Chemistry 605 (Hans J. Reich)

Transcription

1 Chemistry 605 (ans J. Reich) TIRD UR EXAM Mon. May 1, 2012 Question/Points R-11P /15 R-11Q /15 R-11R /20 R-11S /10 R-11T /20 R-11U /20 Total /100 Name 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. Distribution from grade list (average: 7.5; count: 27) Number 2 1 Average 7 i 9 Median 7 AB 80 BC Grade

2 15 Problem R-11P (C 8 8 F 2 Si. Below is a 00 Mz 1 NMR spectrum of vinyl difluorophenylsilane (source: Josh Dykstra/Burke). Analyze the spectrum, and label the spectrum with coupling trees, and label them with g, c and t. Report all coupling in the standard format ( n J X-Y = 00.0 z). Apart from intensities, the spectrum is basically first order. g t c Si F F z 5 pts each t g c J c-g 20.2 z J c-g 2 J t-c J t-g.5 z J t-g 2 J t-c J t-f 1.8 z J g-f 2.9 z J c-si J t-c =.5 z J t-g = 20.2 z J g-f = 2.9 z J t-f = 1.8 z There are also a number of small peaks from 29 Si coupling to the various protons. Pretty indistinct - hard to be sure which might be impurities because usually can't see both satellites. The strange intensities of the middle peak is confirmed by a simulation

3 15 Problem R-11Q (C N 8 ). This problem requires you to determine the stereochemistry of two isomers of sialic acid (A and B). Below is shown a portion of the 126 Mz 1 C NMR spectrum (D 2 solvent) of a 10:1 mixture of two isomers (ori,.; Nakajima, T.; Nishida, Y.; hrui,.; Meguro,. Tetrahedron Lett. 1988, 29, 617). Spectrum 1 is the fully proton decoupled. Spectrum 2 has the decoupler turned off N A 175. C 2 z Scale applies to main spectrum, not expansion N B C Spectrum 2 Spectrum (a) Which carbons of the sialic acid are being shown here? Mark the shifts on the structures. Carbonyl peaks - amide at and 177.9, carboxylic acid at and 175. (b) Interpret the multipicity of the signal at in the coupled spectrum (2). Estimate coupling constants, and assign them. D qd, J = 6, z C N Thus this must be the amide carbonyl - coupled to C (6 z) and the N-C- proton ( z). Coupling cannot be to N- proton since in D 2 this would be N-D. J = 6 J = (c) Which is the major isomer (A or B)? B Give your reasoning below. Be specific and brief N N 175. C C 2 J = 6 z J < 2 z The minor isomer (δ 175.) has a J C coupling between the vicinal axial C and the axial of 6 z. The major isomer has no significant J coupling to either of the vicinal C- protons - Karplus angle is ca 60

4 20 Problem R-11R (C Se 2 ) You are given the structure, and asked to interpret the spectrum (complete spectrum on next page). (a) Analyze the multiplet at δ 2.1 and report couplings. Septet of doublets, J - = 6.5, z ( D ) No Se satellites due to J D-Se can be seen here (satellites of the larger inner peaks are buried under the outer ones) Se A D B C Se (b) Analyze the multiplet at δ.7. Identify all peaks. btain exact shifts and report all shifts and couplings in the form: δ 0.00, n J XY = 00 z. An enlarged copy of the multiplet is shown below. The z values are from TMS at 0 z. 6 Main feature is the AB quartet ( A, B ) of the diastereotopic Se-C 2 - group. 76, 75, 79, 78 z (marked with and ) J AB = = 11z; = 11 z Solve the AB quartet: ν AB = 10.2 z, ν center = 751 z ν A = = 76 z, δ.7 ν B = = 756 z, δ.78 5 There are selenium satellites on both sides of each peak of the AB quartet. The 2 J Se if slightly larger for B (1.5 z) than for A (11.5 z) B A 2 J A-Se 11.5 z C 2 J C-Se 2 There is also a doublet for C at 75 z (marked with ) δ.67, J = z (to D ) ne satelline only is visible (double intensity because there are 2 Se). It is about 2 z from central peak, so 2 J C-Se = z. 2 J B-Se 1.5 z z z

5 Problem R-11R C Se Mz 1 NMR spectrum. Solvent CDCl. (Source: Bob Dykstra/JR 12/25) Se Se z

6 10 Problem R-11S (C Fe 2 2 P 2 ). Below are the 60 Mz 1 NMR spectra of two stereoisomers (E and Z) of the iron Cp complexes shown (J. Am. Chem. Soc 196, 85, 120). C C C P Fe Fe P C C C E C C P Fe Fe P C C C C Z E Spectrum 1 Z z Spectrum (a) Which isomer corresponds to Spectrum 1 E, and which to Spectrum 2 Z? Explain This is a simple symmetry argument - in the Z isomer the PMe 2 group is diastereotopic, so two triplets are seen. In the E they are identical (related by a C 2 axis). (b) Explain the appearance of the multiplet at δ 1.6 (i.e. why does it look like this). In the absence of other effects, the methyl groups should be coupled to one 1 P nucleus, hence a doublet. If the -bond coupling to the remote P was large enough, one would expect to see a dd. ne sees a triplet because there is a large J between the two 1 P nuclei, so anything coupled to them will show "virtual coupling" effects - i.e. the methyl group appears to be coupled equally to both P. (c) Would you expect the spectrum to look significantly different at 00 Mz (instead of the 60 Mz of the spectra shown)? 2 Apart from a larger separation between the two triplets in Spectrum 2, there would be no difference - the chemical shift between the P nuclei is zero at all fields, so the "virtual couplings" effect will always be there.

7 20 Problem R-11T (C F 2 ). Below are part of the 60 Mz 1 NMR spectra of two stereoisomers (A and B) of the fluorinated steroids shown. To aid in your analysis, a conformational drawing is also provided ( J. Am. Chem. Soc. 196, 85, 08) F 6 A 7 2e 2a 6a F 2e 7e F B 7a 2a 6e F 7a 7e A z Spectrum B 1.00 Spectrum (a) Which protons are being shown here? Analyze the coupling, and report them in the standard format (give δ and identify any couplings you found). Spectrum 1: δ 6.05, broad s, 2 J 6a-F = 8 z J 6a-7a = 11 z J 6a-7e = 6 z 7 δ 5.1, dddd, J = 8, 11, 6, 2 Spectrum 2: δ 5.89, d, J = 5z δ 5.05, dt, J = 51, J 6a- = 2 z 2 J 6e-F = 51 z J 6e-7a = z J 6e-7e = z J 6e- < 2 z (b) Which isomer corresponds to Spectrum 1 A, which to Spectrum 2 B. Explain briefly. The large - coupling in Spectrum 1 requires that the proton at 6 be axial, to get one large ax-ax coupling ( J 6a-7a = 11 z). The vicinal couplings in Spectrum 2 are all small ( z) so only eq-eq and eq-ax coupling, hence 6 must be equatorial.

8 20 Problem R-11U (C 02 0 AsF 7 NXe). This problem requires you to interpret the 129 Xe and 1 N spectra of [C C N-Xe-F] + AsF 6 - (Emara, A. A. A; Schrobilgen, G. J. Chem. Commun. 1987, 16) 1 J Xe-F = 6020 z Spectrum Mz 129 Xe NMR spectrum of [C C N-Xe-F] + AsF 6 - in F at -10 C 129 Xe, I = 1/2, 26.% abundant 1 J Xe-1N = 1 z z J Xe-1C = 79 z Spectrum Mz 129 Xe NMR spectrum of [C 1 C N-Xe-F] + AsF 6 - in F at -10 C Spectrum Mz 1 N NMR spectrum of [C C N-Xe-F] + AsF 6 - in F at -10 C 1 J 1N-Xe = 00 z z (a) Analyze Spectrum 1 and 2. Spectrum 2 is of a compound labeled >99% with 1 C at the CN carbon. Report coupling constants. Use the form n J X-Y = 00.0 z. 12 δ Xe = J Xe-F = 6060 z 1 J Xe-1N = 1 z 2 J Xe-1C = 79 z Spectrum 1: a doublet of 1:1:1 triplets, J = 6020, 1 z Spectrum 2: a doublet of 1:1:1 triplets of doublets, J = 6020, 1, 79 z (b) Analyze Spectrum. Make sure you understand and explain the origin of all peaks. Why are the signals somewhat broadened? 5 Central peak: all of the isotopes of Xe except 129 Xe uter peaks: 129 Xe satelites due to 129 Xe - 1 N coupling, J ca 00 z The signals are broadened because 1 N T 1 is quite short due to quadrupolar relaxation, so signals are boadened. There could also be broadening effects from unresolved coupling to F and C.