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exam3-2009 - Name Exam 3 CHEM/BCMB 4190/6190/8189(114...

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Name _________________________ Exam 3, 2009: CHEM/BCMB 4190/6190/8189 1 Exam 3: CHEM/BCMB 4190/6190/8189 (114 points) Thursday, 29 October, 2009 1 ). For a simple two spin ( 1 H- 13 C) coupled system (i.e. 13 CHCl 3 ), the population diagram is shown (below, right). Here “H 1 ” and “H 2 ” represent the 1 H transitions, and “C 1 ” and “C 2 represent the 13 C transitions. We’ll consider the effect of the pulse sequence shown (below, left) assuming the Larmor frequency is equal to the reference frequency ( ± L = ± ref ). Assume ² =1/(4J) and ignore the effects of magnetic field inhomogeneity. a. What is the name of the pulse sequence shown. If it is an acronym, state what the acronym stands for? ( 2 points ) INEPT (Insensitive Nuclei Enhanced by Polarization Transfer) b. Complete the vector diagrams below for points ‘c’, ‘d’, ‘e’, and ‘f’ in the pulse sequence. Be sure to label the vectors (M H C ³ , M H C ´ ), to include arrows indicating the direction of precession for the vectors in the rotating frame, and to indicate the angle ( µ ) between the vectors at each point. In this example, the precession frequency of M H C ´ is slightly faster than the Larmor frequency. ( 8 points ) e ² a b c d f g 90x 180x 90y 180x 90x 13 C ² 1 H g h 1 H 13 C N 2 N 1 N 3 N 4 C 1 ³´ ´³ ³³ ´´ C 2 H 1 H 2 x µ = 0° M H C ´ M H C ´ M H C ³ transverse plane y z x 90 ° x y x ² M H C ³ M H C ´ x µ = 90° y a b c 180 ° x 1 H M H C ´ M H C ³ µ = 90° y 180 ° x 13 C ² M H C ³ M H C ´ µ = 180° x y d e f x M H C ³ M H C ´ µ = 90° y M H C ³
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Name _________________________ Exam 3, 2009: CHEM/BCMB 4190/6190/8189 2 c. Which of the hydrogen transitions, H 2 (N 1 -N 3 ) or H 1 (N 2 -N 4 ) corresponds to M H C ± ? ( 2 points ) H 2 d. Draw the vector diagram showing the M H C ± and M H C ² vectors at point ‘g’ in the pulse sequence. Make sure to label each vector (M H C ± , M H C ² ). ( 4 points ). e. Ignoring the effects of relaxation and magnetic field inhomogeneity, write down the populations of the states N 1 , N 2 , N 3 and N 4 at each point to ‘g/g ³ ’ in the sequence. The values at equilibrium (at point ‘a’) are given below (note: point ‘d’ is following the 1 H 180 ° x pulse, and ‘e/f’ is, subsequently, following the 13 C 180 ° x pulse). ( 8 points ) ‘a’ ‘b/c’ ‘d’ N 4 = N N 4 = N + ´ H/2 N 4 = N + ´ H/2 N 3 = N + ´ C N 3 = N + ´ C + ´ H/2 N 3 = N + ´ C + ´ H/2 N 2 = N + ´ H N 2 = N + ´ H/2 N 2 = N + ´ H/2 N 1 = N + ´ C + ´ H N 1 = N + ´ C + ´ H/2 N 1 = N + ´ C + ´ H/2 ‘e/f’ ‘g/g ³ N 4 = N + ´ H/2 N 4 = N + ´ H N 3 = N + ´ C + ´ H/2 N 3 = N + ´ C N 2 = N + ´ H/2 N 2 = N N 1 = N + ´ C + ´ H/2 N 1 = N + ´ C + ´ H M H C ² M H C ± y z x g
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