MCMP 205 instructor_supplement_15

# MCMP 205 instructor_supplement_15 - Instructor Supplemental...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 15 Dienes, Resonance, and Aromaticity Solutions to In-Text Problems 15.2 The delocalization energy is the energy of each MO times the number of electrons in that MO, minus the same energy for three ethylenes: delocalization energy = 2(1.80 b ) + 2(1.25 b ) + 2(0.44 b ) – (3)(2)(1.00 b ) = 0.98 b 15.5 (a) The two enantiomers of an allene: (b) Because enantiomers have specific rotations of equal magnitudes and opposite signs, the other enantiomer has a specific rotation of +30.7°. 15.6 (b) The calculation is identical to that in part (a) with different numbers: E = hc l = (3.99 ± 10 –13 kJ s mol –1 )(3.00 ± 10 8 m s –1 ) 250 ± 10 –9 m = 479 kJ mol –1 (The corresponding value in kcal mol –1 is 114.) 15.7 (b) Use Eq. 15.1 on text p. 685 with ( I 0 / I ) = 2. Thus, A = log (2) = 0.30. 15.8 The piece with greater absorbance transmits less of the incident radiation. Therefore, the thick piece of glass has greater absorbance. 15.9 (b) The absorbance of the isoprene sample in Fig. 15.5 on text p. 684 at 235 nm is 0.225. With the concentration determined from part (a), Beer’s law gives A = 0.225 absorbance units = e (7.44 ± 10 –5 mol L –1 )(1 cm), or e = 3.02 ± 10 3 absorbance units L mol –1 cm –1 . Another way to determine the extinction coefficient at a different wavelength is based on the fact that the ratio of absorbances at different wavelengths equals the ratio of the extinction coefficients. Hence, the extinction coefficient at 235 nm is e 235 = e 225 A 235 A 225 = (10,750 absorbance units L mol –1 cm –1 ) 0.225 absorbance units 0.800 absorbance units ± ² ³ ´ µ or e 235 = 3.02 ± 10 3 = 3023 absorbance units L mol –1 cm –1 . 15.10 (a) The two alkyl substituents contribute +10 nm to the base l max of 217 for a predicted l max value of 227 nm.

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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 3 2 15.11 (b) 15.12 (a) An analysis like that employed in Study Problem 15.1 (text p. 692) suggests two possibilities. Pair A is preferred because, in many cases, the most reactive dienophiles are those with conjugated electronegative substituents. But if your answer was pair B , you have analyzed the problem correctly. 15.13 (a) With 1,3-butadiene as the diene and ethylene as the dienophile, the product would be cyclohexene. 15.14 (b) As in part (a), two possible orientations of the diene and dienophile lead to the following two possible constitutional isomers: 15.16 The triene contains two diene units with one double bond common to both. The dienophile reacts with the diene unit that is locked in an s -cis conformation. 15.17 (b) The two products correspond to the two possibilities in Eq. 15.14 on text p. 699. They result from addition of the diene at either of the two faces of the alkene (or the alkene at either of the two faces of the diene).
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 3 3 Notice that both stereoisomers result from

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MCMP 205 instructor_supplement_15 - Instructor Supplemental...

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