DGD4_NZ

Elaborat ur answer pls thank u

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)

12. Consider the data below to answer the following questions. C₂H₁40 IR: MS: 1715 cm-¹ M* at m/z = 114, a-cleavage fragment at m/z=71 McLafferty rearrangement fragment at m/z = 86 ¹H NMR: 0.928 (6H, triplet) 1.59 8 (4H, sextet) 2.36 8 (4H, triplet) a) Calculate the degree of unsaturation for this compound. b) What functional group is indicated by the IR data? c) Propose a structure consistent with the spectral data presented above. d) Interpret the mass spectral data by showing both fragmentation patterns.

Spectrum No.1 The infrared spectrum has no interesting features except aliphatic C-H stretching and bending. Give the mechanism that lead to the formation of ion fragments at m/2 = 83 & 55. Relative Intensity 100- Relative Intensity 80 60 9 20- 0-tm 10 100- 80 09 MS-NA-3722 9 20 20 MS-NU-3100 0-t 10 30 20 41 40 Spectrum No.2 The infrared spectrum has a medium intensity peak at about 1650 cm³¹. There is also a C-H out-of- plane bending peak near 800 cm³. Give the mechanism that lead to the formation of ion fragments at m/z = 69. 30 50 55 41 40 60 m/z 70 50 m/z 56 83 60 90 69 M(98) 70 100 110 M(84) 80

Chapter 5: INTEGRATED SPECTROSCOPY PRO Compound 9 is a high-boiling liquid (boiling point 201'C), that reacts with alkoxide to yield spectra, along with C NMR data, are given below Elemental Analysis: C, 49.16; H, 4.13, contains manes an ether. The Mass, IR, and 'H NMR Mass Spectrum 91 65 172-170 170 180 190 200 100 I0 120 130 140 Is0 160 30 40 60 70 90 Infrared Spectrum Wave Number, cm 1 4000 3000 2500 1300 1200 1100 B00 700 2000 1500 1000 10 11 12 13 ' Η ΝMR Wavelength, microns 14 15 2. ppm, 5 "C Spectral Data: Structure: singlet, 130.1 ppm; doublet, 129.0 ppm, doublet, 127.8 ppm; doublet, 127.7 ppm; triplet, 29.3 ppm C2000 Brooks/Cole Publishing Company Absorbance

CHEM 241L. Exp#12 NMR Spectre 3. Compound 3 is a low boiling liquid (bp-58°C) that reacts with Br. in CCl. to form a 1,1,2,2 tetra-bromide. Elemental Analysis: C-48.36%, II - 4.06%; contains halogen "C Spectral Data: 80.4ppm (doublet), 68.3ppm (singlet), 34.7ppm (triplet) Mass Spectrom Absorbence Intonally 48 'H NMR Infrared Spectrom 4000 3000 2500 2000 74 70 80 90 100 110 120 130 140 150 160 170 100 mie Wave Number, cm .1 1300 1200 1100 1000 mon 1600 Wavelength, microne Compound 3 Structure Ppin

k. N,N-diisopropylethylamine 100 80 60 40 20- 25 50 75 100 125 m/z Identify the structure that give rise to the peak at m/z 72 and 114. I. Butyramide (C4H9NO) 100- 'NH2 80 60 40 20 10 20 30 40 50 60 70 80 90 m/z Identify the structure and provide the mechanism that give rise to the peak at m/z 59. Relative Intensity Relative Intensity

The iodination of salicylamide was carried out analogous to Experiment 12 in your "Green Organie Chemistry" textbook: Nal NaOCI `NH lodosalicylamide EIOH OH Salicylamide The 'H NMR spectra of the starting material and product are posted on Blackboard. Processed data are posted in the document A2-spectra.pdf. If you would rather process the data yourself, the raw FID's are also posted on Blackboard with the filenames shown below. In cach case the solvent is de-DMSO (if you process the data yourself, be sure to reference the center peak of DMSO to 2.5 ppm). Salicylamide (the starting material): The recrystallized iodosalicylamide (product) salicylamide.mrc iodo salicylamide xtal.mrc 8. In both the starting material and product the chemical shift of the hydroxyl proton is approximately 13 ppm. Table 21.2 in your “Techniques in Organic Chemistry" textbook gives 4.0 - 8.0 ppm as the typical range for phenolic protons. Explain this apparent discrepancy.

Annotate the Mass Spec, 1H NMR, and 13C NMR for C4H11N or Butylamine Annotate the Mass Spec, 1H NMR, and 13C NMR for C4H8O or Butraldehyde Annotate the Mass Spec, 1H NMR, and 13C NMR for C5H8O or Cyclopentanone Thanks :)

can u please explain ur answers thank u

The following Grignard reaction gave I as a major product. The mass spectrum of H is given below. Note that the molecular ion at m/z 88 is not detected in the spectrum. Determine the structure of I. Explain your answer.

Im a bit confused with organic chemistry spectra material. How do I solve this step by step? How do I rationalize this?

a. 2,2,3-trimethylbutane 100 MS-NJ-3103 80 40 20 10 20 30 40 50 60 70 80 90 100 m/z Identify the peaks that give rise to the peaks at m/z 57 and 85. e.g., m/z 85 m/z 57 Relative Intensity

Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 20 NaоH 0103 Br (B) H2504 → (c) (A) 100- MS-NU-0547 80 40 20 31 10 20 100- MS2016-05353CM 80 60 100 MS-NJ-09-3 80 60 40 20 45 J.L 80 S1 84 M+ absent राग 135 137 S2 62 164 166 11 S3 25 50 75 100 125 150 175 m/z

What are the errors in the following answer?

[References] The mass spectrum of octanoic acid, CH3(CH2)6COOH, exhibits a series of peaks differing by 14 amu at m/z 29, 43, 57, 71, 85, and 99. Draw the structure of the fragment resulting in the m/z 71 peak. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one. • Do not use the square brackets tool in your answer. opy aste H,C earch 84%

Fragmentation A.26 Label each of the following in the mass spectrum of hexan-2-ol [CH3CH(OH)CH2CH2CH2CH3]: the molecular ion, the base peak, the fragment resulting from the loss of H2O, and a cleavage fragments. 100 45 Relative abundance 50 69 8487 102 ↓ 0 0 10 20 30 40 50 60 70 80 90 100 110

Use the spectral data given below to determine the structural formulas for reactant A and product B

9 C 2 H, d 2 H, d 8 6 Solution Process: 5 PPM 4 3 H, s 3 3 H, s 2 1 Analysis compound C: Structure of C (paste in chemdraw): 09 Proton NMR: report in this format shift (splitting, integration); e.g., 3.20 (t, 2H) alevlan (webmerlo ni eresa) O to entou de (naiisipeni pritiige) Hirle temot air ni noger:AMM motors com-Ingh ari tot poitilga ni 10123 stol

please explain and answer

What is the compound?

[References] All ethyl esters of long-chain aliphatic acids (for example, ethyl tetradecanoate, C13H27COOCH₂CH3) show significant fragment ions at m/z 88, 73, and 45. Draw the structure of the ion that gives the m/z 88 peak. star 14

Spectroscopy Unknown. The spectra and data provided were obtained from a pure organic molecule. For ¹H NMR Spectra, the integral is given in number of hydrogens (#H) or as a relative ratio. Important coupling constants (J-values) are listed next to the peaks for some examples. For some spectra, an inset (grey box) is also given showing a "zoom-in" on an important part of the spectrum. Mass Spectrum (not shown): [M] = 177 (100%) m/z IR Spectrum (not shown): 3200 (sh m), 3060, 2981, 1685, 1600, 1495 cm³¹ (all listed are strong (s) unless otherwise indicated) ¹H NMR Spectrum (400 MHz, CDCl3, 25 °C) br s 180 10 1H (C) 8 160 1³C NMR Spectrum, proton-decoupled (125 MHz, CDCI3, 25 °C) 2H 2H (CH) (CH) (C) (C) 140 6 120 PPM 100 PPM 80 60 E 1H 3H S 2 (CH3) 40 (CH) 6H (CH3) 20