exam4-2010

exam4-2010 - Name Exam 4 CHEM/BCMB 4190/6190/8189(118...

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Name _________________________ Exam 4, 2010: CHEM/BCMB 4190/6190/8189 1 Exam 4: CHEM/BCMB 4190/6190/8189 (118 points) Thursday, 9 December, 2010 1 ). Consider the HETCOR pulse sequence, and the energy diagram for the simple 1 H- 13 C spin system (i.e. CHCl 3 ). Just before the second 1 H 90 ° pulse, for a particular t 1 value, the M H C α and M H C β vectors happen to be positioned along the –y and y axes, respectively. a. Define the populations for the individual states N 1 -N 4 and the population differences for the A 1 , A 2 , X 1 and X 2 transitions both before the beginning of the pulse sequence and after the second 1 H 90 ° pulse (assuming the particular value of t 1 mentioned above, and no relaxation losses during t 1 ). Assume that N 4 =N at equilibrium, and Δ H is the difference in the number of spins in α and β states for 1 H, and Δ X is the difference in the number of spins in α and β states for 13 C. Also, define the relationship between Δ H and Δ X. ( 8 points ) After the second 1 H pulse, the M H C β vector will be inverted from its equilibrium position (- z), whereas the M H C α vector will be in the same position as it was at equilibrium (z). Thus, the N 2 and N 4 populations are inverted after the second 1 H pulse. Also, the ratio of Δ H and X is the ratio of their gyromagnetic ratios, which is ~4: equilibrium (beginning after second of pulse sequence) 1 H pulse N 4 = N N 4 = N + H N 3 = N + X N 3 = N + X N 2 = N + H N 2 = N N 1 = N + H + X N 1 = N + H + X A 1 = N 2 – N 4 = H A 1 = N 2 – N 4 = - H A 2 = N 1 – N 3 = H A 2 = N 1 – N 3 = H X 1 = N 3 – N 4 = X X 1 = N 3 – N 4 = X – H = -3 X X 2 = N 1 – N 2 = X X 2 = N 1 – N 2 = X + H = 5 X 13 C 1 H N 2 N 1 N 3 N 4 X 1 αβ βα αα ββ X 2 A 1 A 2 13 C 1 H 90x 90x 90x t 1 t 2 M H C β M H C α y z x
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Name _________________________ Exam 4, 2010: CHEM/BCMB 4190/6190/8189 2 b. Draw the vectors M C H α and M C H β that are present just before the 13 C pulse. Make sure to label each with the correct name (M C H α or M C H β ) and with the proper magnitude (expressed in terms of Δ X). ( 4 points ) c. In the modified HETCOR experiment (right) an additional pulse ( 13 C 180 ° x) and an additional delay ( Δ ) have been added in order to remove the 1 J CH splitting in the t 1 dimension (decouple). For the simple 1 H- 13 C spin system (i.e. CHCl 3 ), at point “a” in the sequence, the M H C α and M H C β vectors are positioned as shown below (in the transverse plane). For an arbitrary value of t 1 /2, the vectors have moved into the positions shown in “b”. Show the results at points “c” and “d”. Then explain why decoupling is achieved. ( 8 points ) Decoupling is achieved because the 1 J CH coupling is refocused during the t 1 /2- 13 C 180x- t 1 /2 period. At “d”, the M H C α and M H C β vectors are refocused, and the net precession to this point for each vector is the same. d. The delay Δ has to be set appropriately in order for the pulse sequence to operate properly.
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exam4-2010 - Name Exam 4 CHEM/BCMB 4190/6190/8189(118...

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