Intensity (PA) Intensity (PA) 400000 350000 300000 250000 200000 150000 100000 50000 0 -50000 700000 600000 500000 400000 300000 200000 100000 0 -100000 0 0 0.5 0.5 1 Reaction of 1-hexene with HBr 1 1 2 1.5 2 Reaction of 1-hexene with HBr Peak Number Retention Time (min) Peak Area %) 2.231 48.611 3.428 51.389 1 2 3 1.5 2.5 Time (min) Reaction of 3,3-dimethyl-1-butene with HBr 2 2.5 3 Time (min) Reaction of 3,3-dimethyl-1-butene with HBr Peak Number Retention Time (min) Peak Area %) 2.214 2.967 3.876 3 26.996 23.451 49.553 3.5 3.5 4 4 4.5 4.5

As solvent polarity increases, the Retention Factor (Rf) of compounds on the TLC plate increases. Group of answer choices True False

Table 1: Selected phytocomponents identified in the plant extract. Retention time, min Name of Compound   10.767 Benzaldehyde   15.790 Azulene   16.002 1-Piperazinecarboxaldehyde   17.935 Phthalan   20.731 2-methoxy-1,3,4-trimethybenzene   24.028 Methylesterpentanoic acid   26.993 Tetradecamethylcycloheptasiloxane   31.753 Hexadecamethyl Cyclooctasiloxane   If GC is combined with MS, what data you will add as the 3rd column in Table 1 above.  Fill in that data for each of the select identified phytocomponents.

3) 3-Methylbutyramide 100 relative intensity .8888 relative abundance .88 CHINO 101 298 888 mass charge (m2) CaH₁N 60 4) n-Butylamine 100 60 CHOCHOCHS R98 8 mass charge (m/z) MW=101.15 MW-73.13

Retention Time 40 20 25 Mntes VWD: Signal A, 284 nm Retention Area Identity Time (min) 7.867 3214697 Vitamin A 8.493 15687 19.410 140354 19.710 64346 20.803 884315 Vitamin E 21.640 36745 22.100 63054 What is the %composition of the Vitamin E peak in the sample? O 21.57% O 20.01% O 22.27% O 20.80% E8.403 89%花

use the following chromatogram and table as your data. VWD: Sgrad A 284 nm Retention Time 40- -40 20 20 5 10 15 20 25 30 35 Minuteo VWD: Signal A, 284 nm Retention Area Identity Time (min) 7.867 3214697 Vitamin A 8.493 15687 19.410 140354 19.710 64346 20.803 884315 Vitamin E 21.640 36745 22.100 63054 8.403 L98 ÞI 20.803 ; 84年 81

1) Study the chromatograph (below) of a mixture of Compounds A and B, run on the GCs in the teaching labs at CU Boulder. Compound A has the shorter retention time. STAAT 61 1.11 227 RT TYPE AREA XXXX XXXX XXXX AREAS 0.009 55874 44.117 ARIHT 0.61 XX XX XX 1.11 2.27 XX XX XX TOTAL AREA=XX MUL FACTOR=XX 1. What is the retention time of compound A? Compound B? 2. Which compound is present in a larger amount? 3. Which compound has the lower boiling point? 4. What would happen to the retention times of compounds A and B if the column temperature were raised? 5. You suspect that compound B is octane. What can you do to provide supporting evidence for this hypothesis?

C3H4O 11 10 HSP-00-252 9 - 00 8 7 6 5 ppm ₁2H 4 3 1H 1H N

6. CH₁00 11 HSP-04-4492 LOD D 10 4000 9 5000 8 T 5 7 T 6 T 2005 5 ppm VES 4 2 2 T 3 1309 1 T 2 1 1000 0 580 200 180 160 140 CDS-06-431 DEPT-90 DEPT-135 2 422 peaks 120 100 pom 80 60 40 20 0

In a Bradford assay, 11 µl of a protein isolate sample was diluted by adding 21 µl of water prior to the addition 2.0 mL Bradford reagent. Calculate the protein concentration (in mg/mL) of the original protein isolate sample whose absorbance reading at 595 nm is 0.304 after dilution. The protein standard solutions used for the calibration curve of the Bradford assay was constructed using bovine serum albumin (BSA) stock solution whose concentration is 200 µg/mL. The concentration and the corresponding absorbance reading at 595 nm of the protein standard solutions are summarized in the table below: Standard soln # 1 2 13 4 6 conc of BSA 12.00 24.00 36.00 48.00 60.00 (pg/mL) Absorbance at 595 nm 0.000 0.156 0.249 0.328 0.411 0.528 zNote: Final answer format must be x.xxx (three decimal places). Round off only in the final answer. Do not round off in the middle of the calculation.