PAGE 1 of 7 UNIVERSITY F MANITBA DEPARTMENT F CEMISTRY 2.339 STRUCTURAL TRANSFRMATINS IN RGANIC CEMISTRY FINAL EAMINATIN Dr. Phil ultin Thursday December 14, 2000. NAME: ANSWERS STUDENT NUMBER: 1) (15 Marks). TEN of the terms or concepts in the left hand column match definitions or properties in the right hand column. FIVE do not match. Indicate the correct definition for each term in the space provided. For those terms that do not have a matching definition, write an in the space. TERMS DEFINITINS Felkin-Anh model I A. Promotes the formation of amides from acids and amines. Sodium cyanoborohydride B. A highly hindered reducing agent. Fourier transform C. Relates the structures of transition states to reaction energetics. Product development control K D. Selectively converts benzylic alcohols to carbonyl groups. ammond s postulate C E. Method for solving the Schrödinger equation. Diimide F. Compounds that differ in their connectivity. Structural isomers F G. Enhances the intensity of certain NMR signals. Manganese dioxide D. A method for extracting individual frequencies from a complex oscillation. Disiamylborane B I. Predicts the stereochemistry of nucleophilic additions to acyclic ketones or aldehydes. Burgi-Dunitz angle J. The influence of orbital interactions on conformation or reactivity. Anomeric effect K. An explanation for the stereochemistry of ketone reductions based on torsional effects. Stereospecific reactions N L. Reduces esters to aldehydes at low temperatures. Stereoelectronic effects J M. Relates 1 NMR coupling constants to dihedral angles. Molecular mechanics N. Form distinct stereoisomeric products from stereoisomeric reactants. Nuclear verhauser effect G. Provides an approximate measure of steric bulk of substituent groups.
PAGE 2 of 7 2) a) (5 Marks) Briefly explain why 1-methyl-1-phenylcyclohexane slightly favors a conformation that places the bulkier phenyl group in the axial position rather than the smaller methyl group. C 3 Ph Ph C 3 G o = -0.32 kcal mol -1 The phenyl ring and the methyl group interfere sterically with one another. Rotation of the phenyl is hindered due to close contacts between the ortho hydrogens and the C3. This forces the phenyl to lie roughly perpendicular to the C-C3 bond. When the phenyl is equatorial, its ortho hydrogens therefore come into conflict with the equatorial hydrogens on C-2 and C-6 of the cyclohexyl ring. In the axial orientation, there are no severe interactions of this kind. C 3 C 3 Steric conflict b) (5 Marks) Consider the most stable conformer of the structure shown below. What are the approximate values (in degrees) of the indicated torsional (dihedral) angles? Torsion angle 8-7-2-3 0 9 C 3 10 3 C 6 1 8 2 7 C 3 Torsion angle 9-1-2-3 Torsion angle 10-6-5-11 90-100 60 3 C 11 5 4 3 C 3 Torsion angle 1-2-3-4 0 Torsion angle 3-4-5-11 180
PAGE 3 of 7 c) (5 Marks) Briefly explain the differences in rate and in the structure of the products obtained when menthyl chloride (A) or neomenthyl chloride (B) are treated under identical conditions with sodium ethoxide in ethanol. NaEt Et fast reaction NaEt Et slow reaction ipr A B NTE RIGINAL ARD CPIES AD STRUCTURES F A AND B REVERSED! ipr C 3 C 3 ipr Compound A has a trans-diaxial arrangement of and in its most stable conformation. Moreover, E2 elimination from this geometry gives the favored Saytseff (Zaitsev) alkene. Compound B has no favorable E2 elimination pathway available from its most stable conformation, and in order to eliminate it must flip into a high-energy chair conformer. Note that the only diaxial arrangement leads to the non-saytseff alkene in this case. C 3
PAGE 4 of 7 3) (35 Marks total) Supply the structures of the major product obtained from each of the following reactions. Be sure to indicate stereochemistry clearly and accurately where it is appropriate. 1) g( 2 CCF 3 ) 2 (1 equiv.) 2 /TF 2) NaB 4 a) b) LiAl 4 TF c) C 3 N(C 3 ) 2 I 2 TF/ 2 I C 3 (4 Marks) d) I 2 2 AcAg Ac/ 2 90-95 o C then Na/ 2 e) Bz 3 C 3 C C 3 3 C C 3 C 3 C 3 C 2 2 Bz 3 C
PAGE 5 of 7 f) (C 3 ) 2 C (6 Marks) C 2 C C(C 3 ) 2 Na CS 2 C 3 I (C 3 ) 2 C S C 3 S C 2 C C(C 3 ) 2 n-bu 3 Sn (C 3 ) 2 C AIBN (cat.) toluene C 2 C C(C 3 ) 2 1) B 2 6, TF 2) 2 2, Na g) C 2 C 2 + Ts (cat.) C 2 2 h) C 3 C 3 PCC C 2 2 1 NMR of product shows no peaks at δ > 4.0 ppm. i) (4 Marks) j) (C 3 ) 2 C C 2 C C(C 3 ) 2 Ac 2 DMS (C 3 ) 2 C C 2 C C(C 3 ) 2
PAGE 6 of 7 4) (20 Marks) The following 2-step transformation has similarities to several reactions encountered in class and in the lab. Provide a detailed stepwise mechanism to explain the formation of the product. Note that the Li( 4 ) reagent acts as a source of a mild electrophile (Li + ) while 4 is a nonnucleophilic counter-ion. Ts N.B. Ts- = 3 C Li 4 TF then 2N (aq) Et This is a pinacol rearrangement followed by acid-catalyzed isomerization of an alkene into conjugation with the resulting ketone. At the same time, the acetal is hydrolyzed by the aqueous acid. Note that the isomerization of the alkene is similar to the one we saw as an alternative workup procedure for Birch reductions of anisole derivatives. S Ar S Li + Ar S Li This compound is stable and can be isolated, although it is not necessary to do so. + 2 + +
PAGE 7 of 7 5) (15 Marks). When compound A (C 4 7 Br) is heated with a mixture of NaC 3 /C 3, a new product (compound B) is obtained. Compound B contains no halogen. When either A or B is heated in Na/ 2, the same product (compound C) is obtained after acidifying the reaction mixture. From the spectra below, what are the structures of A, B and C? Assign the NMR signals to appropriate nuclei in each structure. q, J = 6. 8 z d, J = 6. 8 z Compound A Br quartet ~4.4 ppm singlet ~2.4 ppm doublet ~1.75 ppm h e p t. J = 6. 9 z d, J= 6.9 z Compound B doublet ~1.1 ppm heptet, ~2.5 ppm C 3 singlet ~3.6 ppm hept., J=6.8 z d, J=6.8 z Compound C doublet ~1.5 ppm heptet, ~2.3 ppm singlet, ~ 12 ppm