02 - NMR 1 - Last time: UV spectroscopy and degrees of...

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1 Last time: UV spectroscopy and degrees of unsaturation Cubane has only 5: C 8 H 8 2x8=16; 16+2= 18 that is, a saturated hydrocarbon would have C 8 H 18 as a molec.form. 18-8=10 10/2=5 DU’s Halogens take the place of H; O do nothing to m.f., and N subtract 1 from # of H’s Another way: DU= carbons-(H’s+X’s)/2+N’s/2+1 Examples: •C 5 H 12 BrN: 5-13/2+1/2+1=0 DU’s 6 H 6 (benzene) = 4 DU’s 6 H 12 (cyclohexane, e.g.)= 1DU
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2 Nuclear Magnetic Resonance Spectroscopy (Introduction) • NMR depends on nuclei having spin angular momentum (described by quantum number I) • Nuclei with odd mass number or odd atomic number (or both) will have quantized spin angular momentum. The q.n. I determines the number of spin states that are allowed by the nucleus. Spin states allowed are given by the m I (magnetic q.n.) values : -I, -I+1… I-1, I (2I + 1 states allowed) Nuclear Spin 1, 2, 3 … Odd Even 0 e.g. 12 C, 16 O, 32 S Even Even ½, 3/2, 5/2… Even or odd Odd Nuclear spin I Atomic number Mass number
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3 The resonance condition See appendix 1 The spin angular momentum, P, is quantized: its magnitude and direction can only take certain values. m I is the magnetic quantum number P = (h/2 π ) [I(I+1)] 1/2 Pz = (h/2 π ) m I m I = -I, -I+1, … , I-1, I
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This note was uploaded on 10/08/2009 for the course CHEM 439 taught by Professor Falzone,c during the Spring '08 term at Penn State.

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02 - NMR 1 - Last time: UV spectroscopy and degrees of...

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