OC 2/e Ch 8 S & E

OC 2/e Ch 8 S & E

8 Organic Chemistry William H. Brown & Christopher S. Foote 8-1 8 Nucleophilic Substitution and -EliminationElimination Chapter 8 Chapter 8 8-2 8 Nucleophilic Substitution Nu:- +

C X nucleophilic substitution C Nu + :X- Nucleophile leaving group Nucleophilic substitution: any reaction in which one nucleophile is substituted for another at a tetravalent carbon Nucleophile: a molecule or ion that donates a pair of electrons to another molecule or ion to form a new covalent bond; a Lewis base 8-3 8 Nucleophilic Substitution

An important reaction of alkyl halides Reaction:Nu: + CH3 Br : : CH3 OH HO : :: : : RO CH3 OR :: : : : CH3 SH HS : :: - CH3 Nu + : Bran alcohol an ether a thiol (a mercaptan) HC C :

CH3 C CH an alkyne -: : :C N :I: :- CH3 C N: : CH3 I: : :NH3 : HOH : CH3 NH3 :+ CH3 O-H H + a nitrile an alkyl iodide

an alkylammonium ion an alcohol (after proton transfer) 8-4 8 Solvents Protic solvent: a solvent that is a hydrogen bond donor the most common protic solvents contain -EliminationOH groups Aprotic solvent: a solvent that cannot serve as a hydrogen bond donor nowhere in the molecule is there a hydrogen bonded to an atom of high electronegativity 8-5 8 Dielectric Constant Solvents are classified as polar and nonpolar the most common measure of solvent polarity is

dielectric constant Dielectric constant: a measure of a solvents ability to insulate opposite charges from one another the greater the value of the dielectric constant of a solvent, the smaller the interaction between ions of opposite charge dissolved in that solvent polar solvent: dielectric constant > 15 nonpolar solvent: dielectric constant < 15 8-6 8 Protic Solvents Solvent Water Dielectric Constant (25C) Structure H2 O 79 Formic acid HCOOH

59 Methanol CH3 OH 33 Ethanol CH3 CH2 OH 24 Acetic acid CH3 COOH 6 8-7 8 Aprotic Solvents Solvent Structure Polar

Dimethyl sulfoxide (DMSO) (CH3 ) 2S=O Acetonitrile Dielectric Constant CH3 C N 48.9 37.5 N,N-Dimethylformamide (DMF) (CH3 ) 2NCHO 36.7 Acetone (CH3 ) 2C=O 20.7 Nonpolar Dichloromethane CH2 Cl2

9.1 Diethyl ether CH3 CH2 OCH2 CH3 4.3 Toluene C6 H5 CH3 2.3 Hexane CH3 (CH2 ) 4CH3 1.9 8-8 8 Mechanisms Chemists propose two limiting mechanisms for nucleophilic displacement

a fundamental difference between them is the timing of bond breaking and bond forming steps At one extreme, the two processes take place simultaneously; designated SN2 S = substitution N = nucleophilic 2 = bimolecular (two species are involved in the rate-Elimination determining step) 8-9 8 Mechanism -Elimination SN2 both reactants are involved in the transition state of the rate-Eliminationdetermining step H H - HO: + C

H H Br HO C H - Br HO HH + : Br- C H H Transition state with simultaneous

on reaking an on forming 8-10 8 Mechanism -Elimination SN2 QuickTime and a Photo - JPEG decompressor are needed to see this picture. 8-11 8 Mechanism -Elimination SN1 Bond breaking between carbon and the leaving group is entirely completed before bond forming with the nucleophile begins This mechanism is designated SN1 where S = substitution N = nucleophilic 1 = unimolecular (only one species is involved in the rate-Eliminationdetermining step) 8-12

8 Mechanism -Elimination SN1 Step 1: ionization of the C-EliminationX bond gives a carbocation intermediate H3 C C H3 C H3 C Br slow, rate determining CH3 C+ + :Br- H3 C CH3 A carbocation intermediate; its shape is trigonal planar 8-13

8 Mechanism -Elimination SN1 Step 2: reaction of the carbocation with methanol gives an oxonium ion. Attack occurs with equal probability from either face of the planar carbocation CH3 C+ + :OCH3 fast H H3 C CH3 Lewis acid Nucleophile (Lewis base) CH3 H3 C O H H3 C

+ C CH3 CH3 CH3 C H3 C H3 C O H Oxonium ions Step 3: proton transfer completes the reaction H3 C C H3 C H3 C H + CH3 O + :O

CH3 H fast H CH3 + + C O H O : CH3 H3 C H3 C H3 C 8-14 8 Mechanism -Elimination SN1 QuickTime and a Photo - JPEG decompressor are needed to see this picture. 8-15

8 Evidence of SN reactions 1. What is the rate of an SN reaction affected by: the structure of Nu? the structure of RX? the structure of the leaving group? the solvent? 2. What is the stereochemistry of the product if the Nu attacks at a stereocenter? 3. When and how does rearrangement occur? 8-16 8 Kinetics For an SN1 reaction, the rate of reaction is first order in haloalkane and zero order in nucleophile CH3 CH3 CBr

CH3 CH3 + CH3 OH CH3 COCH3 + HBr CH3 2-Bromo-2-methyl- Methanol 2-Methoxy-2-methylpropane propane (tert-Butyl bromide) (tert-Butyl methyl ether) Rate = - d[ (CH3 ) 3 CBr] = k[ (CH3 )3 CBr] dt 8-17 8 Kinetics For an SN2 reaction, the rate is first order in

haloalkane and first order in nucleophile CH3 Br + Na+OHBromomethane rate = d[ CH3 Br] dt CH3 OH + Na+BrMethanol - = k[ CH3 Br] [OH ] 8-18 8 Nucleophilicity Nucleophilicity: a kinetic property measured by the rate at which a Nu causes a nucleophilic substitution under a standardized set of experimental conditions Basicity: a equilibrium property measured by the position of equilibrium in an acid-Eliminationbase reaction

Because all nucleophiles are also bases, we study correlations between nucleophilicity and basicity 8-19 8 Nucleophilicity EffectivenessNucleophile Br- , ICH S , RS Good 3 HO-, CH3 O-, ROCN- , N3 CH3 COO- , RCOOModerate CH3 SH, RSH, R2 S NH3 , RNH2 , R2 NH, R3N Poor H2 O CH3 OH, ROH CH3 COOH, RCOOH 8-20 8 Nucleophilicity

Relative nucleophilicities of halide ions in polar aprotic solvents are quite different from those in polar protic solvents Solvent Increasing Nucleophilicity Polar aprotic I- < Br- < Cl- < F- Polar protic F- < Cl- < Br- < I- How do we account for these differences? 8-21 8 Nucleophilicity A guiding principle is the freer the nucleophile,

the greater its nucleophilicity Polar aprotic solvents (e.g., DMSO, acetone, acetonitrile, DMF) are very effective in solvating cations, but not nearly so effective in solvating anions. because anions are only poorly solvated, they participate readily in SN reactions, and nucleophilicity parallels basicity: F-Elimination > Cl-Elimination > Br-Elimination > I-Elimination 8-22 8 Nucleophilicity Polar protic solvents (e.g., water, methanol) anions are highly solvated by hydrogen bonding with the solvent the more concentrated the negative charge of the anion, the more tightly it is held in a solvent shell the nucleophile must be at least partially removed from its solvent shell to participate in SN reactions because F-Elimination is most tightly solvated and I-Elimination the least, nucleophilicity is I-Elimination > Br-Elimination > Cl-Elimination > F-Elimination 8-23 8 Nucleophilicity

Generalizations within a period, nucleophilicity increases from left to right; that is, it increases with basicity Period Increasing Nucleophilicity Period 2 F- < Period 3 Cl- OH- < < SH- < NH2 - < CH3 - PH2 -

8-24 8 Nucleophilicity Generalizations in a series of reagents with the same nucleophilic atom, anionic reagents are stronger nucleophiles than neutral reagents Increasing Nucleophilicity - H2 O < OH ROH < RO NH3

< NH2 RSH < RS - - 8-25 8 Nucleophilicity when comparing groups of reagents in which the nucleophilic atom is the same, the stronger the base, the greater the nucleophilicity - Nucleophile RCOO Carboxylate

ion - - Conjugate acid RCOOH 4-5 HOH ROH pKa HO RO Hydroxide Alkoxide ion ion Increasing Nucleophilicity 16-18 15.7 Decreasing Acidity

8-26 8 Stereochemistry For an SN1 reaction at a stereocenter, the product is almost completely racemized - C H Cl Cl -Cl C+ CH3 OH + -H H

Cl R Enantiomer Planar carbocation (achiral) + CH3 O C H Cl C OCH3 H Cl S Enantiomer R Enantiomer A racemic mixture 8-27 8 Stereochemistry

For SN1 reactions at a stereocenter examples of complete racemization have been observed, but partial racemization with a slight excess of inversion is more common Approach of the nucleophile from this side is less hindered R1 C+ H R2 - Cl Approach of the nucleophile from this side is partially blocked by chloride ion, which remains associated with the carbocation as an

ion pair 8-28 8 Stereochemistry For SN2 reactions at a stereocenter, there is inversion of configuration at the stereocenter Experiment of Hughes and Ingold 131 I + 131 I - 2-Iodooctane SN2 acetone I + I

- 8-29 8 Hughes-EliminationIngold Expt the reaction is 2nd order, therefore, SN2 the rate of racemization of enantiomerically pure 2-Elimination iodooctane is twice the rate of incorporation of I-Elimination131 C6 H13 131 - + I: C H H3 C I SN2 acetone (S)-Elimination2-EliminationIodooctane 131 I

C6 H13 C + H CH3 :I- (R)-Elimination2-EliminationIodooctane 8-30 8 Structure of RX SN1 reactions governed by electronic factors; the relative stabilities of carbocation intermediates SN2 reactions governed by steric factors; the relative ease of approach of the nucleophile to the site of reaction Governed by electronic factors

SN1 Carbocation stability R3 CX (3) R2 CHX (2) RCH2 X CH3 X (1) (methyl) SN2 Access to the site of reaction Governed by steric factors 8-31 8 Effect of -EliminationBranching Alkyl Bromide -Branches Br Br

0 1 Relative Rate 1.0 -1 4.1 x 10 Br 2 Br 3 -3 -5 1.2 x 10 1.2x 10

8-32 8 Effect of -EliminationBranching CH3 CH3 CH2 Br CH3 CCH2 Br CH3 free access Bromoethane (Ethyl bromide) blocked access 1-Bromo-2,2-dimethylpropane (Neopentyl bromide) 8-33 8 Allylic Halides

Allylic cations are stabilized by resonance delocalization of the positive charge a 1 allylic cation is about as stable as a 2 alkyl cation + CH2 =CH-CH2 + CH2 -CH=CH2 Allyl cation (a hybrid of two equivalent contributing structures) 8-34 8 Allylic Cations 2 & 3 allylic cations are even more stable + + CH2 =CH-CH-CH3 CH2 =CH-C-CH3 CH3

A 2 allylic carbocation A 3 allylic carbocation As also are benzylic cations CH2 + C6 H5 -CH2 + Benzyl cation The benzyl cation is also written (a benzylic carbocation) in this abbreviated form 8-35 8 The Leaving Group The more stable the anion, the better the leaving ability the most stable anions are the conjugate bases of strong acids Reactivity as a leaving group I- > Br- > Cl- >> F-

O > CH3 CO- > HO- > CH3 O- > NH2- Stability of anion; strength of conjugate acid 8-36 8 The Solvent -Elimination SN2 The most common type of SN2 reaction involves a negative Nu and a negative leaving group negatively charged nucleophile Nu:- + C X negative charge dispersed in the transition state negatively charged leaving group

Nu C X Nu C + : X- Transition state the weaker the solvation of Nu, the less the energy required to remove it from its solvation shell and the greater the rate of SN2 8-37 8 The Solvent -Elimination SN2 Br + N 3

- SN2 solvent Solvent Type Solvent CH3 C N polar aprotic(CH ) NCHO N3 + Brk(solvent) k(methanol) 5000 3 2 2800 (CH3 ) 2 S=O 1300 polar proticH2 O CH3 OH

7 1 8-38 8 The Solvent -Elimination SN1 SN1 reactions involve creation and separation of unlike charge in the transition state of the rate-Elimination determining step Rate depends on the ability of the solvent to keep these charges separated and to solvate both the anion and the cation Polar protic solvents (formic acid, water, methanol) are the most effective solvents for SN1 reactions 8-39 8 The Solvent -Elimination SN1 CH3

CH3 CCl + ROH solvolysis CH3 CH3 COR + HCl CH3 Solvent CH3 k(solvent) k(ethanol) water 100,000 80% water: 20% ethanol 14,000 40% water: 60% ethanol ethanol 100 1 8-40

8 Rearrangements in SN1 Rearrangements are common in SN1 reactions if the initial carbocation can rearrange to a more stable one Cl + CH3 OH 2-Chloro-3phenylbutane CH3 OH + OCH3 + CH3 OH + Cl H 2-Methoxy-2-phenylbutane 8-41 8 Rearrangements in SN1 Mechanism of a carbocation rearrangement

(1) Cl + + :Cl A 2 carbocation (2) + H + H A 3 benzylic carbocation (3) + H + :O-CH3 H

O+ CH3 An oxonium ion 8-42 8 Summary of SN1 & SN2 Type of SN 2 Alkyl Halide SN 2 is favored. Methyl CH3 X SN 1 SN 1 does not occur. The methyl catio is so unstable, it is never observed in solution. Primary RCH2 X SN 2 is favored. SN 1 rarely occurs. Primary cations are so unstable, that they a

rarely observed in solution. Secondary R2 CHX SN 2 is favored in aprotic solvents with good nucleophiles. SN 1 is favored in protic solvents with poor nucleophiles. Carbocation rearrangements may occur. Tertiary R3 CX SN 2 does not occur because SN 1 is favored because of the ease of of steric hindrance around formation of tertiary carbocations. the reaction center. SubstitutionInversion of configuration. Racemization is favored

. The carbocation at a The nucleophile attacks intermediate is planar, and attack of stereocenterthe stereocenter from the nucleophile occurs with equal side opposite the leaving probability from either side. There i group. some net inversion of configuration. 8-43 8 Neighboring Groups In an SN1 reaction, departure of the leaving group is not assisted by Nu In an SN2 reaction, departure of the leaving group is assisted by Nu These two types are distinguished by their order of reaction; SN2 reactions are 2nd order, and SN1

reactions are 1st order But some reactions are 1st order and yet involve two successive SN2 reactions 8-44 8 Mustard Gases Mustard gases contain either S-EliminationC-EliminationC-EliminationX or N-EliminationC-EliminationC-EliminationX N S Cl Cl Cl Cl Bis(2-chloroethyl)sulfide Bis(2-chloroethyl)methylamine (a sulfur mustard gas) (a nitrogen mustard gas) what is unusual about the mustard gases is that they undergo hydrolysis so rapidly in water, a very poor

nucleophile Cl S Cl + 2H2 O HO S OH + 2HCl 8-45 8 Mustard Gases the reason is neighboring group participation by the adjacent heteroatom Cl Cl

+ S : S + slow, rate + determining S + :Cl Cl an internal Cl A cyclic SN2 reaction sulfonium ion fast :O-H a second Cl H SN2 reaction : S

+ H O H proton transfer to solvent completes the reaction 8-46 8 SN1/SN2 Problems Problem 1: predict the mechanism for this reaction, and the stereochemistry of each product Cl + + CH3 OH/ H2 O R enantiomer OCH3 OH

+ HCl Problem 2: predict the mechanism of this reaction Br + Na+CN- DMSO CN + Na+Br - 8-47 8 SN1/SN2 Problems Problem 3: predict the mechanism of this reaction and the configuration of product Br + CH3 S -Na+

R enantiomer SCH3 + Na+Br- acetone Problem 4: predict the mechanism of this reaction O Br + CH3 COH O acetic acid OCCH3 + HBr 8-48 8 SN1/SN2 Problems

Problem 5: predict the mechanism of this reaction Br + (CH3 ) 3 P toluene + P(CH3 ) 3 Br- 8-49 8 Phase-EliminationTransfer Catalysis A substance that transfers ions from an aqueous phase to an organic phase An effective phase-Eliminationtransfer catalyst must have sufficient hydrophilic character to dissolve in water and form an ion pair with the ion to be transported hydrophobic character to dissolve in the organic phase and transport the ion into it

The following salt is an effective phase-Eliminationtransfer catalysts for the transport of anions (CH3 CH2 CH2 CH2 ) 4 N+ClTetrabutylammonium chloride (Bu4N+Cl- ) 8-50 8 Phase-EliminationTransfer Catalysis QuickTime and a Photo - JPEG decompressor are needed to see this picture. 8-51 8 -EliminationElimination -EliminationElimination: a reaction in which a small molecule, such as HCl, HI, or HOH, is split out or eliminated from a larger molecule H C a C

X A haloalkane + CH3 CH2 O-Na+ CH3 CH2 OH Base C C + CH3 CH2 OH + Na+X - An alkene 8-52 8 -EliminationElimination Zaitsev rule: the major product of a -Eliminationelimination is the more stable (the more highly substituted) alkene

Br CH3 CH2 O-Na+ CH3 CH2 OH 2-Bromo-2methylbutane Br CH3 O-Na+ CH3 OH 1-Bromo-1-methylcyclopentane + 2-Methyl-2-butene 2-Methyl-1-butene (major product) + 1-MethylMethylenecyclopentene cyclopentane (major product) 8-53 8 -EliminationElimination

There are two limiting mechanisms for -Elimination elimination reactions E1 mechanism: at one extreme, breaking of the R-EliminationX bond is complete before reaction with base to break the C-EliminationH bond only R-EliminationX is involved in the rate-Eliminationdetermining step E2 mechanism: at the other extreme, breaking of the R-EliminationX and C-EliminationH bonds is concerted both R-EliminationX and base are involved in the rate-Eliminationdetermining step 8-54 8 E1 Mechanism ionization of C-EliminationX gives a carbocation intermediate CH3 CH3 -C-CH3 slow, rate determining

Br CH3 CH3 -C-CH3 + Br + (A carbocation intermediate) proton transfer from the carbocation intermediate to the base (in this case, the solvent) gives the alkene H CH3 O: + H-CH2 -C-CH3 + H3 C fast H H3 C + CH3

O H + CH2 =C-CH3 8-55 8 E1 Mechanism QuickTime and a Photo - JPEG decompressor are needed to see this picture. 8-56 8 E2 Mechanism QuickTime and a Photo - JPEG decompressor are needed to see this picture. 8-57 8 Kinetics of E1 and E2 E1 is a 1st order reaction; 1st order in RX and zero order is base d[RX]

Rate = = k[RX] dt E2 is a 2nd order reaction; 1st order in base and 1st order in RX Rate = d[RX] = k[RX][Base] dt 8-58 8 Regioselectivity of E1/E2 E1: major product is the more stable alkene E2: with strong base, the major product is the more stable alkene double bond character is highly developed in the transition state thus, the transition state of lowest energy is that

leading to the most stable (the most highly substituted) alkene 8-59 8 Stereoselectivity of E2 E2 is most favorable (lowest activation energy) when H and X are oriented anti and coplanar CH3 O: - CH3 O H H C C C X -H and -X are anti and coplanar

(dihedral angle 180) C - :X 8-60 8 Stereochemistry of E2 Consider E2 of these stereoisomers CH3 O-Na+ Cl cis-1-Chloro-2isopropylcyclohexane + CH3 OH 1-Isopropyl3-Isopropylcyclohexene cyclohexene (major product) CH3 O-Na+ CH3 OH Cl

trans1-Chloro-2isopropylcyclohexane 3-Isopropylcyclohexene 8-61 8 Stereochemistry of E2 in the more stable chair of the cis isomer, the larger isopropyl is equatorial and chlorine is axial CH3 O: - H 2 H 6 H H E2 -

+ CH3 OH + :Cl 1 Cl 1-Isopropylcyclohexene 8-62 8 Stereochemistry of E2 in the more stable chair of the trans isomer, there is no H anti and coplanar with X, but there is one in the less stable chair H 6 H Cl 2 H 1

Cl H More stable chair (no H is anti and coplanar to Cl) H 6 H 1 2 H H Less stable chair (H on carbon 6 is anti and coplanar to Cl)

8-63 8 Stereochemistry of E2 it is only the less stable chair conformation of this isomer that can undergo an E2 reaction Cl H CH3 O: - 6 H 1 2 H H E2

+ CH3 OH + :Cl3-Isopropylcyclohexene 8-64 8 Stereochemistry of E2 Problem: account for the fact that E2 reaction of the meso-Eliminationdibromide gives only the E-Eliminationalkene Br Br C6 H5 CH-CHC6 H5 CH3 O- Na+ meso-1,2-Dibromo1,2-diphenylethane CH3 OH C6 H5 C C6 H5 C Br H (E)-1-Bromo-1,2diphenylethylene 8-65

8 Summary of E2 vs E1 Alkyl halide Primary RCH2 X E1 E2 E1 does not occur. E2 is favored. Primary carbocations are so unstable, they are never observed in solution. Secondary Main reaction with weakMain reaction with strong bases such as2O, H ROH. bases such as OH R2 CHX and OR . Tertiary R3 CX Main reaction with weakMain reaction with strong bases such as2O,

H ROH. bases such as OH and OR . 8-66 8 SN vs E Many nucleophiles are also strong bases (OH-Elimination and RO-Elimination) and SN and E reactions often compete The ratio of SN/E products depends on the relative rates of the two reactions nucleophilic substitutionH C H C C Nu + X- C X + Nu-elimination C C + H-Nu + X-

8-67 8 SN vs E Halide Reaction Comments Methyl CH3 X SN2 SN1 reactions of methyl halides are never observed. The methyl cation is so unstable that it is never formed in solution. Primary RCH2 X SN2 The main reaction with good nucleophiles/weak bases such as I- and CH3COO-. E2

The main reaction with strong, bulky bases such as potassium tert-butoxide. Primary cations are never formed in solution, and, th SN1 and E1 reactions of primary halides are never observed. 8-68 8 SN vs E (contd) SecondarySN2 R2 CHX Tertiary R3 CX The main reaction with bases/nucleophiles wher pK a of the conjugate acid is 11 or less, as for example I - and CH3COO-. E2 The main reaction with bases/nucleophiles where the pKa of the conjugate acid is 11 or greater, as for OH- and CH 3CH2O .

SN1/E1 Common in reactions with weak nucleophiles in protic solvents, such as water, methanol, and eth E2 Main reaction with strong bases such as HO- and RO-. SN1/E1 Main reactions with poor nucleophiles/weak bases. SN 2 reactions of tertiary halides are never obser because of the extreme crowding around the 3 8-69 8 Prob 8.9 Draw a structural formula for the most stable carbocation of each molecular formula. + (a)C4 H9 (b) C3 H7 + (c) C8 H15 +

(d) C3 H7 O+ 8-70 8 Prob 8.11 From each pair, select the stronger nucleophile. (a)H2 O or OH(c)CH3 SH or CH3 S (b)CH3 COO or OH - (e)Cl or I in methanol (d) Cl or I in DMSO (f)CH3 OCH3 or CH3 SCH3 8-71 8 Prob 8.12 Draw a structural formula for the product of each SN2 reaction. (a)CH3 CH2 CH2 Cl + CH3 CH2 ONa (b) (CH3 ) 3 N + CH3 I (c) (d)H3 C CH2 Br

ethanol acetone + NaCN acetone Cl + CH3 SNa ethanol 8-72 8 Prob 8.12 (contd) (e)CH3 CH2 CH2 Cl + CH3 C C-Li+ (f) CH2 Cl + NH3 (g) O NH + CH3 ( CH2 )6 CH2 Cl (h)CH3 CH2 CH2 Br diethyl ether

ethanol + NaCN ethanol acetone 8-73 8 Prob 8.14 Account for the fact that the rate of this reaction is 1000 times faster in DMSO than it is in ethanol. Cl Cl + 2KCN 1,3-Dichloropropane NC CN + 2KCl Propanedinitrile 8-74 8 Prob 8.15 The following reaction involves two successive SN2 reactions. Propose a structural formula for the product.

O NH2 + Br OCH3 R3 N Br 1-Aminoadamantane C1 5 H2 3 NO2 A Methyl 2,4-dibromobutanoate 8-75 8 Prob 8.16 Which member of each pair shows the greater rate of SN2 reaction with KI in acetone? (a) Cl or

(b) Cl or (c) (d) Cl Br Cl Br or Cl Br or 8-76 8 Prob 8.17 Which member of each pair gives the greater rate of SN2 reaction with KN3 in acetone?

Br Br (a) or Br Br (b) or 8-77 8 Prob 8.19 Limiting yourself to a single 1,2-Eliminationshift, suggest a structural formula for a more stable carbocation. (b) (a) + + (d)

(c) (e) + (f) O + + 8-78 8 Prob 8.21 Draw a structural formula for the product of each SN1 reaction. Cl + CH3 CH2 OH (a) ethanol

S enantiomer (b) (c) (d) + CH3 OH methanol Cl O Cl + CH3 COH Br acetic acid + CH3 OH methanol 8-79 8 Prob 8.24 From each pair, select the compound that undergoes S N1 solvolysis in ethanol more rapidly. (a)

(c) Cl or Cl or Cl (b) Cl (d) Cl (e) Cl or Cl or Cl Br

(f) Br or Cl Br or 8-80 8 Prob 8.25 Account for the following relative rates on solvolysis under SN1 conditions. O Relative rate of solvolysis (S N1) Cl Cl 0.2 1

O Cl 109 8-81 8 Prob 8.26 Explain why the following compound is very unreactive under SN1 conditions. I 1-Iodobicyclo[2.2.2]octane 8-82 8 Prob 8.27 Propose a synthesis for each compound from a haloalkane and a nucleophile. CN (a) (f)

(b) SH (d) O CN (c) O O (e) (g) SH 8-83 8 Prob 8.29 Propose a mechanism for the formation of each product. CH3 CH=CHCH2 Cl 1-Chloro-2-butene

H2 O OH CH3 CH=CHCH2 OH + CH3 CHCH=CH2 2-Buten-1-ol 3-Buten-2-ol 8-84 8 Prob 8.30 Propose a mechanism for the formation of this product. If the configuration of the starting material is S, what is the configuration of the product? NaOH N Cl H2 O N OH 8-85

8 Prob 8.31 Propose a mechanism for the formation of the products of this solvolysis reaction. Br CH3 CH2 OH warm + OCH2 CH3 + HBr 8-86 8 Prob 8.32 Propose a mechanism for the formation of each product. OSO2 Ar O CH3 COH H H O OCCH3

+ H H 8-87 8 Prob 8.33 Which compound in each set undergoes more rapid solvolysis when refluxed in ethanol? Br Br (a) or (b) Cl (c) (d)

Br Br or Cl Br or or Cl 8-88 8 Prob 8.34 Account for these relative rates of solvolysis in acetic acid. (CH3 ) 3CBr Br 1 10-2 Br

Br 10-7 10-12 8-89 8 Prob 8.35 On SN1 solvolysis in acetic acid, (1) reacts 1011 times faster than (2). Furthermore, solvolysis of (1) occurs with complete retention of configuration. Draw structural formulas for the products of each solvolysis and account for the difference in rates. OSO2 Ar (1) OSO2 Ar (2) 8-90 8 Prob 8.36 Draw structural formulas for the alkene(s) formed on treatment of each compound with sodium ethoxide in

ethanol. Assume reaction by an E2 mechanism. Br (b) (a) (c) Cl Cl (d) (e) Cl Cl (f) Br 8-91 8 Prob 8.37 Draw structural for all chloroalkanes that undergo

dehydrohalogenation when treated with KOH to give each alkene as the major product. (a) (d) (b) (c) (e) 8-92 8 Prob 8.38 On treatment with sodium ethoxide in ethanol, each compound gives 3,4-Eliminationdimethyl-Elimination3-Eliminationhexene. One compound gives an E alkene, the other gives a Z alkene. Which compound gives which alkene? Me H Et C Me Br

C Et (A) H Me C Et Me Br C Et (B) 8-93 8 Prob 8.39 On treatment with sodium ethoxide in ethanol, this compound gives a single stereoisomer. Predict whether the alkene has the E or Z configuration. H3 C

HC H 6 5 Br - + CH3 CH2 O Na HC H 6 5 1-Bromo-1,2diphenylpropane CH3 CH2 OH CH3 C6 H5 CH=CC6 H5 1,2-Diphenylpropene 8-94 8 Prob 8.40 Elimination of HBr from 2-Eliminationbromonorbornane gives only 2-Eliminationnorbornene. Account for the regiospecificity of this elimination reaction. H

H H Br H H 2-Bromonorbornane H H H H 2-Norbornene H 1-Norbornene 8-95 8 Prob 8.43 Arrange these haloalkanes in order of increasing ratio of E2 to SN2 products on reaction of each with sodium ethoxide in ethanol.

CH3 (a)CH3 CH2 Br (b) CH3 CHCH2 Br CH3 (c)CH3 CCH2 CH3 Cl CH3 (d) CH3 CHCH2 CH2 Br 8-96 8 Prob 8.44 Draw a structural formula for the major product of each reaction and specify the most likely mechanism for its formation. (a) Br + CH3 OH methanol CH3 (b)CH3 CCH2 CH3 + NaOH

Cl Cl (c)(R)-CH3 CHCH2 CH3 80 H2 O O + + CH3 CO Na DMSO 8-97 8 Prob 8.44 (contd) - (d) Cl (e) + + CH3 O Na methanol

Cl + NaI acetone R enantiomer O Cl (f) CH3 CHCH2 CH3 + HCOH R enantiomer formic acid (g) CH3 CH2 ONa + CH2 =CHCH2 Cl ethanol 8-98 8 Prob 8.45 Propose a mechanism for the formation of each product. OH OH NaOH CH3 CH2 OH

the trans isomer Cl OH (1) OH OH NaOH CH3 CH2 OH the cis isomer Cl + (2) O (3) 8-99 8 Prob 8.46 Show how to bring about each conversion.

(a) Cl (c) Cl (b) OH (d) OH (e) Br Br OH Br Br (f)

Br OH Br 8-100 8 Prob 8.46 (contd) Show how to bring about each conversion. Br (g) (h) Br (i) O H H O 8-101 8 Prob 8.47 Which reaction gives the tert-butyl ether in good yield?

What is the product of the other reaction? CH3 + (a)CH3 CO K + CH3 CH2 Cl CH3 (b) DMSO CH3 CH2 O-K+ + CH3 CCl CH3 CH3 COCH2 + KCl CH3 CH3

DMSO CH3 COCH2 + KCl CH3 8-102 8 Prob 8.48 Each ether can, in principle, by synthesized by a Williamson ether synthesis forming bond (1) or bond (2). Which combination gives the better yield? (1) (a) (2) O-CH2 CH3 (1) (2)CH3 (b)CH3 O-CCH3 CH3

(1) (2) CH3 (c)CH2 =CHCH2 -O-CH2 CCH3 CH3 8-103 8 Prob 8.49 Propose a mechanism for this reaction. ClCH2 CH2 OH Na2 CO3, H2 O O H2 C CH2 8-104 8 Prob 8.50 Each compound can be synthesized by an SN2 reaction.

Propose a combination of haloalkane and nucleophile that will give each product. (a)CH3 OCH3 (b) CH3 SH (c) CH3 CH2 CH2 PH2 (d) CH3 CH2 CN (e)CH3 SCH2 C(CH3 ) 3 (f) (CH3 ) 3NH+ Cl - 8-105 8 Prob 8.50 (contd) N3 O (g)C6 H5 COCH2C6 H5 (h)(R)-CH3CHCH2CH2CH3 (i) CH2 =CHCH2 OCH(CH3 ) 2

(j) CH2 =CHCH2 OCH2 CH=CH2 (k) H N Cl H - (l) O O 8-106 8 Nucleophilic Substitution and -EliminationElimination End Chapter 8 8-107

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