What is the IR and HNMRAndCNMR and mass IR Spectrum(quid fim)40001008060ETTE% of base peak3000204013C NMR Spectrum(50.0 MHz, CDC, solution)8091DEPT CH CH₂ CHproton decoupled200¹H NMR Spectrum(200 MHz, CDCI, solution)10982000v (cm¹)120m/e160ww16000.08001200Mass Spectrum0.51.0C₂H,Br2801607M170/1722001206240expansion140expansion1405solvent1301308041.5Problem 24UV spectrum0.917 mg/10 mispath length 0 20 cmsolvent hexaneabsorbance200ppmppm403250 300λ (nm)2013508 (ppm)TMS08 (ppm

Please find your solution below : Spectroscopy is the study of absorption or emission of radiation…

4000 100 во IR Spectrum (quid fim) 2828 60 20 % of base peak 3000 40 80 13C NMR Spectrum (50.0 MHz, CDC), solution) 91 DEPT CH CH CHt proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 v (cm¹) M 170/172 120 ww m 0.0 1600 800 1200 Mass Spectrum 0.5 1.0 C,H,Br 1.5 280 160 160 m/e 7 200 120 6 240 expansion 140 expansion 140 1 solvent 5 130 130 1 80 4 Problem 24 UV spectrum 0.917 mg/10 mis path length 0 20 cm solvent hexane 300 200 ppm ppm 40 3 250 2 (nm) 2 0 350 8 (ppm) TMS L L 0 8 (ppm 1

4000 100 во IR Spectrum (quid fim) 2828 60 20 % of base peak 3000 40 80 13C NMR Spectrum (50.0 MHz, CDC), solution) 91 DEPT CH CH CHt proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 v (cm¹) M 170/172 120 ww m 0.0 1600 800 1200 Mass Spectrum 0.5 1.0 C,H,Br 1.5 280 160 160 m/e 7 200 120 6 240 expansion 140 expansion 140 1 solvent 5 130 130 1 80 4 Problem 24 UV spectrum 0.917 mg/10 mis path length 0 20 cm solvent hexane 300 200 ppm ppm 40 3 250 2 (nm) 2 0 350 8 (ppm) TMS L L 0 8 (ppm 1

Macroscale and Microscale Organic Experiments

7th Edition

ISBN:9781305577190

Author:Kenneth L. Williamson, Katherine M. Masters

Publisher:Kenneth L. Williamson, Katherine M. Masters

Chapter12: Nuclear Magnetic Resonance Spectroscopy

Section: Chapter Questions

Problem 3Q

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1. For each set of spectral data, propose an acceptable structure that is consistent with the data. a. MS: M+ m/z=71, IR: 3340, 2960, 2870 cm-1 , 13C NMR: 47.20 (CH2), 25.64 (CH2) ppm b. MS: M+ m/z= 102, IR: 2960, 2870 cm-1, 13C NMR: 72.73 (t), 23.20 (t), 10.80 (q) ppm c. MS: M+ m/z=74, IR: 3310, 2960, 2870 cm^-1, 13C NMR: 62.44 (down), 34.86 (down), 18.98 (down), 13.88 (up) ppm (DEPT-direction) d. MS: M+ m/z=92 (M:M+2=3:1), IR: 2960, 2870 cm^-1, 13C NMR: 67.29 (C), 34.46 (CH3) ppm e. MS:M+ m/z=70, IR: 3070, 2960, 2870, 1666 cm^-1, 13C NMR: 138.92 (t), 114.26 (d), 35.96 (t), 22.17 (t), 13.59 (q) ppm f. MS: M+ m/z= 72, IR: 2960, 2870 cm^-1, 13C NMR: 31.68 (t), 29.80 (d), 22.20 (q), 11.69 (q) ppm g. MS: M+ m/z= 84, IR: 2960, 2870, 1734 cm^-1, 13C NMR: 220.16 (C), 38.30 (CH2), 23.24 (CH2) ppm h. MS:M+ m/z= 138, (M:M+2=1:1), IR: ~3010(broad), 1715 cm^-1, 13C NMR: 173.61(s), 25.21(t) ppm i. MS:M+ m/z=88, IR: 2960, 2870, 1736cm^-1, 13C NMR: 170.95 (C), 60.34 (CH2), 20.98 (CH3), 14.23 (CH3)…
I have an NMR spectrum. There are 5 peaks. There is a doublet at 9.6 ppm (integrates to 2H) where aldehydes usually arise, don't know why there is a doublet. There is another doublet at 6.8 ppm, again integrating to 2H. There is a singlet at 4.5 ppm, integrating to 1H. There is a quartet at 3.4 ppm integrating to 2H, and lastly there is triplet at 1.2 ppm integrating to 3H. i recofnize the ethyl group but don't understand the peak at 9.6 unless it is not an aldehyde.
Complete the following answer using the data provided here: Molecular formula: C4H8O2 Important IR data (cm-1): 3280 (broad); 2980-2730 (many); 1690; 1254 All 1H NMR data (ppm, splitting, integration): 11.0 ppm (s), 1; 2.57 ppm (m), 1; 1.07 ppm (d), 6 1. Sketch out a 1H NMR spectrum showing peak locations and peak splitting on a ppm scale for the data provided above. Include the data labels above in your spectrum, but do not show integration lines. 2. draw the most proper line bond structure for the data given above
I have problems understanding the results of this experiment, could you help me by explaining the results? Spectrophotometric of a Cobalt Chloride Solution Analysis Table 1 Absorption Spectrum of Cobalt Chloride Using a Spectrophotometer Wavelength Absorbance 400 410 420 430 440 450 460 370 480 490 500 510 520 530 540 550 560 570 580 590 600 610 620 630 640 650 660 670 680 690 700 0.036 0.044 0.063 0.095 0.147 0.215 0.284 0.329 0.362 0.402 0.450 0.479 0.464 0.398 0.314 0.216 0.142 0.087 0.054 0.041 0.036 0.036 0.034 0.028 0.030 0.023 0.023 0.023 0.014 0.014 0.011 ]
A H1 NMR signal is observed at 117.0 Hz (downfield of TMS) on a 1.20×102‑MHz instrument. a) What is the observed chemical shift in parts per million (ppm)? Chemical shift is ppm b) What is the chemical shift, in parts per million (ppm), if the sample is analyzed with a 60.0‑MHz instrument? Chemical shift is ppm c) Where will the proton signal appear, in hertz, if the sample is analyzed with a 60.0‑MHz instrument?
For each set of spectral data, propose an acceptable structure that is consistent with the data 1.. MS: M+ m/z=92 (M:M+2=3:1), IR: 2960, 2870 cm^-1, 13C NMR: 67.29 (C), 34.46 (CH3) ppm 2.. MS:M+ m/z=70, IR: 3070, 2960, 2870, 1666 cm^-1, 13C NMR: 138.92 (t), 114.26 (d), 35.96 (t), 22.17 (t), 13.59 (q) ppm 3.. MS: M+ m/z= 72, IR: 2960, 2870 cm^-1, 13C NMR: 31.68 (t), 29.80 (d), 22.20 (q), 11.69 (q) ppm
Spectra data for Unknown Z are listed below. Deduce it's structure. MS: m/z 134 (Molecular Ion) IR: 1720 (strong); 3062 (weak); 2981 (medium); 1605 (medium) 13C NMR and 1H NMR shown on a separate page.
The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 and are found in the accompanying file. The 1H are also listedbelow. Identify the name of compound 1 and Provide a full analysis of the NMR spectra for compound 1 with table showing peak name and assignment 1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).
Compund AFK has a maximum absorbance of 650 nm. Its molar absorptivity at 650 is 8400 M−1cm−1 and the path length is 1 cm. Using a spectrophotometer, determine the concentration of Compound AFK it is absorbance at 650 nm is 0.700 ? 8.33x10-5 mol/L 1.20 x104 mol/L 1.08 x10-3 mol/L 7.74 x10-2 mol/L
The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 .. The 1H and 13C peaks are also listedbelow. Provide a full analysis of the NMR spectra for compound 1.1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74. Include: correct assignment of NMR spectra of 1H spectra. correct rationalisation of 1H spectrum
The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 The 1H and 13C peaks are also listedbelow. Provide a full analysis of the NMR spectra for compound 1.1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.
values : Experimental values of IR Peaks compared to literature values for Benzoic Acid (ATR method) Bond Stretch Functional group Experimental Peak(cm-1) Literature Peak(cm-1) O-H Carboxylic acid 2800 3200-2200 C=O Carboxylic Acid 1674.9 1677.5 C=C (ring) Aromatic stretch 1579.5 1581-1418.5 C-H Aromatic Sp2 3100 C-O stretch 1285.5 O-H Alcohol 929.89 table 3: Experimental values of IR peaks compared to literature values for 2-Naphthol Bond Stretch Functional group Experimental Peak(cm-1) Literature Peak(cm-1) O-H Alcohol 3219.1 3400-3080 C-C Aromatic (ring) stretch 1507.6 C=C Aromatic stretch 1597.7 1627.3-1377.8 C-H Aromatic Sp2 3050 C-O Secondary alcohol stretch 1168.9 IR results used to prove that the compounds are effectively separated. Discuss on bothdiagnostic peaks and fingerprint region.