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CHAPTER 8 HW SOLUTIONS: ELIMINATIONS REACTIONS
CIS-TRANS ISOMERISM
1. Use a discussion and drawing of orbitals to help explain why it is generally easy to rotate around single bonds at room temperature, while it is difficult to rotate around double bonds.
A sigma bond is a cylindrically symmetrical bond made by a head-to-head overlap of orbitals. In an alkane C-C single bond, each C atom uses an sp3 hybrid orbital to make the sigma bond. Rotation around that bond is possible without breaking the overlap of those sp3 orbitals.
The pi bond of a double bond is made by side-by-side overlap of p orbitals. Rotation around a double bond would result in the p-orbitals being perpendicular, which means rotation breaks the pi bond. This takes too much energy, so rotation about a double bond is difficult at room temperature.
2. Which of the following compounds can exist as cis and trans isomers? When this type of isomerism is possible, draw the other isomer (if trans is shown, draw cis).
no cis/trans possible
no cis/trans possible
no cis/trans possible
no cis/trans possible
trans is too strained
3. Draw all isomers of C2H2BrCl, including stereoisomers.
BrH
CC
Br
Br
H
cisOH
Cl
cis trans
BrC
CBr
H
H
transOH
Cltrans
cis
Cl
Cl Brcis
Br trans
C CH
Br Cl
HC C
H
Br H
ClC C
H
H Cl
Br
Cl
HH
Cl
HH
σ rotate Cl
HH
H
ClH
σ
ClH HCl
π
rotate HCl Cl
H
Page 2
4. Identify the relationship between each pair of compounds. Are they identical, constitutional isomers, or stereoisomers?
Pair
Relationship Constitutional Isomers Stereoisomers
Pair
Relationship Identical Constitutional Isomers
Pair
Relationship Stereoisomers Identical
ALKENE STABILITY
5. Answer the questions while using the following enthalpy data.
DH˚ (kcal/mol)
-30.3
-28.6
a. Use the DH˚ data to determine the relative stability of 1-butene compared to cis-2-butene. Explain
how you used the enthalpy data.
See the diagram at right. Both products are the same (butane) so have the same energy. 1-butene releases more energy than cis-2-butene, so 1-butene must start at a higher energy. This means that cis-2-butene is lower energy (by 1.7 kcal/mol).
b. Would you expect a similar reaction of trans-2-butene to release more or less heat than cis-2-butene? Briefly explain.
Since trans-2-butene is lower in energy than cis-2-butene (methyl groups are further apart, fewer electronic repulsions), it would release less energy in a similar reaction. The ∆H˚ value would be smaller (or less negative; actual value = -27.6 kcal/mol)
ClCl
12
33
21
12
12 3
H3C H3C
transtranstrans
trans
H2
Pd
H2
Pd
30.3 28.6
H2 + H2 +
Page 3
6. Answer the questions about the following alkenes. a. Rank the following alkenes from low to high energy. Explain the trend.
The more substitution around a double bond, the lower energy due to more hyperconjugation / electron delocalization of the pi bond.
b. Which alkene would you expect to release the least amount of heat upon combustion? Explain.
The tetrasubstituted alkene is lower energy than the trisubstituted alkene. This means it starts at a lower energy and so releases less heat upon combustion. (Similar looking diagram to question 5a).
E2 REACTIONS: MECHANISMS + TRANSITION STATES
7. Draw the curved arrow mechanism for these E2 reactions, including all lone pairs and charge. Include any missing products.
C C
BrH
H
H H
C C
H
H
H
H
HC C
H
H
H H
C C
H
H
H
H
a.O H
O H Br
O HH
b. Br NaOH
H O H
Br
O HH
c.Cl
KOCH3
HOH3C
OH3C HCl
monosubstituted disub. trisub.highest energy lowest energy
trisub.tetrasub.
Page 4
8. Draw the energy diagram for this E2 reaction, including the structure of the transition state.
9. For each E2 reaction, give the products then draw the E2 transition state.
10. Draw all possible elimination products for the following reactions, including stereoisomers. Decide which should be the major elimination product and briefly explain why.
Only one alkene is possible (there is only one type of b-H).
There are three possible products. The major product is the lowest energy (disubstituted and trans).
Circled is the lowest energy product. It is trisubstituted and has the largest groups on each carbon of the alkene (methyl and ethyl) anti.
Only one alkene is possible.
IOH-
a.Cl
OH-Cl
HO
H
δ−
δ−
+ Cl- + H2O
OKb.
Br + Br- +OH
Br
HOδ−
δ−
a. NaOC(CH3)3Cl
b. KOCH3Br
c. NaOHI
d. ClKOC(CH3)3
SM
PR
TS I
H OH
δ−
δ−
Page 5
E2 REACTIONS: SUMMARY OF FACTORS
11. In each pair of reactions, decide which will be a faster E2 reaction. Explain.
Bromine is a better leaving group than chlorine. It can better stabilize the developing negative charge in the transition state due to its large size.
OH- is charged while H2O is neutral, which makes it start at a higher energy than water. This makes it more reactive, in this case more basic.
3˚ RX are more reactive than 2˚ because the developing alkene in the transition state is more substituted, meaning lower energy.
Aprotic solvents make 2nd order reactions faster because the anion/base is less stabilized than in protic solvents. DMSO can’t solvate OH- as well as water because the d+ is crowded. This makes OH- start at a higher energy in DMSO (more reactive).
12. When NaOCH3 is used as the base in the following reaction, the major product is M. When
NaOC(CH3)3 is used as the base, the major product is N. Explain these different results.
The transition states leading to the two products are important since E2 is a kinetically controlled reaction.
With a “normal” sized base like NaOCH3, transition state M is lower in energy since it involves a more substituted partial alkene.
With a bulky base like NaOC(CH3)3, there is too much steric hindrance in the M transition state (sterics dominate over alkene stability). There is less hindrance in transition state N, where the base removes the more accessible external beta-hydrogen, so this transition state is lower in energy and the major product is N.
a. NaOH
NaOH
Br
Cl
Br
H OH
δ−δ−
b. NaOH
H2O
Br
Br
c. NaOH
NaOH
I
I
I
H OH H
I
OH
δ−
δ−
δ−
δ−
vs. becoming trisub.becoming
disub.
d. NaOH
NaOH
Cl
ClH2O
DMSO
Brbase +
M N
Br
H
Br
H
δ−
δ−
δ−
δ−M N
BaseBase
Page 6
E2 REACTIONS: ANTI ELIMINATION
13. Which should be the major product (J-M) of this E2 reaction? Explain, including Newman projections with your answer.
The major product is not J because it’s only monosubstituted (higher energy), and it definitely is not K- there’s no way for the stereocenter with the D to invert.
The Newman projection is helpful to decide between products L and M (both are disubstituted alkenes). The beta-hydrogen removed by the base must be anti to the leaving group, which is the H, not the D. Thus, the deuterium remains and the major product is M.
14. Only one product is observed in reaction (1) while a mixture is formed in reaction (2).
a. Use chair conformations to explain why only one product is formed in reaction (1).
In chair conformations, the E2 mechanism occurs when the leaving group is axial and the beta-hydrogen is also axial (it is then anti). There is only one appropriate axial b-H (see chair below).
b. Use chair conformations to explain why two products are formed in reaction (2), then why one product is the major.
In reaction (2), there are two b-H that are axial (and thus anti) to the leaving group, thus two products can form. The major product is the one lower in energy, the one with the more substituted alkene.
D
ClKOH
D D
J L M
D
K
Br
CH3H
H HCH3CH2O not
anti
H
Br
CH3
H antianti
Br NaOCH2CH3
HOCH2CH3(1) 100%
Br NaOCH2CH3
HOCH2CH3(2)
major
+
CH2CH3
H DCH3
ClH
OH
Page 7
15. Give the major organic product for these E2 reactions. Keep in mind that the b-hydrogen must be anti to the leaving group in an E2 reaction.
Rotation involved to get b-H anti
E1 REACTIONS: MECHANISMS + TRANSITION STATES
16. Give the curved arrow mechanism for these E1 reactions.
a.I
NaOCH3
H
f.
Br
D
NaOH
D
b.Br
CH3
OKg.
Br
D
NaOH
I
HD
c. NaOCH3D
CH3
Cl
KOHh.CH3
d.
Cl
NaOH i.CH3
NaOCH3
CH3Cl
CH3
CH3
β-H not anti
Br
CH2CH3KOHe.
CH2CH3NaOH
CH3
I
j.
Br
CH3OH
high heat(major product is still SN1, but
E1 competes significantly)a.
OCH3
HH
b.Cl
OH
heat
HO
H
Page 8
17. Draw the energy diagram for this reaction, including the structure of the transition state for the rate determining step.
E1 REACTIONS: SUMMARY OF FACTORS
18. In each pair of reactions, decide which should have a faster E1 reaction and explain. (Note: in some reactions E1 is not the major pathway, but still decide if an E1 were to occur, which would be faster.)
The transition state of the RDS has the leaving group d-. –OSO2CH3 stabilizes the d- better than Cl because it spreads out the partial charge through resonance.
Equal rates. E1 is first order, so the rate is dependent on the structure of the alkyl halide which is the same for both reactions. The Nu or base is not involved in the RDS so doesn’t affect the rate.
3˚ RX are faster in E1 reactions because the RDS forms a carbocation, and 3˚ carbocations are lower energy than 2˚.
1st order reactions are faster with protic solvents (e.g. water) because they stabilize the LG– formed after the RDS (or they stabilize the d- in the transition state). Cl– is not stabilized in DMSO (naked anion) so the intermediate E is higher, and the reaction slower.
Brheat
H2O
a.H2O
H2O
Cl
OSO2CH3
heat
heat
b.
H2O
O
ONa
heat
H2O, heat
Br
Br
c. I
I
H2O
H2O
heat
heat
d.H2O
H2O/DMSO
heat
heat
Cl
Cl
SM
PR
carbo-cation
Brδ+ δ−
+
Br
Cl DMSOCl H2O
Cl Cl
Page 9
E1 VERSUS E2 REACTIONS
19. For each elimination reaction, decide whether it proceeds through an E1 or E2 mechanism. Then give the major product and draw the appropriate curved arrow mechanism.
Reaction E1/ E2?
Major Product Mechanism
E2 CH2CH3
I
CH2CH3 CH2CH3
O CH3H
E1 Br
CH3OHH
E2
I
HO
E1
CH3 H2O
Cl
CH3 CH3H3CH
ELIMINATION VERSUS SUBSTITUTION REACTIONS
20. Explain why the amount of elimination product (relative to substitution product) increases when the heat of the reaction is increased.
Elimination reactions have a favorable change in entropy because they convert 2 reactants into 3 products. Substitutions have the same number of reactants as products, so typically have a negligible ∆S˚.
Since ∆G˚= ∆H˚–T∆S˚, temperature is linked with the change in entropy term. As temperature increases, it makes the T∆S˚ term more important to the overall ∆G. This has little effect on substitutions (∆S˚~0) but makes eliminations more favorable (∆S˚= +).
a. NaOCH3
I
CH2CH3
b.Br
CH3OH
heat
c. NaOC(CH3)3I
H2Od.
CH3
Cl
high heat
Page 10
21. Should compound A or compound B be expected to react faster in each mechanism? Briefly explain.
A is 1˚ RX, B is 3˚ RX.
SN2 SN1 E2 E1
1˚ RX faster (A)
Least hindered at transition state.
3˚ RX faster (B)
Lower energy carbocation
3˚ RX faster (B)
Partial alkene in TS is more substituted.
3˚ RX faster (B)
Lower energy carbocation
22. Which of reactions A-D should never proceed through an E1 or SN1 mechanism (there may be >1
correct answer)? Explain why these reactions should not react through these mechanisms.
A B C D
C will never go through an E1 or SN1 mechanism because it involves reaction of a 1˚ RX, and 1˚ carbocations are too high energy.
A also won’t undergo E1 or SN1 reactions because the base NaOCH3 is too strong. The base is so reactive it will go through 2nd order reactions (will “go for it”!), and E2 is the major pathway for A. Neutral nucleophiles are more common for 1st order reactions because they are only reactive enough if paired with a carbocation.
23. Substitution and elimination reactions always compete with each other. For each reaction below, draw the probable substitution and elimination products, considering if a 1st or 2nd order mechanism is likely. Then identify the major product.
BrBr
A B
BrNaOCH3
BrCH3OH I CH3CH2OH
IH2O
a.H2OBr
heat
OH
SN1 (racemic) E1
b.KOH
CH3
IDMF
CH3
OH
CH3
E2SN2
c.NaCNBr
acetone
CN
E2SN2
Page 11
24. Explain why reaction 1 and 2 produce different alkenes as their major products.
Reaction 1 proceeds through an E2 mechanism, which requires the b-H and leaving group to be anti (“trans-like”). The alkene can’t form near the isopropyl group because the b-H there is not anti. Reaction 2 proceeds through an E1 mechanism, which involves a carbocation, and there is no special orientation of the b-H in order to eliminate. Reaction 2 therefore forms the lower energy more substituted product.
25. For each reaction, decide whether the major pathway should be SN1, SN2, E1, or E2. Then give the appropriate curved arrow mechanism that produces the major product for each.
Mechanism and Major Product SN1, SN2, E1, E2?
a.
ClCH3OH
OCH3OCH3
H
O
H OCH3H
CH3
SN1
b.
Br NaOCH3
H O CH3
E2
c.
I CH3CH2OH
high heatH OCH2CH3
H
E1
d.
OSO2CF3 NaSH
SH
S H
SN2
NaOCH3
BrCH3OH
Br
Reaction 1 Reaction 2
Page 12
26. Give the major organic product for each reaction. Pay attention to stereochemistry and if a mixture of products is expected (i.e. racemic mixture), draw all products. Identify if the reaction mechanism is SN1, SN2, E1, or E2.
a.Br OH OH3C
SN1 racemic
CH3O
j. Br
CH3
CH3
OH
O
O
CH3
CH3 OSN1mixture
of diasteromers
CH3
O
CH3
O
b. KOCH2CH3I OSN2k.
Br CH3CH2OHheat E1
c. NaNH2Cl E2
l.
CH3
OSO2CF3
KOHacetone
CH3
E2
d.H2O
high heatCl E1
m.Cl (CH3)3CONa
E2
e.KSHI
DMSO
SHSN2 n. CH3Br
OK
O
O
O
CH3
SN2
f.NaOH
BrDMF E2 o.
NaN3CH3CNBr N3
SN2
g.
Cl
ONa
O
O OSN2
p.Cl
D
KOHacetone
H
H
E2
h.
ClCH3
ONa
CH3
E2 q.Cl
D
KOH
acetone
H
D
E2
i. KCNO
SCH3
O OCNSN2
r.excess
F Br CH3NH-F N
CH3
HSN2