Question

13. (16 pts) Compound P was a pleasant smelling, colorless liquid with a boiling point of 218 degrees C. It had no effect on either red or blue litmus paper. A mass spectrum of Compound P showed a molecular ion at 134 with a base peak at 105. When an ir spectrum was taken as a liquid film of Compound P, an intense peak was observed at 1683 em. Both proton nmr and carbon-13 nmr were taken of Compound P, and both are shown on the next page. For the proton nmr spectrum, the number of protons contributing to a resonance is shown above the resonance multiplet. For the carbon-13 nmr spectrum, the multiplicities are NOT given. Note that two carbon-13 peaks are very close to each other. Tables with reference infrared frequencies and nmr chemical shifts are given on the two pages just before the answer sheet. Points Breakout Interpretation of infrared, 2 pts Calculation of molecular formula, 2 pts Calculation of degrees of unsaturation, 2 pts Interpretation of proton nmr (show deductions), 3 pts Interpretation of carbon-13 nmr (show deductions), 3 pts Correct structure of Compound P (including reasoning), 4 pts

Please give a through explanation

Answer 1

From the given data, the compound is 1-phenypropanone or propiophenone with the given structure as shown in the images.

From the IR data, an intense peak at 1683 cm-1 represents carbonyl compound.

From the mass data, the peak at 134 represents the molecular ion peak of propiophenone and base peak at 105 represents the cleavage of -(CO)-CH2 bond of the given structure.

1H NMR and 13C NMR are explained as follows in the image.
Proton NMR al CH3 b -COCH2 a-ptotons ptotons 6 Chemical Shift (ppm) Carbon 13 NMR AROMATIC PROTONS 1283 -CH2 132 5 1277 79CH3 313 CO 198.9 136.7 100 80 60 40 20 200 180 160 140 120 Chemical Shift (ppm)