Thermo Xcalibur Qual Browser- injection101 150512200336 [injection101 1505122003

Question

Thermo Xcalibur Qual Browser- injection101 150512200336 [injection101 150512200336.raw] File Edit View Display Grid Actions Tools Window Help injection101_150512200336 05/12/15 20:03:36 MS RT: 3.71 8.76 LILI injection 101150512200336 no file no file no file no file no file no file no file 100 90 80 70 60 55E9 TIC MS njection101 150512200 336 7.49 G 40 30 20 10 5.60 4.02 8.04 5.85 5,97 6.16 6.38 6.61 6.65 6.81 7.01 7.07 7.16 7,167,29 |7,50 7.79 7.29 3.82 3.994.06 4.35 4.44 4.55 4.79 4.88 5.00 5.18 5.40 8.10 8.34 8.56 3.8 4.04.24.44.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0 6.2 6.6.6 6.8 7.0 7.2 7.4 7.6 7.88.08.28.4 8.6 Time (min) injection101-150512200336 #1601-1615 RT: 7.84-7.89 AV: 15 NL: 6.57E8 T: c El Full ms [50.00-450.00 143.09 100 90 80 70 O 60 115.08 o 40 257.15 187.05 30 20 10 55,03 7101 116.11 144.13 170.10 89.05 113.05 188.11 258.18 259.19 281.07 302.12 156.10 171.12 189.11 214.10 242.12 329.06 355.04 378.20 399.19 407.96 429.16 437.27 60 80 100 120 140 160 180 200 220 240 260 280 300 340 360 380 400 440 Mass: 70.72, Intensity: 6.15e+008, Scan Filter: (none) NUM

Explanation / Answer

The interpretation of Mass spectrum is as below, the assigmnment is with respect to MS peaks,

257 is assigned to molecular ion peak for the addition product formed from 1-naphthylurethane and the alcohol

187 is assigned due to [M+1-71] loss of (CH3)2CH(CH2)2 from the above addition product

143 is assigned due to [M+1-115] loss of (CH3)2CH(CH2)2OCO from the bove addition product

115 is assigned due to fragment formed right above (CH3)2CH(CH2)2OCO from the bove addition product

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