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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 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.Give An introduction to NMR spectroscopy ?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)…Treatment of butan-2-one (CH3COCH2CH3) with strong base followed byCH3I forms a compound Q, which gives a molecular ion in its massspectrum at 86. The IR (> 1500 cm−1 only) and 1H NMR spectra of Q aregiven below. What is the structure of Q?
- 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 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 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.
- a.Calculate the degree of unsaturation of compound E. What information can be obtained from the calculated degree of unsaturation? b.Draw the most likely ion fragment for the signals at m/z 911, 139, and 156. c.Why do these signals at 111, 139, 156, and 184 appear as pairs with the intensities of 1/3 at 113, 141, 158, and 186? d.ldentify and draw the structure of compound ESketch the H NMR spectrum for each compound. Assume first order multiplets where possible.(74) Which of the following structures corresponds to the attached 1H-NMR spectrum?