Problem Set 8: Substitution Reactions-ANSWER KEY. (b) nucleophile NH 3 H C. (d) H 3 N

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1 Problem Set 8: Substitution Reactions-ANSWER KEY hemistry 260 rganic hemistry 1. The answers are (1), (3) and (5). Nucleophiles generally have lone pair and/or negative charge. 2. (a) neither 4 (c) nucleophile N (b) nucleophile N 3 (d) electrophile 2 2 I (e) neither (f) electrophile 3 Br 3. (a) (c) 3 3 N S 3 3 (b) (d) 3 N N The answer is (1). S - is more polarizable than. 5. The answer is (3). I - is the most polarizable of the species. 6. The answer is (1). For alkyl halides, electrophilicity increases in the following order: 3º < 2º < 1º < methyl. 1

2 7. The answer is (1). For alkyl halides, electrophilicity increases in the following order: 3º < 2º < 1º < methyl 8. The answer is (4). (-) 2-bromooctane consists of molecules of one enantiomer of 2-bromooctane, but we cannot tell from the (-) designation whether the enantiomer has the (R) or (S) configuration. The (-) and (+) designate the optical rotation properties of the enantiomers while (R) and (S) designate the absolute configuration. An enantiomer of a particular configuration can rotate light either right (+) or left (-), and this depends on the molecule we are considering. For a molecule being studied, to find out if an enantiomer of a particular configuration corresponds to (+) or (-), one has to actually do the experiment and measure the direction in which plane polarized light is rotated by that enantiomer. Therefore, the (-) isomer that is the reactant in this question has either the (R) or (S) configuration, but we cannot tell which one from the information given. An S N 2 reaction will cause an inversion of configuration, changing a molecule of (R) configuration to (S) configuration (and vice versa), but this may or may not change the optical rotation properties of the sample. Therefore, the 2-bromooctane is changed to 2-octanol in the reaction with the nucleophile, but the product could be either (+) 2-octanol or (-) 2-octanol. 9. The answer is (1). 10. The answer is (1). 11. The answer is (3). 12. The answer is (1). 13. (a) 1 o Br S N 2 NaN 3 S 3 (solvent) N + NaBr (b) 2

3 S N 2 1 o Na Br 3 S 3 (solvent) + NaBr (c) 3 o S N I 3 + I (d) R R 3 o S N 1 2 Br R R + Br 3 R = any alkyl First, find the configuration of the reactant (R). The reaction takes place by the S N 2 mechanism, so it must proceed with inversion of connfiguration to give the product with (S) configuration I = 4 S 1 N 2 3 I 3 S Na S 3 (R) (solvent) = 3 S 2 3 S (S) The answer is (b). The reaction is S N 1 at the iodo-carbon, because it is tertiary (no matter that the nucleophile is a very strong nucleophile). The product will be racemic. 16. The Answer is (a). The reaction is S N 2 at a primary center (no matter that methanol is a weak nucleophile). The stereogenic center is not involved in the reaction, so it does not change. 17. The answer is (a). 3

4 The reaction is S N 2 at the primary center. The configuration at the stereogenic center does not change, as it is not involved in the reaction The answer is (c). An S N 1 reaction will occur at the alkyl bromide carbon. The stereochemistry at that carbon will be mixed 50:50, but the stereochemistry at the other stereogenic center will remain as in the starting material. Thus, the product will be a 50:50 mixture of two diastereomers. This mixture will have some optical activity, but it will not be optically pure. 19. An S N 1 reaction occurs at the one stereogenic center and yields a racemic mixture, which has no optical activity Racemic Mixture After water attacks the carbocation, the ensuing protonated alcohol (not shown) loses a proton and yields the alcohol. Due to the 50:50 attack shown on the right, the product (above) is racemic. This carbocation is attacked by water from either the top or bottom. alf the carbocations are attacked from one side, and half are attacked from the other side. 20. (a) The reaction as drawn indicates an S N 2 reaction where iodide displaces hydroxide. Under these conditions (mild, neutral), hydroxide is a poor leaving group. (b) The reactant also contains a tertiary iodide, which is a good leaving group. Loss of iodide would generate a tertiary carbocation (although formation of a carbocation would be relatively slow in TF). The resulting carbocation can be attacked by iodide from either face of the molecule, which would lead to a racemic (and thus optically inactive) mixture. 4

5 21. For each reaction shown below, draw the structure of the major organic product(s). Show stereochemistry where appropriate. 22. Propose a mechanism for the reaction below: 5

6 23. Propose a mechanism for each reaction below. 6

7 24. Answer the following questions with respect to the reaction below: (a) Using the knowledge you have learned in hem 123, how would you classify this reaction? S N 1 reaction (b) When the concentration of the methylcyclopentanol is doubled, what would be the effect on the reaction rate? The reaction rate would double. (c) Assuming the reaction is reversible, if a dehydrating agent (takes away water) is added to the reaction mixture, which side of the equilibrium favour? The equilibrium will shift to the right (d) If the temperature is increased, what will be the effect on the equilibrium constant? If the temperature is increased, the equilibrium constant will change, but we can t really predict which direction based on the given information (e) What would be the effect on the optical activity as this reaction proceeds? The starting material and final product are both achiral, so we expect no change in the optical activity 7