1. The Substrate: CH3, 1 o, 2 o, 3 o, Allyl or Benzyl

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1 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. To accomplish this goal, four steps are required: 1. Analyze the substrate and determine the expected mechanism(s). 2. Determine the function of the reagent. 3. Consider any relevant regiochemical and stereochemical requirements. 4. Consider any solvent effects. 1. The Substrate: CH3, 1 o, 2 o, 3 o, Allyl or Benzyl The most important factor is the substrate! Make sure that if you are going to do an E2 or E1 process that you have a β- hydrogen to react! Example: Even though the benzyl carbocation is as stable as a 3 o, the example on the left can only undergo S N 1 as there are no β- hydrogens; the example on the right there are β- hydrogens and we would show both S N 1 and E1 products: 1 Fall 2011

2 2. The 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: Once we know what category we are in, we can now decide on the course of the reaction: For Nucleophile (only) do not use ANY of these for E2. They lack the basicity needed to lower the E A. For Base (only) use these ONLY for E2. 2 Fall 2011

3 For Strong Nuc/Strong Base the bimolecular mechanisms dominate. As the substrates get more hindered, S N 2 slows down and E2 speeds up. For Weak Nuc/Weak Base the unimolecular mechanisms predominate and are always in competition. For primary substrates that cannot form carbocations, the reaction will have to be S N 2 and it will be very slow. 3. Stereochemistry If the reaction proceeds by S N 2 or E2 reaction, inspect the substrate. If it is written in a stereochemical form, you probably have to consider stereochemistry to determine the outcome of the reaction. If the reaction is SN1/E1 remember that stereochemistry is usually lost (S N 1) or will allow all possible E1 reactions to occur. 4. The Sovlent Polar aprotic solvents are used for S N 2 and/or E2 reactions of 2o and 3o substrates. Primary or methyl substrates can have any solvent for S N 2. Remember that nucleophilicity of elements within a group on the periodic table increases going upward (stronger base). Polar protic solvents are used for S N 1/E1 reactions and the solvent itself may become the nucleophile/base. Remember that Nucleophilicity of elements within a group increases going downward (weaker bases). 3 Fall 2011

4 Mixed Problems 1. Identify the mechanism when 1- bromobutane is treated with the following: 2. Identify the mechanism when 2- bromopentane is treated with the following: 3. Identify the mechanism when 2- bromo- 2- methylpentane is treated with the following: 4. When 2- chloro- 1,1,2,3,3- pentamethylcyclohexane is treated with sodium hydroxide neither S N 2 nor E2 products are formed. Explain. 5. Complete the following reactions (treat the OTs group as a bromide!): 4 Fall 2011

5 6. Complete the following reactions: a. b. c. d. e. f. g. h. i. j. k. 5 Fall 2011

6 7. Complete the following reactions indicating the MAJOR product: a. b. c. d. e. f. g. h. i. 6 Fall 2011

7 KEY Mixed Problems bromobutane is primary: a) NaOH is a strong nucleophile and strong base. The substrate in this case is primary. Therefore, we expect S N 2 (giving the major product) b) NaSH is a strong nucleophile and weak base. The substrate in this case is primary. Therefore, we expect only S N 2 c) When a primary alkyl halide is treated with t- BuOK, the predominant pathway is expected to be E2. d) DBN is a weak nucleophile and a strong base. Therefore, we expect only E2. e) NaOMe is a strong nucleophile and strong base. The substrate in this case is primary. Therefore, we expect S N bromopentane is secondary: a) NaOEt is a strong nucleophile and strong base. The substrate in this case is secondary. Therefore, we expect S N 2 with a possible minor E2 b) NaI is a strong nucleophile and weak base. DMSO is a polar aprotic solvent. The substrate is secondary. Under these conditions, only S N 2 can occur. c) DBU is a weak nucleophile and a strong base. Therefore, we expect only E2. d) NaOH is a strong nucleophile and strong base. The substrate in this case is secondary. Therefore, we expect SN2 e) t- BuOK is a strong, sterically hindered base. Therefore, we expect only E bromo- 2- methylpentane is tertiary: a) EtOH is a weak nucleophile and weak base. The substrate in this case is tertiary. Therefore, we expect both S N 1 and E1. b) b) t- BuOK is a strong, sterically hindered base. Therefore, we expect only E2. c) c) NaI is a strong nucleophile and weak base. The substrate in this case is tertiary. Therefore, we expect only SN1. d) d) NaOEt is a strong nucleophile and strong base. The substrate in this case is tertiary. Therefore, we expect only E2. e) e) NaOH is a strong nucleophile and strong base. The substrate in this case is tertiary. Therefore, we expect only E2. 4. The substrate is tertiary, so S N 2 cannot occur at a reasonable rate. There are no beta protons, so E2 also cannot occur. 7 Fall 2011

8 5. Complete the following reactions (treat the OTs group as a bromide!): 8 Fall 2011

9 k) 9 Fall 2011

10 6. Complete the following reactions: a. b. c. d. e. f. g. h. i. j. k. 10 Fall 2011

11 7. Complete the following reactions indicating the MAJOR product: a. b. c. d. e. f. g. h. i. 11 Fall 2011