Step 1: Calculate the degree of unsaturation. The degree of unsaturation indicates number of double bonds present in the compound.
Step 2: See the IR spectrum, specially the region of wavenumbers greater than 1500 cm-1 . Look for likely functional groups keeping in mind, the degree of unsaturation and molecular formula.
e.g, if there is one double bond and an oxygen, it could be a carbonyl; if there is no double bond and an oxygen, it will be either an ether or an alcohol; if there are 4 double bonds, there might be an aromatic ring.
Step 3: See the PMRand 13C NMR spectrum.Check that the functional groups infered by IR and NMR are consistent or not.
Step 4: Draw possible structures consistent with the unsaturation, IR, and NMR data. Check …show more content…
The 2 protons ( CH2) split to a quartet by the adjacent CH3 protons at 2-2.7 ppm (downfield).
The 3 protons ( CH3 adjacent to CH2 group ) split into a triplet by the adjacent CH2 protons at 1.00 ppm.
13 C NMR Spectrum: 13 C-NMR shows 4 types of C: C=O (210 ppm), and 3 hydrocarbon C at 37, 29 and 8 ppm.
Degree of unsaturation: 1.
The compound can have a carbon-carbon double bond or a carbon-oxygen double bond (carbonyl group) or a ring.
IR Spectrum: There is a band at 1700 – 1720 cm-1 indicating a saturated aliphatic ketone.
Possible structures It is not an aldehyde, because the carbonyl of saturated aliphatic aldehydes absorbs at 1740-1720 cm-1.Furthermore, an aldehyde would also show a distinct band for H-C=O stretch in the region 2830-2695 cm-1.
Proton NMR Spectrum:
The protons at 2.3-2.4 ppm are adjacent to the carbonyl group because they are in downfield region and since they are split to a triplet, they are adjacent to a carbon that has two protons.
The protons at 1.5-1.6 ppm are away from the carbonyl because they are not so downfield. The peak is a …show more content…
No band around 1700 cm-1 (no C=O) No band around 1600 cm-1( no C=C )
Proton NMR spectrum:
Triplet at 3.6 ppm for two protons (CH2 coupled to 2H, deshielded) The -OH deshields the -CH2- but does not couple to it.
Singlet at 2.3 ppm for one proton (OH)
Multiplet(six lines, since it has a total of 5 neighbour protons) at 1.6 ppm for two protons (CH2 coupled to 5 H) triplet at 0.9 ppm for three protons (CH3 coupled to 2H)

13 C-NMR spectrum: 13 C-NMR shows 3 types of C: 69 ppm (deshielded C) and 2 other hydrocarbon C (27 and 10 ppm)
Mass spectrum: In the MS, the molecular ion occurs at m/z = 60 indicating the MW = 60 1-propanol
Mass fragmentation: PROBLEM 9
Molecular formula C5H10O

IR spectrum 1H NMR spectrum 13C NMR spectrum Positive EI Mass Spectrum
Degree of unsaturation: 1
IR Spectrum : The IR shows band near at 1715 cm-1 which indicates C=O. No band around 3500 cm-1 (no -OH or -NH ) and no band in region 1250-1000 cm-1( no