1398743355-910536

6.1 Writer the following as net ionic reactions and designate the nucleophile, substrate, and leaving group in each reaction. (a) CH3I+CH3CH2ONa CH3OCH2CH3+NaI (b) NaI+CH3CH2Br CH3CH2I+NaBr (c) 2CH3OH+（CH3）3CCl （CH3）3COCH3+CH3OH2

++Cl- (d) CH3CH2CH2Br+NaCN CH3CH2CH2CN+NaBr (e) C6H5CH2Br+2NH3 C6H5CH2NH2+NH4Br ANSWER： （a）CH3I+CH3CH2O- CH3OCH2CH3+I-

nucleophile：CH3CH2O-

substrate：CH3I leaving group：I-

（b）I-+CH3CH2Br CH3CH2I+Br- nucleophile： I-

substrate：CH3CH2Br leaving group：Br-

（c）2CH3OH+（CH3）3CCl （CH3）3COCH3+CH3OH2++Cl-

nucleophile：CH3OH

substrate： （CH3）3CCl leaving group：Cl-

（d）CH3CH2CH2Br+CN- CH3CH2CH2CN+Br- nucleophile： CN-

substrate： CH3CH2CH2Br leaving group：Br- （e）C6H5CH2Br+2NH3 C6H5CH2NH2+NH4

++Br- nucleophile： NH2

-

substrate： C6H5CH2Br leaving group：Br- 6.2 Use chair conformational structures (Sect.4.12) and show the nucleophilic substitution reaction that would take place when trans-1-bromo-4-tert-butylcyclohexane reacts with iodide ion. (Show the most stable conformation of the reactant and the product.) Answer:

Br C(CH3)3

I

C(CH3)3

I

6.3 SN2 reactions that involve breaking a bond to a stereocenter can be used to relate configurations of molecules because the stereochemistry of the reaction is known, (a) Illustrate how this is true by assigning configurations to the 2-chloro-butane enantiomers based on

the following date. [The configuration of (--)-2-butanol is given in Section 5.7C.]

(+)-2-ChlorobutaneOH-

SN2(-)-2-Butanol

[ ]D25α = +36.000 [ ]D

25α = -13.520

Enantiomerically pure Enantiomerically pure (b) When optically pure (+)-2-chlorobutane is allowed to react with potassium iodide in acetone in an

SN2 reaction, the 2-iodobutane that is produced has a minus rotation. What is the configuration of (--)-2-iodobutane? Of (+)-2-iodobutane?

Answer, (a)

Et

C

MeH

IHO C

Et

MeH

OH + I

An inversion of configuration (b)

CH3

HI

CH2CH3

R-(-)-2-iodobutane

CH3

IH

CH2CH3

S-(+)-2-iodobutane

6.4 Keep in mind that carbocations have a trigonal planar structure, (a) write a structure for the carbocation intermediate and (b) write structures for the alcohol (or alcohols) that you would expect from the following reaction:

I

CH3C(H3C)3

H2OSN1

Answer: (a) Structure of the carbocation intermediate:

(b) Structures of the alcohols would be:

CH3

OHC(H3C)3

OH

CH3C(H3C)3

6.5 What product(s) would you expect from the methanolysis of the cyclohexane derivative given as the reaction in Problem 6.4?

I

CH3(H3C)3C

Answer:

The product is

OCH3

CH3(H3C)3C ,

OCH3

CH3

(H3C)3Cthe reaction is as

following: I

CH3(H3C)3C +

OCH3

CH3(H3C)3CCH3OH + OCH3

CH3

(H3C)3C

6.6 The relative rates of ethanolysis of four primary alkyl halides are as follows: CH3CH2Br, 1.0; CH3CH2CH2Br, 0.28; (CH3)2CHCH2Br, 0.030; (CH3)3CCH2Br, 0.00000042. (a). Are each of these reactions likely to be SN1 or SN2? (b) Provide an explanation for the relative reactivities that are observed.

Answer: (a) All the reactions are likely to be SN2. (b) For particles (molecules and ions) to react, their reactive centers must be able to come within

bonding distance of each other. Large and bulky groups can often hinder the formation of the required transition state. So the relative rates of the above alkyl halides are from high to low.

6.7 Classify the following solvents as being protic or aprotic: formic acid, C

O

OHH ; acetone,

C

O

CH3H3C ;acetonitrile, C NH3C ; formamide, C

O

NH2H ; sulfur

dioxide, SO2 ; ammonia, NH3 ; trimethylamine, N

CH3

CH3H3C ; ethylene glycol, HOCH2CH2OH. Answer:

Protic: C

O

OHH , C

O

NH2H , NH3, HOCH2CH2OH.

Aptotic: C

O

CH3H3C , C NH3C , N

CH3

CH3H3C , SO2. 6.8 Would you expect the reaction of propyl bromide with sodium cyanide (NaCN), that is,

CH3CH2CH2Br + NaBr CH3CH2CH2CN + NaBr to occur faster in DMF or in ethanol? Explain you answer.

HC

NCH3

O

CH3 DMF

Answer: The reaction occurs faster in DMF than in ethanol. Because it don’t solvate anions to any appreciable extent. These “naked” anions are highly reactive both as base and nucleophiles. This is the opposite of their strength as nucleophiles in alcohol or water solutions.

The rates of SN2 reactions generally are vastly increased when they are carried out in polar aprotic solvents. The increase in rate can be as large as a millionfold.

6.9 Which would you expect to be the stronger nucleophile in a protic solvent: (a) . CH3CO2

- or CH3O-? CH3O- is the stronger nucleophile. (b). H2O or H2S? H2S is the stronger nucleophile. (c). (CH3)3P or (CH3)3N? (CH3)3P is the stronger nucleophile in a protic solvent.

6.10 When tert-butyl bromide undergoes solvolysis in a mixture of methanol and water, the rate of solvoysis (measured by the rate at which bromide ions from in the mixture) increases when the percentage of water in the water increased. (a) Explain this occurrence. (b) Provide an explanation for the observation that the rate of the SN2 reaction of ethyl chloride

with potassium iodide in methanol and water decreases when the percentage of water in the mixture is increased.

Answer: (a) Because water is the most effective solvent for promoting ionization. It made the intermediate

more stable. (b) Because the charge of the transition state is more dispersed than the charge in the starting

material. Increasing the polarity of the solvent will increase the stability of the starting material. Therefore, it will decrease the rate of the reaction.

6.12 Starting with (S)-2-bromobutane, outline syntheses of each of the following compounds.

(a) (b)

(c) (d)

CH3CHCH2CH3

OCH2CH3

(R)- CH3CHCH2CH3

OCCH3

O

(R)-

(R)- (R)-CH3CHCH2CH3

SH

CH3CHCH2CH3

SCH3 Answer:

CH3

H3CCO H

CH3

H H

O

(c)

(a)

CH3

H3CH2CO H

CH3

H H

CH3

HS H

CH3

H H

CH3

H3CS H

CH3

H H(d)

CH3

H Br

CH3

H H+ CH3CH2O + Br

CH3

H Br

CH3

H H+ HS + Br

+ Br

+ Br(b)

CH3

H Br

CH3

H H+ CH3CO

O

CH3

H Br

CH3

H H+ CH3S

6.13 Show how you might use a nucleophilic substitution reaction of propyl bromide to synthesize each of the following compounds. (You may use any other compounds that are necessary.) (a) CH3CH2CH2OH (b) CH3CH2CH2I (c) CH3CH2OCH2CH2CH3 (d) CH3CH2CH2SCH3 (e) CH3COOCH2CH2CH3 (f) CH3CH2CH2N3

N

CH3

H3C

CH3

CH2CH2CH3 Br(g)

(g) CH3CH2CH2CN (h) CH3CH2CH2SH Answers:

+ CH3CH2CH2 CH3CH2CH2OH + BrBr(a) OH

(b) I + CH3CH2CH2 Br CH3CH2CH2I + Br

(c) CH3CH2O + CH3CH2CH2 Br CH3CH2OCH2CH2CH3 + Br

(d) CH3S + CH3CH2CH2 Br CH3CH2CH2SCH3 + Br

(e) CH3CO

O

+ CH3CH2CH2 Br CH3COCH2CH2CH3

O

+ Br

(f) N3 + CH3CH2CH2 Br CH3CH2CH2N3 + Br

(g) (CH3)3N + CH3CH2CH2 Br (CH3)3NCH2CH2CH2 Br

+ CH3CH2CH2 BrBr(h) CN CH3CH2CH2CN +

(i) SH + CH3CH2CH2 Br CH3CH2CH2SH + Br

6.14 Which alkyl halide would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (a) CH3CH2CH2Br or (CH3)2CHBr (b) CH3CH2CH2CH2Cl or CH3CH2CH2CH2I (c) (CH3)2CHCH2Cl or CH3CH2CH2CH2Cl (d) (CH3)2CHCH2CH2Cl or CH3CH2CH(CH3)CH2Cl (e) C6H5Br or CH3CH2CH2CH2CH2CH2Cl Answer: (a) 1-Bromopropane. Because it is less hindered. (b) 1-Iodobutane. Because I- is a good leaving group. (c) 1-Chlorobutane. Because it is less hindered. (d) 1-Chloro-3-methylbutane. Because the carbon bearing the leaving group is less hindered than in

1-Chloro-2-methylbutane. (e) 1-chlorohexane. Phenyl halides are unreactive in SN2. 6.15 Which SN 2 reaction of each pair would you expect to take place more rapidly

in a protic solvent?

Answer:

(a) (1) ,(b) (2) ,(c) (2) ,(d) (2) are expected to take place more rapidly in a protic solvent in each pair. 6.16 Which SN1 reaction of each pair would you expect to take place more rapidly? Explain your answer. (a)

(b)

(C)

(d)

(e)

Answer: (a) Reaction2 will be more rapidly. Because bromide ion is better leaving group than chloride ion (b) Reaction1 will be more rapidly. Because H2O and ethanol are both solvent and nucleophile, and the

water is stranger polar protic solvent than ethanol. (c) Reaction2 will be more rapidly. Because the substrate has bigger concentration. (d) The two reactions have same reaction rate. Because the Nu has no effect on the reaction. (e) Reaction1 will be more rapidly. Because (CH3)3C+ is more stable.

(a) (1) CH3CH2CH2Cl + CH3CH2O- CH3CH2CH2OCH2CH3 + Cl-

(2) CH3CH2CH2Cl + CH3CH2OH CH3CH2CH2OCH2CH3 + HCl

(b) (1) CH3CH2CH2Cl + CH3CH2O- CH3CH2CH2OCH2CH3 + Cl-

(2) CH3CH2CH2Cl + CH3CH2S- CH3CH2CH2SCH2CH3 + Cl-

(c) (1) CH3CH2CH2Br + (C6H5)3N CH3CH2CH2N(C6H5)3+ + Br-

(2) CH3CH2CH2Br + (C6H5)3P CH3CH2CH2P(C6H5)3+ + Br-

(1) CH3CH2CH2Br (1.0M) + CH3O- (1.0M)(d) CH3CH2CH2OCH3 + Br-

(2) CH3CH2CH2Br (1.0M) + CH3O- (2.0M) CH3CH2CH2OCH3 + Br-

(CH3)3CCl +H2O (CH3)3COH + HCl(1)(2)(CH3)3CBr + H2O (CH3)3COH + HBr

(CH3)3CCl +H2O (CH3)3COH + HCl(1)(2)(CH3)3CBr + CH3OH (CH3)3COH + HBr

(CH3)3CCl (1.0M) + CH3CH2O- (1.0M) (CH3)3COCH2CH3 + Cl-(1)(2)(CH3)3CCl (2.0M) + CH3CH2O- (1.0M) (CH3)3COCH2CH3 + Cl-

(CH3)3CCl (1.0M) + CH3CH2O- (1.0M) (CH3)3COCH2CH3 + Cl-(1)(2)(CH3)3CCl (1.0M) + CH3CH2O- (2.0M) (CH3)3COCH2CH3 + Cl-

(CH3)3CCl +H2O (CH3)3COH + HCl(1)(2)C6H5Cl + H2O C6H5OH + HCl

6.17 With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem. (a) CH3I (b) CH3CH2I (c) CH3OH (d) CH3CH2OH (e) CH3SH (f) CH3CH2SH (g) CH3CN (h) CH3CH2CN (i) CH3OCH3 (j) CH3OCH2CH3 (k) Cyclopentene Answer:

(a) CH3I

CH3Br + I- CH3I + Br-

(b) CH3CH2I

CH3CH2Br + I- CH3CH2I + Br-

(c) CH3OH CH3I + OH- CH3OH + I-

(d) CH3CH2OH

CH3CH2I + OH- CH3CH2OH + I-

(e) CH3SH

CH3I + SH- CH3SH + I-

CH3OH (f) CH3CH2SH

CH3CH2I + SH- CH3CH2SH + I-

CH3OH (g) CH3CN

CH3I + CN- CH3CN + I-

CH3OH (h) CH3CH2CN

CH3CH2I + CN- CH3CH2CN + I-

CH3OH (i) CH3OCH3

2 CH3OH + 2 Na 2 CH3ONa +H2

CH3I + CH3O- CH3OCH3 +I-50oCCH3OH

(j) CH3OCH2CH3

CH3CH2OCH3 +I-50oCCH3OHCH3CH2I +CH3O-

(k) Cyclopentene

CH3CH2OHICH3CH2ONa

+ NaI CH3CH2OH+

6.18 Listed below are several hypothetical nucleophilic substitution reactions. None is synthetically useful because the product indicated is not formed at an appreciable rate. In each case provide an explanation for the failure of the reaction to take place as indicated. (a) CH3CH2CH3+OH- CH3CH2OH+CH3

—

(b) CH3CH2CH3+OH- CH3CH2 CH2OH+ H-

(c) + OH- -CH2CH3CH2 CH2OH

(d)

Br

+CN-

CN

+Br-

(e) NH3+CH3OCH3 CH3NH2+CH3OH (f) NH3+CH3O H2

+ CH3NH3++H2O

Answer (a) Alkanide ion (CH3

-) is very powerful base and it’s virtually never acting as leaving groups. Therefore, reaction (a) is not feasible. (b) Hydride ion (H:-) is very powerful base and it’s virtually never acting as leaving groups. Therefore, reaction (b) is not feasible. (c) Alkanide ion (R:-) is very powerful base and it’s virtually never acting as leaving groups.

Therefore, reaction (c) is not feasible. (d) Tertiary alkyl bromide can’t be the substrate for the SN2 reaction. (e) CH3O- is a strongly basic ion and rarely acts as leaving group. Therefore, reaction e is not feasible. (f) CH3O H2

+ only appear in a strong acid and NH3 can’t exist in a strong acid. 6.19 You have the task of preparing styrene (C6H5CH=CH2) by dehydrohalogenation of either 1-bromo-2-phenylethane or 1-bromo-1-phenylethane use KOH in ethanol. Which halide would you choose as your starting material to give the better yield of the alkene? Explain your answer. Answers:

1-Bromo-1-phenylethane will be used as the starting material, because a 1º alkyl bromide gives mainly SN2 except with a hindered strong base and then gives mainly E2. A 2º alkyl bromide gives mainly E2 with strong bases. 6.20 Your task is to prepare isopropyl methyl ether. CH3OCH(CH3)2, by one of the following reactions. Which reaction would give the better yield? Explain your choice.

(1) CH3ONa + (CH3)2CHI CH3OCH (CH3)2

(2) (CH3)2CHONa + CH3I CH3OCH (CH3)2

Answer: The second reaction would give the better yield because the desired reaction is an SN2 reaction, and the substrate is a methyl halide. Use of reaction (1) would result in considerable elimination by an E2 pathway. 6.21 Which product (or products) would you expect to obtain from each of the following reactions? In each part give the mechanism (SN1, SN2, E1, or E2) by which each product is formed and predict the relative amount of each product (i.e., would the product be the only product, the major product, a minor

product, etc.?)

CH3CH2CH2CH2CH2Br + CH3CH2O- 50 C°CH3CH2OH

CH3CH2CH2CH(OCH2CH3)CH3 + Br-

(a) SN2

major product

minor product

CH3CH2CH2CH2CH2Br + CH3CH2O- 50 C°CH3CH2OH

(b) E2

CH3CH2CH2CH2CH2Br + (CH3)3CO- 50 C°(CH3)3COH

CH3CH2CH2CH2CH2Br + (CH3)3CO- 50 C°(CH3)3COH CH3CH2CH2CH2CH2OC(CH3)3 + Br-

major product

minor product

(c) E2

the only product

(CH3)3CBr + CH3O- 50 C°

CH3OHH3C C CH2

CH3

(d) E1

the only product

(CH3)3CBr + (CH3)3CO- 50 C°

(CH3)3COH

CH3CH2CH2CH=CH2

CH3CH2CH2CH=CH2

H3C C CH2

CH3

(e) SN2Cl

(H3C)3C + I- 50 C°acetone

I(H3C)3C

(f) SN1

Cl

(H3C)3C50 C°

OCH3(H3C)3CCH3

CH3OH

the only product

the only product

CH3

(g) E1

3-Chloropentane + CH3O- 50 C°CH3OH

3-Chloropentane + CH3O- 50 C°CH3OH

CH3CH(OCH3)CH2CH2CH3 + Cl-

(h) SN1

3-Chloropentane + CH3CO2- 50 C°

CH3CO2HCH3CH2CH(O2CCH3)CH2CH3 + Cl-

3-Chloropentane + CH3CO2- 50 C°

CH3CO2H

(i) E2

HO- + (R)-2-bromobutane25 C° + H2O + Br

major product

minor product

HO- + (R)-2-bromobutane 25 C°

OCH3

(H3C)3C

CH3

+

CH3CH2CH=CHCH3

CH3CH=CHCH2CH3

OH

(j) SN1

(S)-3-Bromo-3-methylhexane 25 C°

CH3OHH3CH2C C CH2CH2CH3

CH3

OCH3

(k) SN2

(S)-2-Bromooctane + I- 50 C°

CH3OH(R)-2-Iodooctane + Br-

(+)-

6.22 Write conformational structures for the substitution products of the following deuterium-labeled compounds: (a)

(b)

Cl

H

H

D

I-

CH3OH?

Cl

HI-

CH3OH?

H

D

(c)

Cl

H

H

D

I-

CH3OH?

(d)

CH3OH?

Cl

CH3

D

H

H2O

The answer:

(a)

Cl

H

H

D

I-

CH3OHI

H

D

H

+ Cl-

(b)

Cl

HI-

CH3OH

H

D

H

ID

H

+ Cl-

(c)

Cl

H

H

D

I-

CH3OH

H

I

H

D

+ Cl-

(d) Cl

CH3 H2OCH3OH

CH3

D

H H

D

OCH3

CH3

H

D

OH

OCH3

H

D

CH3

OH

H

D

CH3

+

+ +

6.23 Although ethyl bromide and isobutyl bromide are both primary halides. Ethyl bromide undergoes SN2 reactions more than 10 times faster than isobutyl bromide does. When each compound is treated with a strong base/ nucleophile (CH3CH2O-), isobutyl bromide gives a greater yield of elimination products than substitution products, whereas with ethyl bromide this behavior is reversed. What factor account for these results?

Answer: The reason ethyl bromide undergoes SN2 reactions faster than isobutyl bromide is steric effect.

This steric hindrance causes isopropyl bromide to react more slowly in SN2 reactions and to give relatively more elimination (by an E2) when a strong base is used. 6.24 Consider the reaction of I- with CH3CH2Cl. (a) Would you expect the reaction to be SN1 or SN2?

The rate constant for the reaction at 60°C is 5*10-5L mol-1 s-1 . What is the reaction rate if [I-]=0.1mol L-1 and [CH3CH2Cl]=0.1 mol L-1? (c) If [I-]=0.1mol L-1 and [CH3CH2Cl]=0.2mol L-1? (d) If [I-]=0.2mol L-1 and [CH3CH2Cl]=0.1 mol L-1? (e) If [I-]=0.2mol L-1 and [CH3CH2Cl]=0.2 mol L-1? Answer: Because the substrate is the primary alkyl halide, the reaction would be SN2. V=k*[I-][CH3CH2Cl] (b) v =5*10-5*0.1*0.1=5*10-7 (c) v =5*10-5*0.1*0.2=1*10-6 (d) v =5*10-5*0.2*0.1=1*10-6

(e) v =5*10-5*0.2*0.2=2*10-6

6.25 Which reagent in each pair listed here would be the more reactive nucleophile in a protic solvent? (a) CH3NH- or CH3NH2 (e) H2O or H3O+

(b) CH3O- or CH3CO-

O

(f) NH3 or NH4 (c) CH3SH or CH3OH (g) H2S or HS-

(d) (C6H5) 3N or (C6H5) 3P (h) CH3CO-

O

or OH- Answer:

(a) CH3NH- (b) CH3O- (c) CH3SH (d) (C6H5) 3P (e) H2O (f) NH3

(g) HS- (h) OH- 6.26 write mechanisms that account for the products of the following reaction (a)

HOCH2CH2BrOH

_

H2O O

(b)

H2NCH2CH2CH2CH2BrOH

_

H2ON

H

ANSWER: (a)

HO

Br

OH-

OBr

O

(b)

H2NBr

NH H

OH-

N

H 6.27 Many SN2 reactions of alkyl chlorides and alkyl bromides are catalyzed by the addition of sodium or potassium iodide. For example, the hydrolysis of methyl bromide takes place much faster in the presence of sodium iodide. Explain. Answer: Because Alkyl chlorides and bromides are also easily converted to alkyl iodides by nucleophilic substitution reactions. R—Cl- + I- R—I +Cl-

R—Br + I- R—I + Br-

The effect of the leaving group in SN2 reaction: R—I> R—Br> R—Cl> R—F That is why many SN2 reactions of alkyl chlorides and alkyl bromides are catalyzed by the addition of sodium or potassium iodide take place much faster in the presence of sodium iodide. Iodide ion is a good nucleophile and a good leaving group. 6.28 Explain the following observations: When tert-butyl bromide is treated with sodium methoxide in a mixture of methanol and water, the rate of formation of tert-butyl alcohol and tert-butyl methyl ether does not change appreciably as the concentration of sodium methoxide is increased. However, increasing the concentration of sodium methoxide causes a marked increase in the rate at which tert-butyl bromide disappears from the mixture. Answer: tert-Butyl alcohol and tert-butyl methyl ether are formed via an SN1 mechanism. The rate of the reaction is independent of the concentration of methoxide ion. However, A cpompetetion reaction of this reaction is the E2 elimination reaction that is associated with the concentration of the methoxide ion. This mechanism will cause the disappearance of the tert-butyl bromide. The mechanism is as follows:

CH3 O- C

Br

C CH3

CH3H

H

H

CH3OH + CH2 C(CH3)2 + Br-

6.29 (a) Consider the general problem of converting a tertiary alkyl halide to an alkene, for example, the conversion of tert-butyl chloride to 2-methylpropene. What experimental conditions would you choose to ensure that elimination is favored over substitution? (b) Consider the opposite problem, that of carrying out a substitution reaction on a tertiary alkyl halide. Use as your example the conversion of tert-butyl chloride to tert-butyl ethyl ether. What experimental conditions would you employ to ensure the highest possible yield of the ether? Answer: (a.) It is hard to influence the relative partition between SN1 and E1 products, E2 reaction should be

chosen to ensure that elimination is favored, because SN2 reaction is hard to take place as the result of the steric hindrance. So high temperature, bulky strong base are required. In the conversion of tert-butyl chloride to 2-methylpropene, a base of C2H5ONa solved in C2H5OH and a high temperature should be offered.

(b.) Because of the steric hindrance, SN2 reaction couldn’t take place, so the only way to ensure the substitution is to ensure the SN1 reaction. We should use a neutral nucleophile at low temperature. For example, when (CH3)3CCl and C2H5OH are mixed at a low temperature, they can produce C(CH3)3OC2H5.

6.30 1-Bromobicyclo[2,2,1]heptane is extremely unreactive in either SN2 or SN1 reactions. Provide explanations for this behavior. Answer: 1-Bromobicyclo[2,2,1]heptane

Br

First it is one kind of bicycloalkane, so the ring strain in structure baffles the nucleophile attacking to the substrate from back. So it is unreactive in SN2. Second, it is hard for the bridge carbon atom forming carbocation, because the sp2 hybridization should be in one plane. So it is unreactive in SN1. 6.32 Starting with an appropriate alkyl halide, and using any other needed reagents, outline syntheses of each of the following. When alternative possibilities exist for a synthesis you should be careful to choose the one that gives the better yield (a) Butyl sec-butyl ether (g) (S)-2-Pentanol (b) CH3CH2SC(CH3) (h) (R)-2-Iodo-4-methylpentane (c) Methyl neopentyl ether (i) (CH3)3CCHCH2 (d) Methyl phenyl ether (j) cis-4-Isopropylcyclohexanol (e) C6H5CH2CN (k) (R)-CH3CH(CN)CH2CH3 (f) CH3CO2CH2CN (l) trans-1-Iodo-4-methylcyclohexane Answer:

Butyl sec-butyl etherCl +

OH

O

(a)

NaOH

(b)Cl S+

S

O(c) I + O

(d)O

O+I

(e) Cl CNCN-

(f)

ONa

O+

ClO

O

(g)OHH

Cl

+ NaOH + NaCl

(h) IBr+ I-

+ Br-

(i) CH3CH2ONaCl

cis-4-Isopropylcyclohexanol

HO

Br

+ OH- + Cl-(j)

(k)

(R)-CH3CH(CN)CH2CH3

C

HBrH

CN

+ CN- + Cl-

(l)

trans-1-Iodo-4-methylcyclohexane

I

Br--

+ I-+ Br-

6.33 Give structures for the products of each of the following reactions: (a)

F

H

H

Br

+ NaI(1mol) C5H8FI + NaBracetone

(b) 1,4-Dichlorohexane(1mol) + NaI(1 mol) acetone

C6H12ClI +NaCl

(C) 1,2-Dibromoethane(1mol) + NaSCH2CH2SNa C4H8S2 +2NaBr

(d) 4-Chloro-1-butanol +NaH H2

Et2O-

C4H8ClONa Et2Oheat

C4H8O + NaCl

(e) Propyne + NaNH3 liq.NH3

- NH3

C3H3Na CH3I C4H6 + NaI Answers: (a) C5H8FI

F

H

I

H

(b) C6H12ClI Cl

I

4-Chloro-1-iodo-hexane (c) C4H8S2

S S

(d) C4H8ClONa

ClO-Na+

C4H8O

O (e) C3H3Na

H3C C CNa C4H6

H3C C C CH3 6.34 When tert-butyl bromide undergoes SN1 hydrolysis, adding a “common ion”(e.g.,NaBr) to the aqueous solution has no effect on the rate. On the other hand, when (C6H5)2CHBr undergoes SN1 hydrolysis, adding NaBr retards the reaction .Given that the (C6H5CH)2CH+ cation is known to be much more stable than the (CH3)3C+ cation (and we shall see why in section 15.12A), provide an explanation for the different behavior of the two compounds. Answer:

Br + Br-

Ph

Ph

Br

Ph

Ph

+ Br-

H2OOH

H2OPh

Ph

OH

The rate determining step of the SN1 reaction is the formation of the carbocation, in the hydrolysis of tert-butyl bromide, the carbocation intermediate is relative unstable comparing with the diphenyl methyl cation. Which means that diphenyl methyl cation could combine with the bromine anion, a reversible reaction takes place easier than the first reaction. Therefore, increasing the concentration of bromine anion will increase the reaction speed of the reverse one, and retard the hydrolysis reaction. 6.35 When the alkyl bromides ( listed here ) were subjected to hydrolysis in a mixture of ethanol and water ( 80% C2H5OH/ 20% H2O ) at 55℃, the rates of the reaction showed the following order:

(CH3)3CBr﹥CH3Br﹥CH3CH2Br﹥(CH3)2CHBr Provide an explanation for this order of reactivity. Answer: The relative reactivity of the substrates towards SN2 is as follows:

CH3X﹥CH3CH2X﹥(CH3)2CHX The relative reactivity of the substrates towards SN1 is as follows:

(CH3)3CBr﹥(CH3)2CHBr ﹥CH3CH2Br﹥ CH3Br Combine those two mechanisms, the relative reactivity of the substrates towards the nucleophilic substitution reaction is: (CH3)3CBr﹥CH3Br﹥CH3CH2Br﹥(CH3)2CHBr 6.36 The reaction of 1º alkyl halides with nitrite salts produces both RNO2 and RONO. Account for this behavior. Answer: The nucleophilic sites for the NO2

- are:

N OO N OO

NO-

O

R X NO

O

R XN

O

O

R

X-δδ

δδ

transition state

N

-O

OR X N+

-O

OR X N+

-O

OX-R

δδ δδ

6.37 What would be the effect of increasing solvent polarity on the rate of each of the following nucleophilic substitution reaction? (a) Nu: + R—L→R—Nu+ + :L- (b) R—L+ → R+ + :L Answer:

The transition state for the reaction (a) should be: Nu R Lδ+ δ-

, in which the charge is

developing. The more polar the solvent, the better it can solvate the transition state, thus lowing the free energy of the activation and increasing the reaction rate.

The transition state for the reaction (b) should be: R Lδ+δ+

, in which the charge is becoming dispersed. A polar solvent is less able to solvate this transition state than it is to solvate the reactant. The free energy of activation, therefore, will become somewhat larger as the solvent polarity increases, and the rate will be slower. 6.38 Competition experiments are those in which two reactants at the same concentration (or one reactant with two reactive sites) compete for a reagent. Predict the major product resulting from each of the following competent experiment. Answer: (1)

ClH2C C

CH3

H2C

CH3

H2C Cl + I-- Cl

H2C C

H2C

CH3

CH3

H2C I + Cl--

In the solvent DMF, I— is more reactive than in water or methanol as no hydrogen bonds is formed. So it carries out a reaction SN2. The carbon atom 1 is bearing a bulky group which has a dramatic inhibiting effect. (2)

Cl C

CH3

CH3

H2C

H2C Cl + H2O acetone HO C

CH3

CH3

H2C

H2C Cl + HCl

H2O is a neutral molecule, and it can only carry out a reaction SN1. The tertiary carbocation is more stable than the primary carbocation.

Cl C

CH3

CH3

H2C

H2C Cl + H2O acetone HO C

CH3

CH3

H2C

H2C Cl + HCl

6.39 In contrast to SN2 reactions, SN1 reactions show relatively little nucleophile selectivity. That is, when more than one nucleophile is present in the reaction medium, SN1 reaction show only a slight tendency to discriminate between weak nucleophiles and strong nucleophiles, whereas SN2 reactions show a marked tendency to discriminate. (a) Provide an explanation for this behavior. (b) Show how your answer accounts for the fact that CH3CH2CH2CH2Cl reacts with 0.01 M

NaCN in ethanol to yield primarily CH3CH2CH2CH2CN, whereas under the same conditions (CH3) 3CCl reacts to give primarily (CH3) 3COCH2CH3.

Answer: (a) Reaction of alkyl halides by an SN1 mechanism are favored by the use of substrates that can

from relatively stable carbocation, by the use of weak nucleophiles, and by the use of highly ionizing solvent. SN1 mechanism, therefore, are important in solvolysis reactions of tertiary halides, especially when the solvent is highly polar. If we want to favor the reaction of an alkyl halide by an SN2 mechanism, we should use a relatively unhindered alkyl halide, a strong nucleophile, a polar aprotic solvent, and a high concentration of the nucleophile. So they have different behavior.

(b) CH3CH2CH2CH2Cl and 0.01 M NaCN in ethanol use SN2 reactions, and CH3CH2CH2CH2Cl show a marked tendency to NaCN. So the product is primarily CH3CH2CH2CH2CN. Whereas (CH3) 3CCl show slight tendency to NaCN. Because in 0.01 M NaCN in ethanol solution, ethanol is much more than NaCN. So the product is primarily (CH3) 3COCH2CH3.

6.40 In the gas phase, the homolytic bond dissociation energy (Section 10.2A) for the carbon-chlorine bond of tert-butyl chloride is +328KJ mol-1; the ionization potential for a tert-butyl radical is +715KJ mol-1; and the electron affinity of chlorine is –330KJ mol-1. Using these data, calculate the enthalpy change for the gas phase ionization of tert-butyl chloride to a tert-butyl cation and a chloride ion (this is the heterolytic bond dissociation energy of the carbon –chlorine bond). Answer:

Cl C(CH3)3 C(CH3)3 + Cl

C(CH3)3 +C(CH3)3 e

Cl e+ Cl

+ 328 KJ / mol

+ 715 KJ / mol

- 330 KJ / mol

Cl C(CH3)3 C(CH3)3 Cl+ + 713 KJ / mol

+328KJ +715KJ -330KJ

6.41 The reaction of chloroethane with water in the gas phase to produce ethanol and hydrogen chloride has a ΔH°= + 26.6KJ·mol-1and a ΔS°= + 4.81J·K-1mol-1at 25℃

(a) Which of these terms, if either, favors the reaction going to completion Solution: The entropy term is slightly favorable. (b) Calculate ΔG°for the reaction. What can you now say about whether the reaction will proceed to completion? Solution: ΔG°=ΔH°-TΔS°=26.6-298*4.81*10-3 = 25.17 > 0 So, chloroethane with water in the gas phase will produce ethanol and hydrogen chloride at 25 (c) Calculate the equilibrium constant for the reaction Solution: ΔG = - RT ln K° K°= 3.84*10-5 (d) In aqueous solution the equilibrium constant is very much larger than the one you just

calculated. How can you account for this fact? Answer: The equilibrium is very much more favorable in aqueous solution because solvation of the products takes place and thereby stabilizes them.

6.42 When (S)-2-bromopropanoic acid [ (S)-CH3CHBrCO2H] reacts with concentrated sodium hydroxide the product formed (after acidification) is (R)-2-hydroxypropanoic acid [(R)-CH3CHOHCO2H, commonly known as (R)-lactic acid]. This is, of course, the normal stereochemical result for an SN2 reaction. However, when the same reaction is carried out with a low concentration of hydroxide ion in the presence of Ag2O (where Ag acts as a Lewis acid), it takes place with overall retention of configuration to produce (S)-2-hydroxypropanoic acid. The mechanism of this reaction involves a phenomenon called neighboring group participation. Write a detailed mechanism for this reaction that accounts for the net retention of configuration when Ag and a low concentration of hydroxide are used. Answer: at the first condition, there exists sodium hydroxide, a strong base and a good nucleophile. So SN2 reaction takes place. But at the second condition, there exists Ag+, the reaction takes place in the following way:

OO

BrH3CH

Ag+

O

O

CH3H

OO

OHH3CHOH-

6.43 The phenomenon of configuration inversion in a chemical reaction was discovered in1896 by Paul von Walden. Walden’s proof of configuration inversion was base on the following cycle:

HO2CCH2CHClCO2H(-)-Chlorosuccinic acidAg2O

H2O

HO2CCH2CH(OH)CO2H(-)-Malic acid

HO2CCH2CHClCO2H(+)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(+)-Malic acid

Ag2OH2O

KOHPCl5

PCl5KOH

(a) Basing your answer on the preceding problem, which reaction of the Walden cycle are likely

to take place with overall inversion of configuration and which are likely to occur with overall retention of configuration? (b) Malic acid with a negative optical rotation is now known to have the (s) configuration. What are the configuration of the other compounds in the Walden cycle? (c) Walden also found that when (+)-malic acid is treated with thionyl chloride (rather PCl5), the product of the reaction is (+)-chlorosuccinic acid. How can you explain this result? (d) Assuming that the reaction of (-)-malic acid and thionyl chloride has the same steteochemistry, outline a Walden cycle based on the vse of thionyl chloride instead of PCl5.

ANSWER: (a) The reaction of the Walden cycle are likely to take place with overall inversion of

configuration:

HO2CCH2CH(OH)CO2H(-)-Malic acid

HO2CCH2CHClCO2H(+)-Chlorosuccinic acid

KOH

PCl5

HO2CCH2CHClCO2H(-)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(+)-Malic acid

KOH

PCl5 The reaction which are likely to occur with overall retention of configuration:

HO2CCH2CHClCO2H(-)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(-)-Malic acid

Ag2OH2O

HO2CCH2CHClCO2H(+)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(+)-Malic acid

Ag2OH2O

(b)

HO2CCH2CHClCO2H

(-)-Chlorosuccinic acid is (S)

HO2CCH2CH(OH)CO2H(+)-Malic acid is (R)

HO2CCH2CHClCO2H

(+)-Chlorosuccinic acid is (R) (c)The type of reaction is the retention of the configuration.

HO2CCH2CHClCO2H(+)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(+)-Malic acid

SOCl2

(d) The new cycle is

HO2CCH2CHClCO2H(+)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(-)-Malic acid

HO2CCH2CHClCO2H(-)-Chlorosuccinic acid

HO2CCH2CH(OH)CO2H(+)-Malic acid

thionyl chloride

thionyl chloride KOH

KOH

6.44 (R)-(3-Chloro-2-methylpropyl ) methyl ether ( A ) on reaction with azide ion (N3

-) in aqueous ethanol gives ( S )-( 3-azido-2-methylpropyl ) methyl ether (B). Compound A has the structure ClCH2CH(CH3)CH2OCH3.

( a ) Draw wedge-dashed wedge-line formulas of both A and B. ( b ) Is there a change of configuration during this reaction? Answer: A B

C

CH2N3

CH3HCH2OCH3

There is no change of configuration during this reaction.

6.46 cis-4-Bromocyclohexanol

t-BuO- in

t-BuOHracemic C6H10O

(compound C)

C

CH2Cl

H CH3CH2OCH3

Compound C has infrared absorption in the 1620 to 1680 cm- and in the 3590 to 3650 cm- regions. Draw and label the (R) and (S) enantiomers of produce C Answer