Your Name: Answer Key Question 1. Standard Fragmentations in Mass Spectrometry. (20 points)

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1 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key Question 1. Standard Fragmentations in Mass Spectrometry. (20 points) For (b) (d), draw the complete structure of the substrate (all atoms, lone pairs, etc), draw the complete structure of the EI-generated radical cation, and draw curved arrow mechanisms for the fragmentations. (a) Predict the intensities of the relative intensities of the peaks for M (:= 100%), M+1, M+2, M+3, and M+4 of 1,6-dichloro-1,3,5-hexatriyne 6 l 2. Assume that the 13 abundance is 1.1% and that the abundance of 35 l is three times that of 37 l. Show your work and try to be very, very organized. For future reference: M+1 = [ *6 *100%] = 6.6% M+2 = [ *6*5 *100%] + [ *2 *100%] = 0.36% % = 67.0% M+3 = [ *6*5*4 *100%] + [ *2 * 0.011*6 *100%] = 0.016% % = 4.38% M+4 = [ *6*5*4*3 *100%] + [ *2 * *6*5 *100%] + [ *1*1 *100%] = 0.0% % % = 11.13% (b),-dimethylformamide (EI). Explain peaks with m/z 73, 58, 44 and omol m/z = 73 omol. m/z = 58 m/z = 28 3 eterol m/z = 73 m/z = 44 m/z = 28-1-

2 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key (c) 3-itrobenzyl Alcohol (positive ion FAB). Explain the peaks with m/z 154, 136, and 289. [The m/z values are correct.] ionized: m/z = 153 protonated: m/z = 154 [M+- 2 ] + m/z = 136 M forms complex with [M+- 2 ] + m/z = 289 (d) 5-Decyne, 3 -( 2 ) 3 - -( 2 ) 3-3. Explain the formation of the base peak, m/z 54. ote = 84; start with 10 18, loose 6 12, form 4 6. The trick is to notice that -cleavage(s) alone cannot give the 4 6 -fragment, i.e., center piece plus n 2 -units. Two -shifts are needed. -- Adapted from book by Pretsch et al Shift m/z = Shift

3 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key Question 2. Spectra of a MALDI MS Matrix ompound 3-Aminoquinoline. (20 points) (a) Find the MS spectrum of 3-aminoquinoline in the IST database, print it, and attach it to this page (staple). What ionization method was used to generate the IST spectrum? List the three major peaks and outline a detailed fragmentation mechanism consistent with the MS spectrum. IST shows the EI spectrum. Spectrum provided on the following page in JAMP-DX format. M + : m/z = 144 [M-] + : m/z = 143 [M+] + : m/z = 145 [M-] + : m/z = [M-] + : m/z = 117 [M-- 2 ] + : m/z = 89 [M-2] + : m/z = 90-3-

4 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key The spectrum in JAMP-DX format (from IST): ##TITLE=3-Quinolinamine ##JAMP-DX=4.24 ##DATA TYPE=MASS SPETRUM ##RIGI=Japan AIST/IM Database- Spectrum MS-W-2595 ##WER=IST Mass Spectrometry Data enter ollection () 2007 copyright by the U.S. Secretary of ommerce on behalf of the United States of America. All rights reserved. ##AS REGISTRY = ##$IST MASS SPE = ##MLFRM=982 ##MW=144 ##$IST SURE=MSD ##XUITS=M/Z ##YUITS=RELATIVE ABUDAE ##XFATR=1 ##YFATR=1 ##FIRSTX=17 ##LASTX=146 ##FIRSTY=2 ##MAXX=146 ##MIX=17 ##MAXY=999 ##MIY=1 ##PITS=73 ##PEAK TABLE=(XY..XY) 17,2 26,2 27,3 28,12 32,1 37,6 38,9 39,27 40,5 41,10 42,4 43,2 44,3 45,5 49,2 50,18 51,17 52,11 53,3 54,3 56,1 57,5 58,21 59,2 61,7 62,21 63,51 64,15 65,10 66,6 67,3 71,4 72,28 73,1 74,8 75,11 76,15 77,12 78,9 79,3 85,3 86,5 87,8 88,14 89,164 90,128 91,16 92,1 93,1 99,2 100,3 101,7 102,5 103,6 104,16 105,2 114,8 115,12 116, , ,31 119,2 127,9 128,6 129,8 130,5 131,5 141,1 142,14 143, , , ,5 ##ED= -4-

5 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key (b) The MALDI MS spectrum of 3-aminoquinoline appears in the text by Pretsch et al. on page _380_ and it contains peaks with m/z 145, 289, and 433. Draw reasonable structures of the ions that cause these peaks. M + : m/z = 144 [M+] + : m/z = 145 [2M+] + : m/z = [3M+] + : m/z = These are possible structures. There are other possibilities. The point is simply that MALDI is so soft as to allow non-covalent aggregates. -5-

6 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key Question 3. onjugation Effects on UV/Vis Absorptions. (20 points) (a) Draw all π-ms of ethene, buta-1,3-diene, and hexa-1,3,5-triene below. Make sure to indicate nodes clearly. Label the Ms of each molecule starting with the most binding M. Indicate M and LUM for each alkene. (b) Draw the energy levels of the Ms and LUMs of ethene, butadiene and hexatriene. Explain the effect of conjugation on the energy of absorption. LUM pi2 pi6 Energy M pi1 pi5 pi4 pi4 0 pi3 pi3 pi2 pi2 pi1 pi1-6-

7 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key Question 4. UV/Vis Spectroscopy of an Indicator Dye. (20 points) Methyl range, Me = S 3 a (all para), is used an acid-base indicator that changes color in the p range The color in an alkaline medium is YELLW and absorption occurs in the region(s) 425 nm. The color in acid media (p < 3.2) is RED and absorption occurs in the region(s) 490 nm. Structure if p < 3.2: [areful, this is a bit more interesting than it looks at first.] Structure if p > 4.4: Me Me Me S 3 - Me S 3 - Explain the p-dependence of the color of Methyl range. igh p: Appear yellow, absorb indigo-blue. Pretsch says 425 nm. Low p: Appear red, absorb blue-green. Pretsch says 490 nm. Spectra can be found in J. Sol. hem. 2006, 35, Yellow dye absorbs at 437 and 475 nm. Red dye absorbs at 507 nm. Pretsch is pretty close. The anion has a donor at one end and an acceptor at the other and the conjugation can go over the entire molecule. The neutral compound contains two shorter DA systems (amine-sulfonate and dimethylamine-aminium). For systems with the same donors and/or acceptors, the system with the longer conjugation length wins. ere is a case where the conjugation length is not so important. nce protonated, the conjugation over the entire molecule is no longer possible. But the dimethylamino-aminium system is a very good DA-system. The amino-sulfonate system also is better (amino is a better donor than azo). -7-

8 MU, hemistry 8160, FS08, Dr. Glaser Your ame: Answer Key Question 5. AT-Sensitive Fluorescence Probe. (10 points) The 6-aminoacridizinium bromide (1a) has been shown to serve as a fluorescence probe which lights up in AT-rich regions of DA (rg. Biomol. hem. 2003, 1, ). Read the article. Explain why 1a exhibits fluorescence enhancement. Explain why 1b - 1d do not. [Structural drawing are required with your explanation.] The enhancement of the fluorescence emission of 1a in AT-rich DA is thought to result from selective electron-transfer reactivity towards the base pairs (rather than from selective binding to AT-rich regions). sugar Adenine Thymine sugar 2 Will accept any reasonable explanation as to why 1a allows for intercalation while the others do not. The important logical item is simply that 1a needs to be able to get one ring to stack with A in A=T. You do not even need to intercalate the entire aromatic system. Approach from the major groove and just get one ring over the A s amine, for example, and A s oxidation can proceed. 1a s amino group might help by way of -bonding with A s imidazole; T s carbonyl is another option. sugar Adenine Thymine sugar -8-