Chapter 8 Alkenes: Structure and Preparation via Elimination

33 Pages·2010·0.29 MB·English

Chapter 8
Alkenes: Structure and
Preparation via Elimination Reactions

Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 8. Each of the sentences below appears in the section entitled Review
of Concepts and Vocabulary.

• Alkene stability increases with increasing degree of ________________.
• E2 reactions are said to be regioselective, because the more substituted alkene,
called the ___________ product, is generally the major product.
• When both the substrate and the base are sterically hindered, the less substituted
alkene, called the __________ product, is the major product.
• E2 reactions are stereospecific because they generally occur via the
__________________ conformation.
• Substituted cyclohexanes only undergo E2 reactions from the chair conformation
in which the leaving group and the proton both occupy _________ positions.
• E1 reactions exhibit a regiochemical preference for the ___________ product.
• E1 reactions are not stereospecific, but they are stereo___________.
• Strong nucleophiles are compounds that contain a _____________ and/or are
________________.
• Strong bases are compounds whose conjugate acids are _______________.

Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 8. The answers appear in the section entitled
SkillBuilder Review.

8.1 Assembling the Systematic Name of an Alkene
PROVIDE A SYSTEMATIC NAME FOR THE FOLLOWING COMPOUND
1) IDENTIFY THE PARENT
2) IDENTIFY AND NAME SUBSTITUENTS
3) ASSIGN LOCANTS TO EACH SUBSTITUENT
4) ALPHABETIZE

8.2 Assigning the Configuration of a double bond
HO
ASSIGN THE CONFIGURATION
OF THE DOUBLE BOND IN THE
FOLLOWING COMPOUND
140 CHAPTER 8

8.3 Comparing the Stability of Isomeric Alkenes
CIRCLE THE MOST STABLE ALKENE BELOW

8.4 Drawing the Curved Arrows of an Elimination Reaction
A CONCERTED MECHANISM A STEPWISE MECHANISM
DRAW THREE CURVED ARROWS, SHOWING A DRAW A CURVED ARROW SHOWING THE LOSS OF THE LEAVING GROUP
PROTON TRANSFER ACCOMPANIED BY TO FORM A CARBOCATION INTERMEDIATE, FOLLOWED BY ANOTHER
SIMULTANEOUS LOSS OF A LEAVING GROUP TWO CURVED ARROWS SHOWING A PROTON TRANSFER
H - LG H - LG H Base
Base LG LG

8.5 Predicting the Regiochemical Outcome of an E2 Reaction
DRAW THE ELIMINATION PRODUCTS OBTAINED WHEN THE COMPOUND BELOW IS TREATED WITH A STRONG BASE.
Cl
Strong Base +
ZAITSEV HOFMANN

8.6 Predicting the Stereochemical Outcome of an E2 Reaction
PREDICT THE STEREOCHEMICAL OUTCOME OF THE FOLLOWING REACTION, AND DRAW THE PRODUCT.
Strong Base
Cl

8.7 Drawing the Products of an E2 Reaction
PREDICT THE MAJOR AND MINOR PRODUCTS OF THE FOLLOWING REACTION.
Br
NaOEt
+
MAJOR MINOR

8.8 Predicting the Regiochemical Outcome of an E1 Reaction
PREDICT THE MAJOR AND MINOR PRODUCTS OF THE FOLLOWING REACTION.
OH
H SO
2 4 +
heat
MAJOR MINOR
CHAPTER 8 141

8.9 Drawing the Complete Mechanism of an E1 Reaction
IDENTIFY THE TWO CORE STEPS AND TWO POSSIBLE ADDITIONAL STEPS OF AN E1 PROCESS
TWO CORE STEPS
TWO POSSIBLE ADDITIONAL STEPS

8.10 Determining the Function of a Reagent
IDENTIFY REAGENTS THAT FALL INTO EACH OF THE FOUR CATEGORIES BELOW:
NUCLEOPHILE (ONLY) BASE STRONG STRONG WEAK WEAK
(ONLY) NUC BASE NUC BASE

8.11 Identifying the Expected Mechanism(s)
IDENTIFY THE MECHANISM(S) THAT
OPERATE IN EACH OF THE CASES BELOW: 1 +
MINOR MAJOR
STRONG STRONG
2 +
NUC BASE
MAJOR MINOR
3
ONLY

8.12 Predicting the Products of Substitution and Elimination Reactions
FILL IN THE BLANKS BELOW:
STEP 1 STEP 2 STEP 3
DETERMINE THE FUNCTION ANALYZE THE ________________ CONSIDER ANY RELEVANT REGIOCHEMICAL AND
OF THE _________________ AND DETERMINE THE EXPECTED ____________________ REQUIREMENTS
MECHANISM(S).
142 CHAPTER 8

Review of Synthetically Useful Elimination Reactions
Identify reagents that will achieve each of the transformations below. To verify that your
answers are correct, look in your textbook at the end of Chapter 8. The answers appear in
the section entitled Review of Synthetically Useful Elimination Reactions.

OH

Br

Br

Solutions

8.1.
a) 2,3,5-trimethyl-2-heptene
b) 3-ethyl-2-methyl-2-heptene
c) 3-isopropyl-2,4-dimethyl-1-pentene
d) 4-tert-butyl-1-heptene

8.2.
a) b) c)

8.3. 2,3-dimethylbicyclo[2.2.1]hept-2-ene

8.4.

this substituent counts twice
a) trisubstituted b) disubstituted c) trisubstituted
d) trisubstituted e) monosubstituted
CHAPTER 8 143

8.5.
O
F Cl
a) E b) Z c) Z d) Z

8.6. When using cis-trans terminology, we look for two identical groups. In this case,
there are two ethyl groups that are in the trans configuration:

Cl
trans

However, when using E-Z terminology, we look for the highest priority at each vinylic
position. Chlorine receives a higher priority than ethyl, so in this case, the highest
priority groups are on the same side of the pi bond:
Cl
Z
Below are two other examples of alkenes that have the trans configuration, but
nevertheless have the Z configuration:
F Br

8.7.

a)
increasing stability
di- tri- tetra-
substituted substituted substituted

b)
increasing stability
mono- di- tri-
substituted substituted substituted
144 CHAPTER 8

8.8. In the first compound, all of the carbon atoms of the ring are sp3 hybridized and
tetrahedral. As a result, they are supposed to have bond angles of approximately 109.5º,
but their bond angles are compressed due to the ring (and are almost 90º). In other
words, the compound exhibits angle strain characteristic of small rings. In the second
compound, two of the carbon atoms are sp2 hybridized and trigonal planar. As a result,
they are supposed to have bond angles of approximately 120º, but their bond angles are
compressed due to the ring (and are almost 90º). The resulting angle strain (120º (cid:1) 90º)
is greater than the angle strain in the first compound (109.5º (cid:1) 90º). Therefore, the
second compound is higher in energy, despite the fact that it has a more highly
substituted double bond.

8.9.
H
OH
Cl
a)
H
OTs OEt
b)
OMe
H
Br
c)

8.10.
H
- Cl H O
Cl 2
a)

Br - Br MeOH
b) H

- I EtOH
H
I
c)
CHAPTER 8 145

8.11. This mechanism is concerted:
OTs
EtO
H

8.12.
H
- Br EtOH
Br

8.13.
a) 3x faster b) 2x faster c) 6x faster

8.14.
a)
increasing reactivity towards E2
Br Br
Br
primary secondary tertiary
substrate substrate substrate

b)
increasing reactivity towards E2
Cl Cl
Cl
primary secondary tertiary
substrate substrate substrate

8.15.

Cl
EtO
+
major minor
a) (more substituted) (less substituted)
146 CHAPTER 8

Cl
O
+
major minor
b) (less substituted) (more substituted)
I
NaOH
+
major minor
c) (more substituted) (less substituted)
I O
+
major minor
d) (less substituted) (more substituted)
Br
NaOH
e) only product
Br O
f) only product

8.16.
a) The more substituted alkene is desired, so hydroxide should be used.
b) The less substituted alkene is desired, so tert-butoxide should be used.

8.17.
a)
O
Br
HO
Br
CHAPTER 8 147

b)
O
Br
Br HO

8.18.
+
a) major minor b) the only E2 product
c) the only E2 product d) the only E2 product e) the only E2 product

+
f) the only E2 product g) major minor

+
h) major minor

8.19.
Strong Base
Br
148 CHAPTER 8

8.20. The leaving group in menthyl chloride can only achieve antiperiplanarity with one
beta proton, so only one elimination product is observed. In contrast, the leaving group
in neomenthyl chloride can achieve antiperiplanarity with two beta protons, giving rise to
two possible products:
Cl
H
menthyl chloride
H
+
Cl
H
neomenthyl chloride

8.21. Because of the bulky tert-butyl group, the first compound is essentially locked in a
chair conformation in which the chlorine occupies an equatorial position. This
conformation cannot undergo an E2 reaction because the leaving group is not
antiperiplanar to a proton. However, the second compound is locked in a chair
conformation in which the chlorine occupies an axial position. This conformation rapidly
undergoes an E2 reaction. Therefore, the second compound is expected to be more
reactive towards an E2 process than the first compound.

8.22.
NaOEt
+
Cl
a) major minor

Br
NaOEt
+
b) major minor

Br
NaOEt
c) only E2 product