Solutions_Manual_for_Organic_Chemistry_6th_Ed 136

Solutions_Manual_for_Organic_Chemistry_6th_Ed 136 - 6-70...

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Unformatted text preview: 6-70 The problem is how to explain this reaction: : NEt2 Et2 N : I I HO- � H2C - CH - CH2CH3 H2C - CH - CH2CH3 I Cl 1 OH I + Cl- Clearly, the NEt2 group is i nvolved. The nitrogen is a nucleophile and can do an internal nucleophi lic substitution (S N i), a very fast reaction for entropy reasons because two different molecules do not have to come together. very fast Solution 2 facts 1) second order, but several thousand times faster than similar second order reactions without the NEt2 group 2) NEt 2 group migrates The slower step is attack of HO- on intermediate 3; the N is a good leaving group because it has a positive charge. Where will HO- attack 3? On the less substituted carbon, in typical SN2 fashion. H 2C - CH - CH2CH3 T his overall reaction is fast because of the neighboring group assistance in fonning 3. It is second order because the HO- group and 3 c ollide in the slow step (not the only step, however). And the NEt2 group "migrates", although in two steps. (a) 6-71 The symmetry of this molecule is crucial. 3 :NEt2 I 2 OH I III CH3 - C - C - C - CH3 III H Br H Ph H Ph Regardless of which adjacent H is removed by t-butoxide, the product will be 2,4-diphenyl-2-pentene. (b) Here are a Newman projection, a three-dimensional representation and a Fischer projection of the required diastereomer. On both carbons 2 and 4, the H has to be anti-coplanar with the bromine while leaving the other groups to give the same product. Not coincidentally, the correct diastereomer is a meso structure. H Ph Br H CH3 H 2 H 3C H Ph Br H and Br are anti-coplanar Ph � Ph H3C CH3 Ph Ph Br 3 129 CH3 H H ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.

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