Q2. Using Mass spectrum and NMR data deduce the structure of the following compounds; (a) C,H,BrO₂ 'H NMR, 300 MHz, 2.9 ppm (triplet, 2H), and 3.8 ppm (singlet, 3H) 100 Relative Abundance (percent) 87 duba 59 107 55 80 40 20 0 15 27 10 20 30 40 50 60 70 ВО Scanned with CamScanner 90 100 110 m/z 135 M (166) mm 120 130 140 150 160 170

A compound A of mass 114 has the following elemental analysis: C = 63.14%, H=8.83%. The IR, 1H NMR and 13C NMR spectra are given below. 1) Assign and interpret all the data from the 1H NMR and 13C NMR spectra. Propose a formula for A. 7.0 175 с 3000 1,3; 13 I=1 6.0 150 I=1 с 5.0 CH₂ 125 2000 Explain why the answer is this one : CH,—CH,—0 4,2; 60 mavermy I=2 cm-1 4.0 100 1500 3.0 75 135 167 | CH₂ 6 125 CH 3 1,95 ; 18 I=3 1000 2.0 50 H 5,5 I=3 1.0 CH₂ CH₂ 25 500 Ppm ppm

An unknown compound produces an 1H NMR spectrum with only three signals: 1.8, 2.8, 7.5 (broad) ppm, all singlets that integrate in a 6:3:1 ratio, respectively. The 13C NMR spectrum shows 4 signals. Both spectra are provided below. The mass spectrum shows a molecular ion region with the following peaks: m/z 135 (100%), 136 (5.9%), 137 (33%). Determine the number of carbons, molecular formula, and propose a structure. Assign all 1H and 13C NMR signals to specific chemical environments or specific hydrogens in your structure. 'H NMR Spectrum of Unknown 8 6 PPM 13C NMR Spectrum of Unknown 160 140 80 60 100 PPM 180 120 40 20

Propose a molecular formula for a compound that exhibits the given peaks in its mass spectrum. Draw a possible structure. M*• at m/z = 136, relative height = 12.1% (M+1)*• at m/z = 137, relative height = 1.2%

Elemental analysis determined unknown Compound G to contain 40.0 % carbon, 6.7 % hydrogen. Compound G had an M+ = 60 m/z on its MS. Its IR displayed three absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500. The proton NMR showed an absorbance at 1.7 ppm (lowest whole number ratio of H =3), and an absorbance at 11.5 ppm (lowest whole number ratio of H = 1). Compound G was able to turn blue litmus paper red, and neutralize a base. What is Compound G? O CH3CO2H O Formaldehyde O HO C CH2 HO O CHy(OH)CHO O CHOHHCH(OH)

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)

Use spectrometry information given to identify the formula structure and name one of the following compounds

Analysis of a sweet-smelling, neutral compound of carbon, hydrogen and oxygen produced the following results: %C = 54.5; %H = 9.1. From its mass spectrum, the molecular ion had a mass/charge ratio of 88. Its infra-red spectrum showed a prominent peak at 1735 cm–1. Figure below shows the NMR spectrum of the compound. Suggest a structural Formula for this compound.

Use the 1H NMR and IR spectra given below to identify the structure of compound B (molecular formula C4H8O2).

Using the IR and NMR spectra provided, determine the molecular structure for C6H12O2. Label any diagnostic peaks on the IR and draw fragments you can determine from the NMR.