Transmittance/% 100 50- A B 3000 2500 2000 1500 1000 500 Wavenumbers //cm Figure 2. IR spectrum of C3H5N 2'6 2'4 1'8 1'6 12 o's 0.4 o'2 ppm Figure 3. 'H NMR spectrum of C3H5N recorded in CDCI3

A student prepares a compound that she believes has the molecular formula C4H1002. Give a possible structure for this compound using the information from the following spectra. How many peaks will your proposed structure give in the 13C NMR? 1Η NMR singlet 3.25 (3 H) singlet 3.45 (2 H) IR nothing at 3500 cm-1 nothing at 1700 cm-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. 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.

For the protons labeled Hª and Hb in the structure, predict the characteristics of their signals in the ¹H NMR spectrum: the approximate chemical shift, the splitting pattern, and the integration value. Ha a TOT a нь Нь X

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).

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?     CH3CO2H     Formaldehyde     HO                 \               C = CH2       / HO     CH2(OH)CHO     CH(OH)=CH(OH)

A compound displays a sharp doublet at 3400 cm-1 in its IR spectrum. The mass spectrum has a molecular ion with m/z = 115 and a base peak of m/z=72. Draw a structure that best fits this data.