# Ch14 - 296 S olutions Chapter 13: Nuclear Magnetic...

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PrQ 296 Solutions Chapter 13: Nuclear Magnetic Resonance Spectroscopy Problem'13.28 Sketch the NMRspectrum you would expect from a partial molecule with the following parameters, H. = LO ppm H b =3,0 ppm He = 6,0' ppm Jab = 5,0 Hz J be = 8,0 Hz Joe = 1,0 Hz The J ae coupling of 1 Hz is so small that it can be safely ignored since it will not show up on normal resolution spectra. Assuming the R groups are not involved with signal splitting, expect the signal for H at 1.0 ppm, a integrating to a single hydrogen atom, to be split into a doublet with coupling constant Jab = 5.0 Hz. Expect the signal for Heat 6.0 ppm, integrating to two hydrogen atoms, to be split into a doublet with coupling constant he = 8.0 Hz. Expect the signal for H b at 3.0 ppm, integrating to a single hydrogen atom, to be split into a doublet of triplets with coupling constants Jab = 5.0 Hz and,he = 8.0 Hz. The following is drawn as expanded views around the signals. ' .................... ~ I I ,6.0 ppm 3.0 ppm 1.0 ppm Note that if the J ae = 1.0 Hz coupling is seen, the signal for H a will be split into a doublet of 'closely spaced triplets, and the signal for He will be a doublet of closely spaced doublets. _ 8Hz He H a ' , lit H b ~...u1.l.LL~ ~Jul.~ I I I 6.0 ppni 3.0 ppm, 1.0 ppm Solutions 297 Cbapter 14: Mass Spectrometry CHAPTER 14 Solutions to the Problems 14 I 'Calculate the nominal mass of each ion, Unless otherwise indicated, use the mass of the most abundant hIem ' iSQtQpe Qf each element. + (a) [CH 3 Br] . , + (b) [CH 3 81 Br]' 94 96 95 £.rnblem 14,2 Propose a structural formula Jor the cation Qf m/? 41 observed in the mass spectrum Qf methylcyclopentane, The c!ltion of'mlz 41 must have a molecular formula of C 3 H s . This corresponds to the stable allyl cation: CH 2 =CH-CH 2 + Emblem 143 The low-resolutiQn mass spectrum of 2-pentanQI shows IS peaks, AccQunt fQr the formation of the peaks at m/: 73, 70, 55, 45, 43, and 4], , 2.Pentanol has a molecular formula' ofCsH [20 and thus a nominal mass of 88. OH I CH 3 -GH 2 -CH 2 -CH-CH 3 2·Pentanol Loss of a methyl radical would leave a fragment of mass 73: +OH , II , CH 3 -CH 2 2 -CH Loss of water results in a fragment of mass 70: _, [CH~-CH2-CH2-CH=CH2 ] + . [ CH 3 -CH 2 -CH=CH-CH 3 ]+ " The 'above alkene radical cations could then lose a methyl radical to give the following fragments of mass 55: [CH 2 -CH 2 -CH=CH 2 ] + 2 3 ] + An alkyl radical could break off from the alcohol to give a fragment of mass 45: +OH II CH-CH 3 The original alcohol cation radical could have fragmented in such as way as to generate the following cation with mass 43: The allyl cation has a mass of 41: CH 2 2 + Problem 14.4 Dr~w acc~ptable Lewis structures fQ~ the molecular ion (radical cation) formed from the following molecules when each is bombarded by high-energy electrons 111 a mass spectrometer. , H I H........ ...... C;t. ...... H (a) o C III . . . . C, rC, H C H I H

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298 Solutions Chapter 14: Mass Spectrometry H H / \ H-C 'C-H' ~ + / (b) C-C / '\ H H '0 H :0' H . ~ / \ / ' (c) C-:-N< C~N: / \ / + \ H H H H (d) H-C=C-H ..
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## This note was uploaded on 07/09/2011 for the course CHE 301 taught by Professor Diver during the Spring '08 term at SUNY Buffalo.

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Ch14 - 296 S olutions Chapter 13: Nuclear Magnetic...

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