Chemistry 60 (Reich) SECND UR EXAM Thur. April 4, 20 Practice Exam 2 Answer Question/Points R-0F /2 R-0G /20 R-0 /0 R-0I /2 R-0J /20 Total /00 i 9 Average 6 Median 0 AB BC 0 CD 40 Name Grading Distribution from grade list (average: 6.9; count: 4) 4 Number 3 2 0 0 0 20 30 40 0 60 0 0 90 00 Grade If you place answers anywhere else except in the spaces provided, (e.g. on the spectra or on extra pages) clearly indicate this on the answer sheets.
2 2 4 Problem R-0F (C 2 6 Se). In this problem you are required to determine a structure from the IR and NMR spectra of a compound. The compound contains a Ph-Se group. (a) DBE. (b) Report your analysis of the IR spectrum (C 4 ). List the data and any conclusions you drew from it. 0 cm - Ketone 300 cm - Ar-C No triple bond (c) Interpret the 2-proton multiplet at δ 2 to δ 3. What do these signals tell you about the structure. Draw a coupling tree above it to show you understand the multiplet. B 30 20 0 0 z 6 This AB part of an ABX 3 pattern (an AB quartet of quartets) cannot be a quartet of quartets - separation is wrong X C C 3 B X can't be, but could be Se, Ar, or C= (c) Interpret the remaining multiplets in the NMR spectrum. Give multiplicity, coupling constants and part structures you were able to obtain from the signal. δ.0 2. 2.6 2.4 2.2 6 2 triplets, J = z, 2x C 3 C 2 C 3 of ABX 3 above 6 δ. δ 3. 2 m (actually AB of ABX 3 Y) t (J= z) X-C-C 2 These define this fragment: C C 2 C 3 e} Draw the structure of R-0F below. Label it with chemical shifts. 0.9 or.0 2. 3.. SePh 0.9 or.0 Without the PhSe hint, other structures (from 2004 exam) Se Ph NMR K, except δ..60.3 4.0 (obs 3.) Ph Se IR: 690 cm -
Problem R-0F (C 2 6 Se) 00 Mz NMR Spectrum in C 4 (Source: ans Reich 2/) 30 20 0 z 0 2. 2.6 2.4 2.2 2.0..6. 4.96 SePh 3.6 3.4.0 0. 2.2.93.00..0..0 6. 6.0..0 4. 4.0 3. 3.0 2. 2.0..0 0. 0.0
20 Problem R-0G (C Br 2 ). This problem requires you to analyze the signals at δ 4. and δ.2. You are given the structure. (a) Do a "first order" analysis of the two multiplets shown below. Draw a coupling tree, and estimate couplings. What type of pattern is this? Br B X Br 4. 42.9 3.0 X 32.4 22. "AMX" solution (approximate Solution ) J AB 223.4 2.3 A 22. 6 202.9 B 3 4 92.6 J AB z 2.20..0 4.0 4.0 4.00 3.9 (b) Do an accurate (quantitative) analysis. Use the frequencies shown above. If more than one solution is possible, show them both, and draw the proper coupling tree on the spectra below. Use appropriate criteria to distinguish the two. Show your work, and tabulate your data in an easily readable form. 202.9 2 92.6 3 22. 4 202.9 2.3 6 22. 22. 223.4 Solution Solution 2 J AB 9.90 to 0.6 +4.93 (-) -6.99 (+) +0.4 (-) +22.39 (-) ν A 29.04 23.09 ν B 204.36 20.3 ν AB 4.69 2. i 0 = i 0.99 0.322 δ A 4.063 4.044 δ B 4.0 4.034 22. ν A 26.6 209.6 ν B Solution c- = (+3)/2 = 2. 99. Δνab- = δ- = sqrt((-)(-3)) =.9 c- ± δ-/2 = 22., 209.6 ν A ν B Solution 2 c+ = (6+4)/2 = 20. Δνab+ = δ+ = sqrt((-2)(6-4)) =. c+ ± δ+/2 = 26.6, 99. Solution is correct:. Solution 2 has one negative 3 J, which never happens 2. Intensity calculation fits better for Solution
Problem R-0G (C Br 2 ) 300 Mz NMR Spectrum in CD 3 (Source: Wayne Goldenberg/Reich /3) 30 20 0 z 0 4. 42.9 3.0 32.4 22. 223.4 2.3 22. 202.9 92.6 4.99.20..0 4.0 4.0 4.00 3.9 2.00 0.9 6 4 3 2 0
0 Problem R-0 (C N 4 ). A graduate student thought she had prepared the compound below, but was worried about the NMR spectrum (taken in D 2 ), which seemed more than a little odd. Does the NMR spectrum fit the structure? Analyze and assign each of the multiplets. In particular, provide an explanation for the appearance of the key multiplet δ 3.-3.. Problem R-0 (C N 4 ). 300 Mz NMR Spectrum in D 2. Source: eather Schenk/Gellman 0-26 30 20 0 0 z M, N J MX, J NX B Y D M N X B 4.04 J AB J AB 2 N Y C 2 3.90 3. 3.0 3. 3.0 3.6 3.60 3. 0.93 X.00 2.2 2.20 2. 4. 4.0 3. 3.0 2. 2.0 - The key here is that the two C 2 groups are diastereotopic, and each of the C 2 groups are themselvees diastereotopic. So the protons form an AB MN XY system - ne C 2 group (labelled AB) has a reasonably large AB shift, J AB =, =., =. - The other C 2 group (labelled MN) has no detectable shift between the two protons, and so appears as a doublet, with apparently equal couplings J MX = J NX =. z - The Y proton is a doublet, δ 3., J = 3. z. J XY, J MX, J NX To show you understand the spectrum, draw a coupling tree for the multiplet at δ 2.2 (start with an intensity assignment). 3 4 33 4 3 3 3 Expect this proton (X) to be coupled as follows: =. =. J MX =. J NX =. Thus dqd, J =.,., 3. X 40 30 20 0 0 z
2 Problem R-0I (C 2 4 ). You are provided the NMR spectrum of a compound. Interpret the NMR spectrum, and determine the structure or structures. Use the A, B, etc labels on the spectrum. Show the chemical shift and multiplet structure in the form: 0.0 δ, dtd, J AB = 0.0, 0.0, 0.0 z,. You may use first order analysis. 2 (a) DBE (b) Analyze the multiplets A, B, C. Provide part structure(s) defined by these protons. Note: Do not attempt to distinguish among the several isomers which are consistent with this pattern. A δ.39, d, J = z (J-ortho) B C δ.32, d, J = 2 z (J-meta) δ.0, dd, J =, 2 z (J-ortho + J-meta) These are aromatic protons and define a,2,4-trisubstituted benzene C C (c) Interpret the signals D-. Provide part structure(s) defined by these protons. D δ 6.04, dd, J = 9, 2 z E F G δ.9, dd, J = 0, 4 z δ 3., dddd, J =, 0, 4, 2 z δ 2.9, dd, J =, 9. z D F E G δ 2., dd, J =, z 0 (d) Draw the structure of R-00F below. If more than one structure fits the data, draw them, but circle your first choice. Assign the protons (label them with the letters A-). If any assignments are ambiguous, indicate the basis for your choice. C G D 0 F C E 0 0 D E This question elicitated many answers - not a good problem since there are too many hidden carbons. F G
Problem R-0I (C 2 4 ). 20 Mz NMR Spectrum in CD 3. Source: Ieva Reich 0/30 A B C.4.3.2. F G 3. 3. C G 3.0 2.9 2. 2. F C E 30 20 0 0 z 2.92 D 6. 6.0.9. E D 2.0.9 x.00 x x A B C D E F G 6 4 3 2 0
20 Problem R0J (C 24 2 9 ). This problem requires you to analyze part of the NMR spectrum of a tetrahydropyran, and determine the stereochemistry at three centers. A planar projection and conformational drawing is shown below. Bz C Bz A E 3 4 2 3 4 2 6 D Bz = PhC(=)- C 2 Me 6 B F 3 pts for correct answer 3 or 4 for reasoning (a) Determine the stereochemistry at C-6. Explain what signal(s) you used, give their shift and multiplicity (e.g. δ 0.00, tq, J =0, 0) and briefly describe how you made the stereochemical assignment using the data: A =, B = C 2 Me ( or C 2 Me). δ 4.3 δ 2.24, q, J = 2 z The quartet at δ 2.24 (J = 2 z) is the axial proton at C-. The three couplings must be a J gem and two J ax-ax, thus protons on both sides are axial, and the substituents at C-6 and C-4 must both be equatorial. The couplings of the equatorial proton e (dddd J = 2,, 2., ) also help identify the 4 (.40, J=) and 6 (4.3, J=2.) protons 2 3 4 6 Me 2 C Bz Bz (b) Determine the stereochemistry at C-4. Explain what signal(s) you used, give their shift and multiplicity and briefly describe how you made the stereochemical assignment using the data: C =, D = Bz ( or Bz). δ.40 6 See part (a) The signal at.4 shows J = 2,, 3, so one axial-axial coupling (to ), and two ax-eq couplings to 3 and (this also proves that 3 must be equatorial) (c) Determine the stereochemistry at C-3. Explain what signal(s) you used, give their shift and multiplicity and briefly describe how you made the stereochemical assignment using the data: E =, F = Bz ( or Bz). δ.6 The "d" at.6 has to be 3 - it shows only one obvious small coupling. Since 2 is axial, this means that 3 must be equatorial, or else it would show a large J ax-ax Could also use the axial proton at 4.40, ddd, J = 2,, 3. The 2 z coupling is the J ax-ax to, the two smaller coulings have to be the J ax-eq to 3 and, hence 3 has to be equatorial The proton at 2 ( δ 3.6, dd, J =, 2.) has to be axial, if it were equatorial the ring would flip. The z coupling is to, the 2 z couping must be to 3. Thus 3 must be equatorial.
Problem R-0J (C 24 2 9 ) 300 Mz 3 C NMR Spectrum in CD 3 Source: Geoffrey Sametz/Burke 9/99 3 Bz Bz 3 4 2 6 C 2 Me 4 ddd, J = 2,, 3 30 20 0 0 z e dddd J = 2,, 2., a q, J = 2 z.90..0..0.6.60..0.4.40.3 6 dd, J = 9, 4 2. dd J=2, 2. dd, J =, 2 dd, J = 9, 6 ddd, J =, 6, 4 2.0 2.4 2.40 2.3 2.30 2.2 2.20 2. 4.3 4.30 4.2 4.20 4. 4.0 4.0 4.00 3.0 3.6 3.60 30 20 0 0 z..0.9...6..4.3 6.3 4.04 2.93 2.3 3.0.94.92 2.0.00 0.9 0.6..0..0 6. 6.0..0 4. 4.0 3. 3.0 2. 2.0..0 0. 0.0