Using the spectroscopic data given, determine the structure of the unknown. Thank you 13C NMR Spectrum(50.0 MHz, CDCI, solution)DEPT CH2 CH3 CH↑proton decoupledsolvent20016012080408 (ppm)1H NMR Spectrum(200 MHz, CDCl3 solution)10987650TMS43 2108 (ppm) 10080604020IR Spectrum(liquid film)4000% of base peak3000200016001200800V (cm³)5799M114 (<1%)Problem 8Mass SpectrumNo significant UVabsorption above 220 nmC8H18408080120160200240280m/e

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Answered: 13C NMR Spectrum (50.0 MHz, CDCI,…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter24: Introduction To Spectrochemical Methods

Section: Chapter Questions

Problem 24.2QAP

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Concentration of FeSCN2+ Absorbance (Y axis) Percent Transmittance (%T) (X axis) 100 1.00E-04 0.495 32 2.00E-04 0.955 11.1 3.00E-04 1.398 Calibration Curve (Absorbance vs. Concentration of FESCN²*) 1.6 1.4 y = 4713.6x. 1.2 1 0.8 0.6 0.4 0.2 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003 0.00035 Concentration of FeSCN²+ Absorbance
100 - 80 60 40 20 10 15 20 25 30 35 40 45 50 55 60 m/z Absorbance / % 50- 100+ 4000 3000 2500 2000 1500 1000 500 Wavenumbers / cm Relative Intensity / %
Spectrum 28 100% 75%- 50%- 25% 0% 3500 2500 1750 1250 750 4000 3000 2000 1500 1000 50 Wavenumber (cm) NH2 Transmittance (%)
Calculate the total absorbance at 512 nm Compound Absorptivity at 512 Absorptivity at nm 367nm X (5E-5 M) 200 600 Y (5E-7 M) 400 100 0.03 O0.07 0.04 0.01 0.02 C None 0.05
Intensity (a.u.) 4000 3352 cm¹ 3500 3000 UNKNOWN B 1710 cm¹ 2000 1668 cm¹ 2500 Wavenumber (cm¹¹) 1500 1101 cm 1000 500
опанит 1715 1600 IR Spectrum 100 4000 (liquid film) 43 80 75 60 M 40 20 % of base peak 10 3000 40 C 13C NMR Spectrum (20.0 MHz, CDCI solution) proton decoupled 9 80 200 ¹H NMR Spectrum (100 MHz, CDCI, solution) 101 8 2000 v (cm¹) M** 132 120 160 m/e 160 7 6 1200 120 Mass Spectrum 200 240 280 800 C-H 5 C6H1203 solvent L 80 4 -CH3 No significant UV absorption above 220 nm 3 H-C-H u -CH3 40 2 59 TMS 0 1 8 (ppm) TMS 0 8 (ppm)
Determining the concentration of unknown solution using uv-vis spectroscopy. The absorbance vs concentration has an equation of straight line of y=8.4052x+0.0002 with r2=1. The unknown solution's absorbance is 0.150. From 75 mL sample, A 15 mL aliquot was examined. The 15 mL aliquot was mixed with 15 mL 1% percent by mass over volume HCl and was diluted to 40 mL with water. What is the concentration of the solution in the 75 mL sample?
122 Problem 16 IR Spectrum (iquid film) 776 2960 650 4000 3000 2000 1600 1200 800 V (cm) 100- Mass Spectrum 55 80 60 No significant UV 41 absorption above 220 nm 40 M 126/128/130 < 1% 90 20 92 C4H&Cl2 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (50.0 MHz. CDCI, solution) solvent proton decoupled 80 40 8 (ppm) 200 160 120 H NMR Spectrum (200 MHz, CDCI, solution) expansions TMS 3.8 3.6 ppm 20 18 ppm 5 4 3. 1 8 (ppm) 10 9 8 6 105 s of base peak
4000 100 80 60 IR Spectrum (liquid film) 40 20 $ 2 2960 3000 41 77 79 40 13C NMR Spectrum (50.0 MHz. CDCI, solution) proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) T 40 10 9 80 2000 107/109 1 8 120 v (cm¹) m/e 1600 M** 156/158/160 160 160 3.5 1200 120 200 240 766 800 Mass Spectrum expansions ppm 2.5 1 7 6 5 650 C3H6Br Cl 280 T 2.0 ppm solvent 80 4 H-C-H Problem 18 No significant UV absorption above 220 nm H-C-H 1 3 40 H-C-H 0 1 2 1 124 8 (ppm) TMS 0 8 (ppm) 107
1H 11 10 2H 2h 5 3H 3 An IR spectrum, a 13C NMR spectrum, and a ¹H NMR spectrum were obtained for an unknown structure. Draw the structure of this compound. 2 1 ppm
8 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 6H
Unrounded Rounded ε∗,L/μmol 0.0259825 0.0260 Heres the data if needed: TZ # Concentration Absorbance at 430 nmnm 1 33.6480 0.931 2 25.2360 0.757 3 16.8420 0.210 4 8.41200 0.137 5 4.20600 0.122

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