the compoundwith the lowest pKa.

a) Acetic acid

b) Chloroacetic acid

c) Dichloroacetic acid

d) Trichloroacetic acid


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.


20.8 CH3CH2OHNa2Cr2O7

H2SO4

a) Ethanoic acid

b) Propanoic acid

c) Butanoic acid

d) Ethanal

e) Propanal


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.


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


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.


20.10 CCH2CH3CH3CH2Cwarm, conc

KMnO4

a) Hexane-3,4-diol

b) Hexane-3,4-dione

c) Ethanoic acid

d) Propanoic acid


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.


20.11 CH3CH2MgCl

1. CO2

2. H+

a) Ethanoic acid

b) Propanoic acid

c) Butanoic acid

d) Butanoyl chloride

e) Pentanoyl chloride


20.11 CH3CH2MgCl

1. CO2

2. H+

a) Ethanoic acid

b) Propanoic acid

c) Butanoic acid



Explanation:

Grignard reagents add to carbon dioxide to form a salt of a carboxylic

acid. Acid hydrolysis forms a carboxylic acid.


20.12 CH3CH2Cl

1. NaCN

2. H+

a) Ethanoic acid

b) Propanoic acid

c) Butanoic acid




20.12 CH3CH2Cl

1. NaCN

2. H+

a) Ethanoic acid

b) Propanoic acid

c) Butanoic acid



Explanation:

Halogen is replaced using sodium cyanide. Hydrolysis of the cyanide

group gives the carboxylic acid.


20.13 CH3CH2COOHCH3OH

H+

a) CH3CO2CH3

b) CH3CO2CH2CH3

c) CH3CH2CO2CH3

d) CH3COCH3

e) CH3COCH2CH3


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.


20.14 CH3CH2COOHCH2N2

a) CH3CH2NH2

b) CH3CH2CONH2

c) CH3CH2COOCH3

d) CH3NH2

e) CH3COOCH3


20.14 CH3CH2COOHCH2N2

a) CH3CH2NH2

b) CH3CH2CONH2

c) CH3CH2COOCH3

d) CH3NH2

e) CH3COOCH3

Explanation:

Diazomethane converts carboxylic acids to methyl esters.


20.15 CH3CH2COOH1. CH3NH2

2. heat

a) CH3CH2CONH2

b) CH3CH2COOCH3

c) CH3COOCH3

d) CH3CONHCH3

e) CH3CH2CONHCH3


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.


20.16 CH3CH2COOH

1. LiAlH4

2. H3O+

a) CH3CH2OH

b) CH3CHO

c) CH3CH2CH2OH

d) CH3CH2CHO


20.16 CH3CH2COOH

1. LiAlH4

2. H3O+

a) CH3CH2OH

b) CH3CHO

c) CH3CH2CH2OH

d) CH3CH2CHO

Explanation:

The carboxylic acid is reduced to the alcohol.


20.17 CH3CH2COOH1. SOCl2

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

a) CH3CH2CH2OH

b) CH3CH2COCl

c) CH3CH2CHO

d) CH3CH2COOC(CH3)3

e) CH3CH2COOH



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.


20.18 CH3CH2COOH1. 2 CH3Li

2. H2O

a) 3-Pentanone

b) 2-Pentanone

c) Propanone

d) 2-Butanone

e) Methyl propanoate


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.



2. CH3OH

a) CH3CH2COCl

b) CH3CH2CHO

c) CH3CO2CH3

d) CH3CH2COOCH3

e) CH3CH2CH2Cl



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.


20.20 CH3CH2COOH1. (C OCl)2

2. CH3NH2

a) CH3CH2COCl

b) CH3CH2CONHCH3

c) CH3CONHCH2CH3

d) CH3CH2COOCH3

e) CH3CH2CH2Cl


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.


21.1 Name .

a) Ethyl ethanoate

b) Propyl propanoate

c) Ethyl propanoate

d) Propyl ethanoate

e) Propyl butanoate

O

COCH2CH2CH3CH3CH2


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


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 .


21.3 Name .

a) Pentanamide

b) Butanamide

c) N-Ethylethanamide

d) N-Methylethanamide

e) N-Ethylpropanamide

O

CNHCH2CH3CH3CH2


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


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 .


21.5 Name .

a) Pentanenitrile

b) Butanenitrile

c) Propanenitrile

d) 2-Methylbutanenitrile

e) 3-Methylbutanenitrile

N(CH3)2CHCH2C


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


21.6 Name .

a) 1-Chloroethanoyl chloride

b) 2-Chloroethanoyl chloride

c) 1-Chloropropanoyl chloride

d) 2-Chloropropanoyl chloride

O

CClCH3CH

Cl


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

chem 243 midterm review

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