Download Chapter 16 Aldehydes and Ketones

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
a.
CH3CH2
C
O
H
aldehyde
c.
(CH3)3C
C
b.
16.2
CH3CH2
CH3
d. (CH3CH2)2CH
ketone
O
O
C
C
CH3
ketone
CH3CH2CH2
H
aldehyde
O
aldehyde
H
(CH3)2CH
C
H
aldehyde
Trigonal planar carbons are carbons bonded to three other groups. Each trigonal planar carbon is
labeled with an arrow.
CH3 CH3 H
C
C
H
CH3
CH3
H
C
C
C
H
C
CH3
C
H
16.4
C
Draw the constitutional isomers of molecular formula C4H8O and then label each compound using
the definitions from Answer 16.1.
CH3CH2
16.3
ketone
O
O
C
CH3
H
C
C
H
O
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
CH3CHCH2CH2CH2CH
hexane
(6 C's)
CH3
O
CH3CHCH2CH2CH2CH
Answer: 5-methylhexanal
hexanal
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Chapter 16–2
b.
(CH3)3CC(CH3)2CH2CHO
3,3,4,4-tetramethyl
CH3 CH3 H
CH3 C
C
C
O
CH3 CH3 H
C
H
CH3 C
CH3 CH3 H
pentane
(5 C's)
CH3 CH3 H
C
2
H
Answer:
3,3,4,4-tetramethylpentanal
1
CHO
CH3CHCHCH2CH2CHCH3
CH3
2,5,6-trimethyl
re-draw
CH3
CH3
CH3
CH3CH2CHCHCH2CH2CHCH
CH3
octane
(8 C's)
16.5
4
O
C
pentanal
CH2CH3
c.
5
C
CH3
Answer:
2,5,6-trimethyloctanal
CH3CH2CHCHCH2CH2CHCH
O
6 5
CH3
2 1
O
octanal
Work backwards from the name to draw each structure.
!a. 2-chloropropanal
3 C chain with
a CHO at C1
!c. 3,6-diethylnonanal
2-chloro
Cl
9 C chain with
a CHO at C1
CH3CHCHO
3,6-diethyl
CH2CH3
CH3CH2CH2CHCH2CH2CHCH2CHO
CH2CH3
b. 3,4,5-triethylheptanal
7 C chain with
a CHO at C1
16.6
d. o-ethylbenzaldehyde
CH3CH2
4
CH2CH3
CH3CH2CHCHCHCH2CHO
5
3
CHO
benzene ring
with a CHO
CH2CH3
CH2CH3
o-ethyl
To name an aldehyde using the IUPAC system, use the steps in Example 16.1.
8-methyl
1
a.
nonane
(9 C's)
2
8
8
b.
1
2
decane
(10 C's)
nonanal
decanal
8-methylnonanal
8-methyldecanal
8-methyl
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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
O4
a. CH3CH2CCHCH2CH2CH3
CH3CH2CCHCH2CH2CH3
1
CH3
heptane
(7 C's)
2
heptanone
4-methyl
CH3
b.
CH3
2
O
cyclopentane
(5 C's)
c.
CCH2CH2CH2CH3
CH3 O
CH3C
1
CH3
16.8
Answer: 2-methylcyclopentanone
1
2
heptane
(7 C's)
2-methyl
O
cyclopentanone
CH3 O
CH3C
Answer: 4-methyl-3-heptanone
3 CH
3
heptanone
CCH2CH2CH2CH3
Answer: 2,2-dimethyl-3-heptanone
CH3 3
2,2-dimethyl
Work backwards from the name to draw each structure.
acetophenone
O
a. butyl ethyl ketone
butyl
b. 2-methyl-3-pentanone
5 C chain with
C=O at C3
p-ethyl
ethyl
O
CH3CH2
d. 2-propylcyclobutanone
O
C CH3
O
4 C ring with
C=O at C1
CH3CH2CCHCH3
CH3
16.9
c. p-ethylacetophenone
CH3CH2CH2CH2CCH2CH3
2-methyl
CH2CH2CH3
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.
O
or
ketone
higher boiling point
b.
(CH3CH2)2CO
CH3
c.
hydrocarbon
or
ketone
higher boiling point
(CH3CH2)2C=CH2
hydrocarbon
O
ketone
d.
CH3(CH2)6CH3
hydrocarbon
or
OH
alcohol
higher boiling point
or
CH3(CH2)5CHO
aldehyde
higher boiling point
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2-propyl
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
CH3 C
O
CH3
C
CH3
CH3
CH3
small ketone
acetone
large molecule with two ketones
progesterone
O
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
C
OCH3
O
O
C
C
CH2
CH3O
OH
OH
C
C(CH3)3
oxybenzone
one hydroxyl group
one ketone
one ether
O
avobenzone
two ketones
one ether
OCH3
dioxybenzone
two hydroxyl groups
one ketone
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.
a.
b.
CH3CH2CHO
(CH3CH2)2C=O
CH3
O
K2Cr2O7
CH3CH2
K2Cr2O7
CH3
c. CH3C=CHCH2CH2CHCH2CHO
C
OH
No reaction
K2Cr2O7
CH3
CH3
O
CH3C CHCH2CH2CHCH2C OH
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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.
CH3(CH2)6CHO
NH4OH
Ag2O
b.
O
O
Ag2O
CH3(CH2)6C OH
No reaction
CHO
d.
NH4OH
O
Ag2O
c.
C
NH4OH
OH
Ag2O
NH4OH
OH
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.
O
a.
CH3CH2CH2
C
H2
H
Pd
O
OH
CH3CH2CH2CH2
c.
CH3
C
H2
CH2CH3
Pd
OH
CH3CHCH2CH3
OH
O
H2
b.
d.
Pd
CH3
CHO
H2
Pd
CH2OH
CH3
16.16 Work backwards to determine what carbonyl compound is needed to prepare alcohol A.
(CH3)2CHCH2
O
H2
CCH3
Pd
OH
(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.
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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.
O
a.
CH3
C
+ CH3OH
H
CH3OH
H2SO4
OCH3
H2SO4
CH3COH
OCH3
CH3COCH3
H
H
hemiacetal
acetal
b.
+
(CH3CH2)2C=O
CH3OH
H2SO4
OCH3
H2SO4
(CH3CH2)2COH
CH3OH
OCH3
(CH3CH2)2COCH3
hemiacetal
O
C
c.
acetal
OCH2CH3
H2SO4
H
+
CHOH
CH3CH2OH
H2SO4
OCH2CH3
CHOCH2CH3
CH3CH2OH
acetal
hemiacetal
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.
OH
OCH3
OCH3
a.
b.
c.
OCH3
CH3CH3CH2CH2 C OCH3
O
H
ether
acetal
O
d.
hemiacetal
CH3
CH3
acetal
16.20 Label the acetal or hemiacetal in each compound using the definitions in Example 16.7.
acetal
CH2OH
hemiacetal
HOCH2
HOCH2
O
a.
HO
OH
HO
b.
NH2
O
O
HO
OH
OH
16.21 Draw the products of each reaction using the steps in Example 16.6.
O
a.
OH
+ CH3CH2OH
H2SO4
O
OCH2CH3
Replace OH by OCH2CH3
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Chapter 16–7
O
OH
b.
+
OH
O
H2SO4
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
H2SO4
OCH3
a.
CH3
C
O
CH3
OCH3
C
OCH3
CH2CH2CH3
c.
+
CH2CH2CH3
H2O
H2SO4
O
C
C H
H
OCH3
2 CH3OH
+
2 CH3OH
CH3CH2O OCH2CH3
H2O
H2SO4
O
+ 2 CH3CH2OH
b.
Solutions to End-of-Chapter Problems
16.23
Draw a structure to fit each description.
CH2CH3
a. CH3CH2CH2CHCH2CHO
O
O
b. CH3CH2CCHCH3
c.
O
16.24
ketone
C6H12O
C
H
CH3
aldehyde
C8H16O
d.
ketone
C5H8O
aldehyde
C6H10O
Draw the structure of a constitutional isomer to fit each description.
O
a. CH3CH2CH2CH2CH2CH2CHO
b. CH3CH2CCH2CH2CH2CH3
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.
c. CH3CH2CH2CH2CH CHCH2OH
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.
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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.
2-methyl
1
a.
b.
2
3
3-ethyl
pentane
(5 C's)
1
pentanal
cyclohexane
(6 C ring)
2-methylpentanal
cyclohexanone
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
F
O
O
a.
H
Cl
b.
Br
H
2-chloro-2-fluoro
H
benzene
benzaldehyde
m-fluorobenzaldehyde
16.31
cylcopentanone
cyclopentane
2-chloro-2-fluorocyclopentanone
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.
a.
CH3CH2CH2CHCH2CHO
CH3CH2CH2CHCH2CHO
CH3
CH3
hexane
(6 C's)
hexanal
CH3
b.
CH3
CH3CH2CHCH2CHCH2CHO
CH3
heptane
(7 C's)
heptanal
3
3-methyl
1
CH3CH2CHCH2CHCH2CHO
5
CH3
Answer: 3-methylhexanal
Answer: 3,5-dimethylheptanal
3,5-dimethyl
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Chapter 16–9
c.
O C H
H C H
O C H
H C H
CH3CH2CH2 C CH2CH2CH3
CH3CH2CH2 C CH2CH2CH3
H
H
hexane
(6 C's)
hexanal
CH2CH3
CH3
CH2CH3
e. Cl
6
CH2CH3
2
CH3
1
Answer:
6,6-diethyl-2,2-dimethyloctanal
octanal
CHO
Cl
benzaldehyde
16.32
2,2-dimethyl
CH3
CH3CH2CCH2CH2CH2 C CHO
CH3
octane
(8 C's)
3
3-propyl
d. CH3CH2CCH2CH2CH2 C CHO
CH2CH3
Answer: 3-propylhexanal
Answer: p-chlorobenzaldehyde
CHO
p-chloro
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
butane
(4 C's)
butanal
Answer: 3,3-dimethylbutanal
CH3CCH2CHO
CH3 2 methyls on C3
CH2CH3
b.
(CH3CH2)2CHCH2CH2CHO
hexane
(6 C's)
hexanal
CH3CH2CHCH2CH2CHO
4-ethyl
CH3
c.
CH3CH2CH2CH2CHCHCH3
CH2CHO
octane
(8 C's)
octanal
d. (CH3CH2CH2CH2)2CHCH2CHO
heptane
(7 C's)
heptanal
Answer: 4-ethylhexanal
CH3
CH3CH2CH2CH2CHCHCH3
3,4-dimethyl
CH3CH2CH2CH2CHCH2CHO
3-butyl
Answer: 3,4-dimethyloctanal
CH2CHO
Answer: 3-butylheptanal
CH2CH2CH2CH3
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Chapter 16–10
CH3CH2
m-ethyl
CH3CH2
Answer: m-ethylbenzaldehyde
f.
16.33
CHO
CHO
benzaldehyde
benzaldehyde
Work backwards to draw the structure.
3,3-dichloro
Cl
a. 3,3-dichloropentanal
c. o-bromobenzaldehyde
CHO
CH3CH2CCH2CHO
Cl
5 C chain
b. 3,4-dimethylhexanal
benzene ring with CHO
CH3
d. 4-hydroxyheptanal
3,4-dimethyl
16.34
4-hydroxy
OH
7 C chain
CH3
Work backwards to draw the structure.
2-bromo
a. 2-bromooctanal
Br
c. 3,4-dimethoxybenzaldehyde
CH3CH2CH2CH2CH2CH2CHCHO
benzene ring with CHO
b. 2-propylheptanal
2-propyl
H3CO
4
CHO
3
8 C chain
d. 3,4-dihydroxynonanal
CH2CH2CH3
3,4-dimethoxy
H3CO
OH OH
CH3CH2CH2CH2CH2CH CHCH2CHO
CH3CH2CH2CH2CH2CHCHO
9 C chain
7 C chain
16.35
o-bromo
CH3CH2CH2CHCH2CH2CHO
CH3CH2CHCHCH2CHO
6 C chain
Br
3,4-hydroxy
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
a.
CH3CHCH2
C
O
C 1
CH3CHCH2 2 CH3
CH3
4
CH3
CH3
pentane
(5 C's)
pentanone
4-methyl
O
b.
CH3
O
CH3
CH3
6
cyclohexane
(6 C's)
Answer: 4-methyl-2-pentanone
cyclohexanone
1
CH3
2
Answer: 2,6-dimethylcyclohexanone
2,6-dimethyl
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Chapter 16–11
O
C
c.
O
C
CH3
CH2CH2CH2CH3
o-butyl
O
CH3CH
C
CH3
Answer: o-butylacetophenone
CH2CH2CH2CH3
benzene ring with CH3C=O
acetophenone
d.
CH3
O
1
C 4
CH3CH 3 CHCH2CH3
CH3
CH3
CHCH2CH3
CH3
hexane
(6 C's)
2
hexanone
Answer: 2,4-dimethyl-3-hexanone
2,4-dimethyl
3-chloro
Cl
Cl 3
e.
O
cyclopentane
(5 C's)
16.36
2
1
O
Answer: 3-chlorocyclopentanone
cyclopentanone
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
O
C
CH3CH2CH2CH2CHCH2CH3
Answer: 3-ethyl-2-heptanone
C
O 2 CH3
CH3
heptanone
heptane
(7 C's)
O
O 1
3,3-dichloro
b.
Cl
3
Cl
Answer: 3,3-dichlorocyclobutanone
Cl
Cl
cyclobutane
cyclobutanone
(4 C's)
O
c.
O
C
CH3CH2
(CH2)5CH(CH3)2
C
9
CH3CH2 3 (CH2)5CHCH3
9-methyl
decane
Answer: 9-methyl-3-decanone
CH3
decanone
(10 C's)
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Chapter 16–12
O
O
C
4C
d. CH3CH2CH
3
CHCH2CH3
CH3
CH3CH2CH
CH3
heptane
CH3
5
CHCH2CH3
CH3
Answer:
3,5-dimethyl-4-heptanone
heptanone
3,5-dimethyl
(7 C's)
CH2CH3
e.
CH2CH3
2
O
CH3
cyclopentane
Answer: 2-ethyl-4-methylcyclopentanone
O
4
CH3
2-ethyl
4-methyl
cyclopentanone
(5 C's)
16.37
Work backwards from the name to draw each structure.
a. 3,3-dimethyl-2-hexanone
3
CH3
CH3CH2CH2C
6 C chain
C
CH3
C
2
CH3
1
benzene ring with a
CH3C=O
m-ethyl
3,3-dimethyl
O
b. methyl propyl ketone
two alkyl groups with a
C=O in the middle
CH3
C
CH2CH2CH3
O
1
6 C ring
5
2 CH2CH3
4
CH3CH2
propyl
CH3
CH2CH3
d. 2,4,5-triethylcyclohexanone
methyl
16.38
O
c. m-ethylacetophenone
O
CH2CH3
2,4,5-triethyl
Work backwards from the name to draw each structure.
O
a. dibutyl ketone
CH3CH2CH2CH2
two butyl groups
with a C=O in the middle
C
O
c. p-bromoacetophenone
CH2CH2CH2CH3
butyl
C
benzene ring with a
CH3C=O
CH3
Br
p-bromo
b. 1-chloro-3-pentanone
5 C chain
O
1
C
ClCH2CH2 3 CH2CH3
chloro
d. 3-hydroxycyclopentanone
O
1
3
5 C ring
hydroxy
OH
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–13
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
CH3
CH3
1
16.41
3
CH3CH2CHCHO
2
CH3
4 C chain
2,3-dimethylbutanal
1
4 C chain
2-ethylbutanal
Draw the three ketones and then name them using the steps in Example 16.1.
O
O
C
C
O
CH3CH2CH2 2 CH3
CH3CH2 3 CH2CH3
5 C chain
2-pentanone
5 C chain
3-pentanone
CH3
3-methyl
3
C
CH 2 CH3
CH3
4 C chain
3-methyl-2-butanone
Draw the structure and correct each name.
O
O
O
b. CH3CH2CH2CCH2CH3
c. CH3CH2CH2CHCCH3
O
d. CH3CH2CH2CH2CH2CH2CHCH
CH3
1-pentanone
A ketone cannot be at C1.
It must be an aldehyde.
pentanal
4-hexanone
Re-number to use a
lower number.
3-hexanone
CH3
2-methyl-1-octanal
An aldehyde is always at C1.
Omit the "1."
2-methyloctanal
3-propyl-2-butanone
Find the longest chain.
3-methyl-2-hexanone
Draw the structure and correct each name.
a. CH3CH2CH2CH2CH2CCH2CH3
O
O
O
b. CH3CH2CH2CH2CH2CH2CH
6-octanone
Re-number to use a
lower number.
3-octanone
16.43
CH2CH3
1
CH3CHCHCHO
4 C chain
3,3-dimethylbutanal
a. CH3CH2CH2CH2CH
16.42
CH3
1
CH3
4 C chain
2,2-dimethylbutanal
16.40
CH3
CH3CCH2CHO
CH3CH2CCHO
1-heptanone
A ketone cannot be at C1.
It must be an aldehyde.
heptanal
d.
c.
CH3
O
CH2CH2CH3
3-propyl-1-cyclopentanone
The ketone is always at
C1 on a ring.
3-propylcyclopentanone
5-methylcyclohexanone
Re-number to use a
lower number.
3-methylcyclohexanone
Draw benzaldehyde and then the hydrogen bond.
H
hydrogen bond
H
O
O
C
H
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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.
H
hydrogen bond
H
O
O
CH3
C
H
c. Hydrogen bonding is possible between ethanal and methanol.
CH3
hydrogen bond
H
O
O
CH3
16.45
(CH3)3CCH2CH2CH3
hydrocarbon
or
b.
(CH3)3CCH2CHO
COCH3
ketone
aldehyde
higher boiling point
or
CH2CH2OH
alcohol
higher boiling point
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
aldehyde
16.47
H
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.
16.46
C
or
CH3(CH2)7OH
alcohol
higher boiling point
b.
CH3(CH2)6CHO
aldehyde
higher molar mass
higher boiling point
or
CH3(CH2)2CHO
aldehyde
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.
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Chapter 16–15
16.49
Low molecular weight aldehydes and ketones (less than six carbons) are water soluble.
O
CHO
a.
b.
CH3
7 C aldehyde
insoluble
16.50
C
c.
CH2CH3
4 C ketone
soluble
CH3CH2CH2CH3
hydrocarbon
insoluble
Low molecular weight aldehydes and ketones (less than six carbons) are water soluble.
O
a.
b. CH3CH2CH2CHO
CH3
c. CH3CH2CH2OH
4 C aldehyde
soluble
7 C ketone
insoluble
alcohol
soluble
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).
a. CH3(CH2)4CHO
K2Cr2O7
CH3(CH2)4COOH
K2Cr2O7
CH2CHO
b.
CH2COOH
O
c.
16.54
CH2CH3
d.
No reaction
CH3(CH2)4CH2OH
K2Cr2O7
CH3(CH2)4COOH
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
a.
K2Cr2O7
CH3(CH2)8CHO
OH
Cl
c. CH3CHCH2CH2CH3
COOH
CHO
b.
K2Cr2O7
K2Cr2O7
O
CH3CCH2CH2CH3
O
C
CH3(CH2)8COOH
K2Cr2O7
d.
CH3
K2Cr2O7
No reaction
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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.
O
Ag2O
a. CH3(CH2)4CHO
NH4OH
Ag2O
CH3(CH2)4COOH
c.
CH2CH3
Ag2O
CH2CHO
b.
16.56
Ag2O
CH3(CH2)4CH2OH
No reaction
NH4OH
CH3(CH2)8CHO
Ag2O
c. CH3CHCH2CH2CH3
NH4OH
CHO
OH
Cl
Ag2O
a.
16.57
NH4OH
d.
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.
Cl
b.
CH2COOH
No reaction
NH4OH
COOH
CH3(CH2)8COOH
NH4OH
No reaction
O
C
Ag2O
NH4OH
d.
Ag2O
CH3
NH4OH
No reaction
Answer each question about erythrulose.
O
ketone
c.
a, b.
Tollens reagent
HO
OH
1°
erythrulose
O
K2Cr2O7
d. HO
2°
No reaction
OH
1°
OH
O
O
HO
OH
OH
O
O
16.58
2° ROH
H
a, b, c. 1° ROH
HO
C
H
OH
C
H
H
d. HO
H
C
aldehyde
O
OH
C
H
C
H
H
H
C
O
Ag2O
NH4OH
HO
OH
C
C
H
OH
C
H
O
chirality center
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Chapter 16–17
16.59
a.
Work backwards to determine what aldehyde can be used to prepare each carboxylic acid.
CH3
CH3
CH3CH2CHCH2CO2H
CH3CH2CHCH2CHO
c. CH3CH2CHCH2CH3
CH3CH2CHCH2CH3
CO2H
b. CH3
16.60
CO2H
CH3
CHO
CHO
Work backwards to determine what aldehyde can be used to prepare each carboxylic acid.
Cl
Cl
a.
CH2CH2CO2H
CH2CH2CHO
Br
CO2H
CHO
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
H2
a. CH3CH2
16.62
CHO
Pd
CH3CH2
b.
CH2OH
OH
H2
Pd
CH3
CH3
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
a.
CH3(CH2)8CHCHO
Br
b.
16.61
c. CH3(CH2)8CHCO2H
CH3CH2
C
HO
H2
CH2CH(CH3)2
Pd
CH3CH2CHCH2CH(CH3)2
b.
CH3(CH2)6CHO
H2
CH3(CH2)6CH2OH
Pd
16.63
CH3
a, b:
CH3
CH3CH2 CH (CH2)4CHO
c.
CH3CH2 CH (CH2)4CHO
H2
Pd
CH3
CH3CH2 CH (CH2)4CH2OH
chirality center
16.64
O
O
a.
b, c.
H2
OH
Pd
chirality center
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Chapter 16–18
16.65
Work backwards to determine what carbonyl compound is needed to make each alcohol.
O
a. CH3CH2CH2CH2CH2OH
16.66
C
CH3CH2CH2CH2
OH
O
CH3
CH3
b.
H
Work backwards to determine what carbonyl compound is needed to make each alcohol.
O
OH
a. CH3CH2CHCH2CH2CH3
O
CH3CH2CCH2CH2CH3
b. (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
a.
CH3CH2 O
H H
c.
C O CH2CH3
CH3CH2 O C C O CH3
H
H H
acetal
two ethers
H
b.
CH3 C
H
O CH2CH2CH3
d.
OH
H OH
alcohol
CH3 C O C C CH3
H
hemiacetal
H H
ether
16.70
Locate the two acetals in amygdalin.
CN
acetal
HOCH2
O CH
O
HO
O
O
CH2
HO
OH
OH
HO
16.71
OH
Label the functional groups using the definitions from Example 16.7.
hemiacetal
a.
OCH3
CH3 C H
acetal
OCH3
b.
CH3 C H
OH
ether
ether
OCH2CH3
O
OCH3
c.
HOCH2CHCH2CH3
d.
alcohol
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Chapter 16–19
16.72
Label the functional groups using the definitions from Example 16.7.
hemiacetal
OCH3
OH
ether
a.
ether
acetal
c.
b. CH3 C OCH2CH2CH3
OCH3
16.73
O
OCH2CH2CH3
CH3
d.
O
O
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.
CH3
a.
O
2 CH3OH
H2SO4
CH3
CH3
b.
CH3
CH2 O
2 CH3OH
H2SO4
2 CH3OH
H2SO4
O
OCH3
c.
CH3
OCH3
CH2(OCH3)2
C
CH2CH2CH3
d.
CH2CHO
CH3O
CH3 C OCH3
CH2CH2CH3
2 CH3OH
H2SO4
OCH3
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
a. CH3CH2CH2
O
b.
C
CH2CH3
2 CH3CH2OH
H2SO4
CH3CH2O
CH3CH2CH2 C OCH2CH3
CH2CH3
OCH2CH3
2 CH3CH2OH
H2SO4
OCH2CH3
CH3
CH3
c. (CH3)2CHCH2CH2CHO
2 CH3CH2OH
H2SO4
OCH2CH3
(CH3)2CHCH2CH2CH
OCH2CH3
d. CH3CH2CHO
2 CH3CH2OH
H2SO4
OCH2CH3
CH3CH2CH
OCH2CH3
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–20
16.75
Draw the products of each reaction.
b.
a.
HO
O
CH3
16.76
HO
OH
H2SO4
C
H2SO4
CH3 C O
CH3
O
CH3 C O
CH3
CH3
Draw the products of each reaction.
a.
b.
O
C
OH
HO
H
HO
C O
H2SO4
O
H2SO4
H
C O
H
16.77
Answer each question.
CH3OH
H2SO4
hemiacetal carbon
O
O
OH
a.
b.
CH3
a.
CH3
HOCH2CH2CH2CH
OH
Answer each question.
hemiacetal carbon
CH3
O
c.
OH
CH3
b.
CH3
CH3
CH3CH2OH
H2SO4
O
CH3
O
CH3
CH3
OH
O
OCH2CH3
O
HOCH2CCH2CH2CH
CH3
OH
16.79
OCH3
O
O
16.78
O
OH
c.
Draw the product of cyclization.
O
HOCH2CH2CH2CH
C
C
OH
H
CH3
O
CH3
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–21
16.80
Draw the product of cyclization.
O
HOCHCH2CH2CH2
CH3
16.81
C
OH
H
O
D
CH3
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
H2SO4
OCH2CH2CH3
a.
O
OCH2CH2CH3
OCH3
b.
H C CH2CH2CH3
H2O
H2SO4
+ 2 HOCH2CH2CH3
O
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.
OCH2CH3
H2O
H2SO4
O
CH3CH2 C CH2CH3 + 2 HOCH2CH3
a. CH3CH2 C CH2CH3
OCH2CH3
H2O
H2SO4
OCH3
b. CH3O
16.83
CH3O
OCH3
O
+ 2 CH3OH
Answer each question about compound A.
p-methyl
O
c. CH3
a, b. CH3
CH3
p-methylacetophenone
*
*
*
O
*
*
O
d. CH3
CH3
*
*
seven trigonal
planar C's
CH3
2 CH3OH
(acid)
OCH3
CH3
C
CH3
OCH3
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–22
16.84
H
H
a.
C
H
H
*
*
C O
* C
b. *
H
C* O
c.
H
C
* * H
7 planar carbons
H
C O
2 CH3OH
H
(acid)
OCH3
H C OCH
3
C
H
H
H
d.
C
H
16.85
C O
H2
H
Pd
CH2CH2OH
Draw the products of each reaction.
O
a.
C
H
O
b.
C
O
H2
CH2OH
Pd
d.
C
H
COH
O
C
H
16.86
H2SO4
O 2 CH CH OH
3
2
O
K2Cr2O7
e.
C
H
c.
2 CH3OH
H
OCH2CH3
O
Ag2O
COH
f.
C H
NH4OH
OCH2CH3
H2SO4
H2O
H2SO4
OCH3
C H
OCH3
OCH2CH3
C H
OCH2CH3
O
+ 2 CH3CH2OH
C
H
Draw the products of each reaction.
O
a. CH3O
H2
C
Pd
H
O
b. CH3O
C
K2Cr2O7
CH3O
CH2OH
O
CH3O
COH
H
O
c. CH3O
C
H
O
d. CH3O
C
H
Ag2O
NH4OH
2 CH3OH
H2SO4
O
CH3O
OCH3
CH3O
C
H
H2SO4
OCH2CH3
f.
CH3O
C H
OCH2CH3
H2O
H2SO4
C H
OCH3
OCH2CH3
O 2 CH CH OH
3
2
e. CH3O
COH
CH3O
C H
OCH2CH3
O
CH3O
C
+ 2 CH3CH2OH
H
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–23
16.87
Draw the products of each reaction.
O
a.
CH3
C
(CH2)4CH3
O
b.
CH3
C
CH3
16.88
C
CH3 C
Pd
(CH2)4CH3
(CH2)4CH3
d.
C
CH3
(CH2)4CH3
O
e.
No reaction
C
CH3
Ag2O
OCH3
O
(CH2)4CH3
H
K2Cr2O7
O
c.
OH
H2
2 CH3OH
CH3 C
H2SO4
(CH2)4CH3
2 CH3CH2OH
OCH2CH3
OCH3
(CH2)4CH3
H2SO4
OCH2CH3
f. CH3 C (CH2)4CH3
No reaction
NH4OH
OCH2CH3
CH3 C (CH2)4CH3
OCH2CH3
O
H2O
H2SO4
CH3
C
(CH2)4CH3
+ 2 CH3CH2OH
Draw the products of each reaction.
O
a.
(CH3)2CH
C
H2
(CH2)2CH(CH3)2
Pd
OH
(CH3)2CH C
(CH2)2CH(CH3)2
H
O
b. (CH3)2CH
C
K2Cr2O7
(CH2)2CH(CH3)2
No reaction
O
c. (CH3)2CH
C
Ag2O
(CH2)2CH(CH3)2
No reaction
NH4OH
O
d. (CH3)2CH
C
OCH3
2 CH3OH
H2SO4
(CH2)2CH(CH3)2
(CH3)2CH C
OCH3
O
e. (CH3)2CH
C
(CH3)2CH C
(CH2)2CH(CH3)2
OCH2CH3
16.89
(CH3)2CH C
H2SO4
OCH2CH3
f.
OCH2CH3
2 CH3CH2OH
(CH2)2CH(CH3)2
H2O
H2SO4
(CH2)2CH(CH3)2
(CH2)2CH(CH3)2
OCH2CH3
O
(CH3)2CH C
(CH2)2CH(CH3)2 + 2 CH3CH2OH
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
CH3CH CHCH2CHO
or
H2
CH3CH2CH2CH2CH2OH
Pd
CH3CH2CH CHCHO
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–24
16.90
Work backwards to determine the identity of A–C.
H2
O
H2
H2SO4
OH
Pd
Pd
A
16.91
B
C
Draw the products of each reaction.
CH3 O
a.
CH3
CH3 OH
CH3
H2
Pd
HO
HO
CH3 O
b.
CH3
CH3 O
CH3
K2Cr2O7
HO
O
CH3 O
CH3
CH3 OCH3
OCH3
CH3
2 CH3OH
H2SO4
c.
HO
HO
CH3 O
d.
CH3
CH3 OCH2CH3
OCH2CH3
2 CH3CH2OH
CH3
H2SO4
HO
HO
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 16–25
16.92
Answer each question.
OH
O
a.
HO
excess H2
CCH2NH(CH2)6O(CH2)4
HO
CCH2NH(CH2)6O(CH2)4
Pd
X
O C
H
HO CH
H
H
salmeterol
chirality center
OH
b. HO
CCH2NH(CH2)6O(CH2)4
H
HO CH
H
salmeterol
H
C
c.
H
CH2NH(CH2)6O(CH2)4
OH
(CH2)4O(CH2)6NHCH2
HO
C
OH
HO
enantiomers
CH2OH
16.93
Draw the product of oxidation.
CH CH C OH
enzyme
Answer each question.
ether
a.
O
NAD+
CH CH CHO
16.94
CH2OH
O
b.
O
alkene
H2O
O
H2SO4
O
O
+
O
OH
O
OH
O
benzene
16.95
Label each hemiacetal or alcohol.
O
HOCH2
alcohol
16.96
OH of a hemiacetal
HO
Label the actetal carbons in paraldehyde.
CH3
acetal
acetal
16.97
OH
O
O
CH3
O
acetal
CH3
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.
© 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 16–26
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-cisretinal.
16.99
Identify the alcohol, acetal, hemiacetal, ether, and carboxylic acid functional groups.
HO
alcohol
carboxylic acid
CH2CH3
CH3
CH3
O
O
CH3
HO2CCHCH CH
CH3O
CH3
ether
O
O
CH3
ether
acetal
alcohol
CH3
hemiacetal
O
HOCH2
OH CH3
16.100 Determine the structure of chloral hydrate.
Cl O
Cl
C C H
Cl
H2O
Cl OH
Cl
C C H
Cl OH
© 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.