Ch13 - 280 S olutions Chapter 13 Nuclear Magnetic ResonanCe...

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280 Solutions, Chapter 13: Nuclear Magnetic ResonanCe Spectroscopy CHAPTER 13 Solutions to the Problems, Problem 13.1 Calculate the ratio of nuclei in the higher spin state to those in the lower spin state, NhIN!, for 13C at 2YC in an applied field strength of7.05 T. The difference in energy between the higher and lower nuclear spin states in this applied field is approximately 0.030 J (0.00715 cal)/mol. ' The important equation relates the change in energy oftwo spin states to their equilibrium concentrations: ~Go=-R11n ~ Nt Rearranging this expression in terms of Nh/N, gives: ~ _ -~Go bt Nt - RT Substituting in the appropriate values for R (8.314 Jedeg·temol- I ), T (298 K) and ~Go (0.030 Jemol- I ) gives:. " ·1 ~ - - 0.030 Jemol , 1 21 10- 5 in N, - (8.314 Jedeg-lemorI)(298 deg) = -. x 1.0000000 ~~ = 0.9999879 = 1.0000121 Problem 13.2 State the number of sets of equivalent hydrogeris in each compound and the number of hydrogens in each 'set. (a) 3-Methylpentane , Numbers have been added to the carbon atoms of the structures to aid in referring to specific hydrogens. Use the "test atom" approach if you have trouble understanding the answers. 50 10 2® 3y H 34 ® CH 3 -CH 2 -CH-CH 2 -CH 3 , @ There are four sets of equivalent hydrogens. Sf1.a;. 6·hydrogens from the methyl groups of carbon atoms 1 and 5. Sll.1L. 4 hydrogens from the -CHz- groups of carbon atoms 2'and 4. ~ 3 hydrogens from the methyl group of carbon atom 5. ~ 1 hydrogen from the -CH· group of carbon atom 3. .. I. There are four sets of equivalent hydrogens. Sftal9 hydrogens from the methyl groups of carbon atoms 1, 7, and 8. Sll.1L. 6 hydrogens from the methyl groups of carbon atoms 5-and 6. Sll.c. .:. 2 hydrogens froIJ:l the -CHz- group of carbon atom 3. Sll. .d. .:. 1 hydrogen from the -CH- group of carbon atom 4. Chapter 13: Nuclear Magnetic Resonance Spectroscopy Solutions 281 £.c.obl em 13.3 Each ,:ompound gives only one' signal: in its IH-NMR spectrum. Propose a structural formula for each compound- I order for these molecules to give a single absorption peak, each of the hydrogen nuclei must be in an i~entical environment. This will only occur in symmetrical mole~ules. , ' (b) CSHIO (c) CsH 12 H H \ / H" /C" /H C C H........ \ / . ..... H . . . . C-C---- H H I I H H (d) C.Jl6 C1 4 CH 3 -CCI 2 -CCI 2 3 problem 13.4 The line of integration of the two signals in the IH-NMR spectrum of a keto~e.with the molecu~ar formula ~H140 rises 62 and 10 chart divisions, respectively. Calculate the number of hydrogens giVing nse to each Signal, and propose a structural formula for this ketone_ The ratio of signals is approximately 6:1, which corresponds to a 12:2 ratio of the 14 hydrogens. Thus, the larger signal represents 12 hydrogens and the smaller signal represents 2 hydrogens. A structure consistent with this assignment is 2,4-dimetbyl·3-pentanone as shown below: ' Larger Signal I \ -8~S Smaller Signal -- CBfC-':"C-C~ --- Smaller Signal ,~ 8 \ I Larger Signal Problem 13.5 Following are two constitutional isomers of molecular foimula C 4 HgO Z - ° II ° II CH 3 CH 2 0CCH 3 CH3CH2~OCH3 (1) (2) (a) Predict the number of signals in the IH-NMR spectrum of each isomer.
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Ch13 - 280 S olutions Chapter 13 Nuclear Magnetic ResonanCe...

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