CHEM 3331 1nd Edition Lecture 27Outline of Last Lecture I. Mass SpectrometryII. Electromagnetic spectrumIII. Infrared SpectroscopyOutline of Current Lecture I. NMR SpectroscopyII. Number of different peaksIII. Positions of peaksIV. Peak areaV. Peak splittingCurrent LectureI. NMR SpectroscopyNMR stands for nuclear magnetic resonance. It analyzes the nuclei of H1, C13, N15, F19, and P31. Today we will only be focusing on hydrogen. In all of these atoms they have an odd number of protons or (neutrons+protons). They experience a magnetic spin where they rient themselves inan external magnetic field. There are α and β states. Β is high energy and α is low energy. ΔE=Ɣ(h/2π)B0 Ɣ= gyromagnetic ratio, B0= magnetic fieldB=B0- Bshielding shielding= 50-100Hz measured in ppm. II. Number of different peaksThe number of peaks= the number of kinds of H. CH4=1,CH3CH3=1, CH3CH2CH3=2, H2C=CH2=1, H2C=CHF=3, CH3OH=2, benzyne=2. These hydrogens differ or rather don’t differ due to symmetry, being peripheral or internal, relationship to F, Br, Cl, and I.III. Positions of peaksThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.This depends on shielded or deshielded. The scale goes from 0-12 but runs backwards. The deshielded hydrogens are closest to twelve while the shielded H are closer to 0. We use TMS as our 0 point because it is the most shielded molecule. Chemical shift (δ)= difference from TMS peak (Hz)/ frequency of radiowaves (MHz)δ compound4.3 CH3F3.4 CH3OH3.0 CH3Cl2.7 CH3Br2.2 CH3I1.23 CH3CH2OH0.94 CH3CH2CH2OH3.0 CH3Cl5.3 CH2Cl27.2 CHCl3Alkanes= 0.9-1.4Alkenes= 5-6Alkynes= 2.5-3.5Carbonyl= 2.41IV. Peak areaCH3C=OCH2OC(CH3)3= 3 peaks. 3:2:9. The relative areas under the peaks are known as integrals and are equal to the ratio of different H.(CH3)2CHOCH(CH3)2= 2. 12:2--- 6:1CH2CH2OCH2CH3= 2. 6:4----3:2-----1.5:1V. Peak splittingPeak splitting is determined by the number and kinds of neighbors an H has next to it.There is the n+1 rule.3 neighbors = 4 peaks split2 neighbors= 3 peaks split0 neighbors= 0 peaks splitWe can use pascals triangle to determine the integrals of the areas of each peak in the peak split. The peak split ratio depends on the possibilities of spin either β or α. It is always symmetric.(CH3)2CHCl has two peaks. However, these peaks are split. Each CH3 is acted upon by the CH so they both have two 1:1 peaks. The CH is acted upon by the two CH3 molecules so it has seven peaks with a 1:6:15:20:15:6:1 ratio. When a H is acted upon by two or more neighbors that are not of the same kind then the peas are split differently. They are split one at a time.For example a benzene with Cl and Br on 1 and 3. Each hydrogen is different. We will label them H1-H4 starting with the hydrogen between Cl and Br working clockwise. H3 has two neighbors H2 and H4 but they are not the same because H2 is closer to Br and H4 is closer to Cl. We will have H4 act on H3 first creating a doublet. Then H2 will split both of these peaks into two. Creating a double doublet with a 1:1:1:1 ratio. This is different from a quartet which is a 1:3:3:1 ratio and the space in between peaks is much larger for the double doublet than a quartet.Do some practice problems to really get the hang of
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