These three chapters of the Study Guide and Student Solutions Manual for the Third
Edition of McMurry's Organic Chemistry with Biological Applications are provided
as a preview only. Students can purchase the full book using ISBN 9781305087972
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Chapter 3 – Organic Compounds:
Alkanes and Their Stereochemistry
W
Chapter Outline
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I. Functional Groups (Section 3.1).
A. Functional groups are groups of atoms within a molecule that have a characteristic
chemical behavior.
B. The chemistry of every organic molecule is determined by its functional groups.
C. Functional groups described in this text can be grouped into three categories:
1. Functional groups with carbon–carbon multiple bonds.
2. Groups in which carbon forms a single bond to an electronegative atom.
3. Groups with a carbon-oxygen double bond.
II. Alkanes (Sections 3.2–3.5).
A. Alkanes and alkane isomers (Section 3.2).
1. Alkanes are formed by overlap of carbon sp3 orbitals.
2. Alkanes are described as saturated hydrocarbons.
a. They are hydrocarbons because they contain only carbon and hydrogen.
b. They are saturated because all bonds are single bonds.
c. The general formula for alkanes is CnH2n+2.
3. For alkanes with four or more carbons, the carbons can be connected in more than
one way.
a. If the carbons are in a row, the alkane is a straight-chain alkane.
b. If the carbon chain has a branch, the alkane is a branched-chain alkane.
4. Alkanes with the same molecular formula are isomers.
a. Isomers whose atoms are connected differently are constitutional isomers.
i. Constitutional isomers are always different compounds with different
properties but with the same molecular formula.
b. A given alkane can be drawn in many ways.
5. Straight-chain alkanes are named according to the number of carbons in their chain.
B. Alkyl groups (Section 3.3).
1. An alkyl group is the partial structure that results from the removal of a hydrogen
atom from an alkane.
a. Alkyl groups are named by replacing the -ane of an alkane by -yl.
b. n-Alkyl groups are formed by removal of an end carbon of a straight-chain
alkane.
c. Branched-chain alkyl groups are formed by removal of a hydrogen atom from
an internal carbon.
i. The prefixes sec- and tert- refer to the degree of substitution at the branching
carbon atom.
2. There are four possible degrees of alkyl substitution for carbon.
a. A primary carbon is bonded to one other carbon.
b. A secondary carbon is bonded to two other carbons.
c. A tertiary carbon is bonded to three other carbons.
d. A quaternary carbon is bonded to four other carbons.
3. The symbol R refers to the rest of the molecule.
4. Hydrogens are also described as primary, secondary and tertiary.
a. Primary hydrogens are bonded to primary carbons (RCH3).
b. Secondary hydrogens are bonded to secondary carbons (R2CH2).
c. Tertiary hydrogens are bonded to tertiary carbons (R3CH).
42
Chapter 3
Naming alkanes (Section 3.4).
1. The system of nomenclature used in this book is the IUPAC system.
a. In this system, a chemical name has a prefix, a parent, a locant, and a suffix.
i. The prefix shows the location and identity of substituents.
ii. The parent shows the number of carbons in the principal chain.
iii. The locant identifies the position of the primary functional group.
iv. The suffix identifies the functional group family.
2. Naming an alkane:
a. Find the parent hydrocarbon.
i. Find the longest continuous chain of carbons, and use its name as the parent
name.
ii. If two chains have the same number of carbons, choose the one with more
branch points.
b. Number the atoms in the parent chain.
i. Start numbering at the end nearer the first branch point.
ii. If branching occurs an equal distance from both ends, begin numbering at
the end nearer the second branch point.
c. Identify and number the substituents.
i. Give each substituent a number that corresponds to its position on the parent
chain.
ii. Two substituents on the same carbon receive the same number.
d. Write the name as a single word.
i. Use hyphens to separate prefixes and commas to separate numbers.
ii. Use the prefixes, di-, tri-, tetra- if necessary, but don't use them for
alphabetizing.
e. Name a complex substituent as if it were a compound, and set it off in
parentheses.
i. Some simple branched-chain alkyl groups have common names.
ii. The prefix iso is used for alphabetizing, but sec- and tert- are not.
D. Properties of alkanes (Section 3.5).
1. Alkanes are chemically inert to most laboratory reagents.
2. Alkanes react with O2 and Cl2.
3. The boiling points and melting points of alkanes increase with increasing molecular
weight.
a. This effect is due to weak dispersion forces between molecules.
b. The strength of these forces increases with increasing molecular weight.
4. Increased branching lowers an alkane's boiling point.
III. Conformations of straight-chain alkanes (Sections 3.6–3.7).
A. Conformations of ethane (Section 3.6).
1. Rotation about a single bond produces isomers that differ in conformation.
a. These isomers have the same connections of atoms and can't be isolated.
2. These isomers can be represented in two ways:
a. Sawhorse representations view the C–C bond from an oblique angle.
b. Newman projections represent the two carbons as a circle.
3. There is a barrier to rotation that makes some conformers of lower energy than
others.
a. The lowest energy conformer (staggered conformation) occurs when all C–H
bonds are as far from each other as possible.
b. The highest energy conformer (eclipsed conformation) occurs when all C–H
bonds are as close to each other as possible.
c. Between these two conformations lie an infinite number of other conformations.
P
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W
C.
Organic Compounds: Alkanes and Their Stereochemistry
43
E
V
IE
W
4. The staggered conformation is 12 kJ/mol lower in energy than the eclipsed
conformation.
a. This energy difference is due to torsional strain from interactions between C–H
bonding orbitals on one carbon and C–H antibonding orbitals on an adjacent
carbon, which stabilize the staggered conformer.
b. The torsional strain resulting from a single C–H interaction is 4.0 kJ/mol.
c. The barrier to rotation can be represented on a graph of potential energy vs.
angle of rotation.
B. Conformations of other alkanes (Section 3.7).
1. Conformations of propane.
a. Propane also shows a barrier to rotation that is 14 kJ/mol.
b. The eclipsing interaction between a C–C bond and a C–H bond is 6.0 kJ/mol.
2. Conformations of butane.
a. Not all staggered conformations of butane have the same energy; not all eclipsed
conformations have the same energy.
i. In the lowest energy conformation (anti) the two large methyl groups are as
far from each other as possible.
ii. The eclipsed conformation that has two methyl–hydrogen interactions and a
H–H interaction is 16 kJ/mol higher in energy than the anti conformation.
iii. The conformation with two methyl groups 60° apart (gauche conformation)
is 3.8 kJ/mol higher in energy than the anti conformation.
(a).This energy difference is due to steric strain – the repulsive interaction
that results from forcing atoms to be closer together than their atomic
radii allow.
iv. The highest energy conformations occur when the two methyl groups are
eclipsed.
(a).This conformation is 19 kJ/mol less stable than the anti conformation.
(b).The value of a methyl-methyl eclipsing interaction is 11 kJ/mol.
b. The most favored conformation for any straight-chain alkane has carbon–carbon
bonds in staggered arrangements and large substituents anti to each other.
c. At room temperature, bond rotation occurs rapidly, but a majority of molecules
adopt the most stable conformation.
Solutions to Problems
Notice that certain functional groups have different designations if other functional groups
are present in a molecule. For example, a hydrocarbon containing a carbon–carbon double
bond and no other functional group is an alkene; if other groups are present, the group is
referred to as a carbon–carbon double bond. Similarly, a hydrocarbon containing a
benzene ring and only carbon- and hydrogen-containing substituents is an arene; if other
groups are present, the ring is labeled an aromatic ring.
P
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3.1
(a)
sulfide
O
carboxylic
acid
(b)
CO2H
carboxylic
CH3 acid
CH3SCH2CH2CHCOH
NH2
Methionine
aromatic
ring
amine
Ibuprofen
44
Chapter 3
(c)
ether
amide O
C–C double bond
O
H3C
CH3
N
alcohol
H
CH3
W
HO
aromatic ring
Other structures are possible.
(a)
(b)
CH3OH
(c)
CH3
(d)
O
CH3COH
Methanol
CH3NH2
IE
3.2
Capsaicin
Acetic acid
Methylamine
Toluene
(e)
(f)
O
CH3CCH2NH2
Buta-1,3-diene
V
Aminoacetone
3.3
amine H
E
H3C
H
H
C
N
ester
C
C
C
CH3
O
O
C–C double bond
C
C
H
H
Arecoline
C8H13NO2
R
H
H
We know that carbon forms four bonds and hydrogen forms one bond. Thus, draw all
possible six-carbon skeletons and add hydrogens so that all carbons have four bonds. To
draw all possible skeletons in this problem: (1) Draw the six-carbon straight-chain
skeleton. (2) Draw a five-carbon chain, identify the different types of carbon atoms on the
chain, and add a –CH3 group to each of the different types of carbons, generating two
skeletons. (3) Repeat the process with the four-carbon chain to give rise to the last two
skeletons. Add hydrogens to the remaining carbons to complete the structures.
P
3.4
CH3
CH3CH2CH2CH2CH2CH3
CH3
CH3CH2CCH3
CH3
CH3
CH3CH2CH2CHCH3
CH3CH2CHCH2CH3
CH3
CH3CHCHCH3
CH3
Organic Compounds: Alkanes and Their Stereochemistry
3.5
45
(a) Nine isomeric esters of formula C5H10O2 can be drawn.
O
O
CH3CH2CH2COCH3
O
CH3CHCOCH3
CH3CH2COCH2CH3
O
O CH3
CH3COCH2CH2CH3
O
CH3COCHCH3
O CH3
O
HCOCHCH2CH3
W
CH3
HCOCH2CH2CH2CH3
CH3
O CH3
HCOCH2CHCH3
HCOCCH3
IE
CH3
(b) Two isomers can be drawn.
CH3
CH3CH2CH2C
and
N
CH3CHC
N
(c) Three isomeric disulfides can be drawn for C4H10S2 .
3.6
CH3SSCH2CH2CH3
CH3SSCHCH3
CH3
V
CH3CH2SSCH2CH3
(a) Two alcohols have the formula C3H8O.
OH
and
CH3CHCH3
E
CH3CH2CH2OH
(b) Four bromoalkanes have the formula C4H9Br.
Br
R
CH3CH2CH2CH2Br
CH3CH2CHCH3
CH3
Br
CH3CHCH2Br
CH3CCH3
CH3
(c) Four thioesters have the formula C4H8OS.
O
P
CH3CH2CSCH3
O CH3
O
CH3CH2SCCH3
O
HCSCHCH3
HCSCH2CH2CH3
3.7
CH3CH2CH2CH2CH2
CH3CH2CH2CH
CH3
CH3
CH3CHCH2CH2
CH3
CH3CH2C
CH3
CH3CH2CH
CH2CH3
CH3
CH3CHCH
CH3
CH3CH2CHCH2
CH3
CH3
CH3CCH2
CH3
46
Chapter 3
3.8
(b)
p
p
CH3
CH3CHCH2CH2CH3
p
t s
t
(c)
p
CH3CHCH3
s
CH3CH2CHCH2CH3
p
s
p
t s
p
p
CH3
CH3
CH3CHCH2 C
p
q pCH3
t s
p
CH3
W
(a)
p
p = primary; s = secondary; t = tertiary; q = quaternary
3.9
(a)
(b)
p
p
CH3
t s
s
CH3CH2CHCH2CH3
p
p
s
t s
p
p
CH3
CH3
CH3CHCH2 C
p
t s
p
IE
3.10
(c)
CH3CHCH3
CH3CHCH2CH2CH3
p
t p
CH3
p
CH3
p
Other structures are possible.
(a)
t
(b)
CH3
CH3CHCHCH3
CH3CHCH3
CH3CH2CHCH2CH3
t
V
CH3
3.11
(a)
CH3
CH3CH2CHCH3
E
CH3CH2CH2CH2CH3
Pentane
2-Methylbutane
(c)
q
CH3
CH3CH2CCH3
s
CH3
CH3
CH3CCH3
CH3
2,2-Dimethylpropane
P
R
(b)
Step 1: Find the longest continuous carbon chain and use it as the parent name. In (b), the
longest chain is a pentane.
Step 2: Identify the substituents. In (b), both substituents are methyl groups.
Step 3: Number the substituents. In (b), the methyl groups are in the 2- and 3- positions.
Step 4: Name the compound. Remember that the prefix di- must be used when two
substituents are the same. The IUPAC name is 2,3-dimethylpentane.
CH3
5 4 3
2
CH3CH2CHCH
CH3
1
CH3
2,3-Dimethylpentane
Organic Compounds: Alkanes and Their Stereochemistry
(c)
CH3
CH3
CH3CHCH2CHCH3
CH3CCH2CH2CH
2,4-Dimethylpentane
2,2,5-Trimethylhexane
CH3
CH3
W
3.12
(d)
CH3
CH3
47
When you are asked to draw the structure corresponding to a given name, draw the parent
carbon chain, attach the specified groups to the proper carbons, and fill in the necessary
hydrogens.
(a)
(b)
CH3
CH3CH2CH2C
CH3
CH3 CH2CH3
3,4-Dimethylnonane
(c)
3-Ethyl-4,4-dimethylheptane
(d)
CH2CH2CH3
CH3
2,2-Dimethyl-4-propyloctane
CH3
Pentyl
CH3
R
CH3CH2C
CH3
2,2,4-Trimethylpentane
CH3CH2CH2CH
E
CH3CH2CH2CH2CH2
CH3
CH3CHCH2CCH3
V
CH3CH2CH2CH2CHCH2C(CH3)3
3.13
CHCH2CH3
IE
CH3CH2CH2CH2CH2CHCHCH2CH3
CH3
CH3CH2CHCH2
CH3
1-Methylbutyl
CH3
CH3CHCH
CH3
CH3
2-Methylbutyl
CH3
3-Methylbutyl
(Isopentyl)
CH3
CH3CCH2
CH3CH2CH
CH3
2,2-Dimethylpropyl
(Neopentyl)
CH2CH3
1-Ethylpropyl
P
1,1-Dimethylpropyl 1,2-Dimethylpropyl
(tert-Pentyl)
CH3CHCH2CH2
3.14
3,3,4,5-Tetramethylheptane
48
Chapter 3
3.15
Energy
0°
60°
120°
180°
240°
H3C
H
HH
H
HH H
3.16
(a)
H3C
H
H
H
HH H
H3C
H
H3C
H
H
H HH
H
HH H
H3C
H
H
H HH
H
HH
V
CH3
E
H
H3C
H
H
H
360°
The most stable conformation is staggered
and occurs at 60°, 180°, and 300°.
CH3
H
H HH
300°
IE
Angle of Rotation
H3C
W
14 kJ/mol
(b)
CH3
H
The least stable conformation is eclipsed and
occurs at 0°, 120°, 240°, and 360°.
H CH3
R
HH
6.0 kJ/mol
4.0 kJ/mol
6.0 kJ/mol
P
(c),(d)
16 kJ/mol
Energy
0°
60°
120°
180°
240°
300°
360°
Organic Compounds: Alkanes and Their Stereochemistry
49
3.17
H3C
H
3.18
H
CH3
CH3
W
This conformation of 2,3-dimethylbutane
is the most stable because it is staggered and
has the fewest CH3–CH3 gauche interactions.
CH3
The conformation is a staggered conformation in which the hydrogens on carbons 2 and 3
are 60° apart. Draw the Newman projection.
CH3
CH3
IE
H
3.8 kJ/mol
3.8 kJ/mol
H
CH3
CH3
P
R
E
V
3.8 kJ/mol
The Newman projection shows three interactions, each of which has an energy cost of 3.8
kJ/mol. The total strain energy is 11 kJ/mol.
50
Chapter 3
Visualizing Chemistry
3.19
(a)
(b)
CH3
O
C
NH2
N
carboxylic
acid
OH
aromatic
amine
ring
Phenylalanine C9H11NO2
(a)
O
CH3
aromatic
ring
Lidocaine
(b)
amine
C14H22N2O
(c)
3-Ethyl-2-methylpentane
2,2,4-Trimethylpentane
V
3,3,5-Trimethylheptane
(d)
CH2N(CH2CH3)2
IE
3.20
C
W
CH2CH
amide
H
E
4-Isopropyl-2-methylheptane
3.21
R
CH3
HO
P
H
H
CH3
H
In this conformation, all groups are staggered
and the two methyl groups are 180° apart.
Organic Compounds: Alkanes and Their Stereochemistry
51
Additional Problems
Functional Groups
3.22
(b)
CH2OH
(c)
O
amide
H
W
(a)
N
NHCH3
aromatic
ring
C–C double
bond
C–C double
bond
(e)
aromatic
ring
(f)
IE
carboxylic
acid
O
carboxylic
acid chloride
ketone
CH3CHCOH
NH2
O
amine
C–C triple
bond
C–C double
bond
V
3.23
CH3
amine
(d)
O
C
ketone
alcohol
Cl
C
O
Different answers to this problem and to Problem 3.24 are acceptable.
(a)
(b)
O
(d)
(c)
CH3CNHCH2CH3
E
CH3CH2CCH2CH3
O
(e)
O
C
O
R
CH3COCH2CH2CH3
(f)
O
CH3CCH2COCH2CH3
H
O
H2NCH2CH2OH
3.24
(a)
(b)
P
O
C4H8O:
(c)
CH3CCH2CH3
O
O
C5H9N:
C6H11Br: CH3CH2CH
(f)
C6H14:
CH3CH2CH2CH2CH2CH3
N
(d)
C4H6O2: HCCH2CH2CH
(e)
CH3CH2CH2CH2C
C6H12:
CHCH2CH2Br
52
Chapter 3
(g)
(h)
C5H8: CH3CH
CHCH
CH2
O
C5H8O:
H2C
CHCCH2CH3
(a)
(b)
(c)
sp
O
W
3.25
O
sp2
C
C
N
sp2
C
OH
(a) Although it is stated that biacetyl contains no rings or carbon–carbon double bonds, it is
obvious from the formula for biacetyl that some sort of multiple bond must be present. The
structure for biacetyl contains two carbon-oxygen double bonds.
IE
3.26
(b) Ethyleneimine contains a three-membered ring.
(c) Glycerol contains no multiple bonds or rings.
(a)
(b)
also possible:
CH3C
O
O
O
O
O
V
O
CCH3
HCCH2CH2CH
HCCH2CCH3
Biacetyl
E
Isomers
N
H2C
CH2
OH
HOCH2CHCH2OH
Ethyleneimine
Glycerol
(a) Eighteen isomers have the formula C8H18. Three are pictured.
CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH3
R
3.27
(c)
H
CH3CH2CH2CHCH2CH3
CH3 CH3
CH3C
CCH3
CH3 CH3
P
(b) Structures with the formula C4H8O2 may represent esters, carboxylic acids or many
other complicated molecules. Three possibilities:
O
CH3CH2CH2COH
O
CH3CH2COCH3
HOCH2CH
CHCH2OH
Organic Compounds: Alkanes and Their Stereochemistry
53
3.28
CH3
CH3CH2CH2CH2CHCH3
CH3CH2CH2CHCH2CH3
2-Methylhexane
3-Methylhexane
Heptane
CH3
CH3
CH3CH2CH2CCH3
CH3
CH3CH2CHCHCH3
CH3
CH3
2,2-Dimethylpentane
CH3CHCH2CHCH3
CH3
2,3-Dimethylpentane
2,4-Dimethylpentane
CH3 CH3
IE
CH3
CH3CH2CCH2CH3
CH3CH2CHCH2CH3
CH3
CH3CH
CH2CH3
3,3-Dimethylpentane
3-Ethylpentane
3.29
(a)
W
CH3CH2CH2CH2CH2CH2CH3
CH3
CH3
CH3CHCHCH3
CH3
CH3CHCHCH3
CH3
OH
HO
OH
HO
R
same
(c)
1 CH OH
2
different
5
same
HO
same
CH3
CH3CH2CHCH2CHCH3
P
Br
same
E
(b)
CH3CHCHCH3
Br
same
CH3
2,2,3-Trimethylbutane
V
Br
CCH3
different
CH2CH3
CH3
6
HOCH2CHCH2CHCH3
1
CH3
same
OH
CH3
CH3CH2CHCH2CHCH2OH
6
1
same
Give the number "1" to the carbon bonded to –OH, and count to find the longest chain
containing the –OH group.
54
Chapter 3
3.30
The isomers may be either alcohols or ethers.
OH
CH3CH2CH2CH2
OH
CH3
CH3CH2CHCH3
CH3CHCH2
CH3CCH3
W
3.31
OCH3
OH
OH
OCH3
CH3CH2CH2
CH3
CH3CHCH3
CH3CH2
OCH2CH3
First, draw all straight-chain isomers. Then proceed to the simplest branched structure.
IE
(a) There are four alcohol isomers with the formula C4H10O.
OH
CH3CH2CH2CH2OH
CH3CH2CHCH3
CH3
OH
CH3CHCH2OH
CH3CCH3
CH3
V
(b) There are 17 isomers of C5H13N. Nitrogen can be bonded to one, two or three alkyl
groups.
NH2
CH3CH2CH2CH2CH2NH2
CH3CH2CH2CHCH3
E
NH2
CH3CH2CHCH2NH2
NH2
NH2
CH3CH2CCH3
CH3
CH3CH2CHCH2CH3
CH3CHCHCH3
CH3
CH3
CH3
R
CH3CCH2NH2
H2NCH2CH2CHCH3
CH3
CH3
CH3CH2CH2CH2NHCH3
CH3CH2CHNHCH3
CH3
CH3
P
CH3CHCH2NHCH3
CH3
CH3CNHCH3
CH3
CH3CH2CH2NHCH2CH3
CH3
CH3
CH3CH2CH2NCH3
CH3
CH3CHNCH3
CH3
CH3
CH3CH2NCH2CH3
CH3CHNHCH2CH3
Organic Compounds: Alkanes and Their Stereochemistry
55
(c) There are 3 ketone isomers with the formula C5H10O.
O
O
CH3CH2CH2CCH3
O
CH3CH2CCH2CH3
CH3CHCCH3
W
CH3
(d) There are 4 isomeric aldehydes with the formula C5H10O. Remember that the aldehyde
functional group can occur only at the end of a chain.
O
O
CH3CH2CH2CH2CH
O
CH3CHCH2CH
CH3 O
CH3CH2CHCH
CH3
CH
CH3
IE
CH3
CH3C
(e) There are 4 esters with the formula C4H8O2.
O
CH3CH2COCH3
O
O
CH3COCH2CH3
O CH3
HCOCH2CH2CH3
HCOCHCH3
(f) There are 3 ethers with the formula C4H10O.
V
CH3
CH3CH2OCH2CH3
3.32
.
CH3OCH2CH2CH3
CH3OCHCH3
Other structures are possible
(b)
CH3CH2OH
(c)
CH3
E
(a)
CH3CC
N
(d)
SH
CH3CHCH3
OH
CH3CHCH2OH
CH3
R
(e)
CH3
(f)
P
CH3CHOCH3
CH3
CH3CCH3
CH3
Naming Compounds
3.33
CH3CH2CH2CH2CH2Br
1-Bromopentane
Br
Br
CH3CH2CH2CHCH3
CH3CH2CHCH2CH3
2-Bromopentane
3-Bromopentane
56
Chapter 3
3.34
CH3
CH3
CH3
CH3CHCH2CH2CHCH2Cl
CH3
CH3
CH3CHCH2CH2CCH3
CH3
CH3CHCH2CHCHCH3
Cl
Cl
W
1-Chloro-2,5-dimethylhexane 2-Chloro-2,5-dimethylhexane 3-Chloro-2,5-dimethylhexane
3.35
(a)
(b)
CH3
CH3CH2CH2CH2CH2CHCH3
CH3
CH3CH2CH
CH2CCH3
CH2CH3
(c)
4-Ethyl-2,2-dimethylhexane
IE
2-Methylheptane
(d)
CH3 CH3
CH3CH2CH2CH2C
CHCH2CH3
CH3 CH3
CH3CH2CH2CCH2CHCH3
CH2CH3
CH3
2,4,4-Trimethylheptane
V
4-Ethyl-3,4-dimethyloctane
(e)
CH3
(f)
CH2CH3 CH3
CH3CH2CH2CH2CHCH2C
CHCH3
CH2CH3
E
3,3-Diethyl-2,5-dimethylnonane
3.36
CH3
CH3CHCH3
CH3CH2CH2CHCHCH2CH3
CH3
4-Isopropyl-3-methylheptane
Other structures can be drawn.
CH3
R
(a)
(b)
CH3CHCH3
2-Methylpropane
CH3
CH3CCH3
P
CH3
CH3CH2CH2CH2CH2CH3
CH3
2,2-Dimethylpropane
3.37
(a)
(c)
(b)
CH3CHCH3
CH3CH3
2-Methylpropane
Ethane
Hexane
Organic Compounds: Alkanes and Their Stereochemistry
57
3.38
(b)
CH3
CH3CHCH2CH2CH3
(c)
CH3
CH3CH2CCH3
CH3
2-Methylpentane
(d)
CH2CH3
CH3CHCCH2CH2CH3
CH3
2,2-Dimethylbutane
(e)
CH3
CH3CH2CHCH2CH2CHCH3
H3C CH3
W
(a)
CH3
2,3,3-Trimethylhexane
(f)
CH2CH3
CH3CH2CH2CHCH2CCH3
H3C
CH3C
CH3
3.39
CH3CH2CH2CH2CH2CH3
Hexane
CH3
CH3CH2CH2CHCH3
CH3CH2CHCH2CH3
2-Methylpentane
3-Methylpentane
CH3
V
CH3
CH3CH2CCH3
CH3CHCHCH3
CH3
2,2-Dimethylbutane
CH3
2,3-Dimethylbutane
E
Structure and Correct Name
(a)
CH2CH3
CH3
2,2,3,3-Tetramethylhexane
3,3,5-Trimethyloctane
CH3
3.40
CCH2CH2CH3
IE
5-Ethyl-2-methylheptane
H3C
CH3
Error
CH3
8
CH3CHCH2CH2CH2CCH3
The longest chain is an octane.
1
R
CH3
Correct name: 2,2,6-Trimethyloctane
(b)
CH3
CH3CHCHCH2CH2CH3
P
1
6
CH2CH3
The longest chain is a hexane.
Numbering should start from the
opposite end of the carbon chain,
nearer the first branch.
Correct name: 3-Ethyl-2-methylhexane
(c)
CH3
CH3CH2C
1
CHCH2CH3
6
CH3 CH2CH3
Correct name: 4-Ethyl-3,3-dimethylhexane
Numbering should start from the
opposite end of the carbon chain.
See step 2(b) in Section 3.4.
58
Chapter 3
(d)
CH3 CH3
CH3CH2CH
Numbering should start from the
opposite end of the carbon chain.
CCH2CH2CH2CH3
8
1
CH3
(e)
W
Correct name: 3,4,4-Trimethyloctane
CH3
The longest chain is an octane.
CH3CH2CH2CHCH2CHCH3
8
CH3CHCH3
1
3.41
(a)
CH3 CH2CH3
CH3CCH2CCH2CH3
(b)
CH3
CH2CH2CHCH3
CH3CH2CH2CH2CH2CHCH2CH2CH2CH2CH3
V
CH3 CH2CH3
IE
Correct name: 2,3,5-Trimethyloctane
4,4-Diethyl-2,2-dimethylhexane
6-(3-Methylbutyl)-undecane
E
Remember that you must choose an alkane whose
principal chain is long enough so that the substituent
does not become part of the principal chain.
Conformations
3.42
R
(a), (b)
CH3
3.8 kJ/mol
CH3
H3C
CH3
H
H
6.0 kJ/mol
CH3CH2 CHCH3
C2–C3 bond
H
H
CH3
HH
H3C CH3
P
4.0 kJ/mol
11.0 kJ/mol
2-Methylbutane
Most stable conformation
Least stable comformation
The energy difference between the two conformations is (11.0 + 6.0 + 4.0) kJ/mol –
3.8 kJ/mol = 17 kJ/mol.
Organic Compounds: Alkanes and Their Stereochemistry
59
(c) Consider the least stable conformation to be at zero degrees. Keeping the front of the
projection unchanged, rotate the back by 60° to obtain each conformation.
CH3
CH3
H
H
CH3
6.0 kJ/mol
3.8 kJ/mol
CH3
H3C
CH3
H CH3
3.8 kJ/mol
H
W
H
H
H CH3
6.0 kJ/mol
H
H
CH3
6.0 kJ/mol
IE
at 60°: energy = 3.8 kJ/mol at 120°: energy = 18.0 kJ/mol at 180°: energy = 3.8 kJ/mol
CH3
CH3
11.0 kJ/mol
3.8 kJ/mol
CH3
H
CH3
3.8 kJ/mol
HH
H CH3
H
H
CH3
V
4.0 kJ/mol
6.0 kJ/mol
at 240°: energy = 21.0 kJ/mol at 300°: energy = 7.6 kJ/mol
E
Use the lowest energy conformation as the energy minimum. The highest energy
conformation is 17 kJ/mol higher in energy than the lowest energy conformation.
R
P
Energy
(kJ/mol)
7.6
3.8
60°
21
21
18
3.8
120°
180°
240°
Angle of Rotation
300°
360°
60
Chapter 3
3.43
CH3
H
H
CH3
H3C
CH3
CH3
H
H
CH3
H
3 CH3–CH3 gauche
= 3 x 3.8 kJ/mol
= 11 kJ/mol
CH3
CH3
3 CH3–CH3 gauche
= 3 x 3.8 kJ/mol
= 11 kJ/mol
IE
3.44
CH3
H
2 CH3–CH3 gauche
= 2 x 3.8 kJ/mol
= 7.6 kJ/mol
CH3
W
H3C
CH3
Since we are not told the energy values of the interactions for 1,2–dibromoethane, the
diagram can only be qualitative.
least stable
HH
H
H
Br H
Br H
H
Br
E
H
H
H
H
H
H
H
H
Br
H
Br
H
Br
Br
H
HH
Br Br
Br
H
Br
HH
Br H
V
Br Br
H
H
H
H
R
Energy
0°
60°
120°
most stable
180°
240°
300°
360°
P
The anti conformation is at 180°.
The gauche conformations are at 60°, 300°.
3.45
The anti conformation has no net dipole moment because the polarities of the individual
bonds cancel. Each of the other conformations, however, has a dipole moment. Because a
dipole moment (1.0 D at room temperature) is observed, a mixture of conformations must
be present.
Organic Compounds: Alkanes and Their Stereochemistry
The best way to draw pentane is to make a model, showing all hydrogens in the staggered
position) and to copy it onto the page. A model shows the relationship among atoms, and
its drawing shows how these relationships appear in two dimensions. From your model,
you should be able to see that all atoms are staggered in the drawing.
HH
H
HH
C
C
H
C
C
C
HH
HH
HH
W
3.46
3.47
HH
HH
C
C
C
C
Cl
HH
HH
General Problems
IE
Cl
3.48
Other isomers are possible.
(a)
(b)
CH2OCH3
V
CH3
(c)
CH3CH2CHCH2Br
(d)
CH3
CH3CHC
N
(e) There are no aldehyde
(f)
isomers. However, the
structure below is a ketone
isomer.
O
E
CH2OH
CO2H
CH3
CH3CCH3
(a) Because malic acid has two –CO2H groups, the formula for the rest of the molecule is
C2H4O. Possible structures for malic acid are:
R
3.49
61
O
P
HOC
O
CH
O
COH
HOC
CH2OH
HOC
O
CHCH2 COH
HOC
OH
primary alcohol
O
O
CH2OCH2 COH
O
HOC
C
COH
CH3
secondary alcohol
O
OH O
O
OCH2CH2 COH
tertiary alcohol
O
HOC
O
OCH
COH
CH3
(b) Because only one of these compounds (the second one) is also a secondary alcohol, it
must be malic acid.
62
Chapter 3
3.50
To solve this type of problem, read the problem carefully, word by word. Then try to
interpret parts of the problem. For example:
O
IE
W
1) Formaldehyde is an aldehyde, H C H
2) It trimerizes — that is, 3 formaldehydes come together to form a compound, C3H6O3.
Because no atoms are eliminated, all of the original atoms are still present.
3) There are no carbonyl groups. This means that trioxane cannot contain any –C=O
functional groups. If you look back to Table 3.1, you can see that the only oxygencontaining functional groups that can be present are either ethers or alcohols.
4) A monobromo derivative is a compound in which one of the –H's has been replaced by
a –Br. Because only one monobromo derivative is possible, we know that there can
only be one type of hydrogen in trioxane. The only possibility for trioxane is:
CH2
O
H2C
O
Trioxane
CH2
(a) The highest energy conformation of bromoethane has a strain energy of 15 kJ/mol.
Because this includes two H–H eclipsing interactions of 4.0 kJ/mol each, the value of an
H–Br eclipsing interaction is 15 kJ/mol – 2(4.0 kJ/mol) = 7.0 kJ/mol.
V
3.51
O
E
(b)
15 kJ/mol
R
Energy
0°
60°
120°
180°
240°
300°
360°
Br
Br
Br
Br
Br
Br
Br
P
H
HH
H
HH H
H
H
H HH
H
H
H
HH H
H
H HH
H
H
HH H
H
H
H HH
H
HH
Organic Compounds: Alkanes and Their Stereochemistry
63
3.52
CH3
H
H
(b)
H
CH3
H
H3C
(c)
H
CH3
CH3
CH3
HH
(d)
H3C H
H CH3
CH3
7.6 kJ/mol
CH3
26 kJ/mol
CH3
HH
CH3
CH3
CH3
H
CH3
CH3
CH3 CH
3
CH3
15.2 kJ/mol
R
P
23 kJ/mol
CH3
H
H3C
H3C
H CH3
CH3
7.6 kJ/mol
E
H
21 kJ/mol
CH3
V
H3C
Strain energy
CH3
* 3.8 kJ/mol
CH3
CH3
Least stable
IE
(a)
Strain energy
W
Most stable
H3C H
** 28 kJ/mol
CH3 CH
3
*most stable overall (least strain)
**least stable overall (most strain)
64
Chapter 3
3.53
O C
HO
A
O
B
B
O
OH
A
B
O
O
CH3
CH3
A
HO
IE
H3C
D
Zocor C25H38O5
A
C
OH
W
O
C
A
HO
O
alcohol
carboxylic acid
D
Pravachol C23H36O7
ester
C–C double bond (alkene)
B
D
3.54
V
The carboxylic acid group in Pravachol is an ester group in Zocor.
The alcohol at the bottom left of Pravachol is a methyl group in Zocor.
A puckered ring allows all the bonds in the ring to have a nearly tetrahedral bond angle. (If
the ring were flat, C-C-C bond angles would be 120°.)
E
H
H
H
H
H
H
H
H
H
R
H
H
H
3.55
P
H
H3C
H
CH3
H
H3C
CH3
H
In one of the 1,2-dimethylcyclohexanes the two methyl groups are on the same side of the
ring, and in the other isomer the methyl groups are on opposite sides.