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Reaction Summary : S N 2, E2, S N 1/E1 Rate Competition: S

Reaction Summary : S N 2, E2, S N 1/E1 Rate Competition: S N 2, S N 1/E1, E2 S N 2 S N 1 E1 E2 Optimize E2 rate: Factor 1: 3 o > 2 o >>1 o Factor 2: Strong Base: Factor 3: Good LG-weak CB

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Reaction Summary : S N 2, E2, S N 1/E1 Rate Competition: S






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Reaction Summary : S N 2, E2, S N 1/E1

Rate Competition: S N2, SN1/E1, E2 S N 2 SN1 E1 E2 Optimize E2 rate:Factor 1: 3o >2o>>1oFactor 2: Strong Base: Factor 3: Good LG-weak CBFactor 4: Polar aprotic solventFactor 5: +DS, T increase rateFactor 6: Stereospecific and regiospecific Optimize SN1 rate:Factor 1: 3o >2o; never 1o, CH3Factor 2: Any Nu: Factor 3: Good LG-weak CBFactor 4: Polar protic solventFactor 5: DS = 0Factor 6: Non-stereospecific Optimize SN2 rate:Factor 1: CH3>1o>2o; never 3oFactor 2: Strong, small Nu: Factor 3: Good LG-weak CBFactor 4: Polar aprotic solventFactor 5: DS = 0Factor 6: Stereospecific Optimize E1 rate: Factor 1: 3 o > 2 o ; never 1 o , CH 3 Factor 2: Any Base: Factor 3: Good LG-weak CB Factor 4: Polar protic solvent Factor 5: + D S, T increase rate Factor 6: Regiospecific only

Reasons for the Competition The products of all four mechanisms can be different. The stereochemistry of each unimolecular reaction is different from that of the corresponding bimolecular reaction.

Kinetic Control or Thermodynamic Control Before you decide how to predict the major product of any competition, you must know whether the competition takes place under kinetic control or thermodynamic control.

Rate-Determining Steps Revisited Because substitution and elimination reactions generally take place under kinetic control, predicting the outcome of an S N 2/SN1/E2/E1 competition means we have to know how to predict the relative rates of the competing reactions.The rate-determining step of a reaction dictates the rate of the overall reaction.

Factor 1: Structure of R-X/LG Putting it all together: Substitution and elimination reactions are almost always in competition with each other. In order to predict the products of a reaction, it is necessary to determine which mechanisms are likely to occur. Don’t fall into the trap of thinking that there must always be one clear winner. Sometimes there is, but sometimes there are multiple products. The goal is to predict all of the products and to predict which products are major and which are minor.

Factor 1: Structure of R-X/LG To accomplish this goal, four steps are required: Analyze the substrate and determine the expected mechanism(s ) Determine the function of the reagent.Consider any solvent effects.Consider any relevant regiochemical and stereochemical requirements.

Factor 1: Structure of R-X/LG Substrate : CH 3, 1o, 2o , 3o, allyl or benzyl The most important factor is the substrate! CH3 1 o2o3o SN1E1 E2SN2SN2SN2 E2 E2 S N 1 E1 only very slow

Factor 1: Structure of R-X/LG Substrate : CH 3, 1o, 2o , 3o, allyl or benzylAllyl and benzyl – analyze case-by-caseAct like their substitution counterparts for SN2 and E2They form carbocations one level more stable than their substitution for SN1 and E1Be careful with elimination! A b-hydrogen is needed for E2 SN1 only 1o, benzylSN1 and E1 2o, benzyl

Reagent : Nucleophilicity vs. Basicity After we know what is possible for a substrate, we now inspect the nucleophile/base to see what will happen. We can divide nucleophiles/bases into categories : Factor 2: Strength of the Nu/Base

For “ Nucleophile-only” do not use ANY of these for E1/E2 . They lack the basicity needed to react with a hydrogenFactor 2: Strength of the Nu/BaseCH3 1o2o3o SN1SN2 S N 2 S N 2 S N 1 only

For “Base (only)” use these ONLY for E2 (except CH 3 ). Factor 2: Strength of the Nu/Base CH 3 1o 2o3o E2SN2E2 E2 only

Factor 1: Structure of R-X/LG CH 3 1 o 2 o 3o SN2 SN2SN2E2 E2onlyFor “Strong base/strong Nu” the bimolecular mechanisms dominate. As the substrates get more hindered, SN2 slows and E2 speeds up. major only

Factor 1: Structure of R-X/LG For “Weak Nuc/Weak Base” the unimolecular mechanisms predominate and are always in competition. For primary substrates the reaction will be a very slow SN2 CH 3 1o 2o3o SN1E1SN 2SN2SN1E1 only very slow

Factor 3: Leaving Group Ability A leaving group must leave in the rate-determining step of an SN 2, SN1 , E2, or E1 reaction. The identity of the leaving group has an effect on the rate of each reaction. A good leaving group is necessary for the reaction to be exothermic (and spontaneous) via a – DHLeaving group ability strongly affects E1 reactions

Factor 3: Leaving Group Ability Overall, S N2 , E2, SN1/E1 are the same with regard to leaving group ability: Are never LGs!

Factor 4: Solvent Effects The Solvent Polar aprotic solvents are used for SN2 and/or E2 reactions of 2 o and 3o substrates. Primary or methyl substrates can have any solvent for S N2. Remember that nucleophilicity of elements within a group on the periodic table increases going upward (stronger base ).

Factor 4: Solvent Effects The Solvent Polar aprotic solvents are used for SN2 and/or E2 reactions of 2o and 3o substrates. CH 3 1o 2o3o SN1E1E2 SN2SN2SN2 E2 E2 S N 1 E1 only very slow Polar aprotic solvents only

Factor 4: Solvent Effects The Solvent Polar protic solvents are used for SN1/E1 reactions and the solvent itself may become the nucleophile/base. CH 3 1o 2o3o SN1E1E2 SN2SN2SN2 E2 E2 S N 1 E1 only very slow Polar protic solvents only

Factor 5: Heat When substitution and elimination reactions are both favored under a specific set of conditions, it is often possible to influence the outcome by changing the temperature under which the reactions take place. Heat will accelerate the rate of all reactions; however one reaction may increase more than another for a given DTE1/E2 reactions are more strongly accelerated by heat than S N1 or SN2 ∆S °rxn is more positive for an E1 and E2 than for S N1 or SN2.

Factor 5: Heat You cannot use heat to exclude a product with these reactions Heat can only change the relative concentrations of products

Draw the Products - Substitution Substitution reactions are straightforward. Remove the counterion (if even shown) or the acidic hydrogen from the Nu: and replace the LG with the nucleophilic atom

Draw the Products - Substitution If stereochemistry is shown, S N2 reactions cause inversion of the carbon center:

Draw the Products - Substitution If stereochemistry is shown, S N1 reactions cause racemization of the carbon center:

Draw the Products - Elimination The E1 reaction is regioselective, but not stereoselective The E2 reaction is both regioselective and stereoselective For simple E2 (no stereochemistry shown) or any E1: Locate all possible b-hydrogens and a-carbon (has LG) for each instance: Remove the LG from a-carbon Draw a double bond between a-bIf the alkene can exist as cis/trans-isomers, draw both - ZaitsevLabel the alkene with the greatest substitution as ‘major’ (trans is more stable than cis)

Examples: Draw the Products - Elimination

Draw the Products - Elimination For E2 where stereochemistry is shown, the b -hydrogen removed to give a particular product had to be anti-periplanar in ‡, this ‘locks’ the stereochemistry of the product alkene:

Draw the Products - Elimination For E2 where stereochemistry is shown on a cyclohexane, realize that the b -hydrogen removed will be trans to the LG. If that position is occupied by a non-H group, no E2 will occur.

Summary E1 S N 2 SN1 E1 E2 Optimize E2 rate:Factor 1: 3o >2o>>1oFactor 2: Strong Base: Factor 3: Good LG-weak CBFactor 4: Polar aprotic solventFactor 5: +DS, T increase rateFactor 6: Stereospecific and regiospecific Optimize SN1 rate:Factor 1: 3o >2o; never 1o, CH3Factor 2: Any Nu: Factor 3: Good LG-weak CBFactor 4: Polar protic solventFactor 5: DS = 0Factor 6: Non-stereospecific Optimize SN2 rate:Factor 1: CH3>1o>2o; never 3oFactor 2: Strong, small Nu: Factor 3: Good LG-weak CBFactor 4: Polar aprotic solventFactor 5: DS = 0Factor 6: Stereospecific Optimize E1 rate: Factor 1: 3 o > 2 o ; never 1 o , CH 3 Factor 2: Any Base: Factor 3: Good LG-weak CB Factor 4: Polar protic solvent Factor 5: + D S, T increase rate Factor 6: Regiospecific only