Answers to Hour Examination #3, Chemistry 302-302A, 2005 1. Part (a) better be easy.
Cl Cl OH2
(b) Odd stereochemical result? Must be a Neighboring Group Problem. Whats the internal nucleophile? Got to be that Br. The rest is easy.
Br * OH 1
Answers to Hour Examination #3, Chemistry 302X - 2003
O OCH 3 (1 ) O H2 O OCH 3 (1 ) H3 O+ OH (CH ) CHLi 32 (2 ) (CH 3 )2 CHBr O 1. LiAlH4 OCH 3 (1 ) 2. H2 O CH 3 CH 2 CH 2 OH pyridine CrO3 H Li (CH 3 )2 CHLi O 2 Eq. H2 O HOCH 3 full Eq. NaOCH 3 CH 3 I
Hour Examination #3, Chemistry 302X - 2007 We must never stop lernig. Hilary Lincoln Jones 1 (24 points). Here is a lovely problem encountered some time ago, by your old buddy, Amy Alonza Stagg. Your overall job is to write a mechanism for this change.
Hour Examination #3, Chemistry 302X - 2006 Sometimes the straight and narrow path leads to a nasty place. Alfred North Whitehead
1 (24 points). Here is a problem you have seen before.
CH3 COO Et
O 1. HOEt/ NaOEt 2. H3 O+ to pH = 7 CH3 COOEt
Answers to the Final examination, Chemistry 302X - 2006 1. (a)
unk nown - ei th er or CH O H HO H H OH H OH OH CH 2O H
D-G luc os e
CHO H HO or HO H OH H H CH3 O H OH C H 2O H HO HO + CH3O H2 HOCH2 O OCH3
D-G ala ctose
u nkn own - this is C4 o f ei the
Answers to Chemistry 302X-302A Final Examination, 2005 1.
(a) H O O
In the plane of the ring, does not overlap with the carbonyl pi bondhence, no resonance stabilization, and no acidity of this hydrogen (b) O C H3 C OH +
O OCH 2 CH 3 H3 C C O pKa = 16 + H
F inal Examination, Chemistry 302X-302A, 2005 This Final Examination is different in very few respects from the hour exams with which you are all too familiar. However, there is substantial choice. There is a total of 10 questions, but P LEASE do not do a
Opportunity 5 Answer, Chemistry 302X- 2007
O O rev erse a ldol
O (-) O (-) ro tat e
C H3 OH
d epro tonat e
prot ona te O O Mic ha el O d epro tonat e O (-)
+ L iO H
aldo l O O HO p rot on ate O
What Kevin Bartlett and your ancestors s
Answers to Hour Examination #3, Chemistry 302X, 2006 1. (a). That SN2 shown just cant happen with the required inversion. The back of that methyl group is not reachable by the putative nucleophile, the pi system of the enolate. (b). Although displacement
Answers to Hour Examination #3, Chemistry 302X - 2007 1. Start with an analysis of exactly what molecule must have produced the product. The product is a -hydroxy ketone, so a reverse aldol to give X seems in order:
O H 3C = O C H3 O O X If we had compoun