Chapter 12

# Chapter 12 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 12 Introduction to Spectroscopy. Infrared Spectroscopy and Mass Spectrometry Solutions to In-Text Problems 12.1 (b) Apply Eq. 12.1 and include the conversion factor 10 –10 m Å –1 . 12.2 (b) Multiply the frequency obtained in the solution to Problem 12.1(b) times Planck’s constant: E = h n = (3.99 ± 10 –13 kJ sec mol –1 )(6.25 ± 10 14 sec –1 ) = 249 kJ mol –1 12.3 (a) The energy of X-rays is greater than that of any visible light, including blue light. (In fact, the energy is so much greater that prolonged exposure to X-rays is harmful.) 12.4 (b) Apply Eq. 12.7b on text p. 541: 12.6 Using Eq. 12.8 on text p. 541, convert the wavenumber to a frequency, which is the “times per second” equivalent of wavelength or wavenumber. n = c n ~ = (3 ± 10 10 cm sec –1 )(2143 cm –1 ) = 6.43 ± 10 13 sec –1 12.8 Take the ratio of two equations like Eq. 12.13 on text p. 547, one for the C—H bond, and the other for the C—D bond. Everything cancels except the wavenumbers and the square roots of the masses. n ~ D n ~ H = m H m D or n ~ D = n ~ H m H m D = (3090 cm –1 ) 1 2 = 2185 cm –1 In fact, C—D vibrations appear in the IR at lower energy than the corresponding C—H vibrations (to the right in conventional IR spectra). 12.9 (b) Active. The C A O dipole is increased by the stretch because its length changes. Recall (Eq. 1.4, text p. 11) that dipole moment is proportional to length. (d) Inactive. The zero dipole moment of this alkyne is not changed by stretching the triple bond.

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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 12 2 (f) Active. The dipole moment of the molecule is due mostly to the bond dipole of the C—Cl bond, which is the only bond in the molecule with significant polarity. Increasing the length of this bond dipole will increase the dipole moment of the molecule. 12.10 The carbon skeleton of the hydrogenation product defines the following alkenes as possibilities. The strong 912 and 994 cm –1 C—H bending absorptions are very close to the standard values of 910 and 990 cm –1 , which are typical of —CH A CH 2 groups. Therefore compound A is (5) . The weak or absent C A C stretching absorption in the 1660–1670 cm –1 region suggest that the candidates for B , C , and E are compounds (2) , (3) , and (4) . The strong 967 cm –1 C—H bending absorption is the definitive absorption for trans alkenes. Therefore, compound E is (3) . Compound C must be the cis alkene (4), from the C—H bending absorption at 714 cm –1 , because the C—H bending absorption for cis alkenes occurs at lower wavenumber than the C—H bending absorption of other alkene types. The 1650 cm –1 C A C stretching absorption as well as the 885 cm –1 C—H bending absorption point to (1) as the structure of compound D. This leaves only compound B unassigned, and it must therefore have structure
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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 12 - Instructor Supplemental Solutions to Problems...

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