Chapter 16–1 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. O O C C a. CH3CH2 H aldehyde c. (CH3)3C CH3 ketone O O C C b. CH3CH2 CH3 ketone d. (CH3CH2)2CH H aldehyde 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) © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Chapter 16–2 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 © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 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. 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 © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 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. 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 CH3 CH3 CH3 O K2Cr2O7 c. CH C=CHCH CH CHCH CHO CH3C CHCH2CH2CHCH2C OH 3 2 2 2 © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 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. 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 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. © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 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. 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.