Lecture+15+-+Spectroscopy

Lecture+15+-+Spectroscopy - Final Exam Thu Sep 8 2011 10:00...

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Unformatted text preview: Final Exam Thu, Sep 8, 2011 10:00 – 11:40 am Chapters 11, 12.1 – 12.7, 14.1, 14.3 – 5 Closed Book and Closed Notes No electronic devices except calculator At least one assigned homework problem will appear on the exam. Review session on Wed Sep 7 at 4:00 pm in 179 Chem Practice Final Exam on website. Electronic Spectroscopy Electronic spectroscopy measures energy of electrons in bonds. UV or visible light excites electronic transitions: (UV – visible absorption) Born-Oppenhiemer Approximation: Electronic transition faster than motion of nuclei: Franck-Condon Principle: Vertical transition from v=0 initial state. Most probable transition maximizes overlap between final and initial vibrational wavefunctions. Franck-Condon Factor (selection rule): ) ,... ( ) ,... ( ) ,... , ,..., ( 1 1 1 1 m vib n el m n R R r r R R r r ψ ψ ψ × = 2 2 τ ψ ψ d S i vib f vib × ∫ = replaces Δ v = ±1 v=0 v=6 v=4 v=0 π π∗ Vibronic Transitions for Benzene v = 0 v = 0 v = 1 v = 2 v = 3 v = 4 0 → 0 → 1 0 → 2 0 → 3 π π∗ π π * Franck-Condon Factor: 2 * 2 ) ( ) ( τ ψ ψ π π d v v S f i × = = ∫ Electronic Transitions in Organic Molecules UV absorption in organic molecules due to strongly allowed π-> π * transitions: ∫ ⋅ ⋅ = ∞ ∞ − τ ψ μ ψ π π d Intensity * Electronic Spectroscopy of Proteins UV absorption spectra of aromatic amino acids Trp Trp Tyr Tyr n n 280 280 280 ε ε ε × + × = pathlength conc protein A × × = 280 280 ε π-> π * ++++++ +++--- ++0--0-+0-+0 ++-++- +-+-+- π π * N OH Protein concentration: The extinction coefficient of a protein at 280 nm can be calculated as ε = n tyr *2000 M-1 cm-1 + n trp * 5500 M-1 cm-1 . The protein calmodulin has three tyrosine and zero Trp. Calculate the protein concentration (in mg/ml) of a solution if A 280 is measured to be 0.1 using a 1 cm pathlength cuvette. ε 280 = 3*2000 M-1 cm-1 + 0 = 6000 M-1 cm-1 A 280 = ε 280 *concentration*pathlength M cm cm M pathlength A ion Concentrat 5 1 1 280 280 10 6 . 1 ) 1 ( 6000 1 . − − − × = × ⋅ = ⋅ = ε 16 μ M l l m mg g mol g M ion Concentrat / 267 . / 267 . 1 16700 10 6 . 1 5 = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = − ε 280 = n Tyr * ε Tyr + n Trp * ε Trp n Tyr = 3 and n Trp = 0 UV Absorption by Peptide Bond ) 2 ( ) 2 ( ) 2 ( ) ( 13 12 11 1 y N y C y O p c p c p c ψ ψ ψ π ψ + + = ) 2 ( ) 2 ( ) 2 ( ) ( 23 21 2 y N y C y O p c p p c ψ ψ ψ π ψ − + = ) 2 ( ) 2 ( ) 2 ( ) * ( 33 32 31 3 y N y C y O p c p c p c ψ ψ ψ π ψ + − = ) ( ) ( ˆ π ψ π ψ n n n E H = n → π * π → π * λ (nm) Absorbance π → π * λ = 200 nm n → π * λ = 230 nm O 2p x UV Absorption by Nucleotide Bases...
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This note was uploaded on 09/16/2011 for the course CHE 107B CHE 107B taught by Professor Ames during the Summer '11 term at UC Davis.

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Lecture+15+-+Spectroscopy - Final Exam Thu Sep 8 2011 10:00...

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