Question

Use the IR and NMR spectra to confirm the identity of your final product. Identify all of the important peaks in the IR spectrum. Assign all peaks in the NMR spectrum to their respective hydrogen nuclei.

IR and 1H NMR of 9-Fluorenone: LOD TFRITTEN 9000 2000 LOOD MAVENUISERI

Answer 1

6 14 5 OH B 13 6 5 9 2 1 12 8 13 3 2 10 11 12 3 9H-fluoren-9-one 10 11 9H-fluoren-9-ol

The infrared spectrum of 9-fluoren-9-one shows the following peaks.

Aromatic C=C bonds appear at 1600 cm-1. Carbolyl stretch (C=O) appears at 1700 cm-1. The C-H stretch in C=C-H components appears at 3100 cm-1.

NMR spectrum of 9-Fluorenone shows 3 peaks. Please note the numbers given to the carbon in structure.

Hydrogens in 6 and 13 come as a doublet at 8.34 ppm. Hydrogens in 6th and 13 of the positions are similar. The peak of this hydrogen will split by only one hydrogen in the next carbon. so this peak will come as a doublet

Hydrogens at 3 and 10 are similar, and this gives a peak at 8.44 ppm,2H, Doublet

hydrogens at 1,12 are similar, and give a multiplet. at 7.63. ppm,2H

hydrogens at 2,11 are similar and give peaks at 7.68 ppm,2H, multiplet.

The remaining carbons 4,5,7,8 and 9 doesn't have hydrohens.

Infrared spectrum of 9-Fluorenol shows peaks Hydroxyl stretching (-OH vibration) at 3300 cm-1, Aromatic C=C stretching at 1600 cm-1 and Chydrogen stretch in C=C-H component at 3100 cm-1. The O-H stretching at 3300 cm-1 is little bit broad due to the presence of hydrogen bonding.

NMR spectrum is explained. please correlate with the given numbers in structure.

Hydrogens in Carbon 6, 13 -: 7.55 ppm, (2H) doublet

Hydrogens in Carbon 3,10 :-7.90 ppm,(2H), doublet

Hydrogens in Carbon 1,12 :-7.28,(2H), ppm multiplet.

Hydrogens in Carbon 2,11 :- 7.38 ppm,(2H), multiplet.

Hydroxyl hydrogen value at 6.43 ppm,(1H) It is an exchangeable proton.

Hydrogens in Carbon 9 gives a peak at 5.97 ppm with 1 H.