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CHAPTER 7 231
 
 
O
O
OTs
CO2CH3
OTs
H
O
O
H3CO2C
H
H
NOT antiperiplanar to LG
Antiperiplanar to LG
H NOT antiperiplanar to LG
Antiperiplanar to LG
 
 
Therefore, we expect the following E2 product: 
 
 
 
7.99. The substrate is primary and the reagent is a strong nucleophile, so this substitution reaction must occur via an 
SN2 pathway. SN2 reactions are bimolecular processes, so statement (a) is true. In an SN2 process, the rate is linearly 
dependent on the concentration of the nucleophile, so statement (b) is also true. Polar aprotic solvents enhance the rate 
of SN2 processes, so statement (c) is also true. Therefore, statement (d) must be false, and indeed, it is false. SN2 
reactions do not proceed via carbocation intermediates. Carbocation intermediates are involved in SN1 reactions, but 
the alkyl halide is primary in this case, and a primary carbocation is too unstable to form. 
 
7.100. The reagent is both a strong base and a strong nucleophile. With a secondary substrate, the E2 product is 
favored over the SN2 product. This rules out option (a). Since the base is not sterically hindered, we expect the Zaitsev 
product. That is, we expect a disubstituted alkene, rather than a monosubstituted alkene, so we can rule out option (b). 
Options (c) and (d) are cis/trans stereoisomers. Since the E2 process is stereoselective, we expect the trans isomer to 
predominate, so option (d) is the correct answer. 
 
7.101. Option (a) involves a primary substrate being treated with hydroxide (which is both a strong nucleophile and a 
strong base). Under these conditions, the SN2 product is expected to predominate over the E2 product. Option (b) 
would give the more substituted alkene (trans-2-butene) as the major product, not 1-butene. Option (d) would also give 
the more substituted alkene (trans-2-butene) as the major product. Only option (c) will give 1-butene. 
 
7.102. Compound 1 exhibits a mesylate group (OSO2CH3) as a good leaving group (described in Section 7.12). Upon 
its formation, compound 1 undergoes an intramolecular SN2-type reaction, in which the nitrogen atom functions as the 
nucleophile. This results in the formation of a three-membered nitrogen-containing ring fused to a six-membered ring 
(3). Notice the stereochemistry: the nitrogen atom displaced the mesylate from the back face of the molecule, resulting 
in inversion at the original chiral center. Next, an intermolecular SN2 reaction occurs: the chloride ion attacks the three-
membered ring at the least hindered site, resulting in the formation of the product (2). Notice that the configuration of 
the chiral center (that underwent inversion in the first SN2-type reaction) does not change during this second SN2 
reaction (in this step, chloride attacks a carbon atom that is not a chiral center). 
 
 
 
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