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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

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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 +

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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.

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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−

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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

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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

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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

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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

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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

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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

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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

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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