Z 2 (OMe) = Z 3 (Cl) = 1.4. Calculated: Measured: Deviation:

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1 Question 1. hemical Shifts & Increment Systems. (20 points) 13 -NMR of para-hloroanisol (from SDBS, requires the book by Pretsch et al.) The structure of para-chloroanisol is shown along with the atom numbering we will use here. The chemical shift data and intensities are taken from the SDBS database. Using the book by Pretsch et al., find the base value for the chemical shift of an aromatic -atom and locate the Z i -values and provide them in the table below. Make sure to indicate the i of the Z i -values (where the underline is shown). Then make the additions to obtain the calculated chemical shift, assign the measured peaks and enter their chemical shifts in the respective row, and, finally, compute the difference between calculated and measured values MHz 13 -NMR, Peak Data. ppm Int l O H Z 1 (OMe) = 33.5 Z 4 (l) = -1.9 Z 2 (OMe) = Z 3 (l) = 1.4 Z 3 (OMe) = 1.0 Z 2 (l) = 0.4 Z 4 (OMe) = -7.7 Z 1 (l) = 5.3 alculated: Measured: Deviation: Row 2, 10 points: 0.5 Points for each base value, 0.5 points for each subscript, 0.5 points for each Z-value. Row 3, 4 points: For getting the additions right. Row 4, 4 points: For the correct assignments of the measured shifts to the %. Row 5, 2 points: 0.5 Points for each correct error calculation. -1-

2 Question 2. Decoupling and DEPT. (20 points) The decoupled 13 -NMR spectrum of ethyl phenylacetate is shown as recorded on a 100 MHz spectrometer (e.g. carbon frequencies around MHz). The peak at 134 ppm is caused by the ipso-carbon. (a) Directly in the spectrum, indicate the multiplicities of the peaks at 14, 41 and 61 ppm in the respective proton-coupled spectrum. (b) Schematically draw the DEPT spectra in the boxes below. The baseline is given so that you know where to place the peaks. Just decide whether there is no peak, or whether the peak is up or down. DEPT-135 DEPT-90 DEPT-45-2-

3 Question 3. Nuclear Overhouser Effect (NOE). (20 points) alpha H (1) (2) (3) (4) (5) (6) (7) (8) arbon T 1 NOE 13 -Int. H att. (sec) alpha , ipso , ortho , meta , para This is all about toluene. The numbering of the atoms is shown in the drawing. The T 1 and NOE data are taken from the textbook; the intensities are from the toluene spectrum in the SDBS database. (a) The SPIN - LATTIE relaxation times T 1 are given in column (2). In general, the shorter the relaxation time T 1, the higher the signal intensity. In column (3) provide the rankings for the -peak intensities that you would expect based on T 1 alone (from 1 to 5, 1 highest intensity, 5 lowest intensity). (8 p.) (b) The NOE factors are given in column (4). The factors specify the amount of the increase of the 13 -NMR signal that is achieved when the H-nuclei are irradiated. In column (5) provide the rankings for the -peak intensities that you would expect based on NOE alone (from 1 to 5 as above). (5 points) (c) Measured intensities with rankings are listed in columns (6) and (8). State whether the intensities are dominated by T 1 or by NOE. Provide a very brief explanation as to why this is so; a few good words suffice! (4 p.) Neither. In this specific case it looks like it goes more by NOE but even here that does not really hold. Without decoupling, T 1 matters. With decoupling, it s more complicated because now T 1 as well as other relaxations (w2) matter. (d) The NOE for the ipso- is not zero! The NOE effect of the H 3 -carbon is much less than 2. The H-carbons have the best NOEs. Does it matter for a carbon s NOE how many Hs are attached? Does it matter for a carbon s NOE whether any H is directly attached? What then is required for a -atom to show NOE? (3 p.) It does not matter how many Hs are directly attached! -- It does not matter whether any H is directly atteched! -- The only thing that matters is that there are some Hs not too far away from the that wants to benefit from the NOE. -3-

4 Question 4. Spin-Spin oupling: Mechanism and Multiplicity. (20 points) (a) 1 H-NMR of 4 H 8 O (from SDBS) MHz 1 H-NMR, Peak Data. Draw the structure of the molecule (3 points): Hz ppm Int q q q q s t t Look at the chemical shifts; there are no alkene Hs!! The unsaturation has to be a carbonyl. H 3 H 2 O H t 3 J(H,H) = 7.3 Hz (3 points) (b) Schematic for the Mechanism of 3 J(H,H) Spin-Spin oupling Double-arrows indicate nuclear spins ( 1 H only), single-arrows indicate electron spins. omplete the scheme by adding the directions of the remaining spins and by providing the name of the rule that governs the alignments of the pairs marked by {. (6 points; 6x0.5 for the arrows, 1 each for rule) H Pauli Rule no rule here can be either way (c) Solvents (8 points) Hund Rule II In general, multiplicity depends on the number of equivalent neighbors n and their nuclear spin I and the multiplicity is given by 2nI+1. For spin I = ½ nuclei, this formula simplifies to the n+1_-rule. Acetone-d5 H DMSO-d6 H couples to n =_2_ equivalent _D_-atom(s) with nuclear spin I = _1_. Multiplicity of 1 H signal: 5 couples to n =_3_ equivalent _D_-atom(s) with nuclear spin I = _1_. Multiplicity of signal: 7-4-

5 Question 5. Spin-Spin oupling: Tree Diagrams. (20 points) 1 H-NMR of bromethene, 2 H 3 Br (from SDBS) 300 MHz 1 H-NMR, Peak Data. Parameter ppm Hz D(A) 5.97 D(B) 5.84 D() 6.44 J(A,B) J(A,) 7.12 J(B,) (A)H (B)H H() Br (a) On a 300 MHz spectrometer 10 Hz correspond to ppm. (2 points) (b) onsidering your answer to (a), draw as best as you can the tree diagrams for signals A, B, and. (18 pts.) 6.5 A B