Chapter 16 Aldehydes and Ketones - websites.rcc.eduwebsites.rcc.edu/grey/files/2014/08/Chapter-16-Smith.pdf · Solutions to In-Chapter Problems 16.1 An aldehyde has at least one H

Chapter 16–1

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Chapter 16 Aldehydes and Ketones Solutions to In-Chapter Problems 16.1 An aldehyde has at least one H atom bonded to the carbonyl group.

A ketone has two alkyl groups bonded to the carbonyl group.

aldehyde

ketone

ketone

aldehyde

CO

HCH3CH2

CO

CH3CH3CH2

CO

CH3(CH3)3C

CO

H(CH3CH2)2CH

a.

b.

c.

d. 16.2 Draw the constitutional isomers of molecular formula C4H8O and then label each compound using

the definitions from Answer 16.1.

CO

CH3CH3CH2CO

HCH3CH2CH2CO

H(CH3)2CHketone aldehyde aldehyde 16.3 Trigonal planar carbons are carbons bonded to three other groups. Each trigonal planar carbon is

labeled with an arrow.

CH3 CH3

CH3

CC

CC

CH3C

CC

CCH3

COH

H

H H

HH

H

16.4 To name an aldehyde using the IUPAC system, use the steps in Example 16.1:

[1] Find the longest chain containing the CHO group, and change the -e ending of the parent alkane to the suffix -al.

[2] Number the chain or ring to put the CHO group at C1, but omit this number from the name. Apply all of the other usual rules of nomenclature.

a. (CH3)2CHCH2CH2CH2CHO

CH3CHCH2CH2CH2CHOCH3

hexanalhexane(6 C's)

CH3CHCH2CH2CH2CHOCH3

5-methyl

Answer: 5-methylhexanal

Chapter 16–2



(CH3)3CC(CH3)2CH2CHOb.

pentanalpentane(5 C's)

12

5CH3 C

CH3

CH3

C CCH3

CH3

CH

HH

OCH3 C

CH3

CH3

C CCH3

CH3

CH

HH

O

4

3,3,4,4-tetramethyl

Answer: 3,3,4,4-tetramethylpentanal

c. CH3CHCHCH2CH2CHCH3

CHOCH2CH3

CH3

1

2,5,6-trimethyl

Answer: 2,5,6-trimethyloctanal

CH3CH2CHCHCH2CH2CHCHCH3CH3

CH3 Ooctanaloctane

(8 C's)

CH3CH2CHCHCH2CH2CHCHCH3CH3

CH3 O256

re-draw

16.5 Work backwards from the name to draw each structure.

CH3CHCHOCl

CH3CH2CHCHCHCH2CHOCH2CH3CH3CH2

CH2CH3

CH3CH2CH2CHCH2CH2CHCH2CHOCH2CH3

CH2CH3

CHO

CH2CH3

d. o-ethylbenzaldehyde

!a. 2-chloropropanal

b. 3,4,5-triethylheptanal

!c. 3,6-diethylnonanal

3 C chain witha CHO at C1

7 C chain witha CHO at C1

2-chloro9 C chain witha CHO at C1

3,6-diethyl

benzene ring with a CHO

o-ethyl3

4

5

16.6 To name an aldehyde using the IUPAC system, use the steps in Example 16.1.

a.1

28nonane nonanal(9 C's)

8-methylnonanal

8-methyl

b.decane decanal(10 C's)

8-methyldecanal128

8-methyl

Chapter 16–3



16.7 To name a ketone using IUPAC rules, use the steps in Example 16.2: [1] Find the longest chain containing the carbonyl group, and change the -e ending of the parent

alkane to the suffix -one. [2] Number the carbon chain to give the carbonyl carbon the lower number. Apply all of the other

usual rules of nomenclature.

a. CH3CH2CCHCH2CH2CH3

O

CH31

heptane heptanone2 3

(7 C's)

CH3CH2CCHCH2CH2CH3

O

CH3

4

4-methyl

Answer: 4-methyl-3-heptanone

O

CH3

b.

1cyclopentane cyclopentanone

2

(5 C's)

2-methyl

O

CH3

Answer: 2-methylcyclopentanone

c. CH3CCH3

CH3

CCH2CH2CH2CH3

O

1

heptane heptanone

2

3

(7 C's)

CH3CCH3

CH3

CCH2CH2CH2CH3

O

2,2-dimethyl

Answer: 2,2-dimethyl-3-heptanone


CH3CH2CH2CH2CCH2CH3

CH3CH2CCHCH3

CH3

O

CH3CH2 CO

CH3

O

CH2CH2CH3

O

d. 2-propylcyclobutanone

a. butyl ethyl ketone

b. 2-methyl-3-pentanone

c. p-ethylacetophenone

butyl ethyl

5 C chain withC=O at C3

2-methyl

acetophenone

p-ethyl

4 C ring withC=O at C1

2-propyl 16.9 Aldehydes and ketones have higher boiling points than hydrocarbons of comparable size.

Aldehydes and ketones have lower boiling points than alcohols of comparable size.

O CH3

(CH3CH2)2CO

or

or (CH3CH2)2C=CH2

O or OHa.

CH3(CH2)6CH3 CH3(CH2)5CHOorb.

c.

d.

hydrocarbonketonehigher boiling point

alcoholhigher boiling point

ketone

hydrocarbonketonehigher boiling point

hydrocarbon aldehydehigher boiling point

Chapter 16–4



16.10 Acetone will be soluble in water and organic solvents since it is a low molecular weight ketone (less than six carbons). Progesterone will be soluble only in organic solvents since it has many carbons and only two polar functional groups.

CH3C

O

CH3

OCH3

large molecule with two ketonesprogesterone

CH3C

CH3

O

small ketoneacetone

ketone

ketone 16.11 Hexane is soluble in acetone because both compounds are organic and “like dissolves like.” Water

is soluble in acetone because acetone has a short hydrocarbon chain and is capable of hydrogen bonding with water.

16.12 Compare the functional groups in each sunscreen. Dioxybenzone will most likely be washed off in

water because it contains two hydroxyl groups and is the most water soluble.

CO

CO

CCH2

OOH

OCH3

OH OH

OCH3

CO

C(CH3)3CH3O

dioxybenzonetwo hydroxyl groups

one ketoneone ether

most water soluble

oxybenzoneone hydroxyl group

one ketoneone ether

avobenzonetwo ketones

one ether

16.13 Draw the product of each reaction using the guidelines in Example 16.3. Compounds that contain

a C–H and C–O bond on the same carbon are oxidized with K2Cr2O7. • Aldehydes (RCHO) are oxidized to RCO2H. • Ketones (R2CO) are not oxidized with K2Cr2O7.

CO

OHCH3CH2a. CH3CH2CHO

K2Cr2O7

No reactionb. (CH3CH2)2C=OK2Cr2O7

CH3C CHCH2CH2CHCH2CCH3 CH3 O

OHCH3C=CHCH2CH2CHCH2CHOCH3 CH3

c.K2Cr2O7

Chapter 16–5



16.14 Draw the product of each reaction using Example 16.4 as a guide. Only aldehydes (RCHO) react with Tollens reagent, and they are oxidized to RCO2H. Ketones and alcohols are inert to oxidation.

a.

b.

CH3(CH2)6CHO

OH

CHO

O

c.

d.

Ag2O

NH4OH

Ag2O

NH4OH

Ag2O

NH4OH

Ag2O

NH4OH

CH3(CH2)6CO

OH CO

OH

No reaction No reaction

16.15 Draw the products of reduction using the steps in Example 16.5.

• Locate the C=O and mentally break one bond in the double bond. • Mentally break the H–H bond of the reagent. • Add one H atom to each atom of the C=O, forming new C–H and O–H single bonds.

CH3CH2CH2C

H

O

a.

b.

c.

O

CH3

CH3C

CH2CH3

O

d. CHO

CH3CH2CH2CH2

OH

OH

CH3

CH3CHCH2CH3OH

CH2OH

Pd

H2

Pd

H2

Pd

H2

Pd

H2

16.16 Work backwards to determine what carbonyl compound is needed to prepare alcohol A.

Pd

H2CCH3(CH3)2CHCH2

OCHCH3(CH3)2CHCH2

OH

A 16.17 Recall that stereoisomers differ only in the three-dimensional arrangement of atoms in space, but

all connectivity is identical. Constitutional isomers have the same molecular formula, but atoms are connected differently.

a. All-trans-retinal and 11-cis-retinal are stereoisomers, and differ only in the arrangement of

groups around one double bond. b. All-trans-retinal and vitamin A are not isomers. They have different molecular formulas. c. Vitamin A and 11-cis-retinol are stereoisomers, and differ only in the arrangement of groups

around one double bond.

Chapter 16–6



16.18 To form a hemiacetal and acetal from a carbonyl compound, use the steps in Example 16.6. • Locate the C=O in the starting material. • Break one C–O bond and add one equivalent of ROH across the double bond, placing the OR

group on the carbonyl carbon. This forms the hemiacetal. • Replace the OH group of the hemiacetal by OR to form the acetal.

CH3C

H

OCH3OH+a. CH3COH

OCH3CH3COCH3

OCH3

H H

hemiacetal acetal

H2SO4 H2SO4

CH3OH

CH3OH+b. (CH3CH2)2C=O (CH3CH2)2COHOCH3

(CH3CH2)2COCH3

OCH3

hemiacetal acetal

H2SO4 H2SO4

CH3OH

CH

O

CH3CH2OHc. +CHOHOCH2CH3

CHOCH2CH3OCH2CH3

hemiacetal acetal

H2SO4 H2SO4

CH3CH2OH

16.19 Recall the definitions from Example 16.7 to identify the functional groups:

• An ether has the general structure ROR. • A hemiacetal has one C bonded to OH and OR. • An acetal has one C bonded to two OR groups.

O

OCH3CH3

OCH3 OCH3OCH3a. b. c. d.CH3CH3CH2CH2 C

OHOCH3

Hether acetal hemiacetal acetal

16.20 Label the acetal or hemiacetal in each compound using the definitions in Example 16.7.

CH2OHOOHOCH2

HOOH

OH

OOH

NH2

HOCH2

HO

HO

a. b.

hemiacetalacetal

16.21 Draw the products of each reaction using the steps in Example 16.6.

a.O

OH CH3CH2OH+ H2SO4 O

OCH2CH3

Replace OH by OCH2CH3

Chapter 16–7



OOH OH+ H2SO4b.

OO

Replace OH by O

16.22 To draw the products of hydrolysis, use the steps in Example 16.8.

• Locate the two C–OR bonds on the same carbon. • Replace the two C–O single bonds with a carbonyl group (C=O). • Each OR group then becomes a molecule of alcohol (ROH) product.

CO

CH3 CH2CH2CH3+

2 CH3OH

+ 2 CH3CH2OH

C

+

2 CH3OHO

O

H

H2SO4

CH3CH2O OCH2CH3

CH3 COCH3

OCH3

CH2CH2CH3COCH3

HOCH3

a.

b.

c.

H2SO4

H2SO4

H2O

H2O

H2O

Solutions to End-of-Chapter Problems 16.23 Draw a structure to fit each description.

CH3CH2CH2CHCH2CHOCH2CH3

a. b. CH3CH2CCHCH3CH3

Oc. O d. C

O

HaldehydeC8H16O

ketoneC6H12O

ketoneC5H8O

aldehydeC6H10O

16.24 Draw the structure of a constitutional isomer to fit each description.

CH3CH2CH2CH2CH2CH2CHOa. b. CH3CH2CCH2CH2CH2CH3O

c. CH3CH2CH2CH2CH CHCH2OH 16.25 Compare C=O and C=C bonds.

a. Both are trigonal planar. b. A C=O is polar and a C=C is not polar. c. Both functional groups undergo addition reactions.

16.26 Compare RCHO and RCOR.

a. An aldehyde has at least one hydrogen bonded to the carbonyl group and a ketone has two alkyl groups bonded to the carbonyl group.

b. Both are trigonal planar. c. Both are polar.

Chapter 16–8



16.27 An aldehyde cannot have the molecular formula C5H12O. C5H12 has too many H’s. Since an aldehyde has a double bond, the number of C’s and H’s resembles an alkene, not an alkane. An aldehyde with 5 C’s would have the molecular formula C5H10O.

16.28 A ketone cannot have the molecular formula C4H10O. C4H10 has too many H’s. Since a ketone has

a double bond, the number of C’s and H’s resembles an alkene, not an alkane. A ketone with 4 C’s would have the molecular formula C4H8O.

16.29 To name the aldehyde and ketone, use the IUPAC rules in Examples 16.1 and 16.2.

a. b.12

2-methylpentanal3-ethylcyclohexanone

2-methyl

pentanal(5 C's)

pentanecyclohexanone

(6 C ring)cyclohexane

133-ethyl

16.30 To name the aldehyde and ketone, use the IUPAC rules in Examples 16.1 and 16.2.

a. b.

m-fluorobenzaldehyde

m-fluoro

benzaldehydebenzene

FH

O

H

HH

H

O

Cl

Br2-chloro-2-fluoro

2-chloro-2-fluorocyclopentanonecylcopentanonecyclopentane

16.31 To name an aldehyde using the IUPAC system, use the steps in Example 16.1:


[2] Number the chain or ring to put the CHO group at C1, but omit this number from the name. Apply all other usual rules of nomenclature.


3-methyl

Answer: 3-methylhexanalCH3CH2CH2CHCH2CHOCH3

a. CH3CH2CH2CHCH2CHOCH3

heptanalheptane(7 C's)

Answer: 3,5-dimethylheptanalCH3CH2CHCH2CHCH2CHOCH3

CH3

b. CH3CH2CHCH2CHCH2CHOCH3

CH3

13

53,5-dimethyl

Chapter 16–9



CC

c.

C

3

CH2CH2CH3

HCH3CH2CH2

H HO H


CCC

CH2CH2CH3

HCH3CH2CH2

H HO H

3-propyl

Answer: 3-propylhexanal

CH3CH2CCH2CH2CH2 CCH3

CH3

CHOd.CH2CH3

CH2CH3 1

octanaloctane(8 C's)

CH3CH2CCH2CH2CH2 CCH3

CH3

CHOCH2CH3

CH2CH3 26

2,2-dimethylAnswer: 6,6-diethyl-2,2-dimethyloctanal

Cl CHOe. Answer: p-chlorobenzaldehyde

benzaldehyde

Cl CHO

p-chloro 16.32 To name an aldehyde using the IUPAC system, use the steps in Example 16.1:


[2] Number the chain or ring to put the CHO group at C1, but omit this number from the name. Apply all other usual rules of nomenclature.

butanalbutane(4 C's)

2 methyls on C3Answer: 3,3-dimethylbutanal(CH3)3CCH2CHOa. CH3CCH2CHO

CH3

CH3


4-ethyl

Answer: 4-ethylhexanal(CH3CH2)2CHCH2CH2CHOb. CH3CH2CHCH2CH2CHOCH2CH3

octanaloctane(8 C's)

Answer: 3,4-dimethyloctanalCH3CH2CH2CH2CHCHCH3c.CH3

CH2CHOCH3CH2CH2CH2CHCHCH3

CH3

CH2CHO3,4-dimethyl

heptanalheptane(7 C's)

Answer: 3-butylheptanald. (CH3CH2CH2CH2)2CHCH2CHO CH3CH2CH2CH2CHCH2CHO

3-butylCH2CH2CH2CH3

Chapter 16–10



Answer: m-ethylbenzaldehydef.CH3CH2

CHO

benzaldehyde

CH3CH2

CHO

benzaldehyde

m-ethyl

16.33 Work backwards to draw the structure.

CH3CH2CCH2CHOCl

Cl

CH3CH2CHCHCH2CHOCH3

CH3

CHO

Br

CH3CH2CH2CHCH2CH2CHOOHd. 4-hydroxyheptanal

a. 3,3-dichloropentanal

b. 3,4-dimethylhexanal

c. o-bromobenzaldehyde

5 C chain

6 C chain

benzene ring with CHO

7 C chain

3,3-dichloro

3,4-dimethyl 4-hydroxy

o-bromo

16.34 Work backwards to draw the structure.

CH3CH2CH2CH2CH2CH2CHCHO

CH3CH2CH2CH2CH2CHCHO

CHO

CH3CH2CH2CH2CH2CHd. 3,4-dihydroxynonanal

a. 2-bromooctanal

b. 2-propylheptanal

c. 3,4-dimethoxybenzaldehyde

8 C chain

7 C chain

benzene ring with CHO

9 C chain

2-bromo

2-propyl

3,4-hydroxy

Br

CH2CH2CH3

H3CO

H3CO3

4

3,4-dimethoxy

OHCHCH2CHOOH

16.35 To name a ketone using IUPAC rules, use the steps in Example 16.2:

[1] Find the longest chain containing the carbonyl group, and change the -e ending of the parent alkane to the suffix -one.

[2] Number the carbon chain to give the carbonyl carbon the lower number. Apply all of the other usual rules of nomenclature.

a. CH3CHCH2C

CH3

O

Answer: 4-methyl-2-pentanone1

pentane pentanone

24

(5 C's) 4-methyl

CH3

CH3CHCH2C

CH3

O

CH3

b.

OCH3CH3

Answer: 2,6-dimethylcyclohexanone

cyclohexane cyclohexanone(6 C's)

OCH3CH3 1 26

2,6-dimethyl

Chapter 16–11



c.C

CH3

O

CH2CH2CH2CH3

Answer: o-butylacetophenone

benzene ring with CH3C=Oacetophenone

CCH3

O

CH2CH2CH2CH3

o-butyl

d.CH3CH

CCHCH2CH3

O

CH3

Answer: 2,4-dimethyl-3-hexanone

hexane hexanone(6 C's)

CH3

CH3CHC

CHCH2CH3

O

CH3CH3

1 23

4

2,4-dimethyl

e. OCl

Answer: 3-chlorocyclopentanone

cyclopentane cyclopentanone(5 C's)

3-chloro

OCl

13 2

16.36 To name a ketone using IUPAC rules, use the steps in Example 16.2:

[1] Find the longest chain containing the carbonyl group, and change the -e ending of the parent alkane to the suffix -one.

[2] Number the carbon chain to give the carbonyl carbon the lower number. Apply all of the other usual rules of nomenclature.

a. Answer: 3-ethyl-2-heptanone

heptane heptanone(7 C's)

C

CH3CH2CH2CH2CHCH2CH3

CH3O C

CH3CH2CH2CH2CHCH2CH3

CH3O 2

3 3-ethyl

b. Answer: 3,3-dichlorocyclobutanone

cyclobutane cyclobutanone(4 C's)

3,3-dichloro

ClCl

O

ClCl

O 1

3

c. Answer: 9-methyl-3-decanone

decane decanone(10 C's)

CO

CH3CH2 (CH2)5CH(CH3)2CO

CH3CH2 (CH2)5CHCH3CH3

3 9

9-methyl

Chapter 16–12



d. Answer: 3,5-dimethyl-4-heptanone

heptane heptanone(7 C's)

CO

CH3CH2CH CHCH2CH3

4

CH3CH3

CO

CH3CH2CH CHCH2CH3

CH3CH3

3 5

3,5-dimethyl

e. Answer: 2-ethyl-4-methylcyclopentanone

cyclopentane cyclopentanone(5 C's)

4

2CH2CH3

CH3

O

CH2CH3

CH3

O

2-ethyl

4-methyl


CCH3CH2CH2CCH3

CH3

CO

CH2CH2CH3CH3

CO

CH3

CH2CH3

OCH2CH3

CH2CH3

CH3CH2

O

CH3

d. 2,4,5-triethylcyclohexanone

a. 3,3-dimethyl-2-hexanone

b. methyl propyl ketone

c. m-ethylacetophenone

6 C chain

two alkyl groups with aC=O in the middle

benzene ring with a CH3C=O

6 C ring

3,3-dimethyl

methylpropyl

m-ethyl

2,4,5-triethyl

1

2

3

1 2

45


CO

CH2CH3ClCH2CH2

CO

CH3

d. 3-hydroxycyclopentanone

a. dibutyl ketone

b. 1-chloro-3-pentanone

c. p-bromoacetophenone

two butyl groupswith a C=O in the middle

5 C chain

benzene ring with a CH3C=O

5 C ring

chloro

p-bromo

13

CO

CH3CH2CH2CH2 CH2CH2CH2CH3

butyl

13

Br

OH

O

hydroxy

Chapter 16–13



16.39 Draw the four aldehydes and then name them using the steps in Example 16.1.

CH3CH2CCHOCH3

CH3

CH3CHCHCHOCH3

CH3

4 C chain2,2-dimethylbutanal


CH3CCH2CHOCH3

CH3


CH3CH2CHCHO

4 C chain2-ethylbutanal

CH2CH3

1

2

2,2-dimethyl 3,3-dimethyl 2,3-dimethyl 2-ethyl

1 1

12

3

3

2

16.40 Draw the three ketones and then name them using the steps in Example 16.1.

CO

CH3CH2CH2 CH32

5 C chain2-pentanone

CO

CH3CH2 CH2CH33

5 C chain3-pentanone

CO

CH CH32

4 C chain3-methyl-2-butanone

CH3

CH33

3-methyl

16.41 Draw the structure and correct each name.

CH3CH2CH2CH2CHa.

1-pentanoneA ketone cannot be at C1.

It must be an aldehyde.pentanal

b. CH3CH2CH2CCH2CH3

O

4-hexanoneRe-number to use a

lower number.3-hexanone

c.

3-propyl-2-butanoneFind the longest chain.3-methyl-2-hexanone

CH3CH2CH2CHCCH3

O

CH3

d.

2-methyl-1-octanalAn aldehyde is always at C1.

Omit the "1."2-methyloctanal

CH3CH2CH2CH2CH2CH2CHCHCH3

OO

16.42 Draw the structure and correct each name.

CH3CH2CH2CH2CH2CCH2CH3a.

6-octanoneRe-number to use a

lower number.3-octanone

b. CH3CH2CH2CH2CH2CH2CH

1-heptanoneA ketone cannot be at C1.

It must be an aldehyde.heptanal

c.

3-propyl-1-cyclopentanoneThe ketone is always at

C1 on a ring.3-propylcyclopentanone

d.

5-methylcyclohexanoneRe-number to use a

lower number.3-methylcyclohexanone

O OO

CH2CH2CH3

OCH3

16.43 Draw benzaldehyde and then the hydrogen bond.

CO

H

HOH

hydrogen bond

Chapter 16–14



16.44 Draw the structures and then determine if hydrogen bonding is possible. a. Hydrogen bonding is not possible between two molecules of acetaldehyde. b. Hydrogen bonding is possible between ethanal and water.

CH3CO

H

HOH

hydrogen bond

c. Hydrogen bonding is possible between ethanal and methanol.

CH3CO

H

HOCH3

hydrogen bond

16.45 Aldehydes and ketones have higher boiling points than hydrocarbons of comparable size.


a. b.or COCH3 CH2CH2OHor(CH3)3CCH2CH2CH3 (CH3)3CCH2CHO

hydrocarbon alcoholhigher boiling point

aldehydehigher boiling point

ketone

16.46 Aldehydes and ketones have higher boiling points than hydrocarbons of comparable size.


a. b.or orCH3(CH2)6CHO CH3(CH2)7OH

aldehyde aldehydealcoholhigher boiling point

aldehydehigher molar mass

higher boiling point

CH3(CH2)6CHO CH3(CH2)2CHO

16.47 Aldehydes and ketones have higher melting points than hydrocarbons of comparable size. Aldehydes and ketones have lower melting points than alcohols of comparable size.

O OHCH3

Increasing melting point 16.48 Menthol is a solid at room temperature but menthone is a liquid because menthol has a hydroxy

group attached to the cyclohexane ring, whereas menthone has a ketone. Alcohols will have higher melting points than ketones for compounds of comparable size.

Chapter 16–15



16.49 Low molecular weight aldehydes and ketones (less than six carbons) are water soluble.

CH3CH2CH2CH3

CHO

CH3C

CH2CH3

O

a. b. c.

7 C aldehydeinsoluble

4 C ketonesoluble

hydrocarboninsoluble

16.50 Low molecular weight aldehydes and ketones (less than six carbons) are water soluble.

CH3CH2CH2OHCH3CH2CH2CHOa. b. c.

7 C ketoneinsoluble

4 C aldehydesoluble

alcoholsoluble

O

CH3

16.51 2,3-Butanedione has two carbonyl groups capable of hydrogen bonding whereas acetone has one

carbonyl group. This makes 2,3-butanedione more water soluble than acetone. 2,3-Butanedione would also be soluble in an organic solvent like diethyl ether by the “like dissolves like” rule.

16.52 Acetone has a much higher boiling point than formaldehyde because acetone contains three

carbons (CH3COCH3), whereas formaldehyde contains only one carbon (HCHO). Boiling points increase with the number of carbons in a molecule.

16.53 Draw the product of each reaction using the steps in Example 16.3. Compounds that contain a

C–H and C–O bond on the same carbon are oxidized with K2Cr2O7. • Aldehydes (RCHO) are oxidized to RCO2H. • Ketones (R2CO) are not oxidized with K2Cr2O7. • 1° Alcohols (RCH2OH) are oxidized to RCO2H (Section 14.5B).

CH3(CH2)4CHOa. CH2CHOb.K2Cr2O7

K2Cr2O7CH3(CH2)4COOH CH2COOH

O

CH2CH3 CH3(CH2)4CH2OHc. d.K2Cr2O7 K2Cr2O7 CH3(CH2)4COOHNo reaction

16.54 Draw the product of each reaction using the steps in Example 16.3. Compounds that contain a

C–H and C–O bond on the same carbon are oxidized with K2Cr2O7. • Aldehydes (RCHO) are oxidized to RCO2H. • Ketones (R2CO) are not oxidized with K2Cr2O7. • 1° Alcohols (RCH2OH) are oxidized to RCO2H (Section 14.5B).

a. CH3CHCH2CH2CH3

b.

c.

d.

K2Cr2O7

K2Cr2O7

K2Cr2O7

K2Cr2O7 No reaction

Cl

CHO

Cl

COOH

CH3(CH2)8CHO CH3(CH2)8COOHCO

CH3

OHCH3CCH2CH2CH3

O

Chapter 16–16



16.55 Draw the product of each reaction using Example 16.4 as a guide. Only aldehydes (RCHO) react with Tollens reagent, and they are oxidized to RCO2H. Ketones and alcohols are inert to oxidation.

Ag2O

NH4OHCH3(CH2)4CHOa.

CH2CHO

O

CH2CH3

CH3(CH2)4CH2OHb.

c.

d.Ag2O

NH4OHAg2O

NH4OH

Ag2O

NH4OHCH3(CH2)4COOH

CH2COOH

No reaction

No reaction

16.56 Draw the product of each reaction using Example 16.4 as a guide. Only aldehydes (RCHO)

react with Tollens reagent, and they are oxidized to RCO2H. Ketones and alcohols are inert to oxidation with Tollens reagent.

Ag2O

NH4OH

Ag2O

NH4OH

Ag2O

NH4OHa. CH3CHCH2CH2CH3

b.

c.

d. No reaction

Cl

CHO

Cl

COOH

CH3(CH2)8CHO CH3(CH2)8COOHCO

CH3

OH

Ag2O

NH4OH

No reaction

16.57 Answer each question about erythrulose.

erythrulose

a, b.1°

ketone

2°

1°

c.O

HOOH

OH

d.O

HOOH

OH

Tollens reagentNo reaction

K2Cr2O7O

HOOH

O

O

O 16.58

CC

C

O

HHO

H H

H OH

chirality center

aldehyde

1° ROH

2° ROH

a, b, c. Ag2O

NH4OH

d.C

CC

O

OHHO

H H

H OH

CC

C

O

HHO

H H

H OH

Chapter 16–17



16.59 Work backwards to determine what aldehyde can be used to prepare each carboxylic acid.

a.

b.

c.

CO2HCH3

CH3CH2CHCH2CO2HCH3

CH3CH2CHCH2CH3CO2H

CHOCH3

CH3CH2CHCH2CHOCH3

CH3CH2CHCH2CH3CHO

16.60 Work backwards to determine what aldehyde can be used to prepare each carboxylic acid.

a.

b.

c.

CO2H

CH3(CH2)8CHCO2HCH2CH2CO2H CH2CH2CHO

Br

CHO

Br

ClCH3(CH2)8CHCHO

Cl



Pd

H2CHOCH3CH2

O

CH3a. b.

Pd

H2CH2OHCH3CH2CH3

OH



Pd

H2a. b.

Pd

H2CO

CH3CH2 CH2CH(CH3)2 CH3CH2CHCH2CH(CH3)2HO

CH3(CH2)6CHO CH3(CH2)6CH2OH

16.63

CH3CH2 CH

CH3

(CH2)4CHOa, b:

chirality center

CH3CH2 CH

CH3

(CH2)4CH2OHc. CH3CH2 CH

CH3

(CH2)4CHOH2

Pd

16.64

O

a.

O

b, c.Pd

H2OH

chirality center

Chapter 16–18



16.65 Work backwards to determine what carbonyl compound is needed to make each alcohol.

CH3CH2CH2CH2CH2OH

OH

CH3

a. b.CO

HCH3CH2CH2CH2

O

CH3 16.66 Work backwards to determine what carbonyl compound is needed to make each alcohol.

(CH3)2CHCH2CH2OHCH3CH2CHCH2CH2CH3a. b.CH3CH2CCH2CH2CH3 (CH3)2CHCH2CHOH O O

16.67 1-Methylcyclohexanol is a 3o alcohol and cannot be produced from the reduction of a carbonyl

compound because only 1° or 2° alcohols can be formed in these reactions. 16.68 (CH3)3COH cannot be prepared by the reduction of a carbonyl compound because it is a tertiary

alcohol. A carbonyl group attached to the tertiary carbon would give the carbon five bonds. 16.69 Recall the definitions from Example 16.7 to draw a compound of molecular formula C5H12O2 that

fits each description: • An ether has the general structure ROR. • A hemiacetal has one C bonded to OH and OR. • An acetal has one C bonded to two OR groups.

a. CH3CH2 O C O CH2CH3

b. CH3 C O CH2CH2CH3OH

c. CH3CH2 O C C O CH3H H

H H

CH3 C O C C CH3H

H

H

OH

H

Hd.

H

H

Hacetal

hemiacetal

two ethers

alcohol

ether 16.70 Locate the two acetals in amygdalin.

OHOCH2

HO

HO OH

OCH2

O

HO OH

OH

O CHCNacetal

16.71 Label the functional groups using the definitions from Example 16.7.

CH3 COCH3

HOCH3

CH3 COCH2CH3

HOH

HOCH2CHCH2CH3OCH3a. b. c.

Od.

acetal

hemiacetal

alcohol

ether ether

Chapter 16–19



16.72 Label the functional groups using the definitions from Example 16.7.

CH3 COH

OCH2CH2CH3CH3

a. b. c. O Od.

hemiacetal

ether

etherOCH3

OCH3

O

OCH2CH2CH3

acetal

16.73 To form a hemiacetal and acetal from a carbonyl compound, use the steps in Example 16.6.

• Locate the C=O in the starting material. • Break one C–O bond and add one equivalent of CH3OH across the double bond, placing the

OCH3 group on the carbonyl carbon. This forms the hemiacetal. • Replace the OH group of the hemiacetal by OCH3 to form the acetal.

c.a.

b. CH2 O

OCH3

CH3

CH3C

CH2CH2CH3

O

CH2CHOd.

2 CH3OHH2SO4

CH3

CH3

CH3 CCH2CH2CH3

CH2CH

OCH3

OCH3

OCH3

CH3O

OCH3

OCH3

CH2(OCH3)2

2 CH3OHH2SO4

2 CH3OHH2SO4

2 CH3OHH2SO4

16.74 To form a hemiacetal and acetal from a carbonyl compound, use the steps in Example 16.6.

• Locate the C=O in the starting material. • Break one C–O bond and add one equivalent of CH3CH2OH across the double bond, placing

the OCH2CH3 group on the carbonyl carbon. This forms the hemiacetal. • Replace the OH group of the hemiacetal by OCH2CH3 to form the acetal.

c.

a.

b.

CH3CH2CH2C

CH2CH3

O

CH3CH2CHOd.

CH3CH2CH2 CCH2CH3

OCH2CH3

CH3CH2O2 CH3CH2OHH2SO4

2 CH3CH2OHH2SO4

O

CH3

2 CH3CH2OHH2SO4

CH3

OCH2CH3

OCH2CH3

(CH3)2CHCH2CH2CHO

2 CH3CH2OHH2SO4 (CH3)2CHCH2CH2CH

OCH2CH3

OCH2CH3

CH3CH2CHOCH2CH3

OCH2CH3

Chapter 16–20



16.75 Draw the products of each reaction.

H2SO4OC

OHCH3

CH3OC

OCH3

CH3

a. b.

OC

CH3CH3

HO

H2SO4

HO


H2SO4

a. b.

H2SO4C

H

O HOCOH

OH

HO

CO

OH

16.77 Answer each question.

OOH

hemiacetal carbon

a.

b. HOCH2CH2CH2CHO

OOCH3c.

CH3OHH2SO4O

OH

OOH


a. O

OH

CH3

CH3

hemiacetal carbon

b. O

OH

CH3

CH3 HOCH2CCH2CH2CH

CH3CH2OHH2SO4

CH3

CH3

O

c. O

OH

CH3

CH3

O

OCH2CH3

CH3

CH3

16.79 Draw the product of cyclization.

HOCH2CH2CH2CHC

H

O

CH3C

OCH3

OH

Chapter 16–21



16.80 Draw the product of cyclization.

HOCHCH2CH2CH2C

H

O

DO

CH3

OH

CH3 16.81 To draw the products of hydrolysis, use the steps in Example 16.8.


OCH2CH2CH3

OCH2CH2CH3a. H2SO4

O HOCH2CH2CH3+ 2

H2O

H COCH3

CH2CH2CH3OCH3

b. H2SO4H C

OCH2CH2CH3 + HOCH32

H2O

16.82 To draw the products of hydrolysis, use the steps in Example 16.8.


a.H2SO4

HOCH2CH3+ 2

H2O

COCH2CH3

CH3CH2OCH2CH3

CH2CH3

b.H2SO4

O CH3OH+ 2

H2O

CCH3CH2 CH2CH3O

CH3OOCH3

OCH3CH3O

16.83 Answer each question about compound A.

CH3

O

CH3

a, b. c. d.

p-methylacetophenone seven trigonal planar C's

CH3 C

OCH3

CH3OCH3

p-methyl

CH3

O

CH3

**

* *

**

* CH3

O

CH3

2 CH3OH(acid)

Chapter 16–22



16.84

CC O

H

HH

a. CC O

H

HH

b.*

** *

**

*

7 planar carbons

c.2 CH3OH

(acid)CC O

H

HH

CC OCH3

HH

OCH3H

d. CC O

H

HH

CH2CH2OHH2

Pd 16.85 Draw the products of each reaction.

H2CH

Oa.

b.

c.

d.

e.

f.

CH2OH

COHO

COHO

COCH3

OCH3

H

COCH2CH3

OCH2CH3

H

CH

O

CH

O

CH

O

CH

O

CH

OK2Cr2O7

Ag2O

2 CH3OH

2 CH3CH2OH

H2O

Pd

NH4OH

H2SO4

H2SO4

H2SO4COCH2CH3

OCH2CH3

H + 2 CH3CH2OH


H2CH

Oa.

b.

c.

d.

e.

f.

CH2OH

COHO

COHO

COCH3

OCH3

H

COCH2CH3

OCH2CH3

H

CH

O

CH

O

CH

O

CH

O

CH

O

K2Cr2O7

Ag2O

2 CH3OH

2 CH3CH2OH

H2O

Pd

NH4OH

H2SO4

H2SO4

H2SO4COCH2CH3

OCH2CH3

H + 2 CH3CH2OH

CH3O CH3O

CH3O CH3O

CH3O CH3O

CH3O CH3O

CH3O CH3O

CH3O CH3O

Chapter 16–23




+ 2 CH3CH2OH

H2a.

b.

c.

d.

e.

f.

K2Cr2O7

Ag2O

2 CH3OH

2 CH3CH2OH

H2O

Pd

NH4OH

H2SO4

H2SO4

H2SO4

(CH2)4CH3CCH3H

OH

No reaction

(CH2)4CH3CCH3

OCH3

OCH3

(CH2)4CH3CCH3OCH2CH3

OCH2CH3

(CH2)4CH3C

CH3

O

No reaction (CH2)4CH3CCH3

OCH2CH3

OCH2CH3

(CH2)4CH3C

CH3

O

(CH2)4CH3C

CH3

O

(CH2)4CH3C

CH3

O

(CH2)4CH3C

CH3

O

(CH2)4CH3C

CH3

O


+ 2 CH3CH2OH

H2a.

b.

c.

d.

e.

f.

K2Cr2O7

Ag2O

2 CH3OH

2 CH3CH2OH

H2O

Pd

NH4OH

H2SO4

H2SO4

H2SO4

(CH2)2CH(CH3)2C(CH3)2CHH

OH

No reaction

(CH2)2CH(CH3)2C(CH3)2CHOCH3

OCH3

No reaction

(CH2)2CH(CH3)2C

(CH3)2CH

O

(CH2)2CH(CH3)2C

(CH3)2CH

O

(CH2)2CH(CH3)2C

(CH3)2CH

O

(CH2)2CH(CH3)2C

(CH3)2CH

O

(CH2)2CH(CH3)2C

(CH3)2CH

O

(CH2)2CH(CH3)2C(CH3)2CHOCH2CH3

OCH2CH3

(CH2)2CH(CH3)2C(CH3)2CHOCH2CH3

OCH2CH3

(CH2)2CH(CH3)2C(CH3)2CHO

16.89 Draw the three constitutional isomers that can be converted to 1-pentanol. The starting material

needs a C=O at C1 and a C=C.

H2

CH2 CHCH2CH2CHO

CH3CH CHCH2CHO

CH3CH2CH CHCHOPd

CH3CH2CH2CH2CH2OHor

or

Chapter 16–24



16.90 Work backwards to determine the identity of A–C.

H2Pd

O H2SO4OH

A B C

H2Pd


a.

HO

CH3

CH3 OH

HO

CH3

CH3 O

H2Pd

b.

O

CH3

CH3 O

K2Cr2O7

HO

CH3

CH3 O

c. 2 CH3OHH2SO4

HO

CH3

CH3 OCH3OCH3

HO

CH3

CH3O

d. 2 CH3CH2OH

H2SO4

HO

CH3

CH3 OCH2CH3OCH2CH3

HO

CH3

CH3O

Chapter 16–25




excess H2

PdHO

COH

CCH2NH(CH2)6O(CH2)4

Oa.

X

HO

CHH

CCH2NH(CH2)6O(CH2)4

HO

OH

Hsalmeterol

HO

CHH

CCH2NH(CH2)6O(CH2)4

HO

OH

Hsalmeterol

b.

chirality center

c.

H

C

HO(CH2)4O(CH2)6NHCH2

H

COH

CH2NH(CH2)6O(CH2)4

enantiomersCH2OHHO OH

CH2OH 16.93 Draw the product of oxidation.

CH CH CO

OHCH CH CHONAD+

enzyme 16.94 Answer each question.

O

Oa.

ether

alkenebenzene

b.O

O H2OH2SO4 O

O

O

O+

OHOH

16.95 Label each hemiacetal or alcohol.

OHOCH2

HO

OH

alcohol

OH of a hemiacetal

16.96 Label the actetal carbons in paraldehyde.

O

O O

CH3 CH3

CH3

acetal

acetal

acetal

16.97 The main reaction that occurs in the rod cells in the retina is conversion of 11-cis-retinal to its

trans isomer. The cis double bond in 11-cis-retinal produces crowding, making the molecule unstable. Light energy converts this to the more stable trans isomer, and with this conversion an electrical impulse is generated in the optic nerve.

Chapter 16–26



16.98 All-trans-retinal is converted back to 11-cis-retinal by a series of reactions that involve biological oxidation and reduction. NADH reduces the aldehyde in all-trans-retinal to all-trans-retinol. The trans double bond is isomerized to a cis double bond. NAD+ oxidizes 11-cis-retinol to 11-cis-retinal.

16.99 Identify the alcohol, acetal, hemiacetal, ether, and carboxylic acid functional groups.

O OO

O

OCHHO2CCHCH

CH3

CH3

CH2CH3

HOCH2

HO

CH3

CH3

CH3OH

CH3

CH3O CH3

carboxylic acid

ether

ether

alcohol

acetal

alcohol

hemiacetal

16.100 Determine the structure of chloral hydrate.

CCl

Cl CCl

OH

H2O CCl

Cl CCl

OHH

OH

Documents

Chapter 16 Aldehydes and Ketones - websites.rcc.eduwebsites.rcc.edu/grey/files/2014/08/Chapter-16-Smith.pdf · Solutions to In-Chapter Problems 16.1 An aldehyde has at least one H