Practice Problems, November 27, 2000

Transcription

1 Practice Problems, ovember 27, Why do the following groups all have very similar A-values? R-Group A-Value (kcal mol -1 ) Br Sn( 3 ) Si( 3 ) Ph What is the product of the following mercuration reaction? g( 3 ) 2 3 ab Give a mechanism for its formation. 3. Predict the products of the following epoxidations. nly one equivalent of is present in each case. 2 l l l 2 4. Provide a mechanism to explain the following unexpected transformation. I 2, KI a 3 (aq.) 5. ydroboration of an alkene is usually followed by oxidation. If the oxidation is done using 2 2 /a, the product is an alcohol. What is the product of the following reaction? 1) B 2 6, TF 2) 2 S 3

2 Answers to Practice Problems, ovember 27, Why do the following groups all have very similar A-values? R-Group A-Value (kcal mol -1 ) Br Sn( 3 ) Si( 3 ) Ph 1.68 The A-value (conformational free energy) of an R-group on a cyclohexane ring arises primarily from the steric repulsion between an axial R-group and the -atoms that are synaxial on positions 3 and 5. When the R-group is not symmetrical, it too has conformational freedom. Rotation around the bond linking R to the ring encounters a 3-fold barrier, and the lowest energy minimum for 2 X will be the rotamer having the group X outboard with respect to the ring. ote that this means that the synaxial steric interaction is largely due to the 2 fragment. The bulk of the X fragment will only contribute when the R-group is in one of the higher-energy geometries. For this reason, most groups 2 X have nearly the same A-value as 3. X X

3 2. What is the product of the following mercuration reaction? g( 3 ) 2 3 Give a mechanism for its formation. ab g( 3 ) g( 3 ) 2 g( 3 ) 3 3 g( 3 ) 3 g( 3 ) 3 2 added with the ab 4 workup step g( 3 ) + 3 g( 3 ) 3 g( 3 ) 3 This hydration/tautomerization will probably occur very rapidly. It could proceed by an acid-catalyzed route as shown here, although since solutions of ab 4 in water are basic, a basecatalyzed mechanism is perhaps more likely. ab 4 g 3 The remainder of the mechanism is identical to the general pathway for free-radical reduction of organomercury species that is described in the textbook.

4 3. Predict the products of the following epoxidations. nly one equivalent of is present in each case. 2 l l l All of these epoxidations involve reaction of the more electron rich alkene group. ote that steric effects of the alkene substituents have a minimal effect on reactivity because approach of the peroxyacid is nearly perpendicular to the plane of the alkene. The substituents are all oriented more or less in the plane, except in the last case. otice that in this allene structure, the 3 groups will overhang the other alkene because the molecule is not planar.

5 4. Provide a mechanism to explain the following unexpected transformation. I 2, KI a 3 (aq.) I a 3 (aq.) I 2, KI I 3 2 The intended reaction was an iodolactonization. owever, this particular system is not set up appropriately. Recall that halolactonizations will tend to favor the Markovnikov product. In this system, the Markovnikov pathway would require forming a 4-membered lactone ring. While this is possible, it is a strained situation. The alternative 5-membered ring lactone must form via an intermediate that puts more cationic character on the less-substituted atom. Because these pathways are less favorable, and because the iodonium ion positions a good leaving group β with respect to the carboxylate, it is possible to eliminate 2.

6 5. ydroboration of an alkene is usually followed by oxidation. If the oxidation is done using 2 2 /a, the product is an alcohol. What is the product of the following reaction? 1) B 2 6, TF 2) 2 S 3 2 R R 3 B + 2 S 3 R B R S R 2 B R + S 4 otice the similarity between this reagent and the anion in the conventional oxidative workup. In both cases, we have a nucleophile with a built-in leaving group (i.e. :u X). Thus, when the nucleophile attacks the tri-coordinate boron to form the anionic intermediate, migration of an R-group with expulsion of the hydrogensulfate ion (an excellent leaving group, the conjugate base of a very strong acid) is straightforward. The oxidative workup probably also requires the addition of a base, which was not indicated in the problem.