Given the following NMR data, find the unknown compound. a. Identify the molecular formula; find IHDb. Identify the 1H and 13C NMR signals IR Spectrum(liquid film)400010080604020% of base peak1030004013C NMR Spectrum(50.0 MHz, CDCI, solution)57proton decoupled980DEPT CH₂ CH₂ CH200¹H NMR Spectrum(200 MHz, CDCl3 solution)99182000120V (cm¹)M+-114 (<1%)m/e1601600160I7I61200120200 240Mass Spectrum8005C8H₁8280solvent8043No significant UVabsorption above 220 nm40I2018 (ppm)TMS108 (ppm)

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Answered: Given the following NMR data, find the…

Given the following NMR data, find the unknown compound. a. Identify the molecular formula; find IHD b. Identify the 1H and 13C NMR signals

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter12: Structure Determination: Mass Spectrometry And Infrared Spectroscopy

Section12.SE: Something Extra

Problem 48AP: The infrared spectrum of the compound with the mass spectrum shown below lacks any significant...

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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 analysis of the NMR spectra for compound 1 include rationalisation of COSY1H 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.
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.
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)…
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).
EI-MS of ortho nitrotoluene shows a large fragment ion at m/z 120 .The EI-MS of Tri deuteron ortho nitrotoluene does not show any peak at m/z 120 but does have a peak at m/z 122 .Show the fragmentation pattern that accounts for the above observation.
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. correct assignment of NMR spectra of both 13C spectra. correct rationalisation of 13C spectrum1H 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.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment. you should be able to assign these peaks to one of two carbon atoms in 1.
Imine compounds usually show stretching bands in the IR in the range of 1690-1649cm-1 and aromatic amines show a C-N stretch at 1340-1265cm-1. A related compound, N-benzylideneaniline, does not show tautomerism and has a C-N stretch at 1678cm-1 while H2SB-1(2-(2-hydroxybenzylideneamino)phenol ) shows a C-N stretch of 1590 cm-1. Given this information, does H2SB-1 show some C-N single bond character.
Given the information about the unknown: Unknown: MP: 101-106 oC White solid IR: 2837, 1701, 1602, 1391, 1139 cm-1 ** Using the Reference sheet, The IR and the given H-NMR, and C-NMR. Deduce, what the unknown is. Please give the name of uknown
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
The leaves of the Brazilian Tree Senna multijunga contain a number of pryidine alkaloids that inhibit acetylcholinterinase. Two recentyl isolated isomeric compounds have the strcture have the strcture shown below. (NOTE: M=293) Use the mass spectral data provided to determine the precise location of the hydroxyl group in each isomer. Isomer A: EI-MS, m/z(rel. int): 222(20), 150(10), 136(25), 123(100) Isomer B:EI-MS, m/z(re;. int): 236(20), 150(10), 136(25), 123(100)
The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 The 13C peaks are also listedbelow. Name compound 1 and Provide a full analysis of the NMR spectra for compound 1. with a table showing peak shift ( carbon) and assigment 13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment. you should be able to assign these peaks to one of two carbon atoms in 1.