Question

Identify the compounds that display IR spectra with the following peaks:

a) (C2H3O2Cl): 3200-2500 (broad), 1720, 1410 cm^-1

b) (C8H8O): 3030, 1810, 1760, 1715, 1605, 1595, 1495, 1410, 750, 695 cm^-1

d) (C8H8O): 3020, 2970, 1695, 1600, 1480, 1435, 760, 690 cm^-1

Answer 1

1. M.F. C2H3O2Cl

1. Corresponding hydrocarbon structure without heteroatom will be C2H4 (1H compensation for Cl and no compensation for O). The saturate hydrocarbon formula will be C2H6. Hence there is deficiency of 2H i.e. 1 unsaturation site.
2. Broad absorption peak in the region 2500-3200 cm^-1 indicate strongly H-bonded O-H functionality. May carboxyl group present.
3. Absorption at 1720 cm^-1 indicate carboxylic C=O .
4. Absorption at 1410 cm^-1 indicate C-C=O stretch. Normal value of this stretch is 1100-1300 cm^-1, but this unexpected increase is because of –I effect of Cl substitution.
5. Hence structure proposed is 2-Chloroethanoic acid.

1. M.F. C8H8O:

1. Corresponding hydrocarbon structure without hetero atom will be C8H8, so corresponding saturated hydrocarbon formula is C8H8, which indicates deficiency of 10 H i.e. 5 u.s.
2. Bands at 1595, 1495, 1410 cm^-1 indicate aromatic ring C=C . Doublet at 695 and 750 indicate monosubstituted aromatic ring.
3. 1715 band indicate C=O and it’s overtone at 1760 and 1810 cm^-1 . (may Aldehydic C=O)
4. Band at 3030 cm^-1 indicate aldehydic/ aromatic/ sp3 C-H stretch.
5. Hence structure proposed as 2-phenylethanal.

1. M.F. C8H8O:

1. 5 u.s. (same explanation as above)
2. 1600, 1480, 1435, 760, 690 cm^-1 indicates monosubstituted benzene ring.
3. 1695 cm^-1 indicate phenyl conjugated C=O grouping.
4. 2970 and 3020 indicates aromatic sp C-H and aliphatic sp3 C-H stretches.
5. Hence structure proposed is Acetophenone.