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- SPECTRA FOR HOMEWORK 11, CHE 230 002 This page is not to be submitted to Gradescope. Use these spectra as you answer questions on the Homework 11 document. Spectra for Problem 1 100 %T 80 60 40 20 Relative Intensity 4000 100- 80 8 60- 40- 20- 0 20 2H 2H 3000 m/z 50.0 75.0 76.0 155.0 157.0 183.0 185.0 212.0 214.0 100.0 97.5 11.8 11.7 1686 cm 40 60 80 2000 rel, intensity 11.9 16.9 17.6 28.8 27.8 Wavenumber[cm-1] 100 120 m/z 1500 7.90 7.85 7.80 7.75 7.70 7.65 7.60 7.55 1588 cm² 20 20 ₂ مسلسل..............للمسلسل ppm 140 160 180 200 220 2H 1000 3.12 3.00 2.95 2.90 3H 500 400 A. Propose a molecular formula for this compound B. Propuse. a structure selected 1.25 1.20 1.15 201Each of the molecules shown below generates three different HNMR signals. Using the chemical shift differences, the relative peak intensities, and the splitting pattern of these signals, match each molecule to one of the HNMR spectra given below. HO, (a) (b) (c) Spectrum I 3 2 PPM9 What compound has a formula of C5H120 * ?and a peak at 3300 cm-1 (ähäi 1) Ether Alcohol Aldehyde Ketone
- In order to measure riboflavin in a breakfast cereal, a 5.00-gram sample of the cereal was extracted with acetic acid and filtered. The extract was transferred to a 50.0 mL volumetric flask and diluted to the mark. The sample was transferred to a cuvette and the fluorescence was measured. The instrument read 59 units (Fluorescence units –an arbitrary scale). 24.00 mL of the cereal extract was sample was mixed with 1.0 mL of a riboflavin standard that was 0.500 ug/mL. The fluorescence of this mixture was measured and read 94 fluorescence units. Calculate the concentration of riboflavin per gram of cereal (use µg/gram of cereal).Explain what can be determined from the following mass spectrum. M* = 123 Relative Intensity 100- 80 9 20 0-mmm 25 50 wytrwytatyt||||||qqıbıkm 75 m/7 100 125Following the Woodward-Fieser's rules, determine the Amax for the following molecule: X 382 nm 356 nm 378 nm 371 nm O
- One QN not allowed. Replace not allowed with one that is allowed? a) n =3; l =3; ml = +2 b) n=2; l=1; ml = -2; c) n = 1; l=1; ml = 08 11) Chemical Formula: C₁0H12O2 IR: strong peak 1720cm-¹ 10 1H 8 12) Chemical Formula: C9H200 IR: strong broad 3300cm-1 2H 2H PPM 10x2= 20+2=22=12=10/2=5 6 PPM septet 1H 9x2=18+2=20-20=0 2 0 6HWhat is the significance of the max of a spectrum? What are the other uses of the spectrophotometer?
- The levels of Vitamin B1 in a sample of milk was determined using the Standard Addition method and technique of fluorescence spectroscopy. 18.0 ml of the milk sample was diluted to 20.0 ml using distilled water and then the fluorescence was measured. The resulting signal was 210 units. A spike was made by taking 18.0 ml of the same milk sample, adding 1.0 ml of a 6.0 ppm Vitamin B1 standard solution, and then diluting to 20.0 ml using distilled water. The signal of the spiked milk was 540 units. Calculate the concentration of Vitamin B1 in the original milk sample (in ppm). Report to 3 decimal places.Why does fluorescence tend to be the mirror image of absorption?If two signals differ by 90 Hz in a 300 MHz spectrometer, by how much do they differ in a 500 MHzspectrometer?