Chapter 11 Introduction to Organic Molecules and...

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


bent

(1): three groupstrigonal planar



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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°


120°


bent109.5°


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


bent(2): two groups

linear



bent


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


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



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


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).



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


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


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


b. CH3 O C(CH3)3C CH

H H

C Na.

(1) and (3): four groupstetrahedral




bent


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


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°


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


120°

two atoms and two lone pairsbent109.5°


120°


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°



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


(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


F3C

CH2CHNCH2CH3

CH3

Hthree groupstrigonal planar

three atoms + one lone pairtrigonal pyramidal

four atomstetrahedral


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

=


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==



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


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




CH3C

C

O

HC

N(CH3)3

Oa. b. c.

ketones amide

CO

HOCHCH

HO OHC

OH

O

hydroxyl groups

carboxyl groups

O

CH3


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


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


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



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!+


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

!+



all nonpolar bondsnonpolar

polar C=Opolar

H

CBr H

Br

two polar bondsdipoles reinforce

polar

net dipoleCH3CH2C

CH2CH3

O

a. b. c.



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.



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.


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.



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.


Chapter 11 Introduction to Organic Molecules and...

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