Chapter 11
11.1 What product would you expect to obtain from a nucleophilic substitution reaction of
(S)-2-bromohexane with acetate ion, CH3CO2-? Assume that inversion of configuration occur, and
show the stereochemistry of both reactant and product.
Solution:
Br
H
H
OCOCH3
O
(R)-1-methylpentyl acetate
(S)-2-bromohexane
-O
11.2 What product would you expect to obtain from SN2 reaction of OH- with (R)-2-bromobutane?
Show the stereochemistry of both reactant and product.
Solution:
H
HO
Br
H
OH-
(R)-2-bromobutane
(S)-2-butanol
11.3 Assign configuration to the following substance, and draw the structure of the product that would
result on the nucleophilic substitution reaction with HS-. (reddish-brown=Br-)
Solution:
Br
H
H
SH
SH-
S
R
11.4 What product would you expect from SN2 reaction of 1-bromobutane with each of the following?
(a) NaI
Solution:
H
H
Br
I
H
H
I
(b) KOH
Solution:
H
H
Br
OH
H
H
(c) H
C
HO
C
Li
Solution:
H
H
Br
H
C
C
H
H
C
HC
(d)NH3
Solution:
H
H
Br
NH2
H
H
H3N
H
+
11.5 Which substance in each of the following pairs is more reactive as a nucleophile? Explain.
H 3C
H3C
N
(a) H3C
Solution:
or
NH
H 3C
Br
H3C
N
H3C
Because it’s negative charged.
(b)
H3C
CH3
B
CH3
H3C
or
CH3
N
CH3
Solution:
H3 C
CH3
N
CH3
Because the N atom has a lone pair electron, so it’s more likely to donate electron. As a result, it has
higher basicity and is more reactive as a nucleophile. What’s more, the N is below the B in a column of
the periodic table.
(c)H2O or H2S
Solution:
H2S
Because the S is below the O in a column of the periodic table, so it is more reactive as a nucleophile.
11.6 Rank the following compounds in order of their expects reactivity toward SN2 reaction:
CH3Br, CH3OToS, (CH3)3CCl, (CH3)2CHCl
Solution:
Reactivity toward SN2 reaction:
CH3OToS > CH3Br > (CH3)2CHCl > (CH3)3CCl
11.7 Organic solvents such as benzene, ether, and chloroform are neither protic nor strongly polar.
What effect would you expect these solvents to have on the reactivity of a nucleophile in SN2
reactions?
Solution:
Organic solvents such as benzene, ether, and chloroform are not suitable solvents for typical SN2
reactions.
11.8 What products would you expect from reaction of (S)-3-chloro-3-methyloctane with acetic acid?
Show the stereochemistry of both reactant and product.
Solution:
HO
O
This side shielded from attack
ClCl
spontaneous
dissociation
(S)-3-Chloro-3-methyl-octane
This side open to attack
HO
O
O
O
O
O
+
S
retention
R
inversion(excess)
11.9 Among the numerous examples of SN1 reactions that occur with incomplete racemization is one
reported by Winstein in 1952. The optically pure tosylate of 2,2-dimethyl-1-phenyl-1-propanol
([α]D=-30.3°) was heated in acetic acid to yield the corresponding acetate ([α]D = + 5.3°). If complete
inversion had occurred, the optically pure acetate would have had [α]D = + 53.6°. What percentage
recemization and what percentage inversion occurred in this reaction?
OTos
(H3C)3 C
C
H
OAc
HOAc
AcO-
[α ]D=-30.3°
(H3C)3 C
C
H
+ HOTos
Observed [α ]D=+5.3°
(optically pure [α ]D=+53.6° )
We assume that the enantiomer of [α]D=53.6°make up of the mixture by x percent, so the one of
[α]D=-53.6°constitutes by (1-x), we get:
53.6x-53.6(1-x)=5.3
so x=c%,
than we get the conclusion that the there is x-(1-x)=2x-1=9.888% that occurred inversion of its
configuration, and there is 2(1-x)=1-(2x-1)=2-2x=90.112% percentage takes place in the racemization.
11.10
Assign configuration to the following substrate, and show the stereochemistry and identity of
the product you would obtain by SN1 reaction with water (reddish-brown=Br)
Solution:
The first step(rate limiting step)
CH3
S
Br
CH3
C
CH3
C
H2
C
CH3
C
H2
+
Br
CH3
C
C
H2
CH3
O
H
H
H
H
O
H
CH3
O
H
H2C
CH3
CH3
H2C
O
O
CH3
H
H
H
H
OH
CH3
+
+
3
CH3
H2C
O
CH3
+
H
+
3
H
O
H2C
CH3
OH
11.11 Rank the following substances in order of their expected SN1 reactivity.
Br
Br
H3C
Solution:
H2
C
Br
H2C
C
H
C
H
CH3
H2C
C
H
Br
H3C
C
H
CH3
H
Br
H2C
C
H2C
The least stable
C
H
H
H
C
C
H3C
H3C
Br
H
H
The stability of the carboncation
H
H
C
C
H3C
CH3
H3C
Br
CH3
H
H
C
H2C
C
CH3
C
H Br
H2C
C
H
The most stable
CH3
11.12 3-Bromo-1-butene and 1-bromo-2-butene undergo SN1 reaction at nearly the same rate even
though one is a secondary halide and the other is primary. Explain.
Solution:
H2C
C
H
H
C
CH3
H3C
C
H
C
H
H2
C
Br
Br
CH3
H2C
C
H
C
H
H3C
C
H
C
H
H
C
H
The starting materials are different definitely, however, you can see clearly in the diagram above that
the same allylic carbocation is formed by resonance. And so the rate of the reaction is same.
11.13 1-Chloro-1,2-diphenylethane reacts with the nucleophiles fluoride ion and triethylamine at the
same rate, even though one is charged and one is neutral. Explain.
Solution: Because it is a SN1 reaction. The rate of SN1 reaction is only depending on the
concentration of the substrate. And it is not related to the nucleophiles.
11.14 Predict whether each of the following substitution reactions is likely to be SN1 or SN2.
(a)
HCl
OH
Cl
CH3OH
It is likely to be an SN1 reaction. The substrate is secondary and the nucleophile is weakely
base and the solvent is acidic.
CH3
(b)
CH3
Na+ -SCH3
H2C
CCH2Br
CH3CN
H2C
CCH2SCH3
It is likely to be an SN2 reaction. The substrate is primary and the nucleophile is a reasonably
good one and the solvent is polar aprotic.
11.15 Ignoring double-bond stereochemistry, what products would you expect from elimination
reaction of the following alkyl halides? Which product will be major in each case.
CH3
CH3
Br CH3
(a) CH3CH2CHCHCH3
CH3CH2CH
CCH3
+
CH3 CH
(major)
CH3
Cl
CH3
CH3CHCH2 C
(b)
CHCH3
CH3
CH3CHCH
CH3
(minor)
CH3
C
CHCHCH3
CHCH3
CH3
+
CH3
CH3CHCH2 C
CH3
CCH3
CH3
(minor)
(major)
Br
CHCH3
CHCH3
C
H
+
(c)
(major)
11.16
(minor)
What alkyl halides might the following alkenes have been made from?
Solution:
(a)
Br
CH2
(b)
CH3
CH3
CH3
Br
CH3
11.17 What stereochemistry do you expect for the alkene obtained by E2 elimination of
(1R,2R)-1,2-dibromo-1,2diphenylethane ? Draw a Newman projection of the reacting conformation.
Solution : The product of this E2 elimination is Z-1-bromo-1,2-diphenylethene .The structure of it as
Br
following :
Z-1-bromo-1,2-diphenylethene
Ph
The Newman projection of the reacting conformation:
H
Br
H
Br
Ph
11.18 What stereochemistry do you expect for the trisubstitued alkene obtained by E2 elimination of
the following alkyl halide on treatment with KOH? (Reddish-brown =Br)
Solution:
We will get
H3C
H
H2C CH3
CH3
11.19 Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert
–butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable
chair conformation, and explain your answer.
Solution:
Br
Br
t-Bu
t-Bu
The right (cis) one is more faster to undergo E2.Because Br is axial .It does not need to flip over like
the left one (trans).
11.20: Tell whether each of the following reaction is likely to be SN1, SN2, E1, E2.
Solution:
(a):CH3CH2CH2CH2Br+NaN3
.
CH3CH2CH2CH2N3
SN2
(b): CH3CH2CH(Cl)CH3+KOH
CH3CH2CH==CH3
E2
OCOCH3
Cl
O
+
OH
(c):
SN1
11.21: Write the product you would expect from those reactions.
Solution:
(a): 1.CH3CH2Cl + NaSCH3
2. CH3CH2Cl + NaOH
CH3CH2SCH3
CH3CH2OH
Cl
SCH3
+
NaSCH3
(b): 1.
Cl
+
NaOH
2.
SCH3
Cl
+ NaSCH3
(c): 1.
Cl
Cl
Ring-flip
Cl
2.
+ NaOH
11.22 From what alkyl bromide was the following alkyl acetate made by SN2 reaction? Write the
reaction, showing all stereochemistry.
H
CH3
H
O
CH3
O
(Sawhorse)
Solution:
H
H
Br
H
=
CH3
CH3
CH3
H
Br
CH3
CH3 CO2
H
H
CH3
H
O
=
H
O
CH3
CH3
CH3
O
O
11.23 Assign R or S configuration to the following molecule, write the product you would expect from
SN2 reaction with NaCN, and assign R or S configuration to the product:
H
H
C
H
H
C
H
C
O
C
C
C
H H
H Cl
H H
H
Solution: The configuration of the molecule above is S. The product of SN2 reaction with NaCN is that
H
H
C
H
H
C
C
C
following:
H
C
H H
NC H
O
C
H
H H
. It experiences a Walden Inversion and the
configuration is R.
11.24: Draw the structure and assign Z or E stereochemistry to the product you expect from E2 reaction
of the following molecule with NaOH:
Cl
S
Ph
Me
H
Et
R
H
Strategy: The reaction should follow the anti-co-planar.
Solution: The stereochemistry of the product should be E:
Ph
H
Me
Et
11.25:Describe the effects of each of the following variables on both SN2 and SN1 reaction:
(a):Solvent
(b):Leaving group
(c):Nucleophile
(d): Substrate
Solution:
(a):solvent:
SN1: The good solvent will due largely to stabilize carbocation.
SN2: The good solvent will due largely to stabilize the transition state.
(b): Leaving group:
SN1: The good leaving group should be weaker base.
SN2: The good leaving group should be weaker base.
(c): Nucleophile:
SN1: No relation to the reaction rate.
SN2: The strong nucleophile will benefit for SN2.
(d): Substrate:
SN1: The good substrate will yield the more stable carbocation intermediates.
SN2: The good substrate will be primary, allylic and benzilic halides.
11.26 Which choice in each of the following pairs will react faster in an SN2 reaction with OH ?
(a) CH3Br or CH3I
(b) CH3CH2I in ethanol or in dimethyl sulfoxide
(c) (CH3) 3CCl or CH3Cl
(d) H2C CHBr or H2C CHCH2Br
Solution: (a) CH3I will react faster in an SN2 reaction with OH . Because as a leaving group, I
is
more reactive than Br .
CH
3CH2I in dimethyl sulfoxide will react faster in an SN2 reaction with OH . Because
(b)
SN2 reactions react faster in polar aprotic solvents than in protic solvents.
(c) CH3Cl will react faster in an SN2 reaction with OH . Because methyl halides is a more
reactive substrates than tertiary halides.
(d) H2C CHCH2Br will react faster in an SN2 reaction with OH . Because vinylic
halides are unreactive toward SN2 reaction.
11.27 What effect would you expect the following changes to have on the rate of the SN2 reaction of
1-iodo-2-methylbutane with cyanide ion?
(a) The CN concentration is halved, and the 1-iodo-2-methylbutane concentration is doubled.
(b) Both the CN and the 1-iodo-2-methylbutane concentrations are tripled.
Solution: (a) The rate of the SN2 reaction will not be changed.
(b) The rate of the SN2 reaction will be as 9 times as before.
11.28 What effect would you expect the following changes to have on the rate of the reaction of ethanol
with 2-iodo-2-methylbutane?
(a) The concentration of the halide is tripled?
(b) The concentration of the ethanol is halved by adding diethyl ether as an inert solvent.
Solution: (a) The rate will also be tripled
(b)It would not be changed.
11.29 How might you prepare each of the following molecules using a nucleophilic substitution
reaction at some step?
(a) H3CC CCH(CH3)2
(c) H3C O C(CH3)3
3
(e)
(b) CH3CH2CH2CH2CN
(d) CH3CH2CH2NH2
Br
CH3
PCH3Br
(f)
Solution: (a)
CH3
LiC CCHCH3 + CH3I
(b)
THF
H3CC CCH(CH3)2
LiI
Na+CN-
CH3CH2CH2CH2 Br
CH3CH2CH2CH2CN + NaBr
THF-HMPA
H3C O C(CH3)3 + Br
CH3Br
(c) (CH3)3CO +
NaNH2
(d) CH3CH2 CH2Br
CH3CH2 CH2NH2 + NaBr
P
P
CH3
Br
H3C
Br
(e)
OH
CH3
Br
CH3
PBr3
Ether
(f)
11.30 Which reaction in each of the following pairs would you respect to be faster?
(a) The SN2 displacement by I- on CH3Cl or on CH3OTos
Solution: On CH3OTs.
The SN2 displacement by CH3 CO2 on bromoethane or on bromocyclohexane
Solution: On bromoethane.
(c) The SN2 displacement on 2-bromopropane by CH3CH2 O or by CN
(b)
-
Solution: By CN .
(d) The SN2 displacement by
HC
C
on bromoethane in benzene or in hexamethylphoramide
Solution: By hexamethylphoramide.
11.31 What products would you expect from the reaction of 1-bromopropane with each of the
following?
(a) NaNH2
(d) NaCN
(b) KOC(CH3)3
CH
(e) NaC
(c) NaI
(f) Mg, then H2O
Solution:
CH3CH2CH2 Br
+ NaNH2
CH3CH2CH2 Br
KOC(CH3)3
(c) CH3 CH2CH2 Br +
(d)
(e)
CH2 + CH3 CH2 CHNH2 + NaBr
(major)
(a)
(b)
H3CHC
NaI
CH3 CH2CH2 Br + NaCN
H3CHC
CH2
CH3CH2CH2I + NaBr
CH3CH2CH2CN + NaBr
CH3 CH2 CH2 Br
CH3CH2CH2 Br
(f)
Mg, H2O
CH
CH3CH2CH2C
CH
+ NaBr
CH3CH2CH2H + HBr
Which reactant in each of the following pairs is more nucleophilic? Explain.
11.32
(a)
+ NaC
-NH2
or
NH3
Solution: The first one is more nucleophilic. Because it has a lone pair electrons and it can attack the
nucleons.
(b)
H2O or
Solution:
CH3CO2-
The second one is more nucleophilic.
Because it is more basic than water and
nucleophilicity roughly parallels basicity, it is more nucleophilic.
(c)
BF3
or
F-
Solution: The second one is more nucleophilic. It has a lone pair electrons while BF3 has a empty
orbital(it is electronic)
(d)
(CH3)3P or
(CH3)3N
Solution: The first one is more nucleophilic. Its polarization is stronger than the second one’s.
(e)
I-
or
Cl-
Solution: The first one is more nucleophilic. Nucleophilicity usually increases going down a column
of the periodic table. The first one’s capacity of controlling electrons is weaker(Diameter is longer) and
it can be polarized easily than the second one.
C
N or -OCH3
(f)
Solution: The second one is more nucleophilic. The acidity of its conjugate acid (CH3OH) is weaker
than HCN. So the second one’s basicity is stronger.
11.33
Among the Walden cycles carried out by Kenyon and Phillips is the following series of
reaction reported in 1923. Explain the results, and indicate where Walden inversion is occurring.
HO
TosO
TosCl
H3CHCH2C
H3CHCH2C
[α] = +33.0°
D
CH3 CH2OH
[α] = +31.1°
D
Heat
H3CH2CO
H3CHCH2C
K
[α] = -19.9 °
D
KO
CH3CH2Br
H3CHCH2C
H3CH2CO
H3CHCH2C
[α] = +23.5°
D
Solution: The reaction which is written in red shows that the configuration changes from retention to
inversion. But the configuration in the reaction which is written in blue doesn’t. The products in two
reactions are enantiomers. The amounts of their rotation should be same. But here is not. Maybe the
products are not pure. Walden inversion is occurring in the following step.
H3CH2CO
CH3CH2OH
H3CHCH2C
Heat
[α] = -19.9 °
D
(The product show in red is not pure, but the blue one contains only one enantimer.)
11.34 The synthetic sequences shown below are unlikely to occur as written. Tell what is wrong with
each, and predict the true product.
Br
OC(CH3)3
(CH3)3COK
H3C
(a)
C
H
CH2CH3
H3C
(CH3)3 COH
OH
F
NaOH
(b)
Cl
OH
CH3
(c).
SOCl2
Pyridine
CH3
C
H
CH2CH3
Solution:
–
(a).The (CH3)3CO
is a kind of strong base which is good of the E2 elimination in the polar aprotic
solvent, and at the same time, SN2 substitution also occurs, but the major product is alkene.
Br
CH3
H
(CH3)COK
H3C
C
H
H
OC(CH3)3
+
+
C
C
CH2CH3
H
(CH3)CH2 OH
H3 C
H
H3C
H3C
C
H
CH2 CH3
CH3
major product
(b).The F-C bond is too strong for break so it is not good for the SN1 substitution and SN2 substitution,
and the OH – is a kind of strong base which is good for E2 elimination not E1 elimination.
F
NaOH
(c). The substrate is a Tertiary alkyl alcohol, so the reaction is not good for SN2 and E2 reactions, and
the solvent pyridine is a kind of base which means that it is a strong nucleophlie good for the E1
elimination, not SN1 substitution.
OH
SOCl2
CH3
Pyridine
11.35 Order each of the following sets of compounds with respect to SN1 reaction
H3 C
CH3
CH3
NH2
C
H3C
C
Cl
Cl
H3CH2C
C
H
CH3
CH3
(a)
CH3
H3C
(b)
C
CH3
CH3
Cl
H3 C
CH3
C
H3C
Br
CH3
H
H
OH
CH3
CH3
H
C
C
C
Br
Br
C
Br
(c).
Solution:
H3C
CH3
CH3
NH2
C
Cl
>
H3C
C
Cl
H3CH2 C
>
C
H
CH3
CH3
(a).
CH3
H3C
C
Br
>
H3C
CH3
(b).
CH3
CH3
C
Cl
>
H3C
C
CH3
CH3
CH3
H
C
Br
>
H
H
C
C
OH
Br
>
Br
(c).
11.36 Order each of the following sets of compounds with respect to SN2 reactivity.
(a)
CH3
H2
C
H3C
Cl
C
H2
>
H2
C
H3C
CH3
CH
>
H3C
C
Cl
CH3
Cl
(b)
Br
>
Br
>
Br
(c)
O
Ts
>
Br
>
O
11.37 Predict the product and give the stereochemistry resulting from reaction of each of the
following nucleophiles with (R) –2-bromooctane:
(a) -CN
H
H
H3C
-CN
CH3
C
Br
NC
C
C6 H13
C6H13
R configuration
S configuration
(b) CH3CO2H
O
CH3CO2- H3C
H
CH3
C
Br
H3C
C
O
C
C6 H13
C6H13
R configuration
S configuration
-
(c) CH3S
H
CH3S- H3C
H
CH3
C
Br
H3C
S
C6 H13
C
C6H13
R configuration
S configuration
-
11.38 (R)-2-bromooctane undergoes racemization to give (+_)-2-bromoocatane when treat with NaBr
in dimethyl sulfoxide. Explain.
Solution: In DMSO, SN2 is proper to happen, the mechanism will be following:
Br
H
Br
+
H
R
DMSO
Br
C
Br
R
Because the nucleophiles are the same, so it will be equal to yield S and R product.
11.39 Reaction of the following S tosylate with cyanide ion yields a nitrile product that also has S
stereochemistry. Explain.
H
OTos
NaCN
C
?
CH2OCH3
Solution: The TosO- is quite a good leaving group for SN2 reaction, and the substrate is secondly, so it
will be not very stable to form carbocation, and it is easy to undergo SN2 reaction. The mechanism will
be follow:
H
N
C
+
OTos
N
C
CH2OCH3
H
H
C
C
OTos
CH2OCH3
C
N
+ OTos
CH2OCH3
11.40 Ethers can often be prepared by SN2 reaction of alkoxide ions ,RO- ,with alkyl halides. Suppose
you wanted to prepare cyclohexyl methyl ether, which of the two possible routes shown below would
you choose? Explain.
CH 3I
+
O
OCH 3
CH 3O
+
I
Solution:
The route above is better. Cyclohexoxide anion is a good nucleophile for SN2 reaction and methyl
iodide is a good substrate for SN2 reaction, so SN2 reaction is favored in the above route and the very
product will be obtained. In the route below, methoxide anion is also a good nucleophile, but the
substrate, iodo cyclohexane, is not good for SN2 reaction because of the steric effect, and then the
methoxide anion will act as a strong base and the elimination will occur, so the product is not that we
want.
11.41 The SN2 reaction can occur intramolecularly (within the same molecule). What product would
you expect from treatment of 4-bromo-1-butanol with base?
Solution:
O
Br
H
:B
O
Br
Br
O
O
The base will be more likely to react with the proton of –OH than to react with the carbon next to the
bromine. The reaction shown above will be favored; especially the concentrate of the substrate is low.
11.42: In light of your answer to problem 11.41, propose a synthesis of 1,4-dioxane starting only with
1,2-dibromoehane.
O
1,4-Dioxane
O
OHH
Solution:
Br
O
+
Br
O
O
Br
O
H
OH-
2 OH
+
Br
11.43: The following tertiary alkyl bromide does not undergo a nucleophilic substitution reaction by
either SN1 or SN2 mechanisms. Explain.
Br
Solution:
Steric effects in the SN2 reaction.
is unstable.
On the other hand,
11.44 In addition to not undergoing substitution reactions, the alkyl bromide shown in Problem 11.43
also fails to undergo an elimination reaction when treated with base. Explain.
Solution:
Br
There is no appropriate H for E2 elimination.
On the other hand,
is unstable.
11.45 1-Chloro-1,2-diphenylethane can undergo E2 elimination to gibe either cis-or
trand-1,2-diphenylethylene (stilbene). Draw Newman projections of the reactive conformations
leading to both possible products, and suggest a reason why the trans alkene is the major product.
Cl H
2
C C
H
1-Chloro-1,2-diphenylethane
OCH3
C C
H H
trans-1,2-Diphenylthylene
H
H
Cl
H
H
Cl
H
H
As we can see, to (a) the product will be trans alkene; to (b) it will be cis-alkene. Treat with (b) the
transition state of the reaction will have a high energy which does not favor the reaction, so the trans
will be major.
11.46 Predict the major alkene product of the following E1 reaction:
H3C CH3
CH3CHCBr
HOAc
Heat
?
CH2CH3
Solution:
OAc
H3C CH3
CH3CHC
H
Br
HOAc
Heat
CH2CH3
CH3C
H3C
CH3
H3C
C
CH3C
CH2CH3
CH3
C
CH2 CH3
11.47 The tosylate of (2R, 3S)-3-phenyl-2-butanol undergoes E2 elimination on treatment with sodium
ethoxide to yield (Z)-2-phenyl-2-butene. Explain, using Newman projections.
Na
CH3CHCHCH3
OTos
Solution:
OCH2CH3
CH3C
CHCH3
OAc
OAc
H
H
H3C
H
H
CH3
H3C
H
Ph
Ph
CH3
H3C
OTos
CH3
H
Ph
CH3
OTos
Tosylate of (2R,3S)-3-Phenyl-2-butanol
H3C
Ph
(Z)-2-Phenyl-2-butene
11.48 In light of your answer to Problem 11.47, which alkene, E or Z, would you expect from an E2
reaction on the tosylate of (2R,3R)-3-phenyl-2-butanol? Which alkene would you result from E2
reaction on the (2S,3R) and (2S,3S) tosylates? Explain.
Ph
CH3
H
OTos
H
CH3
(2R,3R)
Solution:
CH3
Ph
, when it undergo E2 reaction, the product will be
CH3
H
, and it
is E configuration.
Ph
CH3
H
OTos
H3C
Ph
CH3
H
(2S,3R)
, when it undergo E2 reaction, the product will be
H3C
H
, and it is Z
configuration.
H3C
Ph
H
OTos
H3C
H3C
Ph
H
(2S,3S)
, when it undergo E2 reaction, the product will be
H3C
H
, and it is E
configuration.
11.49 How can you explain the fact that trans-1-bromo-2-methycyclohexane yields the non-Zaitsev
elimination product 3-methylcyclohexene on the treatment with base?
CH3
CH3
H
KOH
Br
H
CH3
H
H
H
Solution:
as we see in the structure, the elimination reaction can happen when both
Br
the C-Br bond and C-H bond are anti-coplanar, but in this molecular, the C-H bond on the more
substitutive carbon is equatorial, so it doesn’t follow the Zaitsev’s rule.
11.50. Predict the products of the following reaction, indicating stereochemistry where necessary
Br
CH3
H2O
Ethanol
?
H3C
H
Solution:
Br
CH3
OH
CH3
OH
CH3
CH3
H2O
Ethanol
H3C
H3C
H3C
H
H3C
H
H
H
11.51. Draw all isomers of C4H9Br, name them, and arrange them in order of decreasing reactivity in
the SN2 reaction.
Solution:
The isomers of C4H9Br:
1.
1-Bromobutane
H2
H2
H3C
C
C
2. 2-Bromobutane
CH2Br
Br
H3C
3.
H2
C
C
CH3
H
2-Bromo-2-methylpropane
CH3
H3C
4.
C
CH3
Br
1-Bromo-2-methylpropane
CH3
H3C
C
CH2Br
H
The order of decreasing reactivity in the SN2 reaction:
1>2>4>3
11.52 reaction of iodoethane with CN- yields a small amount of isonitrile, CH3CH2NC，along with the
nitrile, CH3CH2CN as the major product. Write Lewis structure for both products, assign formal
charges as necessary, and propose mechanism to amount for their formation.
Solution:
CH3 CH2 C
CH3CH2 N
N
CH3CH2 I
+
C
CH3 CH2 I
+
C
C
CH3CH2 C
N
N
N
CH3CH2 N
+ I
+ I
C
11.53 alkynes can be made by dehydrohalogenation of vinylic halides in a reaction that is essentially an
E2
process.
In
studying
the
stereochemistry
of
this
elimination,
it
was
found
that
(Z)-2-chloro-2-butenedioic acid reacts 50 times as fast as the corresponding E isomer. What conclusion
you can draw about the stereochemistry of eliminations in vinilic halides? How does this result
compare with eliminations of alkyl halides?
Solution: Similar to elimination of alkyl halides, elimination in vinylic halides is more favored when
the X and the H are on opposite sides of the molecule—anti periplanar geometry.
11.54
(S)-2-Butanol slowly racemizes on standing in dilute sulfuric acid. Explain.
H3CH2C
H3C
11.55
H
H
H
C
OH
H3CH2C
H3C
C
OH2
C
H3 CH2C
CH3
Reaction of HBr with (R)-3-methyl-3-hexanol leads to (±)-3-bromo-3-methyl-hexane. Explain.
H3C
H3C
C
H3CH2CH2C
OH
H
Br
H3CH2CH2C
H3CH2C
C
OH2
H3CH2C
CH3
H3 CH2CH2C
C+
CH2CH3
BrH3C
CH3
C
H3CH2CH2C
H3CH2C
Br
Br
S
C
CH2CH2CH3
CH2CH3
R
11.56
Treatment of 1-bromo-2-deuterio-2-phenylethane with strong base leads to a mixture of deuterated and
nondeuterated phenylethylenes in an approximately 7:1 ratio. Explain.
D
D
H
Br
OCH2(CH3)
+
7:1ratio
Solution:
The mechanism of this reaction as follows:
(a)
H
H
D
H
H
Br
Ph
D
H
D
H
Br
H
Ph
Two reasons: conformation favored and isotope effect.
11.57
Although anti periplanar geometry is preferred for E2 reactions, it isn’t absolutely necessary. The
deuterated bromo compound shown here reacts with strong base to yield an undeuterated alkene.
Clearly, a syn elimination has occurred. Make a molecular model of reactant, and explain the result.
Br
Base
H
H
D
H
H
Solution：
The molecular model as follow:
from the figure we can see only the D atom is in the same plan with the Br atom, so only D and Br
could be eliminated together to yield an undeuterated alkene.
11.58 In light of your answer to Problem 11.57, explain why one of the following isomers undergoes
E2 reaction approximately 100 times as fast as the other. Which isomer is more reactive, and
why?
Cl
(a)
H
H
RO
Cl
Cl
RO
H
Cl
(b)
H
Cl
Solution: (a) is more reactive, because Cl and H lie in the same plane in the (a) isomer, which is more
favored in E2 reaction, whereas in the (b) isomer, Cl and H are without a periplanar
geometry, they can be in the same plane by ringflip, but it is hard to occur.
11.59 Propose structures for compounds that fit the following descriptions:
(a) An alkyl halide that gives a mixture of there alkenes on E2 reaction
(b) An organohalide that will not undergo nucleophile substitution
(c) An alkyl halide that gives the non-Zaitzav product on E2 reaction
(d) An alcohol that reacts rapidly with HCl at 0℃
Solution: (a)
Br
+
(b)
Br
The carbocation is hard to form plane structure, so it is hard to undergo SN1.
The carbon is tertiary, so it can’t undergo SN2.
(c)
CH(CH3)2
CH(CH3)2
H
H
H3C
H3C
Cl
(d)
CH3
CH3
H3C
C
+
OH
HCl
H3C
C
Cl
CH3
CH3
Tertiary carbocation is much stable, so even at 0℃ the alcohol reacts with HCl rapidly.
11.60 There are eight diastereomers of 1,2,3,4,5,6-hexachlorocyclohexane. Draw each in its more
stable chair formation. One isomer loses HCl in an E2 reaction nearly 1000 times more slowly than the
others. Which isomer reacts so slowly, and why?
Solution:
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
(1)
(3)
(2)
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
(4)
(5)
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
(7)
Cl
Cl
Cl
Cl
(6)
Cl
(8)
Isomer (1) reacts so slowly because there is no anti-planar hydrogen.
11.61 The tertiary amine quinnuclidine reacts with CH3I 50 times as fast as triethylamine,
H3CH2C
CH2CH3
N
CH2CH3
.explain.
Solution:
N
The lone pair electrons of
is more outwards, and the C-N bonds cannot freely
rotate, so it is stringer nucleophile.
11.62 Methyl esters (RCO2CH3) undergo a cleavage reaction to yield carboxylate ions plus
iodomethane on heating with LiI in dimethylformamide:
O
O
OCH3
O
LiI
Li
+
DMF
CH3I
The following evidence has been obtained: (1) The reaction occurs much faster in DMF than in ethanol.
(2) The corresponding ethyl ester (RCO2 CH2CH3) cleaves approximately 10 times more slowly than
the methyl ester. Propose a mechanism for the reaction. What other kinds of experimental evidence
could you gather to suppose your hypothesis?
Solution:
Mechanism:
O
I-
O
CH3
Li
O
DMF
O
+
H3C
I
The mechanism is SN2
11.63 The reaction of 1-chlorooctane with CH3CO2- to give octyl acetate is greatly accelerated by
adding a small quantity of iodide ion. Explain.
Solution:
O
Cl
(CH2 )7CH3
O
+
H3C
O
After adding a small quantity of iodide ion,
(CH2)7 CH3
H3C
O
Cl
+
(CH2) 7CH3
I
I
(CH2)7CH3
O
I
(CH2)7CH3
+
Cl
O
+
H3C
(CH2)7CH3
O
H3C
O
This reaction is more favored than the former, because 1-Iodooctane is more reactive than
1-chlorooctane.
11.64 Compound X is optically inactive and has the formula C16H16Br2. On treatment with strong
base, X gives hydrocarbon Y, C16H14. Compound Y absorbs 2 equivalents of hydrogen when
reduced over a palladium catalyst and reacts with ozone to give two fragments. One fragment,
Z, is an aldehyde with formula C7H6O. The other fragment is glyoxal, (CHO)2. Write the
reactions involved, and suggest structures for X, Y, and Z. What is the stereochemistry of X?
Solution:
Br
Br
H2
C
H2
C
X
H
C
H
Y
H
C
H
C
H
C
H
O
C
H
Z
Br
C
H
Br
H
C
H
C
H
H
Nu:
C
H
C
H
:Nu
C
H
C
H
C
H
+ 2HBr
O
O
C
H
C
H
C
H
C
H
1. O3
2. Zn/H2O+
2
CH
+
HC
O
CH
X is a Meso compound.
11.65 Propose a structure for an alkyl halide that gives only (E)-3-methyl-2-phenyl-2-pentene on E2
elemination. Make sure you indicate the stereochemistry.
Solution:
CH3
H
Br
CH2CH3
H3C
11.66 When primary alcohols are treated with p-toluenesulfonyl chloride at room temperature in the
presence of an organic base such as pyridine, a tosylate is formed. When the same reaction out at
higher temperature, an alkyl chloride is often formed. Propose a mechanism.
Solution:
CH2OH
Cl
CH2O
CH2OTos
TosCl
higher temperature
CH2Cl
11.67 SN2 reactions take place with inversion of configuration, and SN1 reactions take place with
racemization. The following substitution reaction, however, occurs with complete retention of
configuration. Propose a mechanism.
Solution:
By neighboring group participation:
Br
H
1%NaOH, H2O
Br
H
H
OH
O
O-
C
C
C
O
O
O
OHHO
H
OH
H+
HO
H
C
OC
O
O
11.68 Propose a mechanism for the following reaction, an important step in the lab to synthesis
proteins:
CH3
H3C
C
H3C
O
O
O
H 3C
C
N
R
H3 C
C
+
CH2
HO
C
R
N
H
H
Solutions: Cause the CF3CO2H is a strong acid, there will be E1 elimination, and yield carbocation.
CH3
H3C
O
C
H3C
O
O
H
C
N
O C
O
H+
O
C
H3C
R
CH3
H3C
CF3
C
N
H
H
CH3
H3C
O
C+
H
R
N
H3C
H
HO
CH2
H3C
R
C
O
C
CH2
R
C
N
H
HO
)
11.69 Bromohydrins are converted to cyclic ethers called epoxides when treated with base. Propose a
mechanism, using curved arrows to show the electron flow.
Br
OH
O
C
C
C
CH3
H
H
H
H
H 3C
C
H 3C
CH3
Solutions:
H
O
C
H
H 3C
Br
C
CH3
H
CH3
H
O-
-OH
C
H
H 3C
O
C
C
Br
H
H 3C
C
H
CH3
)
11.70 Show the stereochemistry of the expoxide (see Problem 11.69) you would obtain by formation of
a bromohydrin from trans-2-butene, followed by treatment with base.
Solution:
H3C
CH3
Br
CH3
H
H
CH3
H
H3C
Br
H3C
H3C
OH
H
O
H
H
CH3
H3C
H
H
Br
CH3
H
H3C
OH
H
CH3
HO
O
11.71 Amines are converted into alkens by a two-step process called the Hofmann elimination.
Reaction of the amine with excess CH3I in the first step yield intermediate that undergoes E2 reaction
when treated with basic silver oxide. Pentylamine, for example, yields 1-pentene. Propose a structure
for the intermediate, and explain why it undergoes ready elimination.
H3CH2CH2CH2CH2C
NH2
H3CH2CH2CH2CH2C
NH2
CH3I
excess
H3CH2CH2CH2CH2C
N(CH2)3I
H3CH2CH2CH2CH2C
N(CH2)3I
Solution:
CH3I
excess
Ag2O
H2O,heat
CH3CH2CH2CH2CH
CH2
N(CH3 )3
AgI
-NH2 is a kind of poor leaving group. When it reacts with CH3I, it could be converted into
–N+(CH3)3, which is a quaternary ammonium, a better leaving group. Thus the substrate
can undergo elimination.