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

-O

O

H OCOCH3

(R)-1-methylpentyl acetate(S)-2-bromohexane

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:

Br

(R)-2-bromobutane

H

OH-

HHO

(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

SH-

H HSH

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

I

BrH

I

H

(b) KOH Solution:

H

HBr

H

HO

H

OH

(c) H C C Li Solution:

H

HBr

H

C

H

H C C

HC

(d)NH3 Solution:

H

HBr

H

H3N

H

HNH2 + Br

11.5 Which substance in each of the following pairs is more reactive as a nucleophile? Explain.

(a)

H3C

NH3C

or NH

H3C

H3C

Solution:

H3C

NH3C Because it’s negative charged.

(b)

orB

CH3H3C

CH3

NCH3H3C

CH3 Solution:

NCH3H3C

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:

O

HO

Cl

spontaneousdissociation

O

HO

Cl-

This side shielded from attack

This side open to attack

O

O

retention

+

S

O

O

Rinversion(excess)

(S)-3-Chloro-3-methyl-octane

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?

CH

OTos

(H3C)3CHOAcAcO- C

H

OAc

(H3C)3C + HOTos

[α ]D=-30.3° 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)

CCH2

CH3

CH3

Br

CCH2

CH3

CH3

+ Br

S

CCH2

CH3

CH3

O H

H

H2CCH3

O

CH3

H2CCH3

O

CH3

H

H H

HO H

H

O H

H

H2CCH3

OH

CH3

H2CCH3

OH

CH3

++

H3+O

H3+O

11.11 Rank the following substances in order of their expected SN1 reactivity.

H3CH2C Br H2C C

HCH

CH3

Br

H2C CH

Br H3C CH

CH3

Br

Solution:

C

HH3C Br

H H

CH3C H

H2C CH

C

BrCH3

H

H2C CH

CCH3

H

H2C C

Br

H

H2C C

H

CH3C CH3

Br

H

CH3C CH3

H

The stability of the carboncation

The most stable

The least stable

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 CH C

HCH

HC CH3

Br

H3CH2C Br

H2C CH

C

H

CH3

CH

CH

H3C C

H

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)

OH ClHCl

CH3OH

It is likely to be an SN1 reaction. The substrate is secondary and the nucleophile is weakely base and the solvent is acidic.

(b) H2C CCH2Br

CH3

Na+ -SCH3

CH3CN H2C CCH2SCH3

CH3

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.

(a) CH3CH2CHCHCH3

Br CH3

CH3CH2CH CCH3

CH3

+ CH3CH CHCHCH3

CH3

(major) (minor)

(b)

CH3CHCH2 C

CH3 Cl

CH3

CHCH3

CH3

CH3CHCH C

CH3

CH3

CHCH3

CH3

CH3CHCH2 C

CH3

CH3

CCH3

CH3

+

(minor) (major)

(c)

CHCH3

Br

CHCH3 CH+

CH2

(major) (minor) 11.16 What alkyl halides might the following alkenes have been made from? Solution:

(a)

Br

CH3

CH3

BrCH3

CH3

(b)

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

following : Z-1-bromo-1,2-diphenylethene

Br

The Newman projection of the reacting conformation: H

Ph BrBr

PhH

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 H2C CH3

CH3H 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 t-Bu

Br

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

(c):

Cl

+

O

OH

OCOCH3

SN1 11.21: Write the product you would expect from those reactions. Solution: (a): 1.CH3CH2Cl + NaSCH3 CH3CH2SCH3 2. CH3CH2Cl + NaOH CH3CH2OH

(b): 1.

Cl

+ NaSCH3

SCH3

2.

Cl

+ NaOH

(c): 1.

Cl

+ NaSCH3

SCH3

2.

Cl

+ NaOH

Ring-flip

Cl

Cl

11.22 From what alkyl bromide was the following alkyl acetate made by SN2 reaction? Write the reaction, showing all stereochemistry.

OH

CH3

CH3

H

O (Sawhorse)

Solution: H

CH3CH3

Br H

HBr

CH3

CH3

H

=

CH3CO2

H

CH3CH3

H O

OOH

CH3

CH3

H

=

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:

HC

CC

CC

OH

H H HH Cl H

HHCHH

Solution: The configuration of the molecule above is S. The product of SN2 reaction with NaCN is that

following:

HC

CC

CC

OH

H NC HH H H

HHCHH

. 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:

HPh

Cl

Me

H

Et

S

R

Strategy: The reaction should follow the anti-co-planar. Solution: The stereochemistry of the product should be E:

H

Ph

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 .

(b) CH3CH2I in dimethyl sulfoxide will react faster in an SN2 reaction with OH . Because 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 (b) CH3CH2CH2CH2CN (c) H3C O C(CH3)3 (d) CH3CH2CH2NH2

(e) PCH3Br

3 (f)

CH3Br

Solution: (a)

H3CC CCH(CH3)2LiC CCHCH3

CH3

+ CH3I LiITHF

(b)

CH3CH2CH2CH2Br Na+CN-

THF-HMPACH3CH2CH2CH2CN + NaBr

(c) H3C O C(CH3)3(CH3)3CO + CH3Br + Br-

(d) CH3CH2CH2Br NaNH2 CH3CH2CH2NH2 + NaBr

(e)

P

H3C Br

P

CH3

Br

(f)

CH3OH

PBr3

EtherCH3

Br

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.

(b) The SN2 displacement by CH3CO2- on bromoethane or on bromocyclohexane

Solution: On bromoethane.

(c) The SN2 displacement on 2-bromopropane by CH3CH2O- or by CN- 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 (b) KOC(CH3)3 (c) NaI (d) NaCN (e) NaC CH (f) Mg, then H2O Solution:

(a) CH3CH2CH2 + NaNH2 CH3CH2CHNH2 + NaBrBr H3CHC CH2 +

(major)

(b) CH3CH2CH2 Br KOC(CH3)3

H3CHC CH2

(c) CH3CH2CH2 Br + NaI CH3CH2CH2I + NaBr (d) CH3CH2CH2 Br + NaCN CH3CH2CH2CN + NaBr (e)

NaC CHCH3CH2CH2 Br + CH3CH2CH2C CH + NaBr

(f) CH3CH2CH2 BrMg, H2O CH3CH2CH2H + HBr

11.32 Which reactant in each of the following pairs is more nucleophilic? Explain. (a) -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 CH3CO2- Solution: 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.

(f) C N or -OCH3 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.

H3CHCH2C

HO

TosClH3CHCH2C

TosO

CH3CH2OH

Heat

H3CHCH2C

H3CH2CO

K

H3CHCH2C

KO

CH3CH2BrH3CHCH2C

H3CH2CO

[α] +33.0°D= [α] +31.1°D=

[α] -19.9 °D=

[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.

CH3CH2OH

HeatH3CHCH2C

H3CH2CO

[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.

(a)

CH

CH2CH3H3C

Br

(CH3)3COK

(CH3)3COHCH

CH2CH3H3C

OC(CH3)3

(b)

F

NaOH

OH

(c).

CH3

OH

CH3SOCl2

Pyridine

Cl

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.

CH

CH2CH3H3C

Br

(CH3)COK

(CH3)CH2OHC

H

H3C

C

CH3

H

+ CH

CH2CH3H3C

OC(CH3)3

major product

HH

H3C CH3

+

(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.

CH3

OH

SOCl2

Pyridine

11.35 Order each of the following sets of compounds with respect to SN1 reaction

(a)

H3CH2C CH

CH3

NH2C

H3C

Cl

CH3

H3C C

CH3

CH3

Cl

(b)

C

CH3

BrH3C

CH3

C

CH3

ClH3C

CH3

C

CH3

OHH3C

CH3

(c).

C

H H

Br

C

H CH3

Br

C

Br

Solution:

(a).

H3CH2C CH

CH3

NH2

>

C

H3C

Cl

CH3

>H3C C

CH3

CH3

Cl

(b).

> >C

CH3

BrH3C

CH3

C

CH3

ClH3C

CH3

C

CH3

OHH3C

CH3

(c).

C

H H

Br

C

H CH3

BrC

Br

> >

11.36 Order each of the following sets of compounds with respect to SN2 reactivity. (a)

CH3C

CH3

Cl

CH3

H3C

H2C

CH2

Cl

H3C

H2C

CHCH3

Cl

> >

(b)

Br

Br Br> >

(c)

OTs

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

C Br

C6H13

HH3C-CN

NC C

C6H13

CH3H

R configuration S configuration (b) CH3CO2

-

C Br

C6H13

HH3C

O C

C6H13

CH3H

R configuration S configuration

CH3CO2-

CH3C

O

(c) CH3S-

C Br

C6H13

HH3C

S C

C6H13

CH3H

R configuration S configuration

CH3S-

H3C

- 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:

R

Br

H+Br DMSOC

H

Br 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.

C

H OTos

CH2OCH3

NaCN

?

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:

C

H

C OTos

CH2OCH3

C

H OTos

CH2OCH3

N C + NC

H

CH2OCH3N

+ OTos

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.

O

I

CH I

CH O3

3+

+

OCH 3

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:

BrO :B

H BrO

O

Br

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

O

1,4-Dioxane

Br

Br

+ OH2

O

O

+

H

O

OSolution:

Br

Br

HOH-

OH-

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.

On the other hand, is unstable.

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.

Br Solution: 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.

CH

Cl H2C

OCH3 CH

CH

1-Chloro-1,2-diphenylethane trans-1,2-Diphenylthylene

Cl

H

H

H

Cl

H

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

CH3CHCBr

CH3

CH2CH3

HOAcHeat ?

Solution:

H3C

CH3CHC

CH3

CH2CH3

HOAcHeat

Br

H3C

CH3C C

H CH3

CH2CH3

OAc

H3C

CH3C C

CH3

CH2CH3 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.

CH3CHCHCH3

OTos

Na OCH2CH3

CH3C CHCH3

Solution:

OTos

H CH3

H

PhH3C

OAc

OTos

H CH3

H

PhH3C

OAcH CH3

PhH3C

CH3H

H3C Ph

Tosylate of (2R,3S)-3-Phenyl-2-butanol

(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.

Solution:

OTos

H CH3

H

CH3Ph

(2R,3R) , when it undergo E2 reaction, the product will be

Ph CH3

H CH3, and it

is E configuration.

OTos

H3C H

H

CH3Ph

(2S,3R) , when it undergo E2 reaction, the product will be

Ph CH3

H3C H , and it is Z

configuration.

OTos

H3C H

H

PhH3C

(2S,3S) , when it undergo E2 reaction, the product will be

H3C Ph

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

H

H

Br

KOHCH3

Solution: Br

H

CH3

HH

as we see in the structure, the elimination reaction can happen when both

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

CH3

H3C

Br

H

H2OEthanol

?

Solution:

CH3

H3C

Br

H

H2OEthanol

CH3

H3CH

OHOH

H3C

CH3

H

CH3

H3CH

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

2. 2-Bromobutane

3. 2-Bromo-2-methylpropane

4. 1-Bromo-2-methylpropane

H3CH2C

H2C CH2Br

H3CH2C C

HCH3

Br

H3C C

CH3

CH3

Br

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:

CH3CH2 C N CH3CH2 N C

CH3CH2 + C N

+ C N

I

CH3CH2 I

CH3CH2 C N + I

CH3CH2 N C + I

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.

C

H

OHH3CH2C

H3C

C

H

OH2H3CH2C

H3C

C

H

H3CH2C CH3 11.55 Reaction of HBr with (R)-3-methyl-3-hexanol leads to (±)-3-bromo-3-methyl-hexane. Explain.

H3C CH

CH3

CH2Br

C OH

H3C

H3CH2CH2C

H3CH2C

Br-

HC OH2

H3C

H3CH2CH2C

H3CH2C

C+

CH3H3CH2CH2C

CH2CH3

Br

C Br

H3C

H3CH2CH2C

H3CH2C

CBr

CH3

CH2CH2CH3

CH2CH3RS

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

Br

OCH2(CH3)

D H

+

7:1ratio

Solution: The mechanism of this reaction as follows: (a)

Ph

D HH

BrH

H

Ph

D HH

HBr

D

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.

D

H

Br

HBase

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

H

Cl

H

Cl

H

H

Cl

Cl

(a)

(b)

RO

RO

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)

H

Cl

CH(CH3)2

HH3CH3C

CH(CH3)2

(d)

C

CH3

H3C

CH3

OH + HCl C

CH3

H3C

CH3

Cl

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:

ClCl

Cl

Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl

Cl

ClCl

Cl

Cl

Cl

ClCl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

ClCl

Cl

Cl

ClCl

Cl

Cl

Cl

Cl

(1) (2) (3)

(4) (5) (6)

(7)

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,

N

CH2CH3

H3CH2C CH2CH3

.explain. Solution:

The lone pair electrons of N 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:

OCH3

O

LiI

DMF

O

O Li

+ 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

O

CH3

O

O Li+ H3C IDMF

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:

Cl (CH2)7CH3 +

H3C O

O

H3C

O

O(CH2)7CH3

After adding a small quantity of iodide ion,

Cl (CH2)7CH3 + I I (CH2)7CH3 +

I (CH2)7CH3

Cl

+

H3C O

O

H3C

O

O(CH2)7CH3

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:

C H

O

X

Y

Z

CH

CH

CH

CH

H2C

H2C

Br Br

H H

CH

CH

CH

CH

CH

HC

HC C

H

Br Br

Nu:

H H

:Nu

+ 2HBr

CH

CH

CH

CH

1. O3 CHHC CH2. Zn/H2O+

O

2

O O

+

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:

H3C

CH3

CH2CH3

H Br

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 CH2O

TosClCH2OTos

Cl

CH2Cl

higher temperature

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:

C

O

OH

Br H

1%NaOH, H2O

C

O

O-

Br H

C

O

O

H

C

O

O-

HO H

C

O

OH

HO H

H+

OH-

11.68 Propose a mechanism for the following reaction, an important step in the lab to synthesis proteins:

C

H3CCH3

H3CO C

O

N R

H

CH3C

CH2H3C HO C

O

N R

H

+

Solutions: Cause the CF3CO2H is a strong acid, there will be E1 elimination, and yield carbocation.

C

H3CCH3

H3CO C

O

N R

H

CH3C

H3CCH2

CO

O

CF3H

C

H3CCH3

H3C

H+

O C

O

N R

H

C+

H3C CH3

CH2HO

CO

NR

HHHO

CO

NR

H

)

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.

C C

OH

HH3C

Br

CH3

H

C C

O

HH3C

H

CH3 Solutions:

C C

O

HH3C

Br

CH3

H

-OH

H

C C

O-

HH3C

Br

HCH3

C C

O

HH3C

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

H CH3

H3C H

OH

Br

OH

CH3

HH3C

H

HO

Br

CH3

H

H

H3C

CH3HH3C H

O

CH3HH3C H

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

CH3I

excessH3CH2CH2CH2CH2C N(CH2)3I

Solution:

H3CH2CH2CH2CH2C NH2

CH3I

excessH3CH2CH2CH2CH2C N(CH2)3I

Ag2O

H2O,heatCH3CH2CH2CH2CH 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.