90
Solutions
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
CHAPTER 5
Solutions to the Problems
Problem 5.1 Calculate the index of hydrogen deficiency of cyclohexene, C6H10, and account for this deficiency by
reference to its structural formula.
The molecular formula of the reference acyclic alkane of six carbons is C6H14. The index of hydrogen
deficiency of cyclohexane (14-10)/2 = 2 and is accounted for by the combination of one ring and one pi bond in
cyclohexene.
Cyclohexene
Problem 5.2 The index of hydrogen deficiency of niacin is 5. Account for this index of hydrogen deficiency by reference
to the structural formula of niacin.
O
CNH2
Nicotinamide
(Niacin)
N
The index of hydrogen deficiency of 5 can be accounted for by the presence of the ring plus the three pi bonds in
the ring, as well as the pi bond of the carbonyl group.
Problem 5.3 Write the IUPAC name of each alkane.
(a)
(b)
H
H
C
C
H
H 3C
CH 3
CH
CH 3
C
H 3C
CH 3
3-Methyl-1-butene
C
CH 3
2,3-Dimethyl-2-butene
Problem 5.4 Which alkenes show cis,trans isomerism? For each alkene that does, draw the trans isomer.
(a) 2-Pentene
(b) 2-Methyl-2-pentene
(c) 3-Methyl-2-pentene
CH 3 CH 2
H
C
H
C
CH 3
No cis,trans isomers
because there are two
methyl groups on one
end of the double bond.
trans-2-Pentene
CH 3 CH 2
H
C
CH 3
C
CH 3
trans-3-Methyl-2-pentene
Problem 5.5 Name each alkene and specify its configuration by the E-Z system.
Cl
Cl
(a)
(E)-1-Chloro-2,3-dimethyl2-pentene
(b)
Br
(Z)-1-Bromo-1-chloropropene
(c)
(E)-2,3,4-Trimethyl-3-heptene
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
91
Solutions
Problem 5.6 Write the IUPAC name of each cycloalkene
(a)
(b)
(c)
1,3-Dimethylcyclopentene
Cyclooctene
4-(1,1-Dimethylethyl)cyclohexene (IUPAC)
4-tert-Butylcyclohexene (Common)
Problem 5.7 Draw structural formulas for the other two stereoisomers for 2,4-heptadiene.
1
H 3C
2
6
H
4
5
3
C
C
7
CH 2 CH 3
C C
H
H
H
cis,trans-2,4-Heptadiene
H
4
5
2
3
C
C
C
C
1
H 3C
H
H
CH 2 CH 3
6
7
H
cis,cis-2,4-Heptadiene
Problem 5.8 (10E,12Z)-10,12-hexadecadiene-1-ol is a sex pheromone of the silkworm. Draw a structural formula for this
compound.
H 1 1 1 0 CH 2 (CH 2 ) 7 CH 2 O H
CH 3 CH 2 CH 2 1 3 1 2 C C
C C
H
H
H
(10E,12Z)-10,12-Hexadecadiene-1-ol
Structure of Alkenes
Problem 5.9 Predict all approximate bond angles about each highlighted carbon atom. To make these predictions, use
valence shell electron-pair repulsion (Section 1.4).
109.5o
(a)
120o
120o
(b)
CH2OH
CH 2OH
120o
O
(c)
O
C OH
C OH
(d)
CH2
CH3
12 0 o
C
CH2
CH
00
00
000
000
000
000
00
00
00
00
0000
0000
0000
0000
0000
0000
0000
0000
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Solutions
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
180o
H
(e)
HC C C
H C C C H CH2
C
120o
H
H
Problem 5.10 For each highlighted carbon atom in Problem 5.9, identify which atomic orbitals are used to form each sigma
bond and which are used to form each pi bond.
Each bond is labeled sigma or pi and the orbitals overlapping to form each bond are shown.
σ 2 3
H
H σsp3 -sp3
sp -sp
σ 3
sp -1 s
σ 2 3
sp -sp
CH2 OH
(b)
(a)
π
2 p-2 p
σ
2
1 s-sp
π
σ 2 2
σ 2 2
H
2 p-2p
sp -sp
π
sp -sp
σ
2 p-2p
O
(c)
σ
sp2 -sp2
C OH
σ
σ
sp3 -sp2
σ
π
(d)
sp2 -sp3
sp-sp 2
2p-2p
HC C C
σ
C
sp-sp
π
CH 3
2p-2p
H 2C
σ
π
H
2p-2p
σ
C
C
H
σ
sp 2 -sp 2
CH 2
sp 2 -sp 2
2p-2p
H
(e)
π
sp 2 -sp 3
σ
1s-sp 2
H
σ
1s-sp 2
sp 2 -sp 2
Problem 5.11 Following is the structure of 1,2-propadiene (allene)
CH2
C
CH2
Propadiene
(Allene)
(a) Predict all bond angles in this molecule.
(b) State the orbital hybridization of each carbon.
(c) Show the three-dimensional geometry of allene, and explain it in terms of the orbitals used.
180o
π
H
H
C
C
sp
sp 2
H
o
2p-2p
H
C
H
120
sp 2
H
C
CH2
C
CH2
H
C
C
H
σ
sp 2 -sp
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
93
Solutions
The central carbon atom of allene is sp hybridized with bond angles of 180o about it. The terminal carbons are
sp2 hybridized with bond angles of 120o about each. The planes created by H-C-H bonds at the ends of the
molecule are perpendicular to each other.
H
H
C
C
H
C
H
Problem 5.12 Following are lengths for a series of C-C single bonds. Propose an explanation for the differences in bond
lengths.
Length of C-C
single bond (pm)
Structure
CH3 CH3
153.7
CH2 C H CH3
151.0
CH2 C H C H CH2
146.5
H C C CH3
145.9
The s electrons are on average held closer to atomic nuclei than p electrons. Thus, hybrid orbitals with higher s
character have the electrons held closer to the nucleus and thus make bonds that are shorter. As shown in the
table, a σ bond formed from overlap of an sp3 orbital with an sp2 orbital is shorter than a σ bond formed from
overlap of an sp3 orbital with another sp3 orbital. Similarly, sp3-sp overlap produces a bond that is shorter
than that produced by sp3-sp2 overlap.
Nomenclature of Alkenes
Problem 5.13 Draw structural formulas for these alkenes.
(a) trans-2-Methyl-3-hexene
(b) 2-Methyl-2-hexene
(c) 2-Methyl-1-butene
H
CH 3
CH 2 CH 3
C
CH
C
CH 3
CH 2 CHCCH 2CH 3
CH 2 CH 3
(g) (Z)-1-Chloropropene
CH 3
H
C
H
CH 2 CH 2 CH 3
H
H
H
(e) 2,3-Dimethyl-2-butene
H 3C
CH 3
C
H 3C
C
CH 3
(h) 3-Methylcyclohexene
CH 3
CH 2 CH 3
C
C
H 3C
(d) 3-Ethyl-3-methyl-1-pentene
C
C
H
CH 3
Cl
H 3C
C
CH 3
(f) cis-2-Pentene
H 3C
CH 2 CH 3
C
H
C
H
(i) 1-Isopropyl-4-methylcyclohexene
CH 3
CH(CH 3 ) 2
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Solutions
(j) (E)-2,6-Dimethyl-2,6-octadiene
H 3C
C
CH 2 CH 2
H 3C
C
(k) 3-Cyclopropyl-1-propene
(l) Cyclopropylethene
CH 3
C
H
CH CH 2
CH 2CH CH 2
C
H
H 3C
(m) 2-Chloropropene
H
C
H
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
(n) Tetrachloroethylene
Cl
Cl
CH 3
Cl
C
(o) 1-Chlorocyclohexene
Cl
Cl
C
C
Cl
Problem 5.14 Name these alkenes and cycloalkenes.
Cl
(a)
(b)
(E)-3-(2-methylpropyl)-2-Octene
4-Chloro-1,4dimethylcyclopentene
(c)
Ethenylcyclohexane
Vinylcyclohexane
Cl
(d)
F
(e)
(f)
Cl
(E)-1,4-Dichloro-2-butene
trans-1,4-Dichloro-2-butene
2,4-Dimethyl-2-pentene
F
F
F
Tetrafluoroethene
Cl
(g)
5-Chloro-5-ethyl1,3-cyclopentadiene
(h)
(i)
1,4-Cyclohexadiene
1,7,7-Trimethylbicyclo[2.2.1]-2-heptene
Problem 5.15 Arrange the following groups in order of increasing priority.
(a) -CH3 -H -Br -CH2CH3
(b) -OCH3 -CH(CH3)2 -B(CH2CH3)2 -H
-H < -CH3 < -CH2CH3 < -Br
-H < -B(CH2CH 3)2 < -CH(CH3)2 < -OCH3
(c) -CH3 -CH2OH -CH2NH2 -CH2Br
-CH3 < -CH2NH2 < -CH2OH < -CH2Br
Problem 5.16 Assign an E or Z configuration and a cis or trans configuration to these dicarboxylic acids, each of which is
an intermediate in the tricarboxylic acid cycle. Following each is its common name.
H
COOH
HOOC
C
(a)
C
HOOC
H
Fumaric acid
E / trans
COOH
C
(b)
H
C
CH2COOH
Aconitic acid
Z / cis
The highest priority group on each sp2 carbon atom is circled.
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
95
Solutions
Problem 5.17 Name and draw structural formulas for all alkenes of molecular formula C5H10. As you draw these alkenes,
remember that cis and trans isomers are different compounds and must be counted separately.
Four alkenes of molecular formula C5H10 do not show cis,trans isomerism.
CH 3
CH 3
CH 2 CHCH 2CH 2CH 3
CH 2 CCH 2CH 3
CH 2 CHCHCH3
1-Pentene
2-Methyl-1-butene
3-Methyl-1-butene
CH 3
CH 3C CHCH 3
2-Methyl-2-butene
One alkene of molecular formula C5H10 shows cis,trans isomerism.
C C
C C
CH 2 CH 3
H
CH 2 CH 3
CH 3
H
CH 3
H
trans-2-Pentene
H
cis-2-Pentene
Problem 5.18 For each molecule that shows cis,trans isomerism, draw the cis isomer.
CH3
CH3
CH3
CH3
(a)
(c)
(b)
CH3
(d)
CH3
CH3
CH3
CH 3
CH 3
CH 3
CH 3
Problem 5.19 β-Ocimene, a triene found in the fragrance of cotton blossoms and several other essential oils, has the IUPAC
name (Z)-3,7-dimethyl-1,3,6-octatriene. Draw a structural formula for β-ocimene.
H 4 3 CH 3
8
5 C C 2
1 H
CH 3 7 6 CH 2
C C
H
H
C C
CH 3
H
β-Ocimene
(Z)-3,7-Dimethyl-1,3,6-octatriene
Problem 5.20 Draw the structural formula for at least one bromoalkene of molecular formula C5H9Br that shows:
(a) Neither E,Z isomerism nor chirality.
H
C
C
H
CH 2 CH 2 CH 2 Br
or
H
5-Bromo-1-pentene
H 3C
C
C
H 3C
CH 2 B r
or
H
H
C
H
1-Bromo-3-methyl2-butene
C
CH 2 CH 2 B r
or
CH 3
H
CH 2 B r
C
C
H
4-Bromo-2-methyl1-butene
CH 2 CH 3
2-(Bromomethyl)1-butene
(b) E,Z isomerism but not chirality.
H
CH 2 CH 2 B r
C
CH 3
C
CH 3
H
(E)-5-Bromo-2-pentene
CH 2 CH 2 B r
C
or
H
C
H
H
(Z)-5-Bromo-2-pentene
CH 2 B r
C
or
CH 3 CH 2
C
H
(E)-1-Bromo-2-pentene
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Solutions
CH 3 CH 2
CH 2 B r
C
H
or
C
CH 2 B r
C
H
H
CH 2 B r
CH 3
C
C
or
CH 3
CH 3
(Z)-1-Bromo-2-pentene
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
C
CH 3
H
(E)-1-Bromo-2-methyl-2-butene
(Z)-1-Bromo-2-methyl-2-butene
(c) Chirality but not E,Z isomerism.
*
CH 2 CHBrCH 3
H
C
H
*
CHBrCH 2 CH 3
H
C
C
or
H
H
4-Bromo-1-pentene
*
CHBrCH 3
H
C
C
or
H
H
3-Bromo-1-pentene
C
CH 3
3-Bromo-2-methyl-1-butene
(d) Both chirality and E,Z isomerism.
*
CHBrCH
3
H
C
C
*
CHBrCH
3
CH 3
C
or
H
CH 3
(E)-4-Bromo-2-pentene
C
H
H
(Z)-4-Bromo-2-pentene
Problem 5.21 Following are structural formulas and common names for four molecules that contain both a carbon-carbon
double bond and another functional group. Give each an IUPAC name.
O
O
(a) CH2=CHC OH
Acrylic acid
2-Propenoic acid
H
O
(b) CH2=CHCH
Acrolein
2-Propenal
COH
C
(c)
H3C
C
H
Crotonic acid
(E)-2-Butenoic acid
O
(d) CH3 CCH=CH2
Methyl vinyl ketone
3-Buten-2-one
Problem 5.22 Trans-cyclooctene has been resolved, and its enantiomers are stable at room temperature. Transcyclononene has also been resolved, but it racemizes with a half-life of 4 min at 0°C. How can racemization of this
cycloalkene take place without breaking any bonds? Why does trans-cyclononene racemize under these conditions but
trans-cyclooctene does not? You will find it especially helpful to build molecular models of these cycloalkenes.
Following are enantiomers of trans-cyclooctene and trans-cyclononene. It may be helpful to construct molecular
models and prove to yourself that the two different configurations of the ring are in fact non-superimposable
mirror images of each other. For both trans-cyclooctene and trans-cyclononene, the enantiomers are
configurational isomers. The enantiomers are interconverted by a change in configuration that is analogous to
the chair flipping of chair cyclohexane. The trans double bond adds a considerable degree of rigidity to the
ring, because the other atoms must exist in a slightly "stretched" configuration to accommodate the trans
geometry. Nevertheless, the trans-cyclononene ring has more carbon atoms, so it is more flexible and can
undergo the configurational interconversion more readily.
Enantiomers of trans-cyclooctene
Enantiomers of trans-cyclononene
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
97
Solutions
Problem 5.23 Which alkenes exist as pairs of cis,trans isomers? For each that does, draw the trans isomer.
For an alkene to exist as a pair of cis,trans isomers, both carbon atoms of the double bond must have two
different substituents. Thus, (b), (c), and (e) exist as a pair of cis,trans isomers. The trans isomer for each
alkene is drawn under its respective condensed molecular formula.
(a) CH2=CHBr
(b) CH3CH=CHBr
CH 3
Br
H
C
H
(c) BrCH=CHBr
H
C
C
C
Br
H
Br
(d) (CH3)2C=CHCH3
(e) (CH3)2CHCH=CHCH3
H
(CH 3 ) 2 C H
C
H
C
CH 3
Problem 5.24 Four stereoisomers exist for 3-penten-2-ol.
about this
double bond
OH
C H C H C*H CH3
CH3
3-Penten-2-ol
(a) Explain how these four stereoisomers arise.
There is one double bond that provides for cis,trans isomers, and one chiral center in 3-penten-2-ol.
(b) Draw the stereoisomer having the E configuration about the carbon-carbon double bond and the R configuration at the
chiral center.
HO
H C
H
C
CH 3
C
H 3C
H
Molecular Modeling
These problems require molecular modeling programs such as Chem 3DTM or SpartanTM to solve. Pre-built models can be
found at http://now.brookscole.com/bfi4.
Problem 5.25 Measure the CH3, CH3 distance in the energy-minimized model of cis-2-butene, and the CH3, H distance in
the energy-minimized model of trans-2-butene. In which isomer is the nonbonded interaction strain greater?
2.1 Å
C
H
2.4 Å
H 3C
CH3
H3 C
Greater non-bonded
interaction strain
H
C
C
H
H
cis-2-Butene
C
CH 3
trans-2-Butene
For cis-2-butene, the distance between the hydrogen atoms on adjacent methyl groups varies as the methyl group
rotates, but is 2.1 Å in the energy minimized structure. For trans-2-butene, the distance between the hydrogen
atoms of the methyl group and the alkene hydrogen atom also varies as the methyl group rotates, with a value of
2.4 Å in the energy minimized structure. Clearly, there is greater nonbonded interaction strain in cis-2-butene.
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Chapter 5: Alkenes: Bonding, Nomenclature and Properties
Problem 5.26 Measure the C=C-C bond angles in the energy-minimized models of the cis and trans isomers of 2,2,5,5tetramethyl-3-hexene. In which case is the deviation from VSEPR predictions greater?
Severe non-bonded
interaction strain
135°
C(CH 3 ) 3
(H 3 C) 3 C
C
H
C
H
cis-2,2,5,5-Tetramethyl-3-hexene
127°
H
(H 3 C) 3 C
C
C
H
C(CH 3 ) 3
trans-2,2,5,5-Tetramethyl-3-hexene
VSEPR predicts 120° bond angles around the double bond. The minimized cis isomer has a 135° bond angle
due to the severe non-bonded interaction strain present. The trans isomer has less non-bonded interaction
strain and a 127° bond angle. Thus, the cis isomer is the one with significantly more deviation from the VSEPR
predictions.
Problem 5.27 Measure the C-C-C and C-C-H bond angles in the energy-minimized model of cyclohexene and compare
them with those predicted by VSEPR. Explain any differences.
118°
123°
112°
119°
Cyclohexene
Representative measurements have been added to the above structure. Clearly, the angles predicted by Chem3D
are not exactly 120° as predicted by VSEPR for an sp2 hybridized carbon atom. This difference is the result of
strain introduced by the ring of cyclohexene.
Problem 5.28 Measure the C-C-C and C-C-H bond angles in the energy-minimized models of cis and trans isomers of
cyclooctene. Compare these values with those predicted by VSEPR. In which isomer are deviations from VSEPR
predictions greater?
129°
cis-Cyclooctene
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
99
Solutions
119°
trans-Cyclooctene
Only the key C-C=C bond angles are shown on the above structures. VSEPR predicts 120° bond angles around
the double bond. The angles deviate more in the cis isomer compared to the trans isomer due to increased ring
strain in the cis isomer. Note how the trans isomer can more easily accommodate staggered conformations along
the entire chain.
Terpenes
Problem 5.29 Show that the structural formula of vitamin A (Section 5.3G) can be divided into four isoprene units bonded
head-to-tail and cross-linked at one point to form the six-membered ring.
Isoprene chain cross-linked here
CH2OH
Problem 5.30 Following is the structural formula of lycopene C40H56, a deep-red compound that is partially responsible
for the red color of ripe fruits, especially tomatoes. Approximately 20 mg of lycopene can be isolated from 1 kg of ripe
tomatoes. Lycopene is an important antioxidant that may help prevent oxidative damage in atherosclerosis.
H3C
CH3
CH
3
H3C
CH3
CH3
CH3
CH3
H3C
CH3
Lycopene
(a) Show that lycopene is a terpene, that is, its carbon skeleton can be divided into two sets of four isoprene units with the
units in each set joined head-to-tail.
Head-to-head bond
joining two four
isoprene units
Lycopene
(b) How many of the carbon-carbon double bonds in lycopene have the possibility for cis,trans isomerism? Of these,
which are trans and which are cis?
The double bonds on the two ends of the molecule cannot show cis,trans isomerism. The other 11 double bonds
can show cis,trans isomerism, but they are all trans.
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Chapter 5: Alkenes: Bonding, Nomenclature and Properties
Problem 5.31 As you might suspect, β-carotene, C40H56, precursor to vitamin A, was first isolated from carrots. Dilute
solutions of β-carotene are yellow, hence its use as a food coloring. In plants, it is almost always present in combination
with chlorophyll to assist in the harvesting of the energy of sunlight and to protect the plant against reactive species
produced in photosynthesis. As tree leaves die in the fall, the green of their chlorophyll molecules is replaced by the yellow
and reds of carotene and carotene-related molecules. Compare the carbon skeletons of β-carotene and lycopene. What are
the similarities? What are the differences?
H3C
CH3
H3C
CH3
CH3
CH3
CH3
CH3
H3C
CH3
β-Carotene
The main structural difference between β-carotene and lycopene is that β-carotene has six-membered rings on the
ends, not an open chain. On the other hand, both β -carotene and lycopene can be divided into two sets of four
isoprene units as shown below, and all of the double bonds are E in both molecules.
Isoprene chain cross-linked at these two points
Head-to-head bond
joining two four
isoprene units
Problem 5.32 Calculate the index of hydrogen deficiency for β-carotene and lycopene.
Both molecules have an index of hydrogen deficiency of 13. β -Carotene has eleven pi bonds and two rings.
Lycopene has no rings, but thirteen pi bonds.
Problem 5.33 α-Santonin, isolated from the flower heads of certain species of Artemisia, is an anthelminthic (meaning
against instestinal worms. This terpene is used in oral doses of 60 mg to rid the body of roundworms such as Ascaris
lumbricoides. It has been estimated that over one third of the world's population is infested with these slender, thread-like
parasites.
CH3
O
CH3
CH3
O
O
(a) Locate the three isoprene units in santonin, and show how the carbon skeleton of farnesol might be coiled and then
cross-linked to give santonin. Two different coiling patterns of the carbon skeleton of farnesol can lead to santonin. Try to
find them both.
1
8
7
6
5
4
3
2
9
10
5
CH2OH
Farnesol
6
4
9
*
4
O
8
7
3
2
*
* 10
1
O
11
12
5
or
*
11
12
O
2
*
6
O
3
7
8 9*
1
* 10
O
*
11
12
O
(b) Label all chiral centers in santonin. How many stereoisomers are possible for this molecule?
The four chiral centers of santonin are marked on the structures above. There are 24 = 16 stereoisomers possible
for santonin.
Chapter 5: Alkenes: Bonding, Nomenclature and Properties
Solutions
101
(c) Calculate the index of hydrogen deficiency for santonin.
Santonin has three rings and four pi bonds, so it has an index of hydrogen deficiency of seven.
Problem 5.34 Pyrethrin II and pyrethrosin are two natural products isolated from plants of the chrysanthemum family.
Pyrethrin II is a natural insecticide and is marketed as such.
(a) Label all chiral centers in each molecule and all carbon-carbon double bonds about which there is the possibility for
cis,trans isomerism.
cis,trans Isomerism
possible
H
cis,trans Isomerism
O
O
H
possible
O
*
H3 COC
H CH
O
*
*
3
O
CH3 *
CH2
C C
H
CH3
*
*
*
O
CH3 O H CH2
CH3
H
C
*
CH3
CCH3
cis,trans Isomerism
H
O
possible
O
Pyrethrin II
Pyrethrosin
(b) State the number of stereoisomers possible for each molecule.
For pyrethrin II there are two double bonds capable of cis,trans isomerism and three chiral centers for a total of
25 = 32 possible stereoisomers. For pyrethrosin there is one double bond capable of cis,trans isomerism and five
chiral centers for a total of 26 = 64 possible stereoisomers.
(c) Show that the bicyclic ring system of pyrethrosin is composed of three isoprene units.
O
O
O
CH 2
O
CCH 3
O
(d) Calculate the index of hydrogen deficiency for each of these two natural products.
Pyrethrin II has two rings and seven pi bonds for a total index of hydrogen deficiency of 9, and pyrethrosin has
three rings and four pi bonds for a total index of hydrogen deficiency of 7.
Problem 5.35 Limonene is one of most common inexpensive fragrances. Two isomers of limonene can be isolated from
natural sources. They are shown below. The one on the left has the odor of lemons, and the one on the right has the odor
of oranges.
S
R
(a) What kind of isomers are they?
The two isomers differ only in the configuration of the single chiral center so these are enantiomers.
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Chapter 5: Alkenes: Bonding, Nomenclature and Properties
(b) Are E,Z isomers possible in limonene?
There are no E,Z isomers possible for limonene. One alkene is part of a cyclohexene ring, which is too small to
accommodate a trans geometry. The other is a terminal alkene which also does not have E,Z isomer
possibilities.
(c) Why do they smell different?
Many of the molecules of our body are chiral, present as single enantiomers. This includes the receptor
molecules present in our noses that are responsible for our sense of smell. Thus, different enantiomers often
smell different to us, because they interact differently with the chiral receptor molecules within our noses.
Looking Ahead
Problem 5.36 Bromine adds to cis and trans-2-butene to give different diastereomers of 2,3-dibromobutane. What does
this say about the mode of addition of bromine to this alkene.
CH3
Br
Br2
H
CH3
Br
CH3
Br2
H
Br
H
H
Br
CH3
In this reaction, one Br atom of Br2 adds to each of the carbon atoms of the alkene. The interesting feature of
this process is that two new chiral centers are created in the product. Because different diastereomers are
created from different isomeric starting materials, this reaction is classified as being stereospecific. In Section
6.3D you will learn that this occurs because the two bromine atoms add to opposite faces of the original alkene.
This so-called anti-stereoselective addition is caused by formation of a three-membered ring intermediate. The
point to make now is that when new chiral centers are created in reactions, it is important to keep track of how
the mechanism of the reaction dictates which of the possible stereoisomers are produced.