Question

3. There are some additional "rules" that help in the in terpret at ion of mass spectra a. Alcohols typically lose the hydrogen on the oxygen and often water (mass of 18) and have a very small parent ion. b. Compounds that contain an odd number of nitrogen have an odd mo lecular mass C.Carbonyl compounds tend to cleave next to the C O. Based on these new rules determine which of the following spectra correspond to n-butyl amine, butanol, and diet hyl ketone. Explain your reasoning 100 80 40 20 0.0 00 20 40 60 80 100 m/z 40 20 10 20 406070 80 m/z 100 80 60 40 20 0.0 15 30 5 60 7590 m/z

Answer 1

Spectra 1:

It corresponds to diethyl ketone as we can easily see the molecular ion peak at 86. Base peak is seen at m/z 57 which occurs due to fragmentation of ethyl group from compound (M-C2H5,M-29) i.e clevage next to carbonyl group. Further fragmentation of carbonyl group from base peak compound gives peak at 29.

Spectra 2 :

It corresponds to butylamine. We can see small molecular ion peak at 73 which is odd number. Alpha cleavage is dominant in aliphatic amines i.e C-C bond next to C-N bond. Base peak is seen at m/z = 30 (M-CH2CH2CH3)

Spectra 3 :

It corresponds to butanol. Identification of molecular ion peak is difficult in alcohols as it easily loses H-atom from alpha carbon. The peak at m/z 56 corresponds due to loss of water molecule (M-18). Further alcohols can easily loss methyl groups alongwith H2O to m/z peak at 41,loss of ethyl group m/z peak at 43. The peak at m/z 31 is really prominent and characteristric peak of primary alcohols,which occurs due to beta cleavage of compound to resonanace stabilised cation.

Spectra 1 alpha-clcavage z-57 diethylketone m Z base peak Spectra 2 im z - 30 basc peak butylamine m z-73 Spectra 3 M-H20 butanol beta-cleavage miz- 31 m z-43