Provide the structure of the molecule that would give rise to these spectra. Include degree of unsaturation, functional groups present, assignment of 1H and 13C NMR peaks by labeling each peak with specific atoms in the structure (a,b,c,d) and show which atoms give rise at which peaks IR Spectrum(liquid film)400010080604020% of base peak3000401013C NMR Spectrum(500 MHz, CDCI, solution)7320080proton decoupled¹H NMR Spectrum(200 MHz, CDCI, solution)45L9-exchangeswith D₂ODEPT CH₂ CH₂ CH1401817202000115138V (cm1¹)120140m/e160160expansions160020A7m1200M+ 194/196 (<1%)120L6Mass Spectrum240200C6H₁10₂Br280M8001.0 ppm5solvent1.1804No significant UVabsorption above 220 nm4013208 (ppm)1TMS08 (ppm)

Broad band in IR shows OH, NH2, COOH Presence of M and M+2 peaks with 1:1 ratio in Ms shows 1Br atom…

IR Spectrum (liquid film) 4000 100 80 8 8 8 8 % of base peak 60 40 20 3000 40 10 13C NMR Spectrum (500 MHz, CDCI, solution) 73 200 80 proton decoupled ¹H NMR Spectrum (200 MHz, CDCI, solution) 45 9 exchanges with D₂0 DEPT CH₂ CH, CHI J M 1720 40 2000 8 138 116 V (cm¹) 140 120 160 m/e 1600 L 160 expansions ì 20 L 7 1200 M+ 194/198 (<1%) 120 Mass Spectrum 240 200 ill 6 C6H₁10₂Br 280 800 1.0 ppm 5 solvent 1 1 80 1 4 No significant UV absorption above 220 nm 40 L 3 1 2 08 (ppm) 1 TMS 0 8 (ppm)

IR Spectrum (liquid film) 4000 100 80 8 8 8 8 % of base peak 60 40 20 3000 40 10 13C NMR Spectrum (500 MHz, CDCI, solution) 73 200 80 proton decoupled ¹H NMR Spectrum (200 MHz, CDCI, solution) 45 9 exchanges with D₂0 DEPT CH₂ CH, CHI J M 1720 40 2000 8 138 116 V (cm¹) 140 120 160 m/e 1600 L 160 expansions ì 20 L 7 1200 M+ 194/198 (<1%) 120 Mass Spectrum 240 200 ill 6 C6H₁10₂Br 280 800 1.0 ppm 5 solvent 1 1 80 1 4 No significant UV absorption above 220 nm 40 L 3 1 2 08 (ppm) 1 TMS 0 8 (ppm)

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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3 The 1H- and 13C-NMR data of an ester of molecular formula C6H10O2 are given below. Also shown are the COSY and HETCOR NMR spectra of the ester. Identify the ester, explaining how you reach your conclusion. 1H-NMR: 7.20-6.90 (1H), 5.85 (1H), 4.16 (2H), 1.88 (3H), 1.31 (3H) ppm 13C-NMR: 166.7, 144.5, 123.0 , 60.2, 18.0, 14.3 ppm
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.
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
The 1H- and 13C-NMR data of an ester of molecular formula C6H10O2 are given below. Also shown are the COSY and HETCOR NMR spectra of the ester. Draw the structure of the ester, explaining how you reach your conclusion. 1H-NMR: 7.20-6.90 (1H), 5.85 (1H), 4.16 (2H), 1.88 (3H), 1.31 (3H) ppm 13C-NMR: 166.7, 144.5, 123.0 , 60.2, 18.0, 14.3 ppm
The infrared spectrum of the compound with the mass spectrum shown below lacks any significant absorption above 3000 cm-1. There is a prominent peak near 1740 cm-1 and another strong peak near 1200 cm-1. Propose a structure consistent with the data.
Please provide the IR, 1H NMR and UV Spectra values of Benzoic Acid, 3-bromo-, methyl ester. And discuss how the results were obtain. For example the result for 3-(3-methoxyphenyl) propanoic acid would be written like this: 1 H NMR spectrumd (ppm in CDCI3 at 60 MHZ: 11.75 (Singlet, 1H, #7), 7.40-7.23 (Multiplet, 1H, #2), 6.85 (quartet, 3H, #1-3-6), 3.86 (Singlet, 3H, #5), 3.15-2.53 (Multiplet, 4H, #9-8). The discussion would be like this: the unknown molecule contained 12H. Revealing the presence of a peak of medium intensity at 11.75 ppm that we can associate with a carbocyclic acid. It also showed a high intensity peak at 3.86 ppm that we can associate with an ether, an alcohol or an amino group. Since the signal also shows that 3 protons are attached to it, it is therefore considered an ether (OCH3) in the unknown molecule. We also see two average peaks around 7.23 ppm and 6.85 ppm which show the presence of an aromatic.
The 1H-NMR spectrum of lycopene is provided below. Comment on the general features of the NMR spectrum.
Benzoic Acid V end point = 22 ml of NaOH V 1/2 = 11 ml of NaOH pKa = 5 Substituted Acid - P- bromobenzoic Acid V end point = 17 ml of NaOH V 1/2 = 8.5 ml of NaOH pKa = 4.9 Briefly explain the electronic effect of the subsistent present in your acid compared to benzoic acid
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.
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.
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.