hw9 - - Chemistry 2311 Law Section I: Circle the number...

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Unformatted text preview: - Chemistry 2311 Law Section I: Circle the number which corresponds to the answer to each of the following questions There is only one correct answer for each question. 1. Which C4HgBr compound gives aflflgtapproximately 3.5 ppm in the 1H NMR spectrum? 3 Br M /Y Br )4 Br I 3 Br ’ 2 3 4 1 ‘r. V. i r . -2 {Nu 1* €11 .3 .4 noun: 2. Which feature in the 1H NMR spectrum provides information about the electronic environment of the protons in a compound? a. number of signals b. integral c. splitting . 'dichemical shift 3. What is the principle that allows us to use mass spectrometry to determine the molecular weight of a compound? a. higher molecular weight compounds are less volatile b.,,higher molecular weight compounds are more dense . a‘beam of higher molecular weight cations is deflected less by a magnetic field d. ions with higher molecular weight absorb higher frequencies of light 4. Which of the following is not true regarding 1H NMR spectroscopy? \. affa’“downfield” peak appears at a lower value of 6 b. on a 300 MHz instrument, a proton that adsorbs irradiation at a frequency 900 Hz higher than the adsorption of TMS appears at 6 3 ppm. 0. 6 for a particular proton is independent of the magnetic field strength of the instrument used. d. “deshielding” leads to peaks at higher values of 6 5.\Which of the following is the definition of the base peak of a mass spectrum? 5 'a. the peak corresponding to the most abundant ion “fof'the peak corresponding to the ion with lowest m/e peak corresponding to the molecular ion peak d. the peak corresponding to the ion arising to loss of a proton from the molecular ion 6. Which C4HgBr compound(s) gives a 130 NMR spectrum consisting of four signals? 2 Br /\/\Br x/Br )4 Br 1 1 3 a. only 1 g/bii‘only 1 and 2 \cfonly 2 and 3 d. 1, 2, 3 and 4 10.. Which CQH1oO compound gives the following 1H NMR spectrum? 3H 2H 2H 3 PPM O O H / /\/ In, LL 0 I P ,‘g H p “I, | O __. / “K {2 \3'; “2:: a. 1 ¥”/ xgbfl 0' 3 A] 1 i’g;¢l// k f? “a d. 4 Section II. For the following problems, draw the structure that corresponds to the preceding spectra. c l A. CsHloO, MW = 86 Own) v ma— «— a , #73 J W ‘3‘ ' 1 L ‘ l l w,- w ‘l A . I. 1 k e. E = g “" “‘“ g l i ~ I x m f .... _, fl I i: ..._.. .sl— .. g I § ac “_M__..A .... .. _..;____ ....... .. l g I " l 3"“ ‘WflWT """ "Bl l ! Tastfi '1‘"?- '1“ T 1 Ilquld um.» um pm“ W Cwloh? 19v: 809-0 B. C6H14O, = . m m M M i. W E, .9 [+1 whom ” M 1365 ram W. i II I Elfin? Shzkfisfll, r E x a M \\\ v \ u L) 13.. “1} v3.33 ii: 3!;3, ,itii ineligfix. “ w w _ M m . M W H w N wit :Jlg: 7.5T «IIJ‘i;WslmVixnlJ|4 T I! 1,? n m N w a 1 i 258? 7 ' magmas MA!" E i 3533 can 1 law H In buka so" .. E g a- woe 0.5 1.0 PW 4 2.5 2.0 3.0 5:5 4 ‘ o I‘m!st NMR C. C8H14, MW :110 ...
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This note was uploaded on 12/01/2009 for the course 123 123 taught by Professor Alpha during the Spring '09 term at Georgia Institute of Technology.

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hw9 - - Chemistry 2311 Law Section I: Circle the number...

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