Interpretation Introduction To determine: The energies in kilojoules per mole Explanation Given: Infrared radiation with wave length λ=1 .55×10 -6 m Ultra violet radiation with wave length λ=250 nm X-ray radiation with wave length 5 .49 nm Wave length ( λ ): Wave length is a distance between two consecutive crests or troughs of a wave. Frequency ( ν ): Frequency is the number of consecutive crests or thoughts (wave cycle) of that pass a given point in unit time. ν = c λ Formula used: E=hν E=h c λ where, E=energy, h = Plank's constant (6 .626 ×10 -34 J×s) ν = frequency, λ = Wave length c = speed of light (2 .998 × 10 8 m/s) Calculation: Energy for one photon: E = hc λ E= 6 .626 ×10 -34 J× s ×2 .998 × 10 8 m / s 1 .55×10 -6 m E=1 .28×10 -19 J The energy in kilojoules per mole: The energy in joule per photon = 1.28 × 10 − 19 J 1 mol photon = 6.022 × 10 23 photons 1 kJ = 1000 J To calculate the energy in kilojoules per mole we use the following solution: E= 1 .28×10 -19 J 1 photon × 6 .022×10 23 photons 1 m o l p h o t o n × 1kJ 1000 J E=77 .1 k J m o l Convert wavelength in meters: 1 nm = 1×10 -9 m 250 nm × 1×10 -9 m 1 .00 nm =2 .50×10 -7 m Energy for one photon: E = hc λ E= 6 .626 ×10 -34 J× s ×2 .998 × 10 8 m / s 2

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ISBN: 9780135204634

Author: Robinson

Publisher: PEARSON

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Chapter 5, Problem 5.3P

Interpretation Introduction

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The energies in kilojoules per mole

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Chapter 5, Problem 5.2A

Chapter 5, Problem 5.4A

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Part I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400

• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peak

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Ch. 5 - What is the wavelength in meters of an FM radio...Ch. 5 - Conceptual APPLY 5.2 Two electromagnetic waves are...Ch. 5 - PRACTICE 5.3 The biological effects of a given...Ch. 5 - Prob. 5.4A Ch. 5 - Prob. 5.5P Ch. 5 - Conceptual APPLY 5.6 Compare the two elements Rb...Ch. 5 - PRACTICE 5.7 The Balmer equation can be extended...Ch. 5 - APPLY 5.8 (a) What is the longest-wavelength line...Ch. 5 - Prob. 5.9P Ch. 5 - Prob. 5.10A

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