CFQ & PP: Infrared Spectroscopy 97 Reading Brown and Foote: Chapter 12, Section 18.3A Lecture Supplement Infrared (IR) Spectroscopy (page 18 of this Thinkbook) Suggested Text Exercises Brown and Foote Chapter 12: 3 – 12 Optional Interactive Organic Chemistry CD and Workbook Supporting Concepts: IR Spectroscopy Tutorial (p. 76) Spectroscopy: IR (p. 58) Optional Web Site Exercises Webspectra (http://www.chem.ucla.edu/~webspectra)Practice Five Zone analysis. Optional Software IR Tutor (Available on the computers in the Science Learning Center, Young Hall, 4 th floor) You do not need to memorize the typical stretching frequencies for each functional group, as given in the lecture handout. This data will be given on an exam if needed. However you are expected to be intimately familiar with the distribution of functional groups in the Five Zone Analysis. Concept Focus Questions 1. Briefly explain the molecular events that result in an infrared spectrum. 2. What molecular structure features control the intensity of an infrared absorption? 3. Explain why similar functional groups absorb infrared photons of similar energies. 4. Briefly explain the Five Zone approach to analysis of an infrared spectrum. 5. The infrared absorption bands of most common functional groups have characteristic features (e.g., two absorptions, a broad peak, etc.) in addition to the photon energies. Briefly describe the important examples of this in the Five Zone analysis. 6. What is the effect of conjugation on the energy of an infrared absorption band? 7. What is the fingerprint region of the infrared spectrum? Why is it usually ignored?
98 Concept Focus Questions Solutions 1. Absorption of an infrared photon results in the excitation of the molecule to a higher vibrational quantum state. 2. For a stretching vibration, a photon will be absorbed and the molecule excited to a higher vibrational quantum state only if that vibration results in a change in bond dipole. The bond dipole is a product of bond length and charge difference of the bonded atoms. As a bond vibrates, the bond length changes, so this criterion is met. The charge difference is determined by the electronegativity of the bonded atoms. If these atoms are not identical then they have a difference in electronegativity and thus a difference in charge. If these atoms are identical, they will have equal electronegativity and thus no charge difference. In this case, the product of bond length change and charge difference (zero) is zero, so no photon is absorbed. If the atoms are even slightly different, then a small change in bond dipole will occur, along with the corresponding absorption of an infrared photon. Thus, all of the bond stretches of H 2 C=O will show up on the IR spectrum because all of the bonds consist of unequal atoms. For H 2 C=CH 2 , the C-H bond stretches will give IR bands, but the C=C bond (which is made up of two identical carbon atoms) will not. The C=C
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