Name Key 5 W-Exam No. Page I. (6 points) Identify the indicated pairs of hydrogens in each of the following compounds as (i) homotopic, (ii) enantiotopic, or (iii) diastereotopic s. Write the answers as (i), (ii), or (iii). Also write the number of unique s and C s for each compound. (a) a and d : (ii) b and e : (iii) b c d b and c : (iii) b and f : (ii) C e and f : (iii) c and f : (iii) (b) C c d a C a b C e a and b : (i) f c and d : (i) Number of unique 's: 5 Number of unique C's: Number of unique 's: 5 Number of unique C's: (c) c d C a b C a and b : (iii) c and d : (ii) Number of unique 's: 6 Number of unique C's: II. (6 points) An unknown compound, C 8 Cl, has the following spectroscopic properties. NMR (in CDCl ) δ.95 (, quintet, J = 7.0 z),.0 (, quintet, J = 7.0 z),.5 (, triplet, J = 7.0 z), and.58 ppm (, triplet, J = 7.0 z); C NMR (in CDCl ) δ 9.9,.0,.6, and.97 ppm. () In the box below, provide the structure of this dihalide and assign chemical shifts by writing chemical shift values next to the corresponding nuclei. Note: Electronegativities of and Cl are, respectively,.8 and.0. *Assignments could be reversed. *.5 ppm.95 ppm *.58 ppm Cl.0 ppm Structure: points Chemical shift assignments: points () The mass spectrum of the dihalide shows a number of peaks corresponding to its molecular ions due to the isotopes of and Cl atoms. Show below the isotope distribution of the molecular ions. The natural abundance ratio of the two isotopes, 79 and 8, is :, and the natural abundance ratio of the two Cl isotopes, 5 Cl and 7 Cl, is :. Ignore the presence of C isotopes. Show in the box below how you calculated the peak intensity ratios. Make sure to write m/z values. relative intensity Molecular ion peaks 70 7 7 m/z Calculations: C 8 Cl = 56 Cl m/z 70 peak = 56 5 79 m/z 7 peak = 56 5 8 and 56 7 79 m/z 7 peak = 56 7 8 Probability = / x / = /8 Probability = / x / = /8 Probability = / x / = /8 Probability = / x / = /8 Therefore, the peak intensity ratio among m/z 70, 7, and 7 peaks is: /8 : (/8 /8) : /8 = : :
Name Key 5 W-Exam No. Page III. (0 points) An unknown compound, C 6, possesses the following spectroscopic properties: IR (neat): 089-065 (several weak peaks), 950-80 (many small peaks), and 76 cm - (strong). No peaks in the 000-00 cm - region; NMR (CDCl ): δ 0.89 (6, doublet, J = 7.0 z),.88 (, nonet [i.e., 9 peaks], J = 7.0 z),.6 (, s),.89 (, d, J = 7.0 z), 7.0-7. ppm [5, multiplet (overlapping peaks)]; C NMR (CDCl ): δ 9.0 (q), 7.77 (d),.5 (t), 70.86 (t), 7.00 (d), 8.5 (d), 9.8 (d),. (s), and 7.7 ppm (s) [note: s, d, t, and q refer to the splitting patterns based upon one-bond C- couplings]. () ( point) What is (are) the unit(s) of unsaturation of this compound? Five () ( points) Is this compound a ketone or an ester? Ester () ( point) Does this compound a phenyl (C 6 5 ) group? Yes or No (circle the one that applies) () (6 points) n the basis of the spectroscopic data shown above, propose the structure of this compound in the box below. 6 IV. (8 points) The mass spectrum of -methyl--pentanone (C 6 ) shows the base peak at m/z and a strong peak at m/z 58 (50% relative intensity to the base peak). In the boxes below, provide a balanced fragmentation mechanism for the formation of the base peak (m/z ) and the peak at m/z 58 from the molecular ion. For m/z ion: C m/z -methyl--pentanone For m/z 58 ion: or even with
Name Key 5 W-Exam No. Page V. ( points) The thermal reaction of carbocation with cyclopentadiene () proceeds to yield cyclic carbocation. This carbocation then loses a proton with a base to produce bicycyclic alkene. ' ' ' ' enantiomer Explain why the thermal cycloaddition reaction between and takes place to produce by: (a) placing nodal points if necessary (exclude those on the atomic nuclei) and (b) darkening the appropriate portion of all of the atomic orbital lobes of FMs. Indicate which M you are using for each of the two compounds and, and provide in the box below a brief explanation as to why bonds can be formed between certain carbons in and. M or LUM? Circle the one that applies. Explanation: For : M or LUM? Circle the one that applies. ' For : ' ' ' Compound acts as a nucleophile and compound as an electrophile. The M of attacks the LUM of. The signs of A lobes match at the terminal positions (i.e., C-/C-' and C-/C-') in a face to face interaction. VI. (0 points) () Draw in the box below an M diagram for. Label (a) all A s used as sp, sp, sp, p, or s, (b) all M s as σ, σ*, π, π*, or n (non-bonding electrons), (c) M and LUM and fill in the electrons of this molecule. You do not need to use s electrons of the oxygen. s σ Ο Η n LUM M sp As σ Ο Η As () ( points) When is treated with Na, an electron is transferred from Na to. To which M, would an electron be transferred? What would be the chemical consequence of such transfer of an electron to the molecule of? Show the balanced chemical equation of the reaction that would result from the treatment of with Na. To which M? to LUM (σ*-) What chemical reaction will result? Show the reaction equation. -- Na - Na ne electron from Na goes into the antibonding M (LUM), thus breaking up an - bond. radical forms. 7
Name Key 5 W-Exam No. Page 5 VII. (7 points) Complete the following cycloaddition reactions. Predict the major stereochemical and regiochemical outcomes as appropriate. If a racemate is formed, only one of an enantiomeric pair needs to be shown. () [J. Am. Chem. Soc. 00,, 00] room temperature 80% () [J. Am. Chem. Soc. 00,, 00] enantiomer endo product TBS note: TBS = Si 5 C 55% TBS enantiomer () [J. rg. Chem. 00, 75, ] note: N Boc C -78 -> 0 C 0 h 50% Boc = enantiomer Boc N C () [J. Am. Chem. Soc. 00,, 9567] (C ) Si (C ) Si C C C C enantiomer 5 endo product
Name Key 5 W-Exam No. Page 6 VIII. (9 points) Draw in the boxes below the expected product for each of the following Diels-Alder reactions. () [J. rg. Chem. 00, 75, 7] most electron-deficient alkene exo product () [J. rg. Chem. 00, 75, 69] C C C C ~00% C C C C enantiomer exo product 5