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Solutions to Problems 133 36. To form the desired product requires stereochemical inversion. The substrate is secondary, so some care should be taken to ensure that the SN2 mechanism is favored. Use a good nucleophile such as acetate (as opposed to acetic acid, which would be a poor nucleophile) in a polar aprotic solvent like dimethylformamide (DMF): CI OCCH₃ 0 DMF CH₃ CH₃ 37. The first reaction is an uncomplicated displacement. The second is as well, but there is a complication. Let's take a look at how its product can react further, and then perhaps we can see why it hap- pens in the second case but not in the first. The pathway to the byproduct, (CH₃CH₂CH₂CH₂)₂S, must involve reaction between the product of the first displacement, and another molecule of the initial starting material CH₃CH₂CH₂CH₂Br: This is the only reasonable way to get a second butyl group on the sulfur. Since butyl is primary, we are limited to the mechanism. The simplest way to get there is to use the product molecule itself as a nucleophile: CH₃CH₂CH₂CH₂ + S-H CH₃CH₂CH₂CH₂ Then loss of from sulfur completes the sequence. Why doesn't the same thing happen in the first reaction, the formation of the alcohol? What do you know about the differences in nucleophile strength between S and O? Sulfur is far better, especially in protic solvent (Section 6-8). Alcohols are too weak as nucleophiles to carry out SN2 displacements, while thiols can achieve this transformation readily. Did you consider an alternative mechanism, one in which the SH group of the thiol is deprotonated before nucleophilic attack? That is, CH₃CH₂CH₂CH₂S: followed by CH₃CH₂CH₂CH₂ + CH₃CH₂CH₂CH₂ This mechanism is qualitatively plausible, but because the of the thiol SH bond is around 10, in the absence of base the equilibrium of the initial deprotonation is too unfavorable for this sequence to be competitive: The concentration of the conjugate base is too low. 38. (1a) (CH₃)₂C=CH₂ (1b) (1c) CH₃CH₂ (1d) (CH₃)₂C= CH₂=C(CH₃) (1e), (1f) Same as