From the given data
Elemental analysis
mols of C = 84.8/12 = 7.03
mols of H = 7.1/1 = 7.1
mols of O = 8.1/16 = 0.50
divide by smallest fraction,
C = 7.03/0.50 = 14
H = 7.1/0.50 = 14
H = 0.50/0.50 = 1
Empirical formula = C14H14O
empirical formula = molecular formula
UV analysis
lambda-max = 217 nm : we have aromatic system
IR analysis
3100 cm-1 for alkynic linkage
3000-3050 cm-1 for sp2 =C-H stretch
2850-2950 cm-1 for aliphatic C-H stretch
1500-1600 cm-1 for C=C stretch
Mass analysis
M+ 198
m/z 108 formed by loss of Ph and -CCH fragments
m/z 92 formed by loso of oxygen from 108
Molecular formula
C14H14O
on the spectrum , label the major absorption peaks with the structrual fragment that causes the...
chapt
Name 3. Chapter 10 covered the synthesis of alkynes by a double dehydrohalogenation of dihalides. A student tried to convert trans-hex-2-ene to hex-2-yne by adding bromine across the double bord then doing a double elimination. The infrared and mass spectra are shown below. Bra 2 NaNH2 100 Relative Intensity Or 75 m/z a Do the spectra confirm the right product? If not what is it? b. Explain the important peaks in the IR. - Determine the structure using MS...