Question

26. CH10 units of unsaturation :: m 2x11 12-14 2 Sunits RE Darbo 13

Answer 1

C₁₁H₁₄O :

D.B.E or Unsaturation units = 5 (cyclic rings + pi -bonds)

IR : frequencies are poorly visible , only it can be said than carbonyl group (keto ) is present .

Peak ~ 1700 cm-1 (strong ) is due to C=O stretching.

Peaks ~ 2900 cm-1 due to (sp3)C-H stretching and above 3000cm-1 is due to (sp2)C-H stretching.

1H-NMR :

• 3H (0.9 ppm) -CH₃(it probably triplet ; may be part of -CH₂CH₃ group)
• 2H (1.4 ppm) : multiplet (may be part of -CH₂CH₂- group)
• 2H (1.7 ppm) : mutiplet   (may be part of -CH₂CH₂- group)
• 2H (2.9 ppm) : probably triplet : de-sheilded due to -(C=O)
• 3H (7.4  ppm ) ; aromatic protons (ortho to -(C=O) ).
• 2H (7.9 ppm ) : aromatic protons.

13C-NMR : - 4 peaks in sp3 C region : 4 types of carbon.

- 2/3 peaks in sp2 C region : 2 types of aromatic Carbons.

- 200 ppm is due to  -(C=O)

Mass Spectra : Most of data is non readable : - m/z : 105 (base peak ) due to Ph-CO+.

- m/z : 120 ( - 42 (propene) ) : maclafferty rearrangement (Ph-(CO)-CH₃)^+..

Most probable structure is , 1-phenylpentan-1-one