CHAPTER 8 NUCLEOPHILIC SUBSTITUTION SOLUTIONS TO TEXT PROBLEMS 8.1 Identify the nucleophilic anion in each reactant. The nucleophilic anion replaces bromine as a sub- stituent on carbon. (b) Potassium ethoxide serves as a source of the nucleophilic anion CH 3 CH 2 O H11002 . (c) (d) Lithium azide is a source of the azide ion. It reacts with methyl bromide to give methyl azide. (e) The nucleophilic anion in KCN is cyanide ( ). The carbon atom is negatively charged and is normally the site of nucleophilic reactivity. H11001NCH 3 Br CH 3 C Br H11002 Bromide ionMethyl cyanide (product) Methyl bromide NC H11001 Cyanide ion (nucleophile) H11002 NC H11002 N H11002 N H11002 N H11001H11001 H11001 N H11002 N H11001 CH 3 Br CH 3 N Br Bromide ionMethyl azide (product) Methyl bromide Azide ion (nucleophile) H11002 N H11002 N H11002 N H11001 CH 3 Br Br Benzoate ion Methyl bromide Methyl benzoate Bromide ion C H11001O O C H11001OCH 3 O H11002 H11002 H11001H11001CH 3 CH 2 O CH 3 CH 2 OCH 3 CH 3 Br Br Bromide ionEthyl methyl ether (product) Methyl bromideEthoxide ion (nucleophile) H11002 H11002 184 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) The anion in sodium hydrogen sulfide (NaSH) is . (g) Sodium iodide is a source of the nucleophilic anion iodide ion, . The reaction of sodium iodide with alkyl bromides is usually carried out in acetone to precipitate the sodium bromide formed. 8.2 Write out the structure of the starting material. Notice that it contains a primary bromide and a pri- mary chloride. Bromide is a better leaving group than chloride and is the one that is displaced faster by the nucleophilic cyanide ion. 8.3 No, the two-step sequence is not consistent with the observed behavior for the hydrolysis of methyl bromide. The rate-determining step in the two-step sequence shown is the first step, ionization of methyl bromide to give methyl cation. 1. 2. In such a sequence the nucleophile would not participate in the reaction until after the rate- determining step is past, and the reaction rate would depend only on the concentration of methyl bromide and be independent of the concentration of hydroxide ion. Rate H11005 k[CH 3 Br] The predicted kinetic behavior is first order. Second order kinetic behavior is actually observed for methyl bromide hydrolysis, so the proposed mechanism cannot be correct. 8.4 Inversion of configuration occurs at the stereogenic center. When shown in a Fischer projection, this corresponds to replacing the leaving group on the one side by the nucleophile on the opposite side. 8.5 The example given in the text illustrates inversion of configuration in the S N 2 hydrolysis of (S)-(H11001)-2-bromooctane, which yields (R)-(H11002)-2-octanol. The hydrolysis of (R)-(H11002)-2-bromooctane exactly mirrors that of its enantiomer and yields (S)-(H11001)-2-octanol. Hydrolysis of racemic 2-bromooctane gives racemic 2-octanol. Remember, optically inactive reactants must yield optically inactive products. 8.6 Sodium iodide in acetone is a reagent that converts alkyl chlorides and bromides into alkyl io- dides by an S N 2 mechanism. Pick the alkyl halide in each pair that is more reactive toward S N 2 displacement. NaOH S N 2 CH 3 CH 2 (CH 2 ) 4 CH 3 BrH CH 3 CH 2 (CH 2 ) 4 CH 3 HHO (S)-(H11001)-2-Bromooctane (R)-(H11002)-2-Octanol CH 3 H11001 CH 3 OHH11001 HO H11002 fast CH 3 Br CH 3 H11001 H11001 Br H11002 slow ClCH 2 CH 2 CH 2 Br 1-Bromo-3-chloropropane 4-Chlorobutanenitrite NaCN ethanol–water ClCH 2 CH 2 CH 2 CN H11001H11001CH 3 Br Br H11002 CH 3 Bromide ionMethyl iodideMethyl bromideIodide ion I I H11002 acetone I H11002 HS H11001H11001CH 3 Br CH 3 SH Br H11002 Bromide ionMethanethiolMethyl bromideHydrogen sulfide ion H11002 SH H11002 NUCLEOPHILIC SUBSTITUTION 185 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 186 NUCLEOPHILIC SUBSTITUTION (b) The less crowded alkyl halide reacts faster in an S N 2 reaction. 1-Bromopentane is a primary alkyl halide and so is more reactive than 3-bromopentane, which is secondary. (c) Both halides are secondary, but fluoride is a poor leaving group in nucleophilic substitution reactions. Alkyl chlorides are more reactive than alkyl fluorides. (d) A secondary alkyl bromide reacts faster under S N 2 conditions than a tertiary one. (e) The number of carbons does not matter as much as the degree of substitution at the reaction site. The primary alkyl bromide is more reactive than the secondary. 8.7 Nitrite ion has two potentially nucleophilic sites, oxygen and nitrogen. Thus, an alkyl iodide can yield either an alkyl nitrite or a nitroalkane depending on whether the oxy- gen or the nitrogen of nitrite ion attacks carbon. Both do, and the product from 2-iodooctane is a mixture of CH 3 CH(CH 2 ) 5 CH 3 CH 3 CH(CH 2 ) 5 CH 3 NO 2 ONO and NRH11001 N H11001 R H11001 Nitrite ion Alkyl iodide Nitroalkane Iodide ion I H11002 I O O H11002 O H11002 O ON RH11001O I ON RH11001O Nitrite ion Alkyl iodide Alkyl nitrite Iodide ion I H11002H11002 BrCH 2 (CH 2 ) 8 CH 3 CH 3 CHCH 3 Br 2-Bromopropane (secondary; less reactive in S N 2) 1-Bromodecane (primary; more reactive in S N 2) CH 3 CHCH 2 CH 2 CHCH 3 Br CH 3 CH 3 CCH 2 CH 2 CH 2 CH 3 Br CH 3 2-Bromo-2-methylhexane (tertiary; less reactive in S N 2) 2-Bromo-5-methylhexane (secondary; more reactive in S N 2) CH 3 CHCH 2 CH 2 CH 3 F 2-Fluoropentane (less reactive) CH 3 CHCH 2 CH 2 CH 3 Cl 2-Chloropentane (more reactive) BrCH 2 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CHCH 2 CH 3 Br 3-Bromopentane (secondary; less reactive in S N 2) 1-Bromopentane (primary; more reactive in S N 2) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.8 Solvolysis of alkyl halides in alcohols yields ethers as the products of reaction. The reaction proceeds by an S N 1 mechanism. 8.9 The reactivity of an alkyl halide in an S N 1 reaction is dictated by the ease with which it ionizes to form a carbocation. Tertiary alkyl halides are the most reactive, methyl halides the least reactive. (b) Cyclopentyl iodide ionizes to form a secondary carbocation, and the carbocation from 1-methylcyclopentyl iodide is tertiary. The tertiary halide is more reactive. (c) Cyclopentyl bromide ionizes to a secondary carbocation. 1-Bromo-2,2-dimethyl-propane is a primary alkyl halide and is therefore less reactive. (d) Iodide is a better leaving group than chloride in both S N 1 and S N 2 reactions. (CH 3 ) 3 CI (CH 3 ) 3 CCl tert-Butyl chloride (less reactive) tert-Butyl iodide (more reactive) BrH Cyclopentyl bromide (secondary; more reactive in S N 1) 1-Bromo-2,2-dimethylpropane (primary; less reactive in S N 1) (CH 3 ) 3 CCH 2 Br IH 3 C 1-Methylcyclopentyl iodide (tertiary; more reactive in S N 1) IH Cyclopentyl iodide (secondary; less reactive in S N 1) H11001H11001 tert-Butyloxonium ion (CH 3 ) 3 C O H CH 3 H11001 tert-Butyl methyl ether (CH 3 ) 3 C OCH 3 Bromide ion Br H11002 Hydrogen bromide H Br tert-Butyloxonium ion tert-Butyl cation Methanol fast (CH 3 ) 3 C H11001 H11001 O O(CH 3 ) 3 C H11001 CH 3 H CH 3 H (CH 3 ) 3 C H11001 Br H11002 Bromide ion tert-Butyl bromide tert-Butyl cation (CH 3 ) 3 C Br slow H11001 H11001 H11001(CH 3 ) 3 CBr (CH 3 ) 3 COCH 3 CH 3 OH HBr Hydrogen bromide tert-Butyl methyl ether Methanoltert-Butyl bromide NUCLEOPHILIC SUBSTITUTION 187 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.10 The alkyl halide is tertiary and so undergoes hydrolysis by an S N 1 mechanism. The carbocation can be captured by water at either face. A mixture of the axial and the equatorial alcohols is formed. The same two substitution products are formed from trans-1,4-dimethylcyclohexyl bromide be- cause it undergoes hydrolysis via the same carbocation intermediate. 8.11 Write chemical equations illustrating each rearrangement process. Hydride shift: Methyl shift: Rearrangement by a hydride shift is observed because it converts a secondary carbocation to a more stable tertiary one. A methyl shift gives a secondary carbocation—in this case the same carbocation as the one that existed prior to rearrangement. 8.12 (b) Ethyl bromide is a primary alkyl halide and reacts with the potassium salt of cyclohexanol by substitution. (c) No strong base is present in this reaction; the nucleophile is methanol itself, not methoxide. It reacts with sec-butyl bromide by substitution, not elimination. CH 3 OH sec-Butyl bromide CH 3 CHCH 2 CH 3 Br CH 3 CHCH 2 CH 3 OCH 3 sec-Butyl methyl ether H11001 OK Potassium cyclohexanolate OCH 2 CH 3 Cyclohexyl ethyl ether CH 3 CH 2 Br Ethyl bromide Secondary carbocation CC H H CH 3 CH 3 H 3 C H11001 CC H CH 3 H 3 C H11001 H CH 3 Tertiary carbocation C CCH 3 H CH 3 H H 3 C H11001 CC H H H 3 C H11001 CH 3 CH 3 cis-1,4-Dimethylcyclohexyl bromide CH 3 Br H 3 C trans-1,4-Dimethylcyclohexanol OH CH 3 H 3 C cis-1,4-Dimethylcyclohexanol CH 3 OH H 3 C Carbocation intermediate CH 3 H 3 C H 2 O H 2 O H11001 188 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Secondary alkyl halides react with alkoxide bases by E2 elimination. 8.13 Alkyl p-toluenesulfonates are prepared from alcohols and p-toluenesulfonyl chloride. 8.14 As in part (a), identify the nucleophilic anion in each part. The nucleophile replaces the p-toluene- sulfonate (tosylate) leaving group by an S N 2 process. The tosylate group is abbreviated as OTs. (b) (c) (d) (e) 8.15 The hydrolysis of (S)-(H11001)-1-methylheptyl p-toluenesulfonate proceeds with inversion of configura- tion, giving the R enantiomer of 2-octanol. In Section 8.14 of the text we are told that optically pure (S)-(H11001)-1-methylheptyl p-toluenesulfonate is prepared from optically pure (S)-(H11001)-2-octanol having a specific rotation [H9251] 25 D H110019.9°. The conversion of an alcohol to a p-toluenesulfonate proceeds with complete retention of configuration. Hydrolysis of this p-toluenesulfonate with inversion of configuration therefore yields optically pure (R)-(H11002)-2-octanol of [H9251] 25 D H110029.9°. H CH 3 (CH 2 ) 5 H 3 C COTs H HO (CH 2 ) 5 CH 3 CH 3 C H 2 O (S)-(H11001)-1-Methylheptyl p-toluenesulfonate (R)-(H11002)-2-Octanol CH 3 (CH 2 ) 16 CH 2 OTsCH 3 CH 2 CH 2 CH 2 S H11002 Butanethiolate ion H11001 CH 3 (CH 2 ) 16 CH 2 SCH 2 CH 2 CH 2 CH 3 H11001 TsO H11002 Octadecyl p-toluenesulfonate Butyl octadecyl thioether p-Toluenesulfonate anion HS H11002 Hydrogen sulfide ion H11001 CH 3 (CH 2 ) 16 CH 2 OTs CH 3 (CH 2 ) 16 CH 2 SH Octadecyl p-toluenesulfonate 1-Octadecanethiol TsO H11002 p-Toluenesulfonate anion H11001 C Cyanide ion N CH 3 (CH 2 ) 16 CH 2 OTs Octadecyl p-toluenesulfonate CH 3 (CH 2 ) 16 CH 2 Octadecyl cyanide TsO H11002 p-Toluenesulfonate anion H11001H11001 C N H11002 CH 3 (CH 2 ) 16 CH 2 OTs CH 3 (CH 2 ) 16 CH 2 I TsO H11002 Octadecyl p-toluenesulfonate Octadecyl iodide p-Toluenesulfonate anion H11001I H11002 Iodide ion H11001 H11001H11001CH 3 (CH 2 ) 16 CH 2 OH 1-Octadecanol p-Toluenesulfonyl chloride Octadecyl p-toluenesulfonate Hydrogen chloride SCl O O H 3 C HClCH 3 O O CH 3 (CH 2 ) 16 CH 2 OS pyridine sec-Butyl bromide CH 3 CHCH 2 CH 3 Br CH 3 CH 2-Butene (major product; mixture of cis and trans) NaOCH 3 CH 3 OH CHCH 3 CHCH 2 CH 3 1-Butene H11001 H 2 C NUCLEOPHILIC SUBSTITUTION 189 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.16 Protonation of 3-methyl-2-butanol and dissociation of the alkyloxonium ion gives a secondary carbocation. A hydride shift yields a tertiary, and thus more stable, carbocation. Capture of this car- bocation by chloride ion gives the major product, 2-chloro-2-methylbutane. 8.17 1-Bromopropane is a primary alkyl halide, and so it will undergo predominantly S N 2 displacement regardless of the basicity of the nucleophile. (a) (b) (c) (d) (e) ( f ) (g) 8.18 Elimination is the major product when secondary halides react with anions as basic as or more basic than hydroxide ion. Alkoxide ions have a basicity comparable with hydroxide ion and react with CH 3 CH 2 CH 2 Br NaSCH 3 ethanol Methyl propyl sulfide CH 3 CH 2 CH 2 SCH 3 CH 3 CH 2 CH 2 Br NaSH ethanol 1-Propanethiol CH 3 CH 2 CH 2 SH CH 3 CH 2 CH 2 Br NaN 3 ethanol–water 1-Azidopropane CH 3 CH 2 CH 2 N 3 CH 3 CH 2 CH 2 Br NaCN DMSO Butanenitrite CH 3 CH 2 CH 2 CN CH 3 CH 2 CH 2 Br NaOCH 2 CH 3 ethanol Ethyl propyl ether CH 3 CH 2 CH 2 OCH 2 CH 3 CH 3 CH 2 CH 2 Br CH 3 CONa acetic acid Propyl acetate CH 3 CH 2 CH 2 OCCH 3 O O CH 3 CH 2 CH 2 Br CH 3 CH 2 CH 2 I NaI acetone 1-Bromopropane 1-Iodopropane CH 3 CHCH(CH 3 ) 2 OH 3-Methyl-2-butanol HCl H11001H11001 CH 3 CHCH(CH 3 ) 2 Secondary carbocation; less stable CH 3 CH 2 C(CH 3 ) 2 Tertiary carbocation; more stable Cl H11002 Cl H11002 Cl CH 3 CHCH(CH 3 ) 2 Cl 2-Chloro-3-methylbutane (trace) CH 3 CH 2 C(CH 3 ) 2 2-Chloro-2-methylbutane (major product; H1101597%) hydride shift 190 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website secondary halides to give predominantly elimination products. Thus ethoxide ion [part (c)] will react with 2-bromopropane to give mainly propene. 8.19 (a) The substrate is a primary alkyl bromide and reacts with sodium iodide in acetone to give the corresponding iodide. (b) Primary alkyl chlorides react with sodium acetate to yield the corresponding acetate esters. (c) The only leaving group in the substrate is bromide. Neither of the carbon–oxygen bonds is susceptible to cleavage by nucleophilic attack. (d) Hydrolysis of the primary chloride yields the corresponding alcohol. (e) The substrate is a primary chloride. ( f ) Primary alkyl tosylates yield iodides on treatment with sodium iodide in acetone. (g) Sulfur displaces bromide from ethyl bromide. CH 3 CH 2 BrH11001 O CH 2 SNa CH 2 SCH 2 CH 3O Sodium (2-furyl)- methanethiolate Ethyl bromide Ethyl (2-furyl)methyl sulfide (80%) TsOCH 2 CH 3 CH 3 O O ICH 2 CH 3 CH 3 O O NaI acetone (2,2-Dimethyl-1,3-dioxolan-4-yl)- methyl p-toluenesulfonate 2,2-Dimethyl-4-(iodomethyl)- 1,3-dioxolane (60%) NaN 3 acetone–water tert-Butyl chloroacetate tert-Butyl azidoacetate (92%) ClCH 2 COC(CH 3 ) 3 O N 3 CH 2 COC(CH 3 ) 3 O NC CH 2 Cl NC CH 2 OH H 2 O, HO H11002 p-Cyanobenzyl chloride p-Cyanobenzyl alcohol (85%) NaCN ethanol–water 2-Cyanoethyl ethyl ether (52–58%) 2-Bromoethyl ethyl ether CH 3 CH 2 OCH 2 CH 2 CNCH 3 CH 2 OCH 2 CH 2 Br O 2 NCH 2 Cl p-Nitrobenzyl chloride O 2 NCH 2 OCCH 3 O p-Nitrobenzyl acetate (78–82%) CH 3 CONa O acetic acid NaI acetone Ethyl iodoacetate (89%)Ethyl bromoacetate ICH 2 COCH 2 CH 3 O BrCH 2 COCH 2 CH 3 O CH 3 CHCH 3 NaOCH 2 CH 3 Propene Br CH 3 CH CH 2 NUCLEOPHILIC SUBSTITUTION 191 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (h) The first reaction is one in which a substituted alcohol is converted to a p-toluenesulfonate ester. This is followed by an S N 2 displacement with lithium iodide. 8.20 The two products are diastereomers of each other. They are formed by bimolecular nucleophilic sub- stitution (S N 2). In each case, a good nucleophile (C 6 H 5 S H11002 ) displaces chloride from a secondary carbon with inversion of configuration. (a) The trans chloride yields a cis substitution product. (b) The cis chloride yields a trans substitution product. 8.21 The isomers of C 4 H 9 Cl are: The reaction conditions (sodium iodide in acetone) are typical for an S N 2 process. 2-Chlorobutane (sec-butyl chloride) 2-Chloro-2-methylpropane (tert-butyl chloride) CH 3 CCl CH 3 CH 3 CH 3 CHCH 2 CH 3 Cl 1-Chlorobutane (n-butyl chloride) 1-Chloro-2-methylpropane (isobutyl chloride) CH 3 CH 2 CH 2 CH 2 Cl CH 3 CHCH 2 Cl CH 3 cis-4-tert-Butylcyclohexyl chloride C(CH 3 ) 3 Cl trans-4-tert-Butylcyclohexyl phenyl sulfide S C(CH 3 ) 3 H11001 Sodium benzenethiolate NaS H11001 SNa Sodium benzenethiolate trans-4-tert-Butylcyclohexyl chloride C(CH 3 ) 3 Cl cis-4-tert-Butylcyclohexyl phenyl sulfide S C(CH 3 ) 3 CH 2 CH 2 CH 2 CH 2 OTsCH 3 O CH 3 O CH 2 CH 2 CH 2 CH 2 ICH 3 O CH 3 O OCH 3 OCH 3 TsCl pyridine (62%) LiI, acetone (88%) OCH 3 CH 2 CH 2 CH 2 CH 2 OHCH 3 O CH 3 O 4-(2,3,4-Trimethoxyphenyl)-1-butanol 4-(2,3,4-Trimethoxyphenyl)-1-butyl iodide 192 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The order of S N 2 reactivity is primary H11022 secondary H11022 tertiary, and branching of the chain close to the site of substitution hinders reaction. The unbranched primary halide n-butyl chloride will be the most reactive and the tertiary halide tert-butyl chloride the least. The order of reactivity will therefore be: 1-chlorobutane H11022 1-chloro-2-methylpropane H11022 2-chlorobutane H11022 2-chloro-2- methylpropane. 8.22 1-Chlorohexane is a primary alkyl halide; 2-chlorohexane and 3-chlorohexane are secondary. Primary and secondary alkyl halides react with potassium iodide in acetone by an S N 2 mechanism, and the rate depends on steric hindrance to attack on the alkyl halide by the nucleophile. (a) Primary alkyl halides are more reactive than secondary alkyl halides in S N 2 reactions. 1-Chlorohexane is the most reactive isomer. (b) Substituents at the carbon adjacent to the one that bears the leaving group slow down the rate of nucleophilic displacement. In 2-chlorohexane the group adjacent to the point of attack is CH 3 . In 3-chlorohexane the group adjacent to the point of attack is CH 2 CH 3 . 2-Chlorohexane has been observed to be more reactive than 3-chlorohexane by a factor of 2. 8.23 (a) Iodide is a better leaving group than bromide, and so 1-iodobutane should undergo S N 2 attack by cyanide faster than 1-bromobutane. (b) The reaction conditions are typical for an S N 2 process. The methyl branch in 1-chloro- 2-methylbutane sterically hinders attack at C-1. The unbranched isomer, 1-chloropentane, re- acts faster. (c) Hexyl chloride is a primary alkyl halide, and cyclohexyl chloride is secondary. Azide ion is a good nucleophile, and so the S N 2 reactivity rules apply; primary is more reactive than secondary. (d) 1-Bromo-2,2-dimethylpropane is too hindered to react with the weakly nucleophilic ethanol by an S N 2 reaction, and since it is a primary alkyl halide, it is less reactive in S N 1 reactions. tert-Butyl bromide will react with ethanol by an S N 1 mechanism at a reasonable rate owing to formation of a tertiary carbocation. (CH 3 ) 3 CBr tert-Butyl bromide; more reactive in S N 1 solvolysis (CH 3 ) 3 CCH 2 Br 1-Bromo-2,2-dimethylpropane; relatively unreactive in nucleophilic substitution reactions CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 Cl Hexyl chloride is primary, therefore more reactive in S N 2. Cyclohexyl chloride is secondary, therefore less reactive in S N 2. Cl CH 3 CH 2 CHCH 2 Cl CH 3 1-Chloro-2-methylbutane is more sterically hindered, therefore less reactive. CH 3 CH 2 CH 2 CH 2 CH 2 Cl 1-Chloropentane is less sterically hindered, therefore more reactive. 1-Chlorohexane (primary) CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 Cl 2-Chlorohexane (secondary) Cl CH 3 CHCH 2 CH 2 CH 2 CH 3 3-Chlorohexane (secondary) Cl CH 3 CH 2 CHCH 2 CH 2 CH 3 NUCLEOPHILIC SUBSTITUTION 193 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Solvolysis of alkyl halides in aqueous formic acid is faster for those that form carbocations readily. The S N 1 reactivity order applies here: secondary H11022 primary. ( f ) 1-Chlorobutane is a primary alkyl halide and so should react by an S N 2 mechanism. Sodium methoxide is more basic than sodium acetate and is a better nucleophile. Reaction will occur faster with sodium methoxide than with sodium acetate. (g) Azide ion is a very good nucleophile, whereas p-toluenesulfonate is a very good leaving group but a very poor nucleophile. In an S N 2 reaction with 1-chlorobutane, sodium azide will react faster than sodium p-toluenesulfonate. 8.24 There are only two possible products from free-radical chlorination of the starting alkane: As revealed by their structural formulas, one isomer is a primary alkyl chloride, the other is sec- ondary. The problem states that the major product (compound A) undergoes S N 1 hydrolysis much more slowly than the minor product (compound B). Since secondary halides are much more reactive than primary halides under S N 1 conditions, the major (unreactive) product is the primary alkyl halide 1-chloro-2,2,4,4-tetramethylpentane (compound A) and the minor (reactive) product is the secondary alkyl halide 3-chloro-2,2,4,4-tetramethylpentane (compound B). 8.25 (a) The two most stable Lewis structures (resonance forms) of thiocyanate are: (b) The two Lewis structures indicate that the negative charge is shared by two atoms: S and N. Thus thiocyanate ion has two potentially nucleophilic sites, and the two possible products are (c) Sulfur is more polarizable than nitrogen and is more nucleophilic. The major product is butyl thiocyanate and arises by attack of sulfur of thiocyanate on butyl bromide. 8.26 Using the unshared electron pair on its nitrogen, triethylamine acts as a nucleophile in an S N 2 reaction toward ethyl iodide. The product of the reaction is a salt and has the structure shown. The properties given in the prob- lem (soluble in polar solvents, high melting point) are typical of those of an ionic compound. H11001(CH 3 CH 2 ) 3 N (CH 3 CH 2 ) 4 N H11001 CH 2 CH 3 I Triethylamine Ethyl iodide Tetraethylammonium iodide I H11002 CH 3 CH 2 CH 2 CH 2 Br H11001CH 3 CH 2 CH 2 CH 2 S CN CH 3 CH 2 CH 2 CH 2 N CS KSCN DMF 1-Bromobutane Butyl thiocyanate Butyl isothiocyanate SCN H11002 H11002 SCN 2,2,4,4-Tetramethylpentane (CH 3 ) 3 CCH 2 C(CH 3 ) 3 1-Chloro-2,2,4,4- tetramethylpentane (primary) (CH 3 ) 3 CCH 2 C(CH 3 ) 2 CH 2 Cl 3-Chloro-2,2,4,4- tetramethylpentane (secondary) (CH 3 ) 3 CCHC(CH 3 ) 3 Cl Cl 2 hH9263 H11001 (CH 3 ) 2 CHCH 2 Br Isobutyl bromide; primary, therefore less reactive in S N 1 sec-Butyl bromide; secondary, therefore more reactive in S N 1 CH 3 CHCH 2 CH 3 Br 194 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.27 This reaction has been reported in the chemical literature and proceeds as shown (91% yield): Notice that the configuration of the product is the same as the configuration of the reactant. This is because the stereogenic center is not involved in the reaction. When we say that S N 2 reactions pro- ceed with inversion of configuration we refer only to the carbon at which substitution takes place, not a stereogenic center elsewhere in the molecule. \ 8.28 (a) The starting material incorporates both a primary chloride and a secondary chloride. The nucleophile (iodide) attacks the less hindered primary position. (b) Nucleophilic substitution of the first bromide by sulfur occurs in the usual way. The product of this step cyclizes by way of an intramolecular nucleophilic substitution. (c) The nucleophile is a dianion (S 2H11002 ). Two nucleophilic substitution reactions take place; the sec- ond of the two leads to intramolecular cyclization. 8.29 (a) Methyl halides are unhindered and react rapidly by the S N 2 mechanism. (b) Sodium ethoxide is a good nucleophile and will react with unhindered primary alkyl halides by the S N 2 mechanism. (c) Cyclohexyl bromide is a secondary halide and will react with a strong base (sodium ethoxide) predominantly by the E2 mechanism. (d) The tertiary halide tert-butyl bromide will undergo solvolysis by the S N 1 mechanism. (e) The presence of the strong base sodium ethoxide will cause the E2 mechanism to predominate. ( f ) Concerted reactions are those which occur in a single step. The bimolecular mechanisms S N 2 and E2 represent concerted processes. (g) In a stereospecific reaction, stereoisomeric reactants yield products that are stereoisomers of each other. Reactions that occur by the S N 2 and E2 mechanisms are stereospecific. S 2H11002 CH 2 CH 2 CH 2 CH 2 Cl ClH11001 SCH 2 CH 2 Cl CH 2 CH 2 S H11002 Thiolane (C 4 H 8 S) CH 2 CH 2 CH 2 Br CH 2 H11002 S S SS 1,4-Dithiane (C 4 H 8 S 2 ) SCH 2 CH 2 SCH 2 Br CH 2 Br H11001 H11002H11002H11002 SCH 2 CH 2 SCH 2 CH 2 Br 1,3-Dichloropentane NaI acetone ClCH 2 CH 2 CHCH 2 CH 3 Cl 3-Chloro-1-iodopentane (C 5 H 10 ClI) ICH 2 CH 2 CHCH 2 CH 3 Cl CC (S)-1-Bromo-2-methylbutane (S)-1-Iodo-2-methylbutane NaI acetone H CH 3 CH 2 CH 2 Br H 3 C H CH 3 CH 2 CH 2 I H 3 C NUCLEOPHILIC SUBSTITUTION 195 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (h) The unimolecular mechanisms S N 1 and E1 involve the formation of carbocation intermediates. (i) Rearrangements are possible when carbocations are intermediates in a reaction. Thus reac- tions occurring by the S N 1 and E1 mechanisms are most likely to have a rearranged carbon skeleton. ( j) Iodide is a better leaving group than bromide, and alkyl iodides will react faster than alkyl bromides by any of the four mechanisms S N 1, S N 2, E1, and E2. 8.30 (a) Cyclopentyl cyanide can be prepared from a cyclopentyl halide by a nucleophilic substitution reaction. The first task, therefore, is to convert cyclopentane to a cyclopentyl halide. An analogous sequence involving cyclopentyl bromide could be used. (b) Cyclopentene can serve as a precursor to a cyclopentyl halide. Once cyclopentyl bromide has been prepared, it is converted to cyclopentyl cyanide by nucleophilic substitution, as shown in part (a). (c) Reaction of cyclopentanol with hydrogen bromide gives cyclopentyl bromide. Then cyclopentyl bromide can be converted to cyclopentyl cyanide, as shown in part (a). (d) Two cyano groups are required here, both of which must be introduced in nucleophilic substi- tution reactions. The substrate in the key reaction is BrCH 2 CH 2 Br. 1,2-Dibromoethane is prepared from ethylene. The overall synthesis from ethyl alcohol is therefore formulated as shown: (e) In this synthesis a primary alkyl chloride must be converted to a primary alkyl iodide. This is precisely the kind of transformation for which sodium iodide in acetone is used. Isobutyl chloride (CH 3 ) 2 CHCH 2 Cl Sodium iodide NaI acetone H11001 Isobutyl iodide (CH 3 ) 2 CHCH 2 I Sodium chloride NaClH11001 H 2 SO 4 heat Br 2 NaCN CH 3 CH 2 OH Ethyl alcohol BrCH 2 CH 2 Br 1,2-Dibromoethane NCCH 2 CH 2 CN 1,2-Dicyanoethane CH 2 CH 2 Ethylene H11001 ethanol–water or DMSO BrCH 2 CH 2 Br 1,2-Dibromoethane 2NaCN Sodium cyanide NNCCH 2 CH 2 C 1,2-Dicyanoethane HBrH11001 OH Br HBrH11001 Cyclopentene Hydrogen bromide Cyclopentyl bromide Br NaCN CN NaClH11001 ethanol–water or DMSO Sodium cyanide Cyclopentyl cyanide Sodium chloride H11001 Cyclopentyl chloride Cl Cl 2 Cl H11001 HClH11001 light or heat Cyclopentane Chlorine Cyclopentyl chloride Hydrogen chloride 196 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) First convert tert-butyl chloride into an isobutyl halide. Treating isobutyl bromide with sodium iodide in acetone converts it to isobutyl iodide. A second approach is by way of isobutyl alcohol. Isobutyl alcohol is then converted to its p-toluenesulfonate ester, which reacts with sodium iodide in acetone in a manner analogous to that of isobutyl bromide. (g) First introduce a leaving group into the molecule by converting isopropyl alcohol to an iso- propyl halide. Then convert the resulting isopropyl halide to isopropyl azide by a nucleophilic substitution reaction (h) In this synthesis 1-propanol must be first converted to an isopropyl halide. After an isopropyl halide has been obtained, it can be treated with sodium azide as in part (g). (i) First write out the structure of the starting material and of the product so as to determine their relationship in three dimensions. The hydroxyl group must be replaced by azide with inversion of configuration. First, however, a leaving group must be introduced, and it must be introduced in such a way that the configu- ration at the stereogenic center is not altered. The best way to do this is to convert (R)-sec- butyl alcohol to its corresponding p-toluenesulfonate ester. COH H CH 3 CH 2 CH 2 (R)-sec-Butyl alcohol (R)-sec-Butyl p-toluenesulfonate p-toluenesulfonyl chloride pyridine COS O O CH 3 H CH 3 CH 3 CH 2 COH H CH 3 CH 3 CH 2 N 3 H CH 3 CH 2 CH 3 C (R)-sec-Butyl alcohol (S)-sec-Butyl azide H 2 SO 4 heat HBr 1-Propanol CH 3 CH 2 CH 2 OH Propene CH 3 CH CH 2 Isopropyl bromide CH 3 CHCH 3 Br HBr NaN 3 Isopropyl alcohol CH 3 CHCH 3 OH Isopropyl bromide CH 3 CHCH 3 Br Isopropyl azide CH 3 CHCH 3 N 3 tert-Butyl chloride (CH 3 ) 3 CCl Isobutyl alcohol (CH 3 ) 2 CHCH 2 OH NaOCH 3 1. B 2 H 6 2. H 2 O 2 , HO H11002 2-Methylpropene (CH 3 ) 2 C CH 2 Isobutyl bromide (CH 3 ) 2 CHCH 2 Br Isobutyl iodide (CH 3 ) 2 CHCH 2 I NaI acetone tert-Butyl chloride (CH 3 ) 3 CCl Isobutyl bromide (CH 3 ) 2 CHCH 2 Br NaOCH 3 HBr peroxides 2-Methylpropene (CH 3 ) 2 C CH 2 NUCLEOPHILIC SUBSTITUTION 197 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Next, convert the p-toluenesulfonate to the desired azide by an S N 2 reaction. ( j) This problem is carried out in exactly the same way as the preceding one, except that the nucleophile in the second step is HS H11002 . 8.31 (a) The two possible combinations of alkyl bromide and alkoxide ion that might yield tert-butyl methyl ether are 1. 2. We choose the first approach because it is an S N 2 reaction on the unhindered substrate, methyl bromide. The second approach requires an S N 2 reaction on a hindered tertiary alkyl halide, a very poor choice. Indeed, we would expect that the reaction of methoxide ion with tert-butyl bromide could not give any ether at all but would proceed entirely by E2 elimination: (b) Again, the better alternative is to choose the less hindered alkyl halide to permit substitution to predominate over elimination. An attempt to prepare this compound by the reaction gave cyclopentyl methyl ether in only 24% yield. Cyclopentene was isolated in 31% yield. (c) A 2,2-dimethylpropyl halide is too sterically hindered to be a good candidate for this synthesis. The only practical method is H11001 Potassium 2,2-dimethylpropoxide (CH 3 ) 3 CCH 2 OK Bromoethane CH 3 CH 2 Br Ethyl 2,2-dimethylpropyl ether (CH 3 ) 3 CCH 2 OCH 2 CH 3 H11001 Chlorocyclopentane CH 3 ONa Sodium methoxide Cl Cyclopentyl methyl ether OCH 3 OK H11001 Potassium cyclopentoxide CH 3 Br Methyl bromide OCH 3 Cyclopentyl methyl ether fast CH 3 O H11002 H11001 (CH 3 ) 3 CBr Methanol CH 3 OH H11001 2-Methylpropene CH 2 C(CH 3 ) 2 slow Methoxide ion CH 3 O H11002 H11001 tert-Butyl bromide (CH 3 ) 3 CBr tert-Butyl methyl ether (CH 3 ) 3 COCH 3 fast Methyl bromide CH 3 Br H11001 tert-Butoxide ion (CH 3 ) 3 CO H11002 tert-Butyl methyl ether (CH 3 ) 3 COCH 3 (R)-sec-Butyl p-toluenesulfonate COS O O CH 3 H CH 3 CH 3 CH 2 NaSH HS H CH 3 CH 2 CH 3 C (S)-2-Butanethiol COH H CH 3 CH 3 CH 2 (R)-sec-Butyl alcohol p-toluenesulfonyl chloride pyridine (R)-sec-Butyl p-toluenesulfonate COS O O CH 3 H CH 3 CH 3 CH 2 NaN 3 N 3 H CH 3 CH 2 CH 3 C (S)-sec-Butyl azide 198 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.32 (a) The problem states that the reaction type is nucleophilic substitution. Sodium acetylide is therefore the nucleophile and must be treated with an alkyl halide to give the desired product. (b) The acidity data given in the problem for acetylene tell us that HC>CH is a very weak acid (K a H11005 10 H1100226 ), so that sodium acetylide must be a very strong base—stronger than hydroxide ion. Elimination by the E2 mechanism rather than S N 2 substitution is therefore expected to be the principal (probably the exclusive) reaction observed with secondary and tertiary alkyl halides. The substitution reaction will work well with primary alkyl halides but will likely fail for secondary and tertiary ones. Alkynes such as (CH 3 ) 2 CHC>CH and (CH 3 ) 3 CC>CH could not be prepared by this method. 8.33 The compound that reacts with trans-4-tert-butylcyclohexanol is a sulfonyl chloride and converts the alcohol to the corresponding sulfonate. Reaction of compound A with lithium bromide in acetone effects displacement of the sulfonate leav- ing group by bromide with inversion of configuration. 8.34 (a) To convert trans-2-methylcyclopentanol to cis-2-methylcyclopentyl acetate the hydroxyl group must be replaced by acetate with inversion of configuration. Hydroxide is a poor leaving group and so must first be converted to a good leaving group. The best choice is p-toluenesulfonate, because this can be prepared by a reaction that alters none of the bonds to the stereogenic center. Treatment of the p-toluenesulfonate with potassium acetate in acetic acid will proceed with in- version of configuration to give the desired product. KOCCH 3 acetic acid cis-2-Methylcyclopentyl acetate trans-2-Methylcyclopentyl p-toluenesulfonate H 3 C OTs H 3 C OCCH 3 O O trans-2-Methylcyclopentanol p-toluenesulfonyl chloride, pyridine H 3 C OH trans-2-Methylcyclopentyl p-toluenesulfonate H 3 C OTs LiBr acetone OSO 2 (CH 3 ) 3 C NO 2 Compound A cis-1-Bromo-4-tert-butylcyclohexane Compound B (CH 3 ) 3 C Br H11001OH (CH 3 ) 3 C O 2 NSO 2 Cl pyridine OSO 2 (CH 3 ) 3 C NO 2 Compound A H11001H11001 Ethyl bromide CH 3 CH 2 Br Sodium bromide NaBr 1-Butyne CH 3 CH 2 CCH Sodium acetylide Na H11001 CCH H11002 NUCLEOPHILIC SUBSTITUTION 199 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) To decide on the best sequence of reactions, we must begin by writing structural formulas to determine what kinds of transformations are required. We already know from part (a) how to convert trans-2-methylcyclopentanol to cis-2-methyl- cyclopentyl acetate. So all that is really necessary is to design a synthesis of trans-2-methyl- cyclopentanol. Therefore, Hydroboration–oxidation converts 1-methylcyclopentene to the desired alcohol by anti- Markovnikov syn hydration of the double bond. The resulting alcohol is then converted to its p-toluenesulfonate ester and treated with acetate ion as in part (a) to give cis-2-methyl- cyclopentyl acetate. 8.35 (a) The reaction of an alcohol with a sulfonyl chloride gives a sulfonate ester. The oxygen of the alcohol remains in place and is the atom to which the sulfonyl group becomes attached. (b) Sulfonate is similar to iodide in its leaving-group behavior. The product in part (a) is attacked by NaSCH 2 CH 3 in an S N 2 reaction. Inversion of configuration occurs at the stereogenic center. (c) In this part of the problem we deduce the stereochemical outcome of the reaction of 2-butanol with PBr 3 . We know the absolute configuration of (H11001)-2-butanol (S) from the statement of the problem and the configuration of (H11002)-sec-butyl ethyl sulfide (R) from part (b). We are told that the sulfide formed from (H11001)-2-butanol via the bromide has a positive rotation. It must there- fore have the opposite configuration of the product in part (b). NaSCH 2 CH 3 PBr 3 CH 3 CH 2 CH 3 SCH 2 CH 3 H (S)-(H11001)-sec-Butyl ethyl sulfide CH 3 CH 2 CH 3 OHH (S)-(H11001)-2-Butanol 2-Bromobutane CH 3 CHCH 2 CH 3 Br NaSCH 2 CH 3 CH 3 CH 2 CH 3 HCH 3 CH 2 S (R)-(H11002)-sec-Butyl ethyl sulfide CH 3 CH 2 CH 3 OSO 2 CH 3 H (S)-sec-Butyl methanesulfonate CH 3 SO 2 Cl pyridine CH 3 CH 2 CH 3 OHH (S)-(H11001)-2-Butanol CH 3 CH 2 CH 3 OSO 2 CH 3 H (S)-sec-Butyl methanesulfonate H 2 SO 4 heat 1. B 2 H 6 2. H 2 O 2 , HO H11002 1-Methylcyclopentanol 1-Methylcyclopentene H 3 C trans-2-Methylcyclopentanol H 3 C OH H 3 C OH ? cis-2-Methylcyclopentyl acetate 1-Methylcyclopentanol H 3 C OCCH 3 O H 3 C OH 200 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Since the reaction of the bromide with NaSCH 2 CH 3 proceeds with inversion of configuration at the stereogenic center, and since the final product has the same configuration as the starting alcohol, the conversion of the alcohol to the bromide must proceed with inversion of configu- ration. (d) The conversion of 2-butanol to sec-butyl methanesulfonate does not involve any of the bonds to the stereogenic center, and so it must proceed with 100% retention of configuration. Assuming that the reaction of the methanesulfonate with NaSCH 2 CH 3 proceeds with 100% inversion of configuration, we conclude that the maximum rotation of sec-butyl ethyl sulfide is the value given in the statement of part (b), that is, H1100625°. Since the sulfide produced in part (c) has a rotation of H1100123°, it is 92% optically pure. It is reasonable to assume that the loss of optical purity occurred in the conversion of the alcohol to the bromide, rather than in the re- action of the bromide with NaSCH 2 CH 3 . If the bromide is 92% optically pure and has a rota- tion of H1100238°, optically pure 2-bromobutane therefore has a rotation of 38H208620.92, or H1100641°. 8.36 (a) If each act of exchange (substitution) occurred with retention of configuration, there would be no observable racemization; k rac H11005 0. Therefore k rac H20862k exch H11005 0. (b) If each act of exchange proceeds with inversion of configuration, (R)-(H11002)-2-iodooctane will be transformed to radioactively labeled (S)-(H11001)-2-iodooctane. Starting with 100 molecules of (R)-(H11002)-2-iodooctane, the compound will be completely racemized when 50 molecules have become radioactive. Therefore, H11005 2 (c) If radioactivity is incorporated in a stereorandom fashion, then 2-iodooctane will be 50% racemized when 50% of it has reacted. Therefore, H11005 1 In fact, Hughes found that the rate of racemization was twice the rate of incorporation of radioactive iodide. This experiment provided strong evidence for the belief that bimolecular nucleophilic substitution proceeds stereospecifically with inversion of configuration. k rac H5007 k exch k rac H5007 k exch (R)-(H11002)-2-Iodooctane C H H 3 C I H11001 [I * ] H11002 CH 3 (CH 2 ) 5 [(S)-(H11001)-2-Iodooctane] * H11001 I H11002 I * H CH 3 (CH 2 ) 5 CH 3 C CI H11001 [I * ] H11002 H H 3 C CH 3 (CH 2 ) 5 (R)-(H11002)-2-Iodooctane ( * indicates radioactive label) CI * H11001 I H11002 H H 3 C CH 3 (CH 2 ) 5 [(R)-(H11002)-2-Iodooctane] * k exch PBr 3 CH 3 CH 2 CH 3 HBr (R)-(H11002)-2-Bromobutane CH 3 CH 2 CH 3 OHH (S)-(H11001)-2-Butanol NUCLEOPHILIC SUBSTITUTION 201 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.37 (a) Tertiary alkyl halides undergo nucleophilic substitution only by way of carbocations: S N 1 is the most likely mechanism for solvolysis of the 2-halo-2-methylbutanes. (b) Tertiary alkyl halides can undergo either E1 or E2 elimination. Since no alkoxide base is present, solvolytic elimination most likely occurs by an E1 mechanism. (c, d) Iodides react faster than bromides in substitution and elimination reactions irrespective of whether the mechanism is E1, E2, S N 1, or S N 2. (e) Solvolysis in aqueous ethanol can give rise to an alcohol or an ether as product, depending on whether the carbocation is captured by water or ethanol. ( f ) Elimination can yield either of two isomeric alkenes. Zaitsev’s rule predicts that 2-methyl-2-butene should be the major alkene. (g) The product distribution is determined by what happens to the carbocation intermediate. If the carbocation is free of its leaving group, its fate will be the same no matter whether the leaving group is bromide or iodide. 8.38 Both aspects of this reaction—its slow rate and the formation of a rearranged product—have their origin in the positive character developed at a primary carbon. The alcohol is protonated and the carbon–oxygen bond of the resulting alkyloxonium ion begins to break: As positive character develops at the primary carbon, a methyl group migrates. Rearrangement gives a tertiary carbocation, which is captured by bromide to give the product. H11002H 2 O CH 3 CCH 2 O H H H9254H11001H9254H11001 CH 3 CH 3 CH 3 CCH 2 CH 3 H11001 CH 3 CH 3 CCH 2 CH 3 Br Br H11002 CH 3 2-Bromo-2-methylbutane CH 3 CCH 2 OH CH 3 CH 3 CH 3 CCH 2 O H H H11001 CH 3 CH 3 CH 3 CCH 2 O H H H9254H11001H9254H11001 CH 3 CH 3 HBr 2,2-Dimethyl-1-propanol (CH 3 ) 2 CCH 2 CH 3 X H11001 2-Methyl-1-butene CCH 2 CH 3 CH 3 CH 2 2-Methyl-2-butene CHCH 3 (CH 3 ) 2 C (CH 3 ) 2 CCH 2 CH 3 X H11001(CH 3 ) 2 CCH 2 CH 3 OH 2-Methyl-2-butanol (CH 3 ) 2 CCH 2 CH 3 OCH 2 CH 3 Ethyl 1,1- dimethylpropyl ether CH 3 CH 2 OH H 2 O CH 3 CCH 2 CH 3 X CH 3 2-Halo-2-methylbutanes are tertiary alkyl halides. 202 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.39 The substrate is a tertiary alkyl bromide and can undergo S N 1 substitution and E1 elimination under these reaction conditions. Elimination in either of two directions to give regioisomeric alkenes can also occur. 8.40 Solvolysis of 1,2-dimethylpropyl p-toluenesulfonate in acetic acid is expected to give one substitu- tion product and two alkenes. Since five products are formed, we are led to consider the possibility of carbocation rearrangements in S N 1 and E1 solvolysis. Since 2-methyl-2-butene is a product common to both carbocation intermediates, a total of five dif- ferent products are accounted for. There are two substitution products: and three elimination products: 8.41 Solution A contains both acetate ion and methanol as nucleophiles. Acetate is more nucleophilic than methanol, and so the major observed reaction is: Methyl iodide CH 3 I H11001 Acetate CH 3 CO H11002 O Methyl acetate CH 3 OCCH 3 O methanol 2-Methyl-2-butene CHCH 3 (CH 3 ) 2 C 3-Methyl-1-butene CH 2 (CH 3 ) 2 CHCH 2-Methyl-1-butene CCH 2 CH 3 CH 2 CH 3 1,2-Dimethylpropyl acetate OCCH 3 O (CH 3 ) 2 CHCHCH 3 1,1-Dimethylpropyl acetate OCCH 3 O (CH 3 ) 2 CCH 2 CH 3 H11001H11001 2-Methyl-2-butene CHCH 3 (CH 3 ) 2 C 1,2-Dimethylpropyl cation (secondary) (CH 3 ) 2 CHCHCH 3 H11001 1,1-Dimethylpropyl cation (tertiary) (CH 3 ) 2 CCH 2 CH 3 H11001 2-Methyl-1-butene CCH 2 CH 3 CH 2 CH 3 1,1-Dimethylpropyl acetate OCCH 3 O (CH 3 ) 2 CCH 2 CH 3 H11001 H11001 2-Methyl-2-butene CHCH 3 (CH 3 ) 2 C 3-Methyl-1-butene CH 2 (CH 3 ) 2 CHCH 1,2-Dimethylpropyl p-toluenesulfonate (CH 3 ) 2 CHCHCH 3 OTs 1,2-Dimethylpropyl acetate OCCH 3 O (CH 3 ) 2 CHCHCH 3 CH 3 CO 2 H CH 3 CO 2 K CH 3 CO 2 H CH 3 CCH 2 CH 3 CH 3 Br H11001H11001CH 2 CCH 2 CH 3 CH 3 CH 3 H 3 C H 3 C H CCCH 3 CCH 2 CH 3 CH 3 OCCH 3 O 2-Bromo-2- methylbutane 1,1-Dimethylpropyl acetate 2-Methyl-1-butene 2-Methyl-2-butene NUCLEOPHILIC SUBSTITUTION 203 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Solution B prepared by adding potassium methoxide to acetic acid rapidly undergoes an acid–base reaction: Thus the major base present is not methoxide but acetate. Methyl iodide therefore reacts with acetate anion in solution B to give methyl acetate. 8.42 Alkyl chlorides arise by the reaction sequence: The reaction proceeds to form the alkyl p-toluenesulfonate as expected, but the chloride anion formed in this step subsequently acts as a nucleophile and displaces p-toluenesulfonate from RCH 2 OTs. 8.43 Iodide ion is both a better nucleophile than cyanide and a better leaving group than bromide. The two reactions shown are therefore faster than the reaction of cyclopentyl bromide with sodium cyanide alone. 8.44–8.47 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. SELF-TEST PART A A-1. Write the correct structure of the reactant or product omitted from each of the following. Clearly indicate stereochemistry where it is important. (a) (b) ? NaCN CN CH 3 CH 3 CH 2 CH 2 CH 2 Br ? CH 3 CH 2 ONa CH 3 CH 2 OH H Br Cyclopentyl bromide H CN Cyclopentyl cyanide H I Cyclopentyl iodide NaI NaCN RCH 2 Cl H11001H11001 CH 3 SO H11002 O O RCH 2 OS CH 3 O O Cl H11002 Primary alkyl chloride Primary alkyl p-toluenesulfonate RCH 2 OH H11001H11001 N CH 3 SCl O O Primary alcohol p-Toluenesulfonyl chloride Pyridine H11001RCH 2 OS CH 3 O O N H Cl H11002 H11001 Primary alkyl p-toluenesulfonate Pyridinium chloride H11001 Methoxide (stronger base) CH 3 O H11002 H11001 Acetic acid (stronger acid) CH 3 COH O Acetate (weaker base) CH 3 CO H11002 O Methanol (weaker acid) CH 3 OH 204 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) (d) (e) ( f ) (g) A-2. Choose the best pair of reactants to form the following product by an S N 2 reaction: (CH 3 ) 2 CHOCH 2 CH 2 CH 3 A-3. Outline the chemical steps necessary to convert: (a) (b) A-4. Hydrolysis of 3-chloro-2,2-dimethylbutane yields 2,3-dimethyl-2-butanol as the major product. Explain this observation, using structural formulas to outline the mechanism of the reaction. A-5. Identify the class of reaction (e.g., E2), and write the kinetic and chemical equations for: (a) The solvolysis of tert-butyl bromide in methanol (b) The reaction of chlorocyclohexane with sodium azide (NaN 3 ). A-6. (a) Provide a brief explanation why the halogen exchange reaction shown is an acceptable synthetic method: (b) Briefly explain why the reaction of 1-bromobutane with sodium azide occurs faster in dimethyl sulfoxide than in water. A-7. Write chemical structures for compounds A through D in the following sequence of reac- tions. Compounds A and C are alcohols. A-8. Write a mechanism describing the solvolysis (S N 1) of 1-bromo-1-methylcyclohexane in ethanol. AB CD HBr, heat NaNH 2 B H11001 D CH 3 CH 2 O [(CH 3 ) 2 S O] H11001 NaICH 3 CHCH 3 Br I H11001 NaBrCH 3 CHCH 3 acetone (S)-2-Pentanol to (R)-CH 3 CHCH 2 CH 2 CH 3 SCH 3 CH 3 CH 2 CH 3 OTsH CH 3 CH 2 CH 3 CNHto NaSH ? HBr HF CH 3 CH 2 CH 3 NaSCH 3 ?H11001 HBr NaN 3 ?H11001 C(CH 3 ) 3 Br ? (major) CH 3 ONa CH 3 OH Cl ?H11001 1-Chloro-3-methylbutane sodium iodide acetone NUCLEOPHILIC SUBSTITUTION 205 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-9. Solvolysis of the compound shown occurs with carbocation rearrangement and yields an alcohol as the major product. Write the structure of this product, and give a mechanism to explain its formation. PART B B-1. The bimolecular substitution reaction is represented by the kinetic equation: (a) Rate H11005 k[CH 3 Br] 2 (b) Rate H11005 k[CH 3 Br][OH H11002 ] (c) Rate H11005 k[CH 3 Br] H11001 k[OH H11002 ] (d) Rate H11005 k/[CH 3 Br][OH H11002 ] B-2. Which compound undergoes nucleophilic substitution with NaCN at the fastest rate? B-3. For the reaction the major product is formed by (a) An S N 1 reaction (c) An E1 reaction (b) An S N 2 reaction (d) An E2 reaction B-4. Which of the following statements pertaining to an S N 2 reaction are true? 1. The rate of reaction is independent of the concentration of the nucleophile. 2. The nucleophile attacks carbon on the side of the molecule opposite the group being displaced. 3. The reaction proceeds with simultaneous bond formation and bond rupture. 4. Partial racemization of an optically active substrate results. (a) 1, 4 (b) 1, 3, 4 (c) 2, 3 (d) All B-5. Which one of the following alkyl halides would be expected to give the highest substitution- to-elimination ratio (most substitution, least elimination) on treatment with sodium ethoxide in ethanol? CH 2 CH 3 Br (CH 3 ) 3 C (CH 3 ) 3 C (CH 3 ) 3 C (CH 3 ) 3 C (a) CH 2 CH 3 Br (b) CHCH 3 Br (c) CH 2 CH 2 Br (d) CH 3 CH 2 O H11002 Na H11001 H11001 Br Br (e) Br Br (c) (d) Br (a) Br (b) CH 3 Br H11001 OH H11002 H11001CH 3 OH Br H11002 H 2 O CH 3 CH 2 CH 2 CHCH(CH 3 ) 2 ? Br 206 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-6. Which of the following phrases are not correctly associated with an S N 1 reaction? 1. Rearrangement is possible. 2. Rate is affected by solvent polarity. 3. The strength of the nucleophile is important in determining rate. 4. The reactivity series is tertiary H11022 secondary H11022 primary. 5. Proceeds with complete inversion of configuration. (a) 3, 5 (b) 5 only (c) 2, 3, 5 (d) 3 only B-7. Rank the following in order of decreasing rate of solvolysis with aqueous ethanol (fastest A slowest): (a)2H11022 1 H11022 3(b)1H11022 2 H11022 3(c)2H11022 3 H11022 1(d)1H11022 3 H11022 2 B-8. Rank the following species in order of decreasing nucleophilicity in a polar protic solvent (most A least nucleophilic): (a)3H11022 1 H11022 2(b)2H11022 3 H11022 1(c)1H11022 3 H11022 2(d)2H11022 1 H11022 3 B-9. From each of the following pairs select the compound that will react faster with sodium iodide in acetone. 2-Chloropropane or 2-bromopropane 12 1-Bromobutane or 2-bromobutane 34 (a) 1, 3 (b) 1, 4 (c) 2, 3 (d) 2, 4 B-10. Select the reagent that will yield the greater amount of substitution on reaction with 1-bromobutane. (a)CH 3 CH 2 OK in dimethyl sulfoxide (DMSO) (b) (CH 3 ) 3 COK in dimethyl sulfoxide (DMSO) (c) Both (a) and (b) will give comparable amounts of substitution. (d) Neither (a) nor (b) will give any appreciable amount of substitution. B-11. The reaction of (R)-1-chloro-3-methylpentane with sodium iodide in acetone will yield 1-iodo-3-methylpentane that is (a) R (c) A mixture of R and S (e) None of these (b) S (d) Meso CH 3 CH 2 CH 2 S H11002 CH 3 CH 2 CH 2 O H11002 321 CH 3 CH 2 CO H11002 O Br Br CH 3 CHCH 2 CH(CH 3 ) 2 Br H 2 C C CH 3 CH 3 12 3 NUCLEOPHILIC SUBSTITUTION 207 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-12. What is the principal product of the following reaction? B-13. Which of the following statements is true? (a)CH 3 CH 2 S H11002 is both a stronger base and more nucleophilic than CH 3 CH 2 O H11002 . (b)CH 3 CH 2 S H11002 is a stronger base but is less nucleophilic than CH 3 CH 2 O H11002 . (c)CH 3 CH 2 S H11002 is a weaker base but is more nucleophilic than CH 3 CH 2 O H11002 . (d)CH 3 CH 2 S H11002 is both a weaker base and less nucleophilic than CH 3 CH 2 O H11002 . B-14. Which of the following alkyl halides would be most likely to give a rearranged product under S N 1 conditions? (e) None of these. Rearrangements only occur under S N 2 conditions. Br (a) Br (b) Br (c) Br (d) NaN 3 H11001 HBr HH CH 3 CH 3 HCl HN 3 HH CH 3 CH 3 HCl N 3 H HH CH 3 CH 3 Cl H N 3 H HH CH 3 CH 3 HCl HN 3 HH CH 3 CH 3 Cl H (a)(b)(c)(d) 208 NUCLEOPHILIC SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website