0243Pb53an-09.cwk (WP)

0243Pb53an-09.cwk (WP) - When L = OH, there can be no...

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Answers to Problem 53, Chemistry 0243 - 2009 To make (a), we need to do an S N 2 displacement (note the inversion). So we need a good nucleophile (HS ), a nonpolar solvent to minimize S N 1, and a good leaving group like OTs. To make (b), we need to do an S N 1. A good leaving group is still necessary, but we want a poor nucleophile such as H 2 O, and a polar solvent to aid ionization. To make (c), the product of a Saytzeff E2 elimination, we want a good base (not nucleophile), and we might as well use a big one to minimize S N 2; hence (CH 3 ) 3 CO . We still need a good leaving group (OTs is fine) and a nonpolar solvent to minimize S N 1. For (d), we need to do a Hofmann elimination, so the conditions of (c) will do as long as we use a Hofmann leaving group such as F.
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Unformatted text preview: When L = OH, there can be no displacement or elimination. Hydride is a poor nucleophile and a poor initiator of elimination reactions. But it is a great Brønsted base when it comes to deprotonations. So the reaction is the formation of the alkoxide by deprotonation of the alcohol. As H 2 is the other product, the reaction is irreversible. O H H H 2 O + In the first S N 2 - the more polar solvent makes the reaction go faster- it will stabilize the polar TS more than the nonpolar starting materials. In the second S N 2, the more polar solvent makes the reaction go slower. The more polar the solvent, the more that starting materials will be stabilized relative to the less polar TS. See Book, pp 298-299 for diagrams....
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This note was uploaded on 03/02/2010 for the course CHEM orgo taught by Professor Jones during the Fall '10 term at NYU.

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