Chapter 11–1
Chapter 11 Introduction to Organic Molecules and Functional Groups Solutions to In-Chapter Problems 11.1 Organic compounds contain the element carbon.
a. C6H12—organic c. KI—inorganic e. CH4O—organic b. H2O—inorganic d. MgSO4—inorganic f. NaOH—inorganic
11.2 Draw in all the H’s and lone pairs as in Example 11.1. Each C and heteroatom must be
surrounded by eight electrons.
a. C C C C C C
O
CC
CO
C C C
OC C
O
N Cb.
c.
d.
e.H
HH
H H
H
HH
H
H
HH
H H
HH
H
HH
H
H
HH
HH
HH
H
HH
H H
H
11.3 To determine the shape around each atom, draw in all the lone pairs on any heteroatom and then
count groups as in Example 11.2.
C CH
H
Br
H
H C C C
H
H
O H
H C
O
O H
CCC
C CC
HH
H
H H
C
H
H
C N
a.
b.
c.
d.
(1) (2)
(2): four groupstetrahedral
(3): two atoms +two lone pairs
bent
(1) and (2): three groupstrigonal planar
(1): two groupslinear
(2): two atoms +two lone pairs
bent
(1): three groupstrigonal planar
(1): three groupstrigonal planar
(2): two groupslinear
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Introduction to Organic Molecules 11–2
11.4 To determine the bond angles, first use the steps in Example 11.2, and then use the values in Table 11.1.
(1), (2), and (3): four groupstetrahedral109.5°
F CF
FCH
BrCl
C CH
CH
HH
HH
CCC
C CC O H
H
H
H H
H
a. b. c.(1)
(2)
(3)
(2)(3)
(2)(1) (1)
(3)
(1): four groupstetrahedral109.5°
(2) and (3): three groupstrigonal planar
120°
(3): two atoms +two lone pairs
bent109.5°
(1) and (2): three groupstrigonal planar
120°
11.5 When drawing three-dimensional structures: • A solid line is used for bonds in the plane. • A wedge is used for a bond in front of the plane. • A dashed line is used for a bond behind the plane.
a. b.C ClH
HH
HC
C
H
Br H
H
H
11.6 Count the groups around each atom to determine the molecular shape as in Example 11.3.
C C C C C CC C
H
H
C C
H
HCC
H
H
H
H
C
H
H
C
O
H
CH
HO
H
H
H
C
H
H
CH
HHH
(6): four groupstetrahedral
(1): two atoms +two lone pairs
bent(2): two groups
linear
(3): three groupstrigonal planar
(5): two atoms +two lone pairs
bent
(4): three groupstrigonal planar
11.7 Convert each compound to a condensed formula as in Example 11.4.
H C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
H
a. = CH3CH2CH2CH2CH3 = CH3(CH2)3CH3
b. Br C
H
H
C
H
Br
H
= BrCH2CH2Br = Br(CH2)2Br
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Chapter 11–3
H C
H
H
C
H
H
O C
H
H
C
H
O
H
Hc. = CH3CH2OCH2CH2OH = CH3CH2O(CH2)2OH
H C C
H
H
H
H
C
C
C
C
HHH
HH
H
HH
H
d. = CH3CH2C(CH3)3= CH3CH2CCH3
CH3
CH3
11.8 Work backwards to convert each condensed formula to a complete structure.
CH
HCHH
HCH
HCH
HC CH
H H
HC CH
HC
H
H H
HCH
HHa. CH3(CH2)8CH3 =
CH
HH C
H
HCH
HCH
HC OH
HH
CH
HH C
Cl
ClCl
CH
HH C
H
HCH
HCH
HCH
HCC
CH
HH
HH
H
H
CCC
HC
HH
HH
HH
H
HN HH
b.
c.
d.
e.
CH3(CH2)4OH
CH3CCl3
CH3(CH2)4CH(CH3)2
(CH3)2CHCH2NH2
=
=
=
=
11.9 To convert each skeletal structure to a complete structure, place a C atom at the corner of each
polygon and add H’s to give each carbon four bonds as in Example 11.5.
a. b.CC
CC C
C
Cl Cl
Cl
Cl Cl
Cl c.CCC C
CC
CC CCCC C
CC
O
C
H HH
H
HH
HHHH
H
HH
H
HH
HHH
HH
H HH
H
HH
HH H HH
CH
C
HH
H HH
HH
HH
11.10 Each carbon must have four bonds. To determine the number of H’s bonded to each C, count the
number of bonds, and then add H’s to equal four.
(CH3)2CHCH2 CHCH3
CO
OH
C1: 4 bonds, no H'sC2: 3 bonds, 1 H
CH2CH2 N NC
OCH2CH3
C3: 2 bonds, 2 H's
C4: 3 bonds, 1 H C5: 3 bonds, 1 H
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Introduction to Organic Molecules 11–4
11.11 Identify the functional groups in each compound. Concentrate on the heteroatoms and multiple bonds.
CH3CHCH3
OHCH=CH2 H2N CH2CH2CH2CH2 NH2a. b. c.
hydroxyl
aromatic ring
alkene
amine amine
11.12 Identify the functional groups as in Example 11.6 and then draw out the complete structures.
CCC
C CC C
CCC
C CC O CC
CC C
CN
O
a.
b.
c.
d.
e.
aldehyde ketoneamide
carboxylic acid ester
H H H HH
HHH
HH
HO
H
H H H H
HHHHH
HH
HHH
H
H H HH H
CH
H
H
CH
HC O HO
CH
HCH
HH C
HH C C
H
OO C C
H
H HH
HH
H
11.13 Identify the functional groups.
ether
hydroxyl group
aromatic ring
amide
CH2CH2N
OO CH2CHCH2NHCH(CH3)2
OH
amine
11.14 Use the common element colors shown on the inside back cover to convert the ball-and-stick model to a shorthand representation, and then identify the functional groups.
ether
a. b.CH3CH2OCH2CH3 C CH
CH3CH2
H
CH2CHO
alkene
aldehyde
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Chapter 11–5
11.15 Draw the skeletal structure as in Sample Problem 11.9.
aromatic ring
CH3O
CH3OCH3O
etherether
ether alkene
11.16 All of the bonds except C–C and C–H bonds are polar. The symbol δ+ is given to the less
electronegative atom, and the symbol δ– is given to the more electronegative atom.
C OH
HH C
F
FO C
FC
F
HF
Cla. b. c. d. C
H
HCH
CH3
NH
CH3
!"
!"
!"
!+ !"!+
!"
!" !" !"
!"
!"
!+ !+ !+
!"!+ !+!"
Cl CCl
HCl
!+
11.17 With organic compounds that have more than one polar bond, the shape of the molecule
determines the overall polarity. • If the individual bond dipoles cancel in a molecule, the molecule is nonpolar. • If the individual bond dipoles do not cancel, the molecule is polar.
C OCH3
CH3
a. b. c. d.
all nonpolar bondsnonpolar polar C=O
polar
Cl
CCl Cl
Cl
four polar bondsDipoles cancel.nonpolar
net dipole
three polar bondsnet dipolepolar
NCH3CH2CH2 H
H
11.18 Draw the molecule in three dimensions around the O atom to determine why dimethyl ether is
polar.
net dipole OCH3 CH3
The two C–O bonds are polar.The molecule is bent, so there is a net dipole. Therefore, the molecule is polar.
11.19 Use the rule “like dissolves like” to determine if the compounds are soluble in water.
a. Octane is a hydrocarbon, and therefore nonpolar. This means it is insoluble in water. b. Acetone is a small polar molecule due to the C=O. This means it is water soluble. c. Stearic acid is a fatty acid with a polar functional group, but since it has a very long
hydrocarbon chain, it is insoluble in water.
11.20 DDT is a nonpolar, fat-soluble compound that is soluble in tissues. Therefore, once it is ingested by birds, it stays in their tissues for long periods of time, making it detectable.
11.21 Niacin is soluble in water because it is a small molecule that contains many polar bonds (C–N,
C=O, C–O, and O–H).
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Introduction to Organic Molecules 11–6
11.22 Vitamin K1 is not water soluble because it has relatively few polar bonds compared to the number of nonpolar bonds. This makes it soluble in organic solvents.
Solutions to End-of-Chapter Problems 11.23 Organic compounds contain the element carbon.
a. H2SO4—inorganic b. Br2—inorganic c. C5H12—organic 11.24 Organic compounds contain the element carbon.
a. LiBr—inorganic b. HCl—inorganic c. CH5N—organic 11.25 Draw in all the H’s and lone pairs as in Example 11.1. Each C and heteroatom must be
surrounded by eight electrons.
C C C C C
C C C O
C C CCl
C
C C C C N
Ob.
c.
d.
H
HH
H HHa.
HH
H H
HH
H
HH
HH
H
H
H
HH
HH
11.26 Draw in all the H’s and lone pairs as in Example 11.1. Each C and heteroatom must be
surrounded by eight electrons.
C C C O N C C O
O C C C Ob.
c.
d.
OH
O Ha. HH
H H
H H
H H
C C C OH
HO OH
H H
H
H
C
O
H HH
H
OH
H
H
H 11.27 To determine the shape around each atom, draw in all the lone pairs on any heteroatom and then
count groups as in Example 11.2.
b. CH3 O C(CH3)3C CH
H H
C Na.
(1) and (3): four groupstetrahedral
(1): three groupstrigonal planar
(2): two groupslinear
(2): two atoms +two lone pairs
bent
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Chapter 11–7
11.28 To determine the shape around each atom, draw in all the lone pairs on any heteroatom and then count groups as in Example 11.2.
OH CH2 CCH2CH3
HNH
CH2CH2 N CH
CH2CH3
HCH2 O H
two atoms + two lone pairsbent
four groupstetrahedral
three groupstrigonal planar
four groupstetrahedral
11.29 To determine the bond angles, first use the steps in Example 11.2 and then use the values in Table
11.1.
CH3 C C Cl CH2 C
H
Cl CH3 C
H
Cl
H
a. b. c.
(1)
(1)
(1)
(2)
(2) (2)
(1) and (2): two groupslinear180°
(1) and (2): four groupstetrahedral109.5°
(1) and (2): three groupstrigonal planar
120° 11.30 To determine the bond angles, first use the steps in Example 11.2 and then use the values in Table
11.2.
a. CH3(CH2)7 CH
CH
(CH2)7 CO
O H
three groupstrigonal planar
120°
two atoms and two lone pairsbent109.5°
three groupstrigonal planar
120°
three groupstrigonal planar
120°
b. NH
H
C
H
H
(CH2)5NH2
three atoms + one lone pair
trigonal pyramidal109.5°
four groupstetrahedral109.5°
three atoms + one lone pair
trigonal pyramidal109.5°
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Introduction to Organic Molecules 11–8
11.31 To determine the shape around each atom, draw in all the lone pairs on any heteroatom and then count groups as in Example 11.3.
O
O CH2CHNCH3
CH3
H(1): two atoms +two lone pairs
bent
(2): three groupstrigonal planar
(4): three atoms + one lone pairtrigonal pyramidal
(3): four atomstetrahedral
11.32 To determine the shape around each atom, draw in all the lone pairs on any heteroatom and then
count groups as in Example 11.3.
F3C
CH2CHNCH2CH3
CH3
Hthree groupstrigonal planar
three atoms + one lone pairtrigonal pyramidal
four atomstetrahedral
four groupstetrahedral
11.33 Each C in benzene is surrounded by three groups, so each carbon is trigonal planar, giving 120°
bond angles.
11.34 See page 347 for the structure of ethane (CH3CH3) drawn using solid lines, dashes, and wedges.
C
CCl
Cl
ClCl
ClCl
11.35 Convert each compound to a condensed structure as in Example 11.4.
H C C
H
C
H
H H
C
H
H
C
C
C
H
H
H
H H
H
H
H
H
a. CH3(CH2)2C(CH3)3CH3CH2CH2CCH3
CH3
CH3
==
H C C
Br
H
C
H
H
Cl
C
H
H
H
C
H
H
Hb. BrCH2CH2CHCH2CH3Cl
= Br(CH2)2CHCH2CH3Cl
=
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Chapter 11–9
C C
C
C O C
H H
H
H
H
C
H
H
C
H
H
H
H
H
H
H
H H
c. (CH3)2CHCH2O(CH2)2CH3CH3CHCH2OCH2CH2CH3
CH3
= =
H C
H
H
C
H
H
C
H
H
C
O
Hd. CH3(CH2)2CHO= =CH3CH2CH2CHO
11.36 Convert each compound to a condensed structure as in Example 11.4.
C C C C C
H H
H
H
H
C
H
H
O H
H
H
Ha. CH3(CH2)5OHCH3CH2CH2CH2CH2CH2OH= =
H
H
H
C C C C N
H H
H H
C
H
H
H
H
H
Hb. CH3(CH2)3NHCH3CH3CH2CH2CH2NHCH3= =
H
H
H
C C
C
C H
H H
H
H
H
H
H
H H
c. (CH3)3CHCH3CHCH3
CH3
= =
C C
C
C
HH
H
H
H
H H
d. (CH3)2CHCO2CH2CH3CH3CHCOCH2CH3
CH3
= =O
O C
H
C H
H
H H O
11.37 To convert each compound to a skeletal structure, remove all C’s and H’s bonded to C’s. Remove
the lone pairs.
a. b.CCCC C
CO
NH
H
C
H
H
H
C
C CC
C CCl
H
HCl
H
HH
H
HH H
H
H H HH
H
H H
H HH
OCH3
Cl
Cl
CH3H2N==
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Introduction to Organic Molecules 11–10
11.38 To convert each compound to a skeletal structure, remove all C’s and H’s bonded to C’s. Remove the lone pairs.
a. b.CC
CC C
C BrH3C=CH
HH
H H
H O H
Br
OHC C
CCH H
H H
H
H H
CO
O H=
CO
OH
11.39 Work backwards to convert each shorthand structure to a complete structure.
a.
b.
(CH3)2CH(CH2)6CH3
(CH3)3COH
CH3CO2(CH2)3CH3
HO OCH(CH3)2
ClCl
c.
d.
C
CCCC C
CO
ClCl
C
C
H
HH
HHH
HC C C C C CH
H
H
H
H
H
H
H
H
HC
H
H
HH
H
CC
C
OC
HHHH
HH
HH H
H
CCH
HH
OO C CC C
H
H H
H
H
HH
H
H
H
HH
HH
H
HHHO C
HC
CH
H H HH
H
=
=
=
=
11.40 Work backwards to convert each shorthand structure to a complete structure.
a.
b.
(CH3)2CHO(CH2)4CH3
(CH3)3C(CH2)3CBr3
(CH3)2CHCONH2
CHO
c.
d.
CC
C
H
HH
HHH
HO C C C C C
H
H
H
H
H
H
H
HH
H
H
C
C
C
CC
HHH
H
HH
HH H
C
CCH
HH C HN
H
=
=
=
=
H
H
H
HCH
HCBr
BrBr
C
H
H HH
O
(CH3)2CH
C
C C
C
CC
C
C
C CO
HH
HH
H H
HHH
H HH
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Chapter 11–11
11.41 Remember that C can have only four bonds.
(CH3)3CHCH2CH3
CH3(CH2)4CH3OH
CH3
CH3
CH3CH3
CH3
a.
b.
c.
d.
e.
(CH3)2C=CH3
Five bonds to C
Five bonds to C
Five bonds to C
Five bonds to C
Five bonds to C
11.42 Remember that C can have only four bonds.
a.
b.
c.
d.
e.
Three bonds to C
Five bonds to C
Three bonds to O
Five bonds to C
HC CCH2C CH2
CH3C C(CH3)2
HC CCH2C(CH3)2
Cl
Cl
Cl
Cl
OHThree bonds to C
11.43 Convert the shorthand structures to complete structures by drawing in C’s, H’s, and lone pairs.
CCCC C
CO CH
H H
HHH
HCH
NH
H
C O H
O
HO CH2CHCO2H
NH2
=
11.44 Convert the shorthand structure to a complete structure by drawing in C’s, H’s, and lone pairs.
CHCH2NHC(CH3)3HO
HOCH2
OH CCCC C
C CO
C
O
H
H H
H
H
HCH
HNH
C
CHH
H
CH
H
CH
H
H
HH
HOH
=
11.45 Identify the functional groups in each molecule.
C CCH3
C CH
H
HH
H
CH3C
(CH2)4CH3
O
CH3C
O CH2CH2CH3
Oa. b. c.
alkenes ketone ester
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Introduction to Organic Molecules 11–12
11.46 Identify the functional groups in each molecule.
CH3C
C
O
HC
N(CH3)3
Oa. b. c.
ketones amide
CO
HOCHCH
HO OHC
OH
O
hydroxyl groups
carboxyl groups
O
CH3
11.47 Identify the functional groups in each molecule.
aldehyde
ester
C C C CCH2=CH CHCH2 CH2 CH=CH (CH2)5CH3
OH
CH
OHO
CH3O
O CH2CH(CH3)2C
H
Oa.
b.
c.
alkene
hydroxyl grouptwo alkynes
alkene
hydroxyl group
ether
aromaticring
11.48 Identify the functional groups in each molecule.
amideNC
HHO
a.
b.
c.
hydroxyl grouparomatic
ring
COH
HCCH3
HNH
CH3
aromaticring
hydroxyl group
amine
OCH3
CO
CHCOOHCH3
carboxyl group
ketone
two aromatic rings
11.49 Functional groups in dopamine: two hydroxyl groups (OH groups with red spheres), aromatic
ring (six-membered ring with three C=C’s), and amine (NH2 group with a blue sphere).
HO
HO CH2CH2NH2
hydroxyl groupsamine
aromatic ring
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Chapter 11–13
11.50 Functional groups in GHB: one hydroxyl group (OH groups with red spheres) and one carboxyl group (COOH group with red spheres).
hydroxyl group carboxyl group
C OHO
CH2CH2CH2HO
11.51
C CHCH2CH2CCH3
CH3
CHCH
CH3 O aldehyde
alkenealkene 11.52
H3CO
CH2CH CH2
etheraromatic ring
alkene
11.53 In a carboxylic acid, the C=O is bonded to a hydroxyl (OH) group. In an ester, the C=O is bonded
to an OR group.
Examples: !CH3CH2COOH (carboxylic acid) and CH3CO2CH3 (ester) 11.54 In an amine, N is bonded to C or H. In an amide, N is bonded to a C=O group.
CH3CNH2 (amide)CH3NHCH3 (amine) andExamples:O
11.55 Draw an organic compound that fits each set of criteria.
a. a hydrocarbon having molecular formula C3H4 that contains a triple bond: HC≡CCH3 b. an alcohol containing three carbons: CH3CH2CH2OH c. an aldehyde containing three carbons: CH3CH2CHO d. a ketone having molecular formula C4H8O: CH3CH2CCH3
O
11.56 Draw an organic compound that fits each set of criteria.
a. an amine containing two CH3 groups bonded to the N atom: CH3NHCH3
! c. an ether having two different R groups bonded to the ether oxygen: CH3CH2OCH3
b. an alkene that has the double bond in a ring:
CH3CH2CNH2!d. an amide that has molecular formula C3H7NO:O
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Introduction to Organic Molecules 11–14
11.57 Draw the three five-carbon structures.
CH3CH2CH2CH2CH3
C5H12
alkane alkene
CH3 C C C CH3
H
H
H H
C5H10
alkyne
CH3 C C C CH3H
H
C5H8 11.58 Identify the functional groups.
CH2CH CO
NH2N CHCH2H C
OCH3O
CHO
O
aromatic ringcarboxylic acid
amine
amide
ester 11.59 Solubility properties can help you differentiate unknown compounds. NaCl dissolves in water but
not in dichloromethane, and cholesterol dissolves in dichloromethane but not in water. Therefore, when you test the solubility, the water-soluble compound is NaCl, and the water-insoluble (dichloromethane-soluble) compound is cholesterol.
11.60 a. KI is an ionic compound and CH3CH2CH2CH2CH3 is a covalent compound. b. KI is soluble in water but CH3CH2CH2CH2CH3 is not. c. KI is insoluble in an organic solvent, but CH3CH2CH2CH2CH3 is soluble in an organic solvent. d. CH3CH2CH2CH2CH3 has a lower melting point than KI. e. CH3CH2CH2CH2CH3 has a lower boiling point than KI. 11.61 Net dipoles and molecular geometry will determine whether a molecule is polar or nonpolar.
When a molecule has polar bonds, one must determine whether the bond dipoles cancel or reinforce. To determine if they cancel or reinforce, one must know the shape of the molecule.
11.62 Your friend is incorrect. 1,1-dichloroethylene is a polar molecule. The dipoles in the two C–Cl
bonds do not cancel.
CCCl
H Cl
!"
!"
!+H
11.63 All of the bonds except the C–C and C–H bonds are polar. The symbol δ+ is given to the less
electronegative atom, and the symbol δ– is given to the more electronegative atom.
CH2CHCH2
OF C
ClCl
Cla. b.!+
!"
!"
!"
!"
!"
!" !"
!+ !+!+O O HH
!+!+H!+
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Chapter 11–15
11.64 All of the bonds except the C–C and C–H bonds are polar. The symbol δ+ is given to the less electronegative atom, and the symbol δ– is given to the more electronegative atom.
a. b.C CHH
HCl!"
!+ !" !+O CH3
!+
11.65 With organic compounds that have more than one polar bond, the shape of the molecule
determines the overall polarity. • If the individual bond dipoles cancel in a molecule, the molecule is nonpolar. • If the individual bond dipoles do not cancel, the molecule is polar.
all nonpolar bondsnonpolar
polar C=Opolar
H
CBr H
Br
two polar bondsdipoles reinforce
polar
net dipoleCH3CH2C
CH2CH3
O
a. b. c.
11.66 With organic compounds that have more than one polar bond, the shape of the molecule
determines the overall polarity. • If the individual bond dipoles cancel in a molecule, the molecule is nonpolar. • If the individual bond dipoles do not cancel, the molecule is polar.
a. b. c.OH3C CH2CH3
two polar bondsdipoles reinforce
polar
NH3CH2C CH2CH3
three polar bondsnet diplepolar
H
net dipole
all nonpolar bondsnonpolar
11.67 Ethylene glycol is more water soluble than butane because it contains two polar hydroxyl groups
capable of hydrogen bonding. Butane has no polar groups, and cannot hydrogen bond, and is therefore less water soluble.
11.68 Vitamin B5 is water soluble because it contains two polar hydroxyl groups, one amide group and
one carboxyl group, all capable of hydrogen bonding. 11.69 Sucrose has multiple OH groups and this makes it water soluble. 1-Dodecanol has a long
hydrocarbon chain and a single OH group so it is insoluble in water. 11.70 Acetic acid is small, with one carbon atom attached to the carboxyl group, so acetic acid is
soluble in water. Palmitic acid, on the other hand, has a long hydrocarbon chain (15 carbons) attached to the carboxyl group, so it is insoluble in water.
11.71 a. Spermaceti wax is an ester since it contains two R groups bonded to –CO2–;
CH3(CH2)14CO2(CH2)15CH3. b. The very long nonpolar hydrocarbon chains make it insoluble in water but soluble in organic
solvents.
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Introduction to Organic Molecules 11–16
11.72 Beeswax is a large molecule (46 carbons) with one carboxyl group. The long nonpolar hydrocarbon chains make it insoluble in water, but soluble in organic solvents. Therefore, beeswax will be the least soluble in water. The molecule is largely nonpolar due to the long hydrocarbon chains, so it will be most soluble in hexane.
11.73 Fat-soluble vitamins will persist in tissues and accumulate, whereas any excess of water-soluble
vitamins will be excreted in the urine. Therefore, if large quantities of fat-soluble vitamins are ingested, they can build up to toxic levels because they are not excreted readily in the urine, whereas large quantities of water-soluble vitamins will be eliminated much more readily.
11.74 The vitamin E found in leafy greens and vitamin E tablets are identical, so neither is better for an
individual. However, leafy greens contain additional nutrients and therefore should be part of a healthy diet.
11.75 Vitamin E is a fat-soluble vitamin, making it soluble in organic solvents and insoluble in water. 11.76 Vitamin B6 is a water-soluble vitamin, making it soluble in water and insoluble in organic
solvents. 11.77 a. C3H8: Yes, there are just enough H’s to give each C four bonds.
b. C3H9: No, there are too many H’s. The maximum number of H’s for 3 C’s is 8 H’s. c. C3H6: Yes, if you put the three C’s in a ring, each C gets two H’s.
C C C HH
H
H
HH
H
HC C C
H
H
H
H
HH
H
H
H CC C
H H
HH
HH
One C has 5 bonds.invalid structure
a. b. c.
Examples:
C3H8 C3H9 C3H6
11.78 Yes, an oxygen-containing organic compound can have the molecular formula C2H6O. Two
examples are ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3. 11.79
a.
N CH3
H
!"
!+!+!+
CH3CH2CH2b.
CH3CH2CH2NHCH3
c. Four groups around N (three atoms and one lone pair); trigonal pyramidal around N. d. This is a polar molecule. The bond dipoles don’t
cancel.
11.80 a.
O H!"
!+!+ CH3CH2CH2CH2b.
CH3CH2CH2CH2OH
c. Four groups around O (two atoms and two lone pairs); bent around O. d. This is a polar molecule. The bond dipoles don’t
cancel.
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Chapter 11–17
11.81 a. C9H11NO2 b. Five lone pairs are drawn in on benzocaine. c. There are seven trigonal planar carbons (*). d. Benzocaine is water soluble due to the multiple polar bonds. e. Polar bonds are drawn in bold.
N
CO
O
CH2CH3H
H
*** **
**
11.82
a. C8H8O3 b. Six lone pairs are drawn in on methyl salicylate. c. There are seven trigonal planar carbons (*). d. Methyl salicylate is water soluble due to the multiple polar bonds. e. Polar bonds are drawn in bold.
CO H
OCH3
O
***
**
*
*
11.83 Answer each question about aldosterone.
O
CH3
HOCHO C
OCH2OH
**
*
**
*12
34
51'2'
3'
4'
5'
6'
ketone alkene
hydroxyl
aldehydeketone
hydroxyl
b. Each O needs two lone pairs.c. There are 21 C's.d. Number of H's at carbon:
O
CH3
OCHO C
OCH2O
!+!+ !+ !+
!+
!"
!"
!"
!"
H
H!+
!+
!"
a.
!1': 0 H's!2': 0 H's!3': 1 H!4': 2 H's!5': 1 H!6': 1 H
e. Shape at each carbon!1: Tetrahedral – 4 atoms!2: Tetrahedral – 4 atoms!3: Bent – 2 atoms, 2 lone pairs!4: Trigonal planar – 3 atoms!5: Tetrahedral – 4 atoms
f.
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Introduction to Organic Molecules 11–18
11.84 Answer each question about levonorgestrel.
O
*
* *
1 31'
2' 3'
ketone 4
hydroxyl
b. There are 21 C's.c. Number of H's at carbon:
a.
!1': 1 H!2': 2 H's!3': 0 H's
f. Levonorgestrel is soluble in organic solvents.
g. Levonorgestrel is slightly soluble in water.
2C
OHCH
CH3CH2
alkene
alkyne
d. Shape at each carbon!1: Trigonal planar – 3 atoms!2: Tetrahedral – 4 atoms!3: Linear – 2 atoms!4: Tetrahedral – 4 atoms
e.
O
COH
CH
CH3CH2
!"
!+
!+!"
!+
11.85 The waxy coating on cabbage leaves will prevent loss of water and keep leaves crisp. 11.86 a. A moisturizer composed mainly of hydrocarbon material will help to keep skin from drying out
because it will form an inert layer on the outside of the skin that prevents water from leaving the skin.
b. A moisturizer composed mainly of propylene glycol will allow water to permeate the skin to keep it hydrated.
11.87.1 THC is insoluble in water. THC is fat soluble and will therefore persist in tissues for an extended
period of time. Ethanol is water soluble and will be quickly excreted in the urine. Therefore, THC is detectable many weeks after exposure, but ethanol is not.
11.88
a.
NCH3
COOCH3
H O CO
amine
ester
ester
aromatic ring
b. Cocaine hydrochloride will have the higher boiling point. c. Cocaine hydrochloride will be more soluble in water. d. Cocaine hydrochloride is a salt and will dissolve once it is injected into the bloodstream, whereas crack is absorbed once it enters the lungs.
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