Chapter 13 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 13 Nuclear Magnetic Resonance Spectroscopy Solutions to In-Text Problems 13.1 (b) Use the same equation and solve for B : n = 900 ± 10 6 Hz = H g 2 p B = 26,753 rad gauss –1 sec –1 2 p rad B Solving, B = 211,373 gauss Because ± and B are proportional, we could also have used the result of the last part and multiplied the value of B in the last part by the ratio of frequencies (= 900 MHz/500 MHz). 13.2 (b) Following the procedure used in part (a), B b B TMS = (10 –6 )(3.35 ppm)( B 0 ) = 0.24 gauss 13.3 The greater the chemical shift, the less shielded are the protons. Therefore, the protons at d 5.5 are least shielded), and those at d 1.3 are the most shielded. 13.5 (a) Subtract the two chemical shifts in ppm and apply Eq. 13.1. That is, chemical-shift difference = d 2 d 1 == ± n n 0 = 45 60 = 0.75 ppm (b) Using the procedure from part (a) gives chemical-shift difference = 45/300 = 0.15 ppm. 13.6 Parts (a)–(c) are answered in the text discussion that follows the problem (p. 587). (d) Because Si is more electropositive (less electronegative) than any of the other atoms in the table, hydrogens near the Si are more shielded . Because chemical shift decreases with increased shielding, this means that (CH 3 ) 4 Si has a smaller chemical shift. A derivative (CH 3 ) x M, in which M is an element more electropositive than Si, should have a negative chemical shift. (CH 3 ) 2 Mg and (CH 3 ) 4 Sn are two of several possible correct answers. 13.7 (a) The order is C < B < A . The protons of methylene chloride (dichloromethane, CH 2 Cl 2 ) have the greatest chemical shift, because chlorine is more electronegative than iodine. The chemical shift of methylene iodide is greater than that of methyl iodide because two iodines have a greater chemical shift contribution than one. 13.8 (b) Because protons H a and H b are constitutionally nonequivalent, their chemical shifts are different. (d) Because protons H a and H b are enantiotopic, their chemical shifts are identical. 13.9 The question asks essentially how many chemically nonequivalent sets of protons there are in each case.
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 13 2 (b) This compound has two chemically nonequivalent sets of protons; hence, its NMR spectrum consists of two resonances. 13.10 (b) 13.12 (b) The (CH 3 ) 3 CBr impurity in CH 3 I is more easily detected, because a given mole fraction of (CH 3 ) 3 CBr gives a resonance that is three times as strong as the resonance for the same amount of CH 3 I, as the solution to part (a) of this problem demonstrated. 13.13 In the following discussions, the integral is not mentioned. It corresponds to the number of protons under observation in each case. (b)
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This note was uploaded on 02/13/2010 for the course EGR 012 taught by Professor Witfield during the Spring '10 term at Aarhus Universitet.

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Chapter 13 - Instructor Supplemental Solutions to Problems...

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