Chapter 7 Substitution Reactions

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1 Chapter 7 Substitution Reactions Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 7. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary. Substitution reactions exchange one for another. Evidence for the concerted mechanism, called S 2, includes the observation of a -order rate equation. The reaction proceeds with of configuration. S 2 reactions are said to be because the configuration of the product is determined by the configuration of the substrate. Evidence for the stepwise mechanism, called S 1, includes the observation of a -order rate equation. The step of an S 1 process is the rate-determining step. An S 1 reaction is a stepwise process with a first-order rate equation. There are four factors that impact the competition between the S 2 mechanism and S 1: 1) the, 2) the, 3) the, and 4) the. solvents favor S 2. Review of Skills Follow the instructions below. To verify that your answers are correct, look in your textbook at the end of Chapter 7. The answers appear in the section entitled SkillBuilder Review. SkillBuilder 7.1 Drawing the Curved Arrows of a Substitution Reaction A CCERTED MECAISM DRAW CURVED ARRWS, SWIG UCLEPILIC ATTACK ACCMPAIED BY SIMULTAEUS LSS F A LEAVIG GRUP A STEPWISE MECAISM DRAW A CURVED ARRW SWIG TE LSS F TE LEAVIG GRUP T FRM A CARBCATI ITERMEDIATE, FLLWED BY ATER CURVED ARRW SWIG TE UCLEPILIC ATTACK uc - LG - LG LG uc uc LG uc SkillBuilder 7.2 Drawing the Product of an S 2 Process DRAW TE MAJR PRDUCT F TE FLLWIG REACTI

2 CAPTER SkillBuilder 7.3 Drawing the Transition State of an S 2 Process DRAW TE TRASITI STATE F TE FLLWIG REACTI as S TRASITI STATE SkillBuilder 7.4 Drawing the Carbocation Intermediate of an S 1 Process DRAW TE CARBCATI TAT WULD BE FRMED IF A CLRIDE I IS EXPELLED FRM TE FLLWIG CMPUD - SkillBuilder 7.5 Drawing the Products of an S 1 Process PREDICT TE PRDUCTS F TE FLLWIG S 1 REACTI ac SkillBuilder 7.6 Drawing the Complete chanism of an S 1 Process IDETIFY TE TW CRE STEPS AD TREE PSSIBLE ADDITIAL STEPS F A S 1 PRCESS TW CRE STEPS TREE PSSIBLE ADDITIAL STEPS SkillBuilder 7.7 Drawing the Complete chanism of an S 2 Process IDETIFY TE E CRE STEP (CCERTED) AD TW PSSIBLE ADDITIAL STEPS F A S 2 PRCESS CRE STEP TW PSSIBLE ADDITIAL STEPS

3 116 CAPTER 7 SkillBuilder 7.8 Determining whether a Reaction Proceeds via an S 1 chanism or an S 2 chanism FILL I TE TABLE BELW, SWIG TE FEATURES TAT FAVR S 2 R S 1 REACTIS S 2 S 1 SUBSTRATE UC LG SLVET SkillBuilder 7.9 Identifying the Reagents ecessary for a Substitution Reaction IDETIFY TE REAGETS ECESSARY T ACIEVE TE FLLWIG TRASFRMATI 1) C 2) Review of Reactions Follow the instructions below. To verify that your answers are correct, look in your textbook at the end of Chapter 7. The answers appear in the section entitled Review of Reactions. S 2 DRAW TE CURVED ARRWS TAT SW TE FLW F ELECTR DESITY DURIG TE FLLWIG S 2 REACTI LG uc uc LG S 1 DRAW TE CURVED ARRWS TAT SW TE FLW F ELECTR DESITY DURIG TE FLLWIG S 1 REACTI LG - LG uc uc

4 CAPTER Solutions chloro-4-ethylheptane 1-bromo-1-methylcyclohexane 4,4-dibromo-1-chloropentane d) (S)-5-fluoro-2,2-dimethylhexane 7.2. S S I I 7.3. I 7.4.

5 118 CAPTER the rate of the reaction is tripled. the rate of the reaction is doubled. the rate of the reaction will be six times faster S F 3 C 2 3 S 1 The reaction does proceed with inversion of configuration. owever, the Cahn-Ingold- Prelog system for assigning a stereodescriptor (R or S) is based on a prioritization scheme. Specifically, the four groups connected to a chirality center are ranked (one through four). In the reactant (above left), the highest priority group is the leaving group (bromide) which is then replaced by a group that does not receive the highest priority. In the product, the fluorine atom has been promoted to the highest priority as a result of the reaction, and as such, the prioritization scheme has changed. In this way, the stereodescriptor (S) remains unchanged, despite the fact that chirality center undergoes inversion. 2 S 1 F C 3 3 4

6 CAPTER d) δ δ Being formed Being broken This step is favorable (downhill in energy) because ring strain is alleviated when the three-membered ring is opened C 3 R S R C 3 C 3 icotine

7 120 CAPTER 7 3 C 3 C C 3 R S R 3 C C 3 C 3 choline The rate of the reaction will be doubled, because the change in concentration of sodium chloride will not affect the rate. The rate of the reaction will remain the same, because the change in concentration of sodium chloride will not affect the rate Draw the carbocation intermediate generated by each of the following substrates in an S 1 reaction: ( ( ( (d) The first compound will generate a tertiary carbocation, while the second compound will generate a tertiary benzylic carbocation that is resonance stabilized. The second compound leads to a more stable carbocation, so that compound will lose its leaving group more rapidly than the first compound S a S d)

8 CAPTER S S Diastereomers o Yes o d) Yes e) Yes f) o o Yes Yes d) Yes e) o f) o g) o h) Yes i) o j) o k) Yes l) o o Yes Yes d) o e) o f) o 7.21.

9 122 CAPTER 7 d) Et Et Et Et e) f) g) S S h) Et Et Et Et

10 CAPTER LG uc Attack - d) - C LG rearr. uc Attack - e) - LG uc Attack - f) - C LG rearr. uc Attack - g) -LG uc Attack h) - LG uc Attack - Problem 7.20c and 7.20h exhibit the same pattern. Both problems are characterized by three mechanistic steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton transfer The chirality center at C2 is lost when the leaving group leaves to form a carbocation with trigonal planar geometry. The chirality center at C3 is lost during the hydride shift in the following step. nce again, the chirality center is converted into a trigonal planar sp 2 hybridized center (which is no longer a chirality center).

11 124 CAPTER Et Et Et d) S 1 S 2 either d) S 1 e) Both f) either g) Both

12 CAPTER S 1 S 2 S 2 d) S 2 e) S Ts I 2 F Ts I 2 F S 1 S 2 S 1 d) S 2 e) S 1 f) S 2 g) S 2 h) S Acetone is a polar aprotic solvent and will favor S 2 by raising the energy of the nucleophile, giving a smaller E a I S 1 MPA S 2 S 1 Racemic d) Ts ac DMF C S 2

13 126 CAPTER 7 I 2 S 1 e) Racemic f) ac DMS C S o. Preparation of this amine via the Gabriel synthesis would require the use of a tertiary alkyl halide, which will not undergo an S 2 process ( I a ( as S ( I I (d) DMS 1) Ts, pyridine (e) DMS (f) 2) a, DMS (g) I (h) 2 (i) 1) Ts, pyridine 2) ac C ) Ts, pyridine S 2) ai, DMS 3) as, DMS (R)- (R)- 2-butanol 2-butanethiol

14 CAPTER melphalan uc 2 uc 2 uc uc 2 uc uc Systematic ame = 2-chloropropane Common ame = isopropyl chloride Systematic ame = 2-bromo-2-methylpropane Common ame = tert-butyl bromide Systematic ame = 1-iodopropane Common ame = propyl iodide d) Systematic ame = 2-chlorobutane Common ame = propyl iodide d) Systematic ame = (R)-2-bromobutane Common ame = (R)-sec-butyl bromide e) Systematic ame = 1-chloro-2,2-dimethylpropane Common ame = neopentyl chloride f) Systematic ame = chlorocyclohexane Common ame = cyclohexyl chloride

15 128 CAPTER Increasing reactivity (S 2) secondary primary primary more sterically hindered primary less sterically hindered I secondary tertiary d) better leaving group o. Preparation of this compound via the process above would require the use of a tertiary alkyl halide, which will not undergo an S 2 process as sodium hydroxide methoxide dissolved in DMS tertiary primary primary tertiary Ts allylic d) better leaving group

16 CAPTER The rate of the reaction is doubled. The rate of the reaction is doubled The rate of the reaction is doubled The rate of the reaction will remain the same aprotic protic aprotic d) protic e) protic R C R The reaction is an S 2 process, and it does proceed with inversion of configuration. owever, the prioritization scheme changes when bromide (#1) is replaced with a cyano group (#2). As a result, the Cahn-Ingold-Prelog system assigns the same configuration to the reactant and the product δ δ Iodide functions as a nucleophile and attacks (S)-2-iodopentane, displacing iodide as a leaving group. The reaction is an S 2 process, and therefore proceeds via inversion of configuration. The product is (R)-2-iodopentane. The reaction continues until a racemic mixture is obtained.

17 130 CAPTER Racemic The chirality center is lost when the leaving group leaves to form a carbocation with trigonal planar geometry. The nucleophile can then attack either face of the planar carbocation, leading to a racemic mixture S S Racemic E Reaction coordinate Increasing stability secondary tertiary primary secondary

18 CAPTER Steps S S 2 Steps 2 3 Steps

19 132 CAPTER 7 d) Ts Et Et Et Et 4 Steps Et Et Ts a d) I ac DMS C Although the substrate is primary, it is still sterically hindered. As a result, S 2 reactions at neopentyl halides do not occur at an appreciable rate.

20 CAPTER The substrate is primary, and therefore, the reaction must proceed via an S 2 process. S 2 reactions are highly sensitive to the strength of the nucleophile, and the nucleophile (water) is a weak nucleophile. As a result, the reaction occurs slowly. ydroxide is a strong nucleophile, which favors the S 2 process a Ts 1) Ts, py 2) ac C d) as S e) a

21 134 CAPTER C d) S e) 2 f) 2 S S C S Et The second method is more efficient because the alkyl halide (methyl iodide) is not sterically hindered. The first method is not efficient because it employs a tertiary alkyl halide, and S 2 reactions do not occur at tertiary substrates ) Ts, pyridine 2) a a

22 CAPTER S S Rate = k as The rate would be slower. E d) Reaction coordinate e) δ S Et δ S 1 (tertiary substrate) Rate = k d) o. The rate is not dependent on the concentration or strength of the nucleophile. E e) Reaction coordinate

23 136 CAPTER S 2 C C Rate = k ac d) Yes. The reaction rate would double. E e) Reaction coordinate I

24 CAPTER I I This reaction occurs via an S 2 process. As such, the rate of the reaction is highly sensitive to the nature of the substrate. The reaction will be faster in this case, because the methyl ester is less sterically hindered than the ethyl ester Base Et Et Et Et When the leaving group leaves, the carbocation formed is resonance stabilized: Ts Resonance stabilized Iodide is a very good nucleophile (because it is polarizable), and it is also a very good leaving group (because it can stabilize the negative charge by spreading the charge over a large volume of space). As such, iodide will function as a nucleophile to displace the chloride ion. nce installed, the iodide group is a better leaving group than chloride, thereby increasing the rate of the reaction.

25 138 CAPTER

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