Chapter 16 Aldehydes and Ketones

Chapter 16–1

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.

O O C C a. CH3CH2 H aldehyde c. (CH3)3C CH3 ketone

O O C C (CH CH ) CH H aldehyde b. CH3CH2 CH3 ketone d. 3 2 2

16.2 Draw the constitutional isomers of molecular formula C4H8O and then label each compound using the definitions from Answer 16.1.

O O O C C C CH3CH2 CH3 ketone CH3CH2CH2 H aldehyde (CH3)2CH H aldehyde

16.3 Trigonal planar carbons are carbons bonded to three other groups. Each trigonal planar carbon is labeled with an arrow.

H CH CH3 CH3 3 H C C C H C C C H H C H CH CH3 3 C C H O

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 5-methyl

CH3 O CH3 O

CH3CHCH2CH2CH2CH CH3CHCH2CH2CH2CH Answer: 5-methylhexanal

hexane hexanal (6 C's)

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b. (CH3)3CC(CH3)2CH2CHO 3,3,4,4-tetramethyl

CH3 CH3 H O CH3 CH3 H O

CH3 C C C C H CH3 C C C C H Answer: 3,3,4,4-tetramethylpentanal CH3 CH3 H 5 CH3 CH3 H 1 4 2 pentane pentanal (5 C's)

CH2CH3 CHO

c. CH3CHCHCH2CH2CHCH3

CH3 re-draw 2,5,6-trimethyl

CH3 CH3 CH3 CH3

CH3CH2CHCHCH2CH2CHCH CH3CH2CHCHCH2CH2CHCH Answer: CH O CH O 2,5,6-trimethyloctanal 3 6 5 3 2 1 octane octanal (8 C's)

16.5 Work backwards from the name to draw each structure.

!a. 2-chloropropanal !c. 3,6-diethylnonanal 3,6-diethyl Cl 2-chloro CH2CH3 3 C chain with CH CHCHO 9 C chain with a CHO at C1 3 a CHO at C1 CH3CH2CH2CHCH2CH2CHCH2CHO

CH2CH3

d. o-ethylbenzaldehyde b. 3,4,5-triethylheptanal CH3CH2 CH CH 2 3 CHO 7 C chain with 4 benzene ring CH3CH2CHCHCHCH2CHO a CHO at C1 5 3 with a CHO CH CH o-ethyl CH2CH3 2 3

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

8-methyl

1 nonane nonanal 8-methylnonanal a. 8 2 (9 C's)

1 decane decanal 8-methyldecanal 8 2 b. (10 C's)

8-methyl

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.

O O 4

a. CH3CH2CCHCH2CH2CH3 CH3CH2CCHCH2CH2CH3 Answer: 4-methyl-3-heptanone 1 2 3 CH3 CH3 heptane heptanone (7 C's) 4-methyl

CH3 CH3 2-methyl

2 b. O O Answer: 2-methylcyclopentanone

cyclopentane cyclopentanone 1 (5 C's)

2 CH3 O CH3 O c. CH3C CCH2CH2CH2CH3 CH3C CCH2CH2CH2CH3 Answer: 2,2-dimethyl-3-heptanone 1 3 CH3 CH3 heptane heptanone 2,2-dimethyl (7 C's)

16.8 Work backwards from the name to draw each structure.

O acetophenone a. butyl ethyl ketone CH CH CH CH CCH CH c. p-ethylacetophenone 3 2 2 2 2 3 O p-ethyl butyl ethyl CH3CH2 C CH3

b. 2-methyl-3-pentanone d. 2-propylcyclobutanone O O 5 C chain with CH CH CCHCH 4 C ring with C=O at C3 3 2 3 C=O at C1 CH3 2-methyl 2-propyl CH2CH2CH3

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.

c. a. O or CH3 O or OH

ketone hydrocarbon ketone alcohol higher boiling point higher boiling point

d. CH (CH ) CH or b. (CH3CH2)2CO or (CH3CH2)2C=CH2 3 2 6 3 CH3(CH2)5CHO ketone hydrocarbon hydrocarbon aldehyde higher boiling point higher boiling point

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.

ketone O C CH3 CH3 O

C CH3 CH3 CH3 small ketone large molecule with two ketones acetone O progesterone

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.

O OH O O OH O OH C C C C CH2

OCH3 CH3O C(CH3)3 OCH3 oxybenzone avobenzone dioxybenzone one hydroxyl group two ketones two hydroxyl groups one ketone one ether one ketone one ether one ether most water soluble

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.

O K2Cr2O7 a. CH3CH2CHO C CH3CH2 OH

K2Cr2O7 b. (CH3CH2)2C=O No reaction

CH3 CH CH3 CH3 O 3 K Cr O 2 2 7 CH C CHCH CH CHCH C OH c. CH3C=CHCH2CH2CHCH2CHO 3 2 2 2

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.

O Ag O O 2 Ag2O a. CH3(CH2)6CHO CH3(CH2)6C OH c. CHO C NH OH 4 NH4OH OH

Ag O 2 Ag2O b. O No reaction d. OH No reaction NH OH 4 NH4OH

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.

O OH O H OH H2 2 C CH CH CH CH C a. CH CH CH H 3 2 2 2 c. CH3CHCH2CH3 3 2 2 Pd CH3 CH2CH3 Pd

O OH H 2 H2 b. d. CHO CH2OH Pd Pd CH3 CH3

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

O H2 OH (CH3)2CHCH2 CCH3 Pd (CH3)2CHCH2 CHCH3

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.

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.

CH OH O OCH 3 OCH H SO 3 H SO 3 a. 2 4 2 4 C + CH3OH CH3COH CH3COCH3 CH3 H H H hemiacetal acetal

CH OH OCH3 3 OCH H2SO4 3 H2SO4 b. (CH3CH2)2C=O + CH3OH (CH3CH2)2COH (CH3CH2)2COCH3 hemiacetal acetal

O OCH2CH3 OCH CH CH3CH2OH 2 3 C H SO H 2 4 CHOH H SO CHOCH CH c. 2 4 2 3 + CH3CH2OH

hemiacetal acetal

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.

OCH3 OCH3 OH O CH3 a. b. OCH3 c. CH3CH3CH2CH2 C OCH3 d. H O CH3 ether acetal hemiacetal acetal

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

acetal hemiacetal CH2OH HOCH2 HOCH O O O 2 a. HO OH b. HO OH HO NH2 OH

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

O H2SO4 O a. OH + CH3CH2OH

OCH2CH3 Replace OH by OCH2CH3

O O H SO b. OH + OH 2 4 O

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.

H2O O OCH H2O 3 H2SO4 C O OCH3 H SO a. CH C OCH CH3 CH2CH2CH3 2 4 C 3 3 c. C H + H CH2CH2CH3 OCH3 2 CH3OH +

2 CH3OH

CH3CH2O OCH2CH3 H2O O H2SO4 b. + 2 CH3CH2OH

Solutions to End-of-Chapter Problems

16.23 Draw a structure to fit each description.

CH2CH3 O O a. CH3CH2CH2CHCH2CHO b. CH3CH2CCHCH3 c. O d. C CH3 H aldehyde ketone ketone aldehyde C8H16O C6H12O C5H8O C6H10O

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

O a. CH3CH2CH2CH2CH2CH2CHO b. CH3CH2CCH2CH2CH2CH3 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.

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.

2-methyl

a. 1 b. 2 1 3 3-ethyl

pentane pentanal (5 C's) cyclohexane cyclohexanone 2-methylpentanal (6 C ring) 3-ethylcyclohexanone

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

m-fluoro H H O F a. O b. Cl

H H Br H 2-chloro-2-fluoro

benzene benzaldehyde cyclopentane cylcopentanone m-fluorobenzaldehyde 2-chloro-2-fluorocyclopentanone

16.31 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 other usual rules of nomenclature.

Answer: 3-methylhexanal a. CH3CH2CH2CHCH2CHO CH3CH2CH2CHCH2CHO CH CH 3 3 3-methyl hexane hexanal (6 C's)

CH3 CH3 3 1

b. CH3CH2CHCH2CHCH2CHO CH3CH2CHCH2CHCH2CHO Answer: 3,5-dimethylheptanal 5 CH3 CH3 3,5-dimethyl heptane heptanal (7 C's)

O C H O C H H C H H C H

c. CH3CH2CH2 C CH2CH2CH3 CH3CH2CH2 C CH2CH2CH3 Answer: 3-propylhexanal H H 3 hexane hexanal 3-propyl (6 C's)

2,2-dimethyl CH2CH3 CH3 CH2CH3 CH3

d. CH3CH2CCH2CH2CH2 C CHO CH3CH2CCH2CH2CH2 C CHO Answer: 6,6-diethyl-2,2-dimethyloctanal CH2CH3 CH3 CH2CH3 CH3 1 6 2 octane octanal (8 C's)

e. Cl CHO Cl CHO Answer: p-chlorobenzaldehyde

benzaldehyde p-chloro

16.32 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 other usual rules of nomenclature.

CH3 a. (CH3)3CCH2CHO CH3CCH2CHO Answer: 3,3-dimethylbutanal

butane butanal CH3 2 methyls on C3 (4 C's)

CH2CH3

b. (CH3CH2)2CHCH2CH2CHO CH3CH2CHCH2CH2CHO Answer: 4-ethylhexanal hexane hexanal 4-ethyl (6 C's)

CH3 CH3

c. CH3CH2CH2CH2CHCHCH3 CH3CH2CH2CH2CHCHCH3 Answer: 3,4-dimethyloctanal

CH2CHO CH2CHO octane octanal 3,4-dimethyl (8 C's)

d. (CH3CH2CH2CH2)2CHCH2CHO CH3CH2CH2CH2CHCH2CHO Answer: 3-butylheptanal

CH2CH2CH2CH3 heptane heptanal 3-butyl (7 C's)

CH3CH2 CH3CH2 m-ethyl f. Answer: m-ethylbenzaldehyde CHO CHO

benzaldehyde benzaldehyde

16.33 Work backwards to draw the structure.

a. 3,3-dichloropentanal Cl 3,3-dichloro c. o-bromobenzaldehyde CHO CH3CH2CCH2CHO Cl 5 C chain benzene ring with CHO Br o-bromo

d. 4-hydroxyheptanal b. 3,4-dimethylhexanal OH 4-hydroxy CH3 3,4-dimethyl CH3CH2CH2CHCH2CH2CHO CH3CH2CHCHCH2CHO 7 C chain 6 C chain CH3

16.34 Work backwards to draw the structure.

2-bromo Br a. 2-bromooctanal c. 3,4-dimethoxybenzaldehyde 4 H3CO CHO CH3CH2CH2CH2CH2CH2CHCHO 3 8 C chain benzene ring with CHO H3CO 3,4-dimethoxy

d. 3,4-dihydroxynonanal b. 2-propylheptanal OH OH 2-propyl CH2CH2CH3 CH3CH2CH2CH2CH2CH CHCH2CHO CH3CH2CH2CH2CH2CHCHO 9 C chain 3,4-hydroxy 7 C chain

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.

O O C 4 C 1 a. CH3CHCH2 CH3 CH3CHCH2 2 CH3 Answer: 4-methyl-2-pentanone

CH3 CH3 pentane pentanone 4-methyl (5 C's)

O O

CH3 CH3 CH3 CH3 b. 1 6 2 Answer: 2,6-dimethylcyclohexanone

cyclohexane cyclohexanone 2,6-dimethyl (6 C's)

O O C C c. CH3 CH3 Answer: o-butylacetophenone CH2CH2CH2CH3 CH2CH2CH2CH3

benzene ring with CH3C=O o-butyl acetophenone

O O d. C 1 2 C 4 CH3CH CHCH2CH3 CH3CH 3 CHCH2CH3 Answer: 2,4-dimethyl-3-hexanone

CH3 CH3 CH3 CH3 hexane hexanone 2,4-dimethyl (6 C's)

3-chloro Cl Cl 3 2 e. 1 O O Answer: 3-chlorocyclopentanone

cyclopentane cyclopentanone (5 C's)

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.

3-ethyl 3 a. CH3CH2CH2CH2CHCH2CH3 CH3CH2CH2CH2CHCH2CH3 Answer: 3-ethyl-2-heptanone C C O CH3 O 2 CH3 heptane heptanone (7 C's)

O O 1 3,3-dichloro b. Answer: 3,3-dichlorocyclobutanone Cl Cl 3 Cl Cl cyclobutane cyclobutanone (4 C's)

O O

c. C C 9 Answer: 9-methyl-3-decanone 3 CH3CH2 (CH2)5CH(CH3)2 CH3CH2 (CH2)5CHCH3 CH 9-methyl 3 decane decanone (10 C's)

O O 4 C 3 C 5 d. CH3CH2CH CHCH2CH3 CH3CH2CH CHCH2CH3 Answer: 3,5-dimethyl-4-heptanone CH3 CH3 CH3 CH3 heptane heptanone 3,5-dimethyl (7 C's)

CH2CH3 CH2CH3 2-ethyl 2 e. Answer: 2-ethyl-4-methylcyclopentanone O 4 O CH3 CH3 4-methyl cyclopentane cyclopentanone (5 C's)

16.37 Work backwards from the name to draw each structure.

a. 3,3-dimethyl-2-hexanone c. m-ethylacetophenone O 3 CH3 O C CH CH CH C C 2 CH3 6 C chain 3 2 2 benzene ring with a CH3 CH3 CH3C=O 1 3,3-dimethyl m-ethyl CH2CH3

b. methyl propyl ketone O d. 2,4,5-triethylcyclohexanone O

C 2 CH2CH3 CH3 CH2CH2CH3 6 C ring 1 two alkyl groups with a 5 C=O in the middle 4 propyl CH3CH2 methyl CH CH 2,4,5-triethyl 2 3

16.38 Work backwards from the name to draw each structure.

a. dibutyl ketone O c. p-bromoacetophenone O C C CH CH CH CH CH CH CH CH CH 3 2 2 2 2 2 2 3 benzene ring with a 3 two butyl groups CH3C=O with a C=O in the middle butyl Br p-bromo

O b. 1-chloro-3-pentanone O d. 3-hydroxycyclopentanone 1 C 1 ClCH2CH2 3 CH2CH3 5 C ring 5 C chain 3 hydroxy OH chloro

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

2,2-dimethyl 3,3-dimethyl 2,3-dimethyl 2-ethyl 2 2 3 CH CH CH3 3 3 1 1 CH2CH3 CH CH CCHO CH3CCH2CHO CH3CHCHCHO 3 2 3 CH3CH2CHCHO 2 1 CH3 1 CH3 CH3

4 C chain 4 C chain 4 C chain 4 C chain 2,2-dimethylbutanal 3,3-dimethylbutanal 2,3-dimethylbutanal 2-ethylbutanal

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

O O O C C 3 C CH3CH2CH2 2 CH3 CH3CH2 3 CH2CH3 CH3 CH 2 CH3 5 C chain 5 C chain 3-methyl CH3 4 C chain 2-pentanone 3-pentanone 3-methyl-2-butanone

16.41 Draw the structure and correct each name.

O O O O

a. CH3CH2CH2CH2CH b. CH3CH2CH2CCH2CH3 c. CH3CH2CH2CHCCH3 d. CH3CH2CH2CH2CH2CH2CHCH

CH3 CH3 1-pentanone 4-hexanone 2-methyl-1-octanal A ketone cannot be at C1. Re-number to use a 3-propyl-2-butanone An aldehyde is always at C1. It must be an aldehyde. lower number. Find the longest chain. Omit the "1." pentanal 3-hexanone 3-methyl-2-hexanone 2-methyloctanal

16.42 Draw the structure and correct each name.

O O O CH3 O a. CH3CH2CH2CH2CH2CCH2CH3 b. CH3CH2CH2CH2CH2CH2CH c. d. 6-octanone 1-heptanone Re-number to use a A ketone cannot be at C1. CH2CH2CH3 lower number. It must be an aldehyde. 3-octanone heptanal 3-propyl-1-cyclopentanone 5-methylcyclohexanone The ketone is always at Re-number to use a C1 on a ring. lower number. 3-propylcyclopentanone 3-methylcyclohexanone

16.43 Draw benzaldehyde and then the hydrogen bond.

H hydrogen bond O H O C H

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.

H hydrogen bond O H O C CH3 H

c. Hydrogen bonding is possible between ethanal and methanol.

CH3 hydrogen bond O H O C CH3 H

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

a. (CH3)3CCH2CH2CH3 or (CH3)3CCH2CHO b. COCH3 or CH2CH2OH

hydrocarbon aldehyde ketone alcohol higher boiling point higher boiling point

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

a. CH3(CH2)6CHO or CH3(CH2)7OH b. CH3(CH2)6CHO or CH3(CH2)2CHO aldehyde aldehyde aldehyde alcohol higher molar mass higher boiling point higher boiling point

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.

CH3 O OH

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.

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

O CHO C a. b. CH3 CH2CH3 c. CH3CH2CH2CH3

7 C aldehyde 4 C ketone hydrocarbon insoluble soluble insoluble

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

a. b. CH3CH2CH2CHO c. CH3CH2CH2OH

CH3 4 C aldehyde alcohol 7 C ketone soluble soluble insoluble

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).

K Cr O K2Cr2O7 2 2 7 CH CHO CH COOH a. CH (CH ) CHO CH3(CH2)4COOH b. 2 2 3 2 4

K2Cr2O7 K2Cr2O7 c. CH2CH3 No reaction d. CH3(CH2)4CH2OH CH3(CH2)4COOH

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).

Cl Cl OH O K2Cr2O7 K2Cr2O7 a. c. CH3CHCH2CH2CH3 CH3CCH2CH2CH3 CHO COOH O C K2Cr2O7 K2Cr2O7 CH3 b. CH3(CH2)8CHO CH3(CH2)8COOH d. No reaction

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. O Ag2O Ag O CH (CH ) COOH 2 a. CH3(CH2)4CHO 3 2 4 CH CH No reaction NH OH c. 2 3 4 NH4OH

Ag O 2 Ag2O CH CHO CH2COOH d. CH (CH ) CH OH No reaction b. 2 NH OH 3 2 4 2 4 NH4OH

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.

OH Cl Cl Ag O Ag2O 2 a. c. CH3CHCH2CH2CH3 No reaction NH OH NH4OH CHO 4 COOH O C Ag2O Ag2O CH3 b. CH3(CH2)8CHO CH3(CH2)8COOH d. No reaction NH OH NH4OH 4

16.57 Answer each question about erythrulose.

O ketone Tollens reagent c. a, b. HO No reaction 1° OH 1° OH

erythrulose O O O K2Cr2O7 d. HO HO 2° OH OH OH O O

16.58

2° ROH

H OH H OH H OH HO a, b, c. 1° ROH C H d. HO C H Ag2O HO C OH C C C C C C aldehyde NH OH H H 4 H H O H O H O chirality center

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

CH3 CH3 a. CH3CH2CHCH2CO2H CH3CH2CHCH2CHO c. CH3CH2CHCH2CH3 CH3CH2CHCH2CH3

CO2H CHO b. CH3 CO2H CH3 CHO

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

Cl Cl a. CH2CH2CO2H CH2CH2CHO c. CH3(CH2)8CHCO2H CH3(CH2)8CHCHO

Br Br b. CO2H CHO

16.61 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.

H O OH 2 H2 a. CH3CH2 CHO CH3CH2 CH2OH b. Pd CH Pd 3 CH3

16.62 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.

O HO H2 C H2 a. CH CH CH CH(CH ) CH3CH2CHCH2CH(CH3)2 b. CH (CH ) CHO CH (CH ) CH OH 3 2 2 3 2 Pd 3 2 6 3 2 6 2 Pd

16.63 CH3 CH3 CH3 H2 a, b: CH CH CH (CH ) CHO c. CH CH CH (CH ) CHO CH CH CH (CH ) CH OH 3 2 2 4 3 2 2 4 Pd 3 2 2 4 2 chirality center

16.64 O O OH H2 a. b, c. Pd chirality center

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

OH O O C a. CH3CH2CH2CH2CH2OH CH3CH2CH2CH2 H b.

CH 3 CH3

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

OH O O a. CH CH CCH CH CH b. CH3CH2CHCH2CH2CH3 3 2 2 2 3 (CH3)2CHCH2CH2OH (CH3)2CHCH2CH

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.

H H H c. a. CH3CH2 O C O CH2CH3 CH3CH2 O C C O CH3 H H H acetal two ethers H H H OH alcohol b. CH3 C O CH2CH2CH3 d. CH3 C O C C CH3 OH hemiacetal H H H ether

16.70 Locate the two acetals in amygdalin.

acetal CN HOCH2 O O CH HO O O CH2 OH HO OH HO OH

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

hemiacetal ether ether OCH3 OCH2CH3 OCH O a. 3 CH C H b. CH C H d. 3 3 c. HOCH2CHCH2CH3 acetal OCH3 OH alcohol

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

hemiacetal ether OCH acetal 3 OH O a. ether b. CH C OCH CH CH c. O O 3 2 2 3 d. OCH CH CH OCH3 CH3 2 2 3

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.

2 CH3OH 2 CH3OH O CH O CH CH 3 3 H SO 3 H2SO4 2 4 OCH3 c. C a. O CH3 C OCH3 CH3 CH2CH2CH3 CH CH CH OCH3 2 2 3 CH3 CH3

2 CH3OH 2 CH3OH H2SO4 OCH3 b. CH2 O CH2(OCH3)2 H2SO4 d. CH CHO 2 CH2CH OCH3

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.

O CH CH O 2 CH3CH2OH 3 2 C H SO 2 4 CH3CH2CH2 C OCH2CH3 a. CH3CH2CH2 CH2CH3 CH2CH3

O OCH2CH3 2 CH3CH2OH OCH2CH3 b. H2SO4 CH3 CH3

2 CH CH OH 3 2 OCH2CH3 H2SO4 c. (CH3)2CHCH2CH2CHO (CH3)2CHCH2CH2CH

OCH2CH3

OCH2CH3 2 CH3CH2OH CH3CH2CH H2SO4 d. CH CH CHO 3 2 OCH2CH3

16.75 Draw the products of each reaction.

a. b. HO HO O OH O H2SO4 H2SO4 C CH3 C O CH C O CH CH 3 3 3 CH 3 CH3

16.76 Draw the products of each reaction.

a. b. O HO OH HO C C O O H H2SO4 H SO H 2 4 C O H

16.77 Answer each question.

hemiacetal carbon CH3OH O O H2SO4 O a. OH c. OH OCH3

O O b. OH HOCH2CH2CH2CH

16.78 Answer each question.

hemiacetal carbon CH3CH2OH CH3 CH CH3 3 H2SO4 O O O a. CH3 c. CH3 CH3 OH OH OCH2CH3

CH3 CH3 O O b. CH3 HOCH2CCH2CH2CH OH CH3

16.79 Draw the product of cyclization.

O OH

C CH3 HOCH2CH2CH2CH H O C CH3

16.80 Draw the product of cyclization.

O OH C HOCHCH2CH2CH2 H O CH3 D CH3

16.81 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.

H2O OCH2CH2CH3 H SO a. 2 4 O + 2 HOCH2CH2CH3 OCH2CH2CH3

H2O OCH3 O H2SO4 b. H C CH2CH2CH3 H C CH2CH2CH3 + 2 HOCH3 OCH3

16.82 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.

H2O OCH CH O 2 3 H2SO4 a. CH3CH2 C CH2CH3 CH3CH2 C CH2CH3 + 2 HOCH2CH3

OCH2CH3

H2O OCH 3 H2SO4 b. CH3O CH3O O + 2 CH3OH OCH3

16.83 Answer each question about compound A.

p-methyl OCH O * * O O 3 2 CH3OH a, b. CH c. CH * * d. CH CH3 C CH3 3 3 * 3 (acid) CH CH CH OCH 3 * * 3 3 3 p-methylacetophenone seven trigonal planar C's

16.84 H H H OCH3 * * C* O H C O C O 2 CH3OH C OCH3 a. C b. * * C c. C C H H H (acid) H H * * H H H 7 planar carbons

H C O H2 d. C CH2CH2OH H Pd H

16.85 Draw the products of each reaction.

O O H OCH3 a. C 2 d. 2 CH3OH CH2OH C C H Pd H SO H H 2 4 OCH3 O O O OCH CH 2 CH3CH2OH 2 3 K2Cr2O7 b. C COH e. C C H H SO H 2 4 H OCH2CH3

OCH CH O O 2 3 H O O c. Ag2O 2 C COH f. C H C + 2 CH3CH2OH H2SO4 NH4OH H OCH2CH3 H

16.86 Draw the products of each reaction.

O H2 a. CH3O C CH3O CH2OH Pd H

O O K2Cr2O7 b. CH3O C CH3O COH H

O O Ag2O c. CH3O C CH3O COH H NH4OH

O OCH3 2 CH3OH d. CH3O C CH3O C H H2SO4 H OCH3

O OCH2CH3 2 CH3CH2OH e. CH3O C CH3O C H H2SO4 H OCH2CH3

OCH2CH3 O H2O f. CH O C H CH3O C + 2 CH3CH2OH 3 H SO 2 4 H OCH2CH3

16.87 Draw the products of each reaction.

O OH OCH H O 3 2 2 CH3OH a. C CH3 C (CH2)4CH3 d. CH C (CH ) CH CH (CH ) CH C 3 2 4 3 3 2 4 3 Pd CH (CH ) CH H2SO4 H 3 2 4 3 OCH3

OCH CH O 2 CH3CH2OH 2 3 K2Cr2O7 O b. e. CH C (CH ) CH C No reaction C H SO 3 2 4 3 CH3 (CH2)4CH3 CH (CH ) CH 2 4 3 2 4 3 OCH2CH3

O OCH CH O 2 3 H O c. Ag2O 2 C C No reaction f. CH3 C (CH2)4CH3 H SO CH3 (CH2)4CH3 CH3 (CH2)4CH3 NH OH 2 4 4 OCH2CH3 + 2 CH3CH2OH

16.88 Draw the products of each reaction.

O OH H2 a. C (CH3)2CH C (CH2)2CH(CH3)2 (CH3)2CH (CH2)2CH(CH3)2 Pd H O C K2Cr2O7 b. (CH3)2CH (CH2)2CH(CH3)2 No reaction

C Ag2O c. (CH3)2CH (CH2)2CH(CH3)2 No reaction NH4OH

O OCH3 C 2 CH3OH d. (CH3)2CH (CH2)2CH(CH3)2 (CH3)2CH C (CH2)2CH(CH3)2 H2SO4 OCH3 O OCH CH C 2 CH3CH2OH 2 3 e. (CH3)2CH (CH2)2CH(CH3)2 (CH3)2CH C (CH2)2CH(CH3)2 H2SO4 OCH2CH3

OCH2CH3 O H2O f. (CH3)2CH C (CH2)2CH(CH3)2 (CH3)2CH C (CH2)2CH(CH3)2 + 2 CH3CH2OH H2SO4 OCH2CH3

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.

CH2 CHCH2CH2CHO or H2 CH3CH CHCH2CHO CH3CH2CH2CH2CH2OH or Pd CH3CH2CH CHCHO

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

H H SO H O 2 OH 2 4 2 Pd Pd A B C

16.91 Draw the products of each reaction.

O CH3 CH3 OH a. CH3 CH3 H2 Pd

HO HO

O O CH3 CH3 b. CH3 CH3 K2Cr2O7

HO O O OCH CH3 CH3 3 OCH3

CH3 CH3 2 CH OH c. 3 H2SO4 HO HO O OCH CH CH3 CH3 2 3 OCH2CH3

CH3 CH3 d. 2 CH3CH2OH

H2SO4

HO HO

16.92 Answer each question.

O OH excess H2 a. HO CCH2NH(CH2)6O(CH2)4 HO CCH2NH(CH2)6O(CH2)4 Pd H O C X HO CH salmeterol H H chirality center OH b. HO CCH2NH(CH2)6O(CH2)4 H HO CH salmeterol H H H

C C c. CH2NH(CH2)6O(CH2)4 (CH2)4O(CH2)6NHCH2 OH HO HO OH enantiomers CH OH CH2OH 2

16.93 Draw the product of oxidation.

O NAD+ CH CH CHO CH CH C OH enzyme

16.94 Answer each question.

ether O O H2O O O a. b. + OH H2SO4 OH O alkene O O O benzene

16.95 Label each hemiacetal or alcohol.

O HOCH2 OH OH of a hemiacetal

alcohol HO

16.96 Label the actetal carbons in paraldehyde.

CH3 O CH3

acetal O O acetal

acetal CH3

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.

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.

alcohol HO CH2CH3 carboxylic acid CH3 CH3 CH O O O 3 CH3 O CH HO2CCHCH CH ether 3 CH O CH hemiacetal 3 3 acetal O

alcohol CH ether HOCH2 OH 3

16.100 Determine the structure of chloral hydrate.

Cl O Cl OH H2O Cl C C H Cl C C H Cl Cl OH