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196 Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE Solutions to Problems 25. To do this, you need to tell the difference between frequencies, v, in units of s⁻¹, and wavenumbers, v, in units of cm⁻¹. Section 10-2 shows how they are related: v = and = so v = or = v/c. For AM radio (v = 10⁶ s⁻¹), S ¹), = ≈ 3 X 10⁻⁵ cm⁻¹; and for FM and TV (v = 10⁸ = X 10¹⁰) ≈ 3 X 10⁻³ cm All these are well to the right end of the chart, very low in energy relative to most of the forms of electromagnetic radiation on the chart. 26. The conversion formulas are = and v = (Section 10-2). (a) = 1/(1050 cm⁻¹) = 9.5 X 10⁻⁴ cm = 9.5 µm (b) 510 nm = 5.1 X 10⁻⁵ cm; v = (3 X 10¹⁰ cm ¹)/(5.1 X 10⁻⁵ cm) = 5.9 X 10¹⁴ s⁻¹ (c) 6.15 µm = 6.15 X 10⁻⁴ cm; = 1/(6.15 X 10⁻⁴ cm) = 1.63 X 10³ cm⁻¹ (d) v = = (3 X 10¹⁰ cm X 10³ cm⁻¹) = 6.75 X 10¹³ s⁻¹ 27. Use E = 28,600/X (Section 10-2), and use the equations = 1/v and = c/v. Be sure to convert the units of to nm before calculating E, though! (a) À = 1/750 = 1.33 X 10⁻³ cm = 1.33 X 10⁴ nm (1 cm = 10⁻² m, and 1 nm = 10⁻⁹ m, or 1 cm = 10⁷ nm), so E = (2.86 X 10⁴)/(1.33 X 10⁴) = 2.15 kcal mol⁻¹ (b) = 1/2900 = 3.45 X 10⁻⁴ cm = 3.45 X 10³ nm, so E = (2.86 X 10⁴)/(3.45 X 10³) = 8.29 kcal mol⁻¹ (c) = 350 nm (given), so = (2.86 X = 82 kcal mol⁻¹ (d) = 3 X X 10⁷) = 3.4 X 10² cm = 3.4 X nm, so E = (2.86 X 10⁴)/(3.4 X = 8.4 X 10⁻⁶ kcal mol⁻¹ (e) = 7 X 10⁻² nm, so E = (2.86 X 10⁴)/(7 X 10⁻²) = 4.1 X 10⁵ kcal mol⁻¹ 28. Only the value of v is needed to calculate E. Use E = 28,600/X, together with À = c/v. (a) = (3 X 10¹⁰ cm X 10⁷ s⁻¹) = 333 cm = 3.33 X nm, so E = (2.86 X 10⁴)/ (3.33 X = 8.59 X 10⁻⁶ kcal mol⁻¹ (b) E = 4.76 X 10⁻⁵ kcal mol⁻¹ 29. (a) Increasing radio frequency-to the LEFT (b) increasing magnetic field strength (moving "upfield")-to the RIGHT (c) increasing chemical shift-LEFT (d) increased shielding-RIGHT 30. (a) ¹⁹F H₀ = 2.11 T CFCl₃ ¹³C 84.6 22.6 8.82 7.34 0 MHz (b) Like (a), but with an additional signal at 90 MHz (¹H).