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Practice Problems for Final Part 1 ANSWER KEY UPDATED

Practice Problems for Final Part 1 ANSWER KEY UPDATED -...

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Unformatted text preview: Angora!“ Keg ,, Praoé‘ca/ @094sz P01!“ 1 0 ~——z,_¢) The silyl protecting group is considerably larger than the benzyl protecting group. Thus its preference for the less hindered alcohol is more pronounced. A more detailed explanation involves the analysis of the respective transition states. [ If: \ 613 {ram}? ‘H‘m $61,446,: $2 ./ Pro (1L 0/) ' \ ‘ \ 6!. M" M C7” 0 vs. 8 .. r l .-...9,0 As depicted above, there is a slight steric clash in the transition state for the benzylation of the secondary alcohol compared to the primary alcohol. This results in a minor preference for the primary alcohol in terms of the product distribution (2:1) of benzyl ether protection. In the second comparison, it is clear that there is a significant steric interaction in the silylation of the secondary alcohol compared to the primary alcohol in the corresponding transition states. This is directly due to the additional steric bulk stemming from the substituents surrounding the silicon. This significant difference in transition state energies affects the product distribution of silylation of the primary and secondary alcohols (10:1). > Any larger silane would have a further enchanced preference for the primary vs. secondary b alcohol. Y bf—fzrf- buffillwaf‘figliflal w. #9: ~D"R .W A LDW 0 _%:~% >rp\/\jl SEE: :KfiVAJXfw “Jaw ‘> <( jR' @cfi '\/\ 9 (OH Pgrg a) MA): we): X0 A anm, Xo/VK ”"4 X0 M I 6r “£9 '5." Pg OH [fl L33 P MONK, H XOAX' k—i Xo/j C XDA/Q/ b lVaU <3 H-OCH U! 5 8’ 5 W mm ‘ ,7,” j < {‘9 ' D 29 (“5 {Di acELa—bas‘. Haw; reCKJJWJ“ at [wfi— («M/Hg!) ”f has“; [70“: 9"" Mama (9‘; 9“,?ng '1: [‘9 ,_ ‘ 5‘} S. g CH3 f I” 5/ .__..———-5 ”H" ________) H ’3 14/ \L‘H} 19w. {Mir 13" fl Lb,“ (a g“ ep3$t1W5o€€ ‘5‘ 3 fi' _ 7 ~ ‘ ‘ Hack, ace-u; a (560— ba 5t. “a“ ()(Jru‘z‘c 1 e. ’44- cwbhm v( rm (-0 hay/i" 85" ha M02467,“ 2e, 9 H (agink 7:)'M-?M. P’fl‘fkcf {1‘05 '5 r549 \4 «G H K ‘D 26 4, If)” .f; (”MT—x "1;: £9“ .I'H HS 83H .- ufi's H839 05‘05H __ a ’ c '3 Wk 3 g, ,4 ; J , M, <9 6"“. 5‘ 2? H5" H #9541 , f " Mat-A H osgapf §‘ ////,/( \\\\\ (H5 1/) a) ‘ ‘59 1: 5“ t /\ 5437 A3:' For the reaction with A, the transition state possesses partial negative character on the oxygen atom, whereas for the reaction with B, the transition state contains a partial negative charge on the nitrogen atom. The oxygen atom is better at stabilizing a negative charge compared to the nitrogen atom due to oxygen’s higher electronegativity. This stabilization of the transition state for A leads to faster reaction and a higher rate constant. EH) ..- >:° ' a. \K I; _,,_ ”j v E CA > /,,/ 99‘- 1 J Y / {Q9 5 /‘S /~ The partial negative charge on the nitrogen atom in the transition state is stabilized by resonance for C as represented in the resonance structures above. This stabilizing effect results in a larger rate constant for the reaction with C. Q ‘ 6 / (j ( f C ‘ 6) F H ”p Aim”: “4‘ : ///// DE ‘ oY 0 O Aziridine B goes through a highly unfavorable transition state that forces a 3—membered ring with an ideal internal bond angle of 60 degrees to adopt the sp2 hybridized geometry that would normally contain a bond angle of 120 degrees. The high energy transition state for this unfavorable geometry causes the rate of pyramidal inversion to be slow at room temperature for B. Aziridine C also goes through the same unfavorable bond angles in the transition state for pyramidal inversion. However, the transition state of aziridine C places the lone pair on the nitrogen atom in a p orbital, which conjugates with the adjacent p orbitals from the pi bond. This overlap and form of resonance stabilizes the transition state for C and leads to pyramidal inversion at room temperature. d) The, W ll" 3"m€6¢7, ”My fire. bet/7f [‘lééfigfip, ém/V M 4/— mD/c l7 (bawqd’e’ ( IL\H MW, llama 9P3 hybrézflyxwhx ch WK Three—membered rings possess bent bond character to accommodate the deviation of bond angle from the idealized sp3 hybridized angle of ~109.5 degrees to 60 degrees. In order to achieve this geometry, the hybridized orbitals that form the bonds Within the ring possess higher p character, contributing to the smaller bond angle. The other two hybridized orbitals that are not involved in the cyclic structure have compensatory s character. Consequently, the lone pair on the nitrogen atom in Aziridine B is held closer to the nucleus due to the increased s character of the hybridized orbital, rendering the lone pair less basic. E I basic E 0% C ( “413+wa 0% The basic solution upon increasing the pH would deprotonate the thiol functionality of DTT leading to the monothiolate anion. The negatively charged sulfur is a significantly better nucleophile compared to the neutral thiol and would therefore enhance the rate of reaction since the first step in the mechanism is the nucleophilic attack of the disulfide. ...
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