L TOPI 7. ELIMINATION EATIONS (chapter 7 and parts of chapters 4,6,20) OBJETIVES 1. Describe mechanisms for elimination of a leaving group and adjacent proton to form a pi-bond. 2. Discuss the effect of starting material ( substrate ), leaving group, and reaction conditions on the course and outcome of a reaction. 3. Describe syntheses of alkenes and alkynes. 4. Use combinations of elimination and substitution reactions in developing two-step syntheses of value-added compounds. D.M. ollard 2007
EVIEW OF ALKENES: STUTUE, NOMENLATUE, AND STABILITY S:7.1-7.4 Prob: 7.18-21 Structure Preview of eactivity Additions (Topic 8) 2 Pt or Pd l l l l 2 O 2 SO 4 O Systematic Nomenclature e.g., 3 F Assign ahn-ingold-prelog priority to substituents on each sp 2 carbon ighest priority on same side: Z (zusammen=same) opposite sides: E (entgegen=opposite) Problem: Name the following compound 3 2 ( 3 ) 2 3 D.M. ollard 2007
Problem: Provide IUPA names for the following alkenes Problem: Determine the E/Z configurations of the double bonds in each of the following alkenes O geraniol elative Stability eats of ydrogenation 4 8 + Pt 2 4 10 or Pd Δ o < 0 + 2 + 2 + 2 Δ o = -127 kj/mol Δ o = -120 kj/mol Δ o = -115 kj/mol + 2 + 2 + 2 Δ > 0 D.M. ollard 2007
Overall Stability number of substituents Number of substituents emember hyperconjugation? yperconjugation accounts for the enhanced stability of cations, radicals and alkenes with higher degrees of substitution. relative stability arbocation adical Alkene stability explains D.M. ollard 2007
ycloalkenes cyclopropene cyclobutene cyclopentene cyclohexene cycloheptene cyclooctene cis cyclooctene trans cyclooctene strain energies: 15 kcal/mol 27 kcal/mol PEVIEW OF FOMATION OF ALKENES & ALKYNES BY ELIMINATION EATIONS S:7.5 base O acid Zn base NaN 2 D.M. ollard 2007
L BASE-POMOTED DEYDOALOGENATION OF ALKYL ALIDES S:6.15-6.17 3 3 3 2 O but: 3 3 3 NaO ELIMINATION MEANISMS Bimolecular Elimination (E2) eaction ate = k[-l][base] B D.M. ollard 2007
Unimolecular Elimination (E1) eaction ate = k[-l], independent of [base] S N 2 SUBSTITUTION VESUS E2 ELIMINATION X S:6.18-6.19 Prob: 6.21,23,29 L X L Substitution and elimination mechanisms might be competitive pathways in a reaction mixture. The outcome of a reaction depends on: i. the structure of the substrate (steric hindrance to nucleophilic attack) ii. the basicity of the nucleophile D.M. ollard 2007
Observations and Explanations 3 3 3 3 ONa or KN 3 2 3 3 ONa 3 2 3 2 O 3 3 3 ONa 3 2 3 3 3 3 KN N 3 3 2 3 3 3 OK heat 2 2 Summary 3 substrates only undergo substitution with weakly basic nucleophiles (O, 2 O, O 2 ). Stronger bases promote elimination. 1 substrates generally undergo substitution unless the base itself is sterically crowded (e.g., t-buo - ). Substrate Nucleophile/Base Mechanism 3 weak bases: O, 2 O, O 2 give solvolysis strong bases: anionic bases (O -, O - N 2- ), N 3 S N 1 (some E1 on heating) E2 No S N 2 with 3 substrates D.M. ollard 2007
Summary 3 substrates only undergo substitution with weakly basic nucleophiles (O, 2 O, O 2 ). Stronger bases promote elimination. 1 substrates generally undergo substitution unless the base itself is sterically crowded (e.g., t-buo ). Substrate Nucleophile/Base Mechanism 2 Less basic than O : e.g.., S, S, N 3, N, O 2 S N 2 O and more basic: O, N 2 weak nucleophiles: O, 2 O E2 S N 1 (some E1 on heating) Summary 3 substrates only undergo substitution with weakly basic nucleophiles (O, 2 O, O 2 ). Stronger bases promote elimination. 1 substrates generally undergo substitution unless the base itself is sterically crowded (e.g., t-buo ). Substrate Nucleophile/Base Mechanism 1 Me All except t-buo t-bu-o (hindered strong base) all S N 2 E2 S N 2 D.M. ollard 2007
STEEOEMISTY AND EGIOEMISTY egiochemistry: Zaitev Elimination EtOK EtO S:7.6; 20.13 Prob: 7.22a-b, 23a-b,29, 30,34,35 20.30h, 40(r,s,t) Zaitsev s ule: substituted (stable) alkene is formed ΔG eaction coordinate Anti-Zaitsev Elimination 1. Use of bulky bases can lead to formation of anti-zaitsev products t-buok t-buo D.M. ollard 2007
2. Formation of anti-zaitsev products via the ofmann Elimination ( 3 ) 3 N N( 3 ) 3 N( 3 ) 3 O Ag 2 O 2 O Ag 2 150 o Mechanistic explanation (+5% 2-isomer) δ O δ O δ δ N( 3 ) 3 oncerted E2 of alkyl bromide Developing alkene character in the transition state is stabilized by hyperconjugation, giving Zaitsev product ofmann elimination Developing carbanionic character in transition state (facilitated by the N + )is destabilized by alkyl substitutents, giving anti-zaitzev product Orientation of E2 Eliminations Antiperiplanar: preferred orientation for elimination l B Synperiplanar: slower elimination l l l l B l D.M. ollard 2007
Explain the observation that cis-4-t-butyl cyclohexyl chloride undergoes elimination with a strong base 500 times faster than the trans isomer l l t-buok t-buok t-bu trans t-bu t-bu cis reacts 500 times faster than trans cis l t-bu l t-bu DEYDOBOMINATION AND DEBOMINATION OF VIINAL DIBOMIDES S:7.9 Prob:7.22e 2 eq. NaN 2 Zn D.M. ollard 2007
Problem. Why does treatment of 1,2-dibromocyclohexane give 1,3- cyclohexadiene, not an alkyne? NaN 2 L AID-POMOTED DEYDATION OF ALOOLS Acid-Promoted Elimination S:7.7-7.8 Prob: 7.22c-d, 23c,26, 32,33,36 O acid Ease of dehydration: 3 > 2 >> 1 Proceeds under relatively mild conditions 85 20% aq. 2 SO 4 equires relatively harsh conditions 180 conc. 2 SO 4 D.M. ollard 2007
Ease of Formation of arbocations: Kinetics vs. Thermodynamics Stability of starting material and products of a reaction relates to thermodynamics (equilibrium). The relative energy of the starting material and transition state relates to the rate of the reaction (kinetics). i-pr +, 2 O t-bu +, 2 O i-pro, + reaction coordinate t-buo, + reaction coordinate Mechanism: Dehydration of 2 and 3 Alcohols by E1 3 3 O 3 O 2 D.M. ollard 2007
Mechanism: Dehydration of 1 Alcohols by E2 2 3 O OSO 3 earrangement of arbocations During Elimination O 3 PO 4 Δ not: -OPO 3 2 arbocations rearrange by hydride or alkyl shifts: 2 o 3 o D.M. ollard 2007
A SUMMAY OF EATIONS We ve added a few more reactions (eliminations) to your chart of reactions..we ll add a few more in the next topic. 2, hν substitutions Elim'n Elim'n 2 /Pt Add'n Elim'n Add'n Elim'n Subst'n Subst'n O substitutions SYNTETI STATEGIES: TWO (O MOE) STEP SYNTESES Prob: 7.24a-c,f,g,j,k, 31 At the end of Topic 6 you were challenged to recognize one-step synthetic transformations. But not all transformations can be achieved in one step. For example, there is no method to dehydrogenate (i.e., remove 2 ) alkanes. So how would you bring about the following transformation???? You need to recognize that there is a single intermediate which can: 1. be made from the starting material, and 2. be transformed into the desired product D.M. ollard 2007
Problem: What two-step process can be used to achieve the following overall transformation? Problem: This process can, of course, be extended to longer sequences of reactions. ow could you prepare decane from bromohexane, bromoethane and ethyne? from D.M. ollard 2007
Problem: eaction of trans-1-bromo-2-methylcyclohexane with sodium ethoxide, a non-bulky base, results in formation of 3-methylcyclohexene, the anti-zaitzev product. Explain. 3 EtONa EtO 3 TOPI 7 ON EXAM 4 Types of Questions - Describe elimination reaction mechanisms - Predict the products of elimination reactions - Describe some simple reactions of alkenes (hydrogenation) and alkynes (acidity, hydrogenation) The problems in the book are good examples of the types of problems on the exam. Preparing for Exam 4: - Work as many problems as possible. - Work in groups. - Do the Learning Group Problem at the end of the chapter. - Work through the practice exam D.M. ollard 2007