6.1 Writer the following as net ionic reactions and designate the nucleophile, substrate, and leaving group in each reaction. (a) 3 + 3 2 Na 3 2 3 +Na (b) Na+ 3 2 3 2 +Na (c) 2 3 +( 3 ) 3 ( 3 ) 3 3 + 3 + 2 + - (d) 3 2 2 +NaN 3 2 2 N+Na (e) 6 5 2 +2N 3 6 5 2 N 2 +N 4 ANSWER: (a) 3 + 3 2-3 2 3 + - nucleophile: 3 2 - substrate: 3 leaving group: - (b) - + 3 2 3 2 + - nucleophile: - substrate: 3 2 leaving group: - (c)2 3 +( 3 ) 3 ( 3 ) 3 3 + 3 + 2 + - nucleophile: 3 substrate: ( 3 ) 3 leaving group: - (d) 3 2 2 +N - 3 2 2 N+ - nucleophile: N - substrate: 3 2 2 leaving group: - (e) 6 5 2 +2N 3 6 5 2 N 2 +N + 4 + - - nucleophile: N 2 substrate: 6 5 2 leaving group: - 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.) ( 3 ) 3 ( 3 ) 3 6.3 S N 2 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) llustrate 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 5.7.]
(+)-2-hlorobutane - S N 2 (-)-2-Butanol 25 [ α ] = +36.00 0 25 [ α ] = -13.52 0 Enantiomerically pure Enantiomerically pure (b) When optically pure (+)-2-chlorobutane is allowed to react with potassium iodide in acetone in an S N 2 reaction, the 2-iodobutane that is produced has a minus rotation. What is the configuration of (--)-2-iodobutane? f (+)-2-iodobutane? Answer, (a) Et Et + Me Me (b) An inversion of configuration 3 3 2 3 2 3 R-(-)-2-iodobutane 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: ( 3 ) 3 3 2 SN1 (a) Structure of the carbocation intermediate: (b) Structures of the alcohols would be:
3 ( 3 ) 3 ( 3 ) 3 3 6.5 What product(s) would you expect from the methanolysis of the cyclohexane derivative given as the reaction in Problem 6.4? (3 ) 3 3 3 3 The product is following: (3 ) 3 3, ( 3 ) 3 3 3 the reaction is as 3 (3 ) 3 3 + 3 (3 ) 3 + 3 3 ( 3 ) 3 6.6 The relative rates of ethanolysis of four primary alkyl halides are as follows: 3 2, 1.0; 3 2 2, 0.28; ( 3 ) 2 2, 0.030; ( 3 ) 3 2, 0.00000042. (a). Are each of these reactions likely to be S N 1 or S N 2? (b) Provide an explanation for the relative reactivities that are observed. (a) All the reactions are likely to be S N 2. (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 assify the following solvents as being protic or aprotic: formic acid, ; acetone, 3 3 ;acetonitrile, 3 N ; formamide, N 2 ; sulfur
3 dioxide, S 2 ; ammonia, N 3 ; trimethylamine, 3 N 3 ; ethylene glycol, 2 2. Protic:, N 2, N 3, 2 2. 3 Aptotic: 3 3, 3 N, 3 N 3, S 2. 6.8 Would you expect the reaction of propyl bromide with sodium cyanide (NaN), that is, 3 2 2 +Na 3 2 2 N + Na to occur faster in MF or in ethanol? Explain you answer. N 3 MF 3 The reaction occurs faster in MF 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 S N 2 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). 3 2 or 3 -? 3 - is the stronger nucleophile. (b). 2 or 2 S? 2 S is the stronger nucleophile. (c). ( 3 ) 3 P or ( 3 ) 3 N? ( 3 ) 3 P 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 S N 2 reaction of ethyl chloride with potassium iodide in methanol and water decreases when the percentage of water in the mixture is increased.
(a) Because water is the most effective solvent for promoting ionization. t made the intermediate more stable. (b) Because the charge of the transition state is more dispersed than the charge in the starting material. ncreasing 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) (R)- (b) 3 2 (R)- 3 3 2 3 2 3 3 (c) (R)- 3 2 3 (d) (R)- 3 2 3 S S 3 (a) 3 + 3 2 3 2 3 + 3 3 (b) 3 + 3 3 3 + 3 3 (c) 3 3 S + S + 3 3 3 3 (d) + 3 S 3 S + 3 3
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) 3 2 2 (b) 3 2 2 (c) 3 2 2 2 3 (d) 3 2 2 S 3 (e) 3 2 2 3 (f) 3 2 2 N 3 3 (g) 3 N 2 2 3 3 (g) 3 2 2 N (h) 3 2 2 S Answers: (a) + 3 2 2 3 2 2 + (b) + 3 2 2 3 2 2 + (c) 3 2 + 3 2 2 3 2 2 2 3 + (d) 3 S + 3 2 2 3 2 2 S 3 + (e) 3 + 3 2 2 3 2 2 3 +
(f) N 3 + 3 2 2 3 2 2 N 3 + (g) ( 3 ) 3 N + 3 2 2 ( 3 ) 3 N 2 2 2 (h) N + 3 2 2 3 2 2 N + (i) S + 3 2 2 3 2 2 S + 6.14 Which alkyl halide would you expect to react more rapidly by an S N 2 mechanism? Explain your answer. (a) 3 2 2 or ( 3 ) 2 (b) 3 2 2 2 or 3 2 2 2 (c) ( 3 ) 2 2 or 3 2 2 2 (d) ( 3 ) 2 2 2 or 3 2 ( 3 ) 2 (e) 6 5 or 3 2 2 2 2 2 (a) 1-omopropane. Because it is less hindered. (b) 1-odobutane. Because - is a good leaving group. (c) 1-hlorobutane. Because it is less hindered. (d) 1-hloro-3-methylbutane. Because the carbon bearing the leaving group is less hindered than in 1-hloro-2-methylbutane. (e) 1-chlorohexane. Phenyl halides are unreactive in S N 2. 6.15 Which S N 2 reaction of each pair would you expect to take place more rapidly in a protic solvent?
(a) (1) 3 2 2 + 3 2-3 2 2 2 3 + - (2) 3 2 2 + 3 2 3 2 2 2 3 + (b) (1) 3 2 2 + 3 2-3 2 2 2 3 + - (2) 3 2 2 + 3 2 S - 3 2 2 S 2 3 + - (c) (1) 3 2 2 + ( 6 5 ) 3 N + 3 2 2 N( 6 5 ) 3 + - (2) 3 2 2 + ( 6 5 ) 3 P + 3 2 2 P( 6 5 ) 3 + - (d) (1) 3 2 2 (1.0M) + 3 - (1.0M) 3 2 2 3 + - (2) 3 2 2 (1.0M) + 3 - (2.0M) 3 2 2 3 + - each pair. (a) (1),(b) (2),(c) (2),(d) (2) are expected to take place more rapidly in a protic solvent in 6.16 Which S N 1 reaction of each pair would you expect to take place more rapidly? Explain your answer. (a) (b) (1)( 3 ) 3 + 2 ( 3 ) 3 + (2)( 3 ) 3 + 2 ( 3 ) 3 + (1)( 3 ) 3 + 2 ( 3 ) 3 + (2)( 3 ) 3 + 3 ( 3 ) 3 + () (1)( 3 ) 3 (1.0M) + 3 2 - (1.0M) ( 3 ) 3 2 3 + - (2)( 3 ) 3 (2.0M) + 3 2 - (1.0M) ( 3 ) 3 2 3 + - (d) (1)( 3 ) 3 (1.0M) + 3 2 - (1.0M) ( 3 ) 3 2 3 + - (2)( 3 ) 3 (1.0M) + 3 2 - (2.0M) ( 3 ) 3 2 3 + - (e) (1)( 3 ) 3 + 2 ( 3 ) 3 + (2) 6 5 + 2 6 5 + (a) Reaction2 will be more rapidly. Because bromide ion is better leaving group than chloride ion (b) Reaction1 will be more rapidly. Because 2 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 ( 3 ) 3 + is more stable.
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) 3 (b) 3 2 (c) 3 (d) 3 2 (e) 3 S (f) 3 2 S (g) 3 N (h) 3 2 N (i) 3 3 (j) 3 2 3 (k) yclopentene (a) 3 3 + - 3 + - (b) 3 2 3 2 + - 3 2 + - (c) 3 3 + - 3 + - (d) 3 2 3 2 + - 3 2 + - (e) 3 S 3 + S - 3 S + - 3 (f) 3 2 S 3 2 + S - 3 3 2 S + - (g) 3 N 3 + N - 3 3 N + - (h) 3 2 N 3 2 + N - 3 3 2 N + - (i) 3 3 2 3 + 2 Na 2 3 Na + 2 3 + 3-50 o 3 3 + - 3 (j) 3 2 3 3 2 + 3 - (k) yclopentene 50 o 3 3 2 3 + - 3 2 Na 3 2 + Na + 3 2
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. n each case provide an explanation for the failure of the reaction to take place as indicated. (a) 3 2 3 + - 3 2 + 3 (b) 3 2 3 + - 3 2 2 + - (c) + - - 2 3 2 2 N (d) +N - + - (e) N 3 + 3 3 3 N 2 + 3 + (f) N 3 + 3 2 3 N + 3 + 2 Answer (a) Alkanide ion ( - 3 ) is very powerful base and it s virtually never acting as leaving groups. Therefore, reaction (a) is not feasible. (b) ydride ion (: - ) 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 S N 2 reaction. (e) 3 - is a strongly basic ion and rarely acts as leaving group. Therefore, reaction e is not feasible. (f) 3 + 2 only appear in a strong acid and N 3 can t exist in a strong acid. 6.19 You have the task of preparing styrene ( 6 5 = 2 ) by dehydrohalogenation of either 1-bromo-2-phenylethane or 1-bromo-1-phenylethane use K in ethanol. Which halide would you choose as your starting material to give the better yield of the alkene? Explain your answer. Answers: 1-omo-1-phenylethane will be used as the starting material, because a 1º alkyl bromide gives mainly S N 2 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. 3 ( 3 ) 2, by one of the following reactions. Which reaction would give the better yield? Explain your choice. (1) 3 Na + ( 3 ) 2 3 ( 3 ) 2 (2) ( 3 ) 2 Na + 3 3 ( 3 ) 2 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? n each part give the mechanism (S N 1, 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.?) (a) S N 2 major product 3 2 2 2 2 + 3 2-50 3 2 3 2 2 ( 2 3 ) 3 + - minor product 3 2 2 2 2 + 3 2-50 3 2 3 2 2 = 2 (b) E2 major product 3 2 2 2 2 + ( 3 ) 3-50 ( 3 ) 3 3 2 2 = 2 minor product 3 2 2 2 2 + ( 3 ) 3-50 ( 3 ) 3 3 2 2 2 2 ( 3 ) 3 + - (c) E2 the only product ( 3 ) 3 + 3-50 3 3 2 3 (d) E1 the only product ( 3 ) 3 + ( 3 ) 3-50 ( 3 ) 3 3 2 3
(e) S N 2 the only product ( 3 ) 3 + - 50 acetone ( 3 ) 3 (f) S N 1 the only product 3 ( 3 ) 3 3 50 3 ( 3 ) 3 3 3 + ( 3 ) 3 3 (g) E1 3-hloropentane + 3-50 3 3 2 = 3 3-hloropentane + 3-50 3 3 ( 3 ) 2 2 3 + - (h) S N 1 3-hloropentane + 3 2-50 3 2 3 2 ( 2 3 ) 2 3 + - 3-hloropentane + 3 2-50 3 2 3 = 2 3 (i) E2 major product - + (R)-2-bromobutane 25 + 2 + minor product - + (R)-2-bromobutane 25
(j) S N 1 (S)-3-omo-3-methylhexane (k) S N 2 25 3 3 3 2 2 2 3 3 (+)- (S)-2-omooctane + - 50 3 (R)-2-odooctane + - 6.22 Write conformational structures for the substitution products of the following deuterium-labeled compounds: (a) - 3? (b) - 3? (c) - 3? (d) 3 2 3? The answer:
(a) - 3 + - (b) - 3 + - (c) - 3 + - (d) 3 3 3 2 3 3 + 3 3 + + 3 6.23 Although ethyl bromide and isobutyl bromide are both primary halides. Ethyl bromide undergoes S N 2 reactions more than 10 times faster than isobutyl bromide does. When each compound is treated with a strong base/ nucleophile ( 3 2 - ), 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? The reason ethyl bromide undergoes S N 2 reactions faster than isobutyl bromide is steric effect. This steric hindrance causes isopropyl bromide to react more slowly in S N 2 reactions and to give relatively more elimination (by an E2) when a strong base is used. 6.24 onsider the reaction of - with 3 2. (a) Would you expect the reaction to be S N 1 or S N 2?
The rate constant for the reaction at 60 is 5*10-5 L mol -1 s -1. What is the reaction rate if [ - ]=0.1mol L -1 and [ 3 2 ]=0.1 mol L -1? (c) f [ - ]=0.1mol L -1 and [ 3 2 ]=0.2mol L -1? (d) f [ - ]=0.2mol L -1 and [ 3 2 ]=0.1 mol L -1? (e) f [ - ]=0.2mol L -1 and [ 3 2 ]=0.2 mol L -1? Because the substrate is the primary alkyl halide, the reaction would be S N 2. V=k*[ - ][ 3 2 ] (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) 3 N - or 3 N 2 (e) 2 or 3 + (b) 3 - or 3 - (f) N 3 or N 4 (c) 3 S or 3 (g) 2 S or S - (d) ( 6 5 ) 3N or ( 6 5 ) 3P (h) 3 - or - (a) 3 N - (b) 3 - (c) 3 S (d) ( 6 5 ) 3P (e) 2 (f) N 3 (g) S - (h) - 6.26 write mechanisms that account for the products of the following reaction (a) 2 2 _ 2 (b) 2 N 2 2 2 2 _ 2 N ANSWER: (a)
- (b) 2 N N - N 6.27 Many S N 2 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. Because Alkyl chlorides and bromides are also easily converted to alkyl iodides by nucleophilic substitution reactions. R - + - R + - R + - R + - The effect of the leaving group in S N 2 reaction: R > R > R > R F That is why many S N 2 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. odide 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. owever, increasing the concentration of sodium methoxide causes a marked increase in the rate at which tert-butyl bromide disappears from the mixture. tert-butyl alcohol and tert-butyl methyl ether are formed via an S N 1 mechanism. The rate of the reaction is independent of the concentration of methoxide ion. owever, 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: 3 3-3 3 + 2 ( 3 ) 2 + -
6.29 (a) onsider 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) onsider 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? (a.) t is hard to influence the relative partition between S N 1 and E1 products, E2 reaction should be chosen to ensure that elimination is favored, because S N 2 reaction is hard to take place as the result of the steric hindrance. So high temperature, bulky strong base are required. n the conversion of tert-butyl chloride to 2-methylpropene, a base of 2 5 Na solved in 2 5 and a high temperature should be offered. (b.) Because of the steric hindrance, S N 2 reaction couldn t take place, so the only way to ensure the substitution is to ensure the S N 1 reaction. We should use a neutral nucleophile at low temperature. For example, when ( 3 ) 3 and 2 5 are mixed at a low temperature, they can produce ( 3 ) 3 2 5. 6.30 1-omobicyclo[2,2,1]heptane is extremely unreactive in either S N 2 or S N 1 reactions. Provide explanations for this behavior. 1-omobicyclo[2,2,1]heptane 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 S N 2. Second, it is hard for the bridge carbon atom forming carbocation, because the sp 2 hybridization should be in one plane. So it is unreactive in S N 1. 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) 3 2 S( 3 ) (h)(r)-2-odo-4-methylpentane (c) Methyl neopentyl ether (i)( 3 ) 3 2 (d) Methyl phenyl ether (j) cis-4-sopropylcyclohexanol (e) 6 5 2 N (k)(r)- 3 (N) 2 3 (f) 3 2 2 N (l) trans-1-odo-4-methylcyclohexane
(a) + Na Butylsec-butyl ether (b) + S S (c) + (d) + (e) N N - (f) + Na (g) + Na + Na (h) + - + - (i) 3 2 Na
(j) + - + - cis-4-sopropylcyclohexanol (k) N + N - + - (R)- 3 (N) 2 3 (l) + - + - -- trans-1-odo-4-methylcyclohexane 6.33 Give structures for the products of each of the following reactions: (a) (b) 1,4-ichlorohexane(1mol) + Na(1 mol) F acetone + Na(1mol) 5 8 F + Na acetone 6 12 +Na () 1,2-ibromoethane(1mol) + NaS22SNa 4 8 S 2 +2Na - 2 heat (d) 4-hloro-1-butanol +Na Et 2 4 8 Na Et 2 4 8 + Na -N 3 (e) Propyne + NaN3 liq.n 3 3 3 Na 3 4 6 + Na Answers: (a) 5 8 F F
(b) 6 12 4-hloro-1-iodo-hexane (c) 4 8 S 2 S S (d) 4 8 Na 4 8 - Na + (e) 3 3 Na 3 Na 4 6 3 3 6.34 When tert-butyl bromide undergoes S N 1 hydrolysis, adding a common ion (e.g.,na) to the aqueous solution has no effect on the rate. n the other hand, when ( 6 5 ) 2 undergoes S N 1 hydrolysis, adding Na retards the reaction.given that the ( 6 5 ) 2 + cation is known to be much more stable than the ( 3 ) 3 + cation (and we shall see why in section 15.12A), provide an explanation for the different behavior of the two compounds. + - 2 Ph Ph + - Ph Ph 2 Ph Ph
The rate determining step of the S N 1 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% 2 5 / 20% 2 ) at 55, the rates of the reaction showed the following order: ( 3 ) 3 > 3 > 3 2 > ( 3 ) 2 Provide an explanation for this order of reactivity. The relative reactivity of the substrates towards SN2 is as follows: 3 X> 3 2 X> ( 3 ) 2 X The relative reactivity of the substrates towards SN1 is as follows: ( 3 ) 3 > ( 3 ) 2 > 3 2 > 3 ombine those two mechanisms, the relative reactivity of the substrates towards the nucleophilic substitution reaction is: ( 3 ) 3 > 3 > 3 2 > ( 3 ) 2 6.36 The reaction of 1º alkyl halides with nitrite salts produces both RN 2 and RN. Account for this behavior. The nucleophilic sites for the N - 2 are: N N N - δ R δ X N δ R δ X N R X - transition state - N δ R δ X - δ N + - δ R X N+ R X -
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 δ + δ - The transition state for the reaction (a) should be: Nu R L, 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. δ + δ + The transition state for the reaction (b) should be: R L, 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 ompetition 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. (1) 3 3 2 2 2 + -- 2 2 2 + -- 3 3 n the solvent MF, is more reactive than in water or methanol as no hydrogen bonds is formed. So it carries out a reaction S N 2. The carbon atom 1 is bearing a bulky group which has a dramatic inhibiting effect. (2) 3 3 2 2 + 2 acetone 2 2 + 3 3 2 is a neutral molecule, and it can only carry out a reaction S N 1. The tertiary carbocation is more stable than the primary carbocation.
3 3 2 2 + 2 acetone 2 2 + 3 6.39 n contrast to S N 2 reactions, S N 1 reactions show relatively little nucleophile selectivity. That is, when more than one nucleophile is present in the reaction medium, S N 1 reaction show only a slight tendency to discriminate between weak nucleophiles and strong nucleophiles, whereas S N 2 reactions show a marked tendency to discriminate. (a) Provide an explanation for this behavior. (b) Show how your answer accounts for the fact that 3 2 2 2 reacts with 0.01 M NaN in ethanol to yield primarily 3 2 2 2 N, whereas under the same conditions ( 3 ) 3 reacts to give primarily ( 3 ) 3 2 3. (a) Reaction of alkyl halides by an S N 1 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. S N 1 mechanism, therefore, are important in solvolysis reactions of tertiary halides, especially when the solvent is highly polar. f we want to favor the reaction of an alkyl halide by an S N 2 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) 3 2 2 2 and 0.01 M NaN in ethanol use S N 2 reactions, and 3 2 2 2 show a marked tendency to NaN. So the product is primarily 3 2 2 2 N. Whereas ( 3 ) 3 show slight tendency to NaN. Because in 0.01 M NaN in ethanol solution, ethanol is much more than NaN. So the product is primarily ( 3 ) 3 2 3. 3 6.40 n 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). ( 3 ) 3 ( 3 ) 3 + + 328 KJ / mol ( 3 ) 3 + ( 3 ) 3 e + 715 KJ / mol + e - 330 KJ / mol ( 3 ) 3 ( 3 ) 3 + + 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 Δ = + 26.6KJ mol -1 and a ΔS = + 4.81J K -1 mol -1 at 25 (a) Which of these terms, if either, favors the reaction going to completion Solution: The entropy term is slightly favorable. (b) alculate ΔG for the reaction. What can you now say about whether the reaction will proceed to completion? Solution: ΔG =Δ -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) alculate the equilibrium constant for the reaction Solution: ΔG = - RT ln K K = 3.84*10-5 (d) n aqueous solution the equilibrium constant is very much larger than the one you just calculated. ow can you account for this fact? 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)- 3 2 ] reacts with concentrated sodium hydroxide the product formed (after acidification) is (R)-2-hydroxypropanoic acid [(R)- 3 2, commonly known as (R)-lactic acid]. This is, of course, the normal stereochemical result for an S N 2 reaction. owever, when the same reaction is carried out with a low concentration of hydroxide ion in the presence of Ag 2 (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. at the first condition, there exists sodium hydroxide, a strong base and a good nucleophile. So S N 2 reaction takes place. But at the second condition, there exists Ag +, the reaction takes place in the following way: 3 Ag + 3-3 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:
Ag 2 2 2 2 2 (-)-hlorosuccinic acid P 5 K 2 2 () 2 (-)-Malic acid 2 2 () 2 (+)-Malic acid P 5 K 2 2 2 (+)-hlorosuccinic acid Ag 2 2 (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 P 5 ), the product of the reaction is (+)-chlorosuccinic acid. ow 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 P5. ANSWER: (a) The reaction of the Walden cycle are likely to take place with overall inversion of configuration: K 2 2 () 2 2 2 2 (-)-Malic acid P (+)-hlorosuccinic acid 5 2 2 2 (-)-hlorosuccinic acid K P 5 2 2 () 2 (+)-Malic acid The reaction which are likely to occur with overall retention of configuration: Ag 2 2 2 2 2 2 () 2 (-)-hlorosuccinic acid 2 (-)-Malic acid 2 2 2 (+)-hlorosuccinic acid Ag 2 2 2 2 () 2 (+)-Malic acid (b) 2 2 2 (-)-hlorosuccinic acid is (S)
2 2 () 2 (+)-Malic acid is (R) 2 2 2 (+)-hlorosuccinic acid is (R) (c)the type of reaction is the retention of the configuration. 2 2 () 2 (+)-Malic acid S 2 2 2 2 (+)-hlorosuccinic acid (d) The new cycle is K 2 2 2 (+)-hlorosuccinic acid thionyl chloride 2 2 () 2 (-)-Malic acid 2 2 () 2 (+)-Malic acid thionyl chloride K 2 2 2 (-)-hlorosuccinic acid 6.44 (R)-(3-hloro-2-methylpropyl ) methyl ether ( A ) on reaction with azide ion (N - 3 ) in aqueous ethanol gives ( S )-( 3-azido-2-methylpropyl ) methyl ether (B). ompound A has the structure 2 ( 3 ) 2 3. ( a ) raw wedge-dashed wedge-line formulas of both A and B. ( b ) s there a change of configuration during this reaction? A B 2 2 N 3 3 2 3 3 2 3 There is no change of configuration during this reaction. 6.46 cis-4-omocyclohexanol t-bu - in t-bu racemic 6 10 (compound )
ompound has infrared absorption in the 1620 to 1680 cm - and in the 3590 to 3650 cm - regions. raw and label the (R) and (S) enantiomers of produce Answer