Question

0 20 40 Transmittance [%] 60 80 100 120 3500 140 BRUKER 3111.11 3000 2952.75 2500 Wavenumber cm-1 2000 1500 1692.77 1643.06 1597.66 1545.30 1479.20 1453.97 1428.43 1401.52 1357.55 1325 67 1284.62 1237.02 1187.98 1071.68 1023.83 972.09 926.61 859.35 1000 758.71 743.18 699.85 644.05 609.17 480.32 443.28 423.06

based off of the presented IR, ID all funtional groups present. based off of the functional groups, what is the compound? benzoic acid, EDTA disodium dihydrate, glucose, ascorbic acid, NH4Cl, boric acid, oxalic acid dihydrate, cystenine, or caffeine?

Answer 1

In the given IR spectrum there is a peak at :

3111 cm-1 (m) which corresponds to C-H stretch of alkenes,

2975 cm-1 which corresponds to C-H stretch of alkenes.

As there is no stretch for O-H at 3500 cm-1 alcohol must not be present.

Hence, glucose is not present.

Another peak is near 1692 cm-1 and 1643 cm-1 which belongs to C=O stretch.

But the stretch for carboxylic acid C=O is present between 1700-1725cm-1.

Thus the carboxylic acid is not present.

But it belongs to amide , thus indicating the presence of two -CO-NH- group.

From the provided compounds we come to conclusion that cysteine or caggeine must be present.

But as cysteine contains amino group ,the stretch of N-H must be present at 3500 cm-1 which is absent in the spectrum.

Thus the spectrum is of caffeine , which contains two amide groups.

Peaks between 1000-1350 cm-1 is of C-N group,

C=N is between 1640-1690 cm-1,

Structure of caffeine is as follows:

NNO