Major Chemistry B.Sc.III ( Semester V)
Name of paper- Spectral analysis
CHE/DSC/T/301
Problems based on Mass Spectroscopy
Exercise 1.Predict the relative abundance of the parent ion in the case of: (a) Propane (b) n-Pentane
Solution
The relative abundance of the propane molecular ion is greater than that of n-pentane. The abundance of the molecular ion peak can be increased with respect to the abundance of fragment ions by recording the spectrum at low ionisation potential, i.e., by bombarding the sample with low-energy electrons.
Exercise 2. Calculate molecular formula of a compound with molecular mass 94 amu.
Solution. n = 7 and r =3
The base formula is C7H10 The index of hydrogen deficiency or DBE = (7-3+2)/2 = 3
Exercise3. Identify the compound having m/e values at 72, 71, 44 (100%), 43, 29 etc.
Solution. Here a signal at 44 (100%) is due to McLafferty rearrangement ion. Clearly, it is a straight chain aldehyde as M⁺ peak at 72 also accompanies M⁺ − 1 peak at 71. Hence formula of the compound is CH₃CH₂CH₂CHO.
Exercise 4. Determine the molecular formula of the compound from the following mass spectrum
Solution. Since the pair of peaks are of equal intensity, it is a bromo compound. Two isotopes of bromine are Br⁷⁹ and Br⁸¹. The pair on the extreme right is due to M⁺ and (M⁺ + 2) peaks. The spectrum corresponds to the molecular formula C₂H₅Br.
Exercise5. Explain its fragmentation pattern of pentanal.
Solution: Fragmentation pattern of pentanal is as follows
Exercise6. Explain its fragmentation pattern of 3- pentanone
Solution: Fragmentation pattern of 3-pentanone is as follows
Exercise7. Predict different peaks in the mass spectrum of ethyl chloride.
Solution. Peaks are seen at:
64 (M⁺)
66 (M⁺ + 2) (intensity ratio 3 : 1)
29 (Base peak)
The loss of Cl• (chlorine radical) occurs.
Exercise8. Explain its fragmentation pattern of 1-bromohexane.
Solution: Various fragmentation modes of 1-bromohexane shown below.
Exercise9. Explain its fragmentation pattern of 1-butanol.
Solution: Various fragmentation modes of 1-butanol shown below.
Exercise10.The following figure shows the mass spectrum of a saturated hydrocarbon (containing only carbon and hydrogen with only single bonds between carbons, not double bonds).Draw five different structures that would have the molecular weight of this compound. Choose three smaller m/z values from the spectrum and draw one structure for each of them. Note that these fragments will not have complete Lewis structures.
Solution: Molecular Weight = 114, which cooresponds to a C8H18 hydrocarbon. There is the possibility of 18 isomers, but here are a few isomers:
Exercise11. What are the masses of all the components in the following fragmentations?
Exercise 12: 5-Chloro-2-pentanone has the mass spectrum shown. Which peak represents the M+? Which is the base peak? Why is there a peak at 122? Explain what the fragment for the base peak would be.
Solution: M+ = 120
Base peak = 43
The m/z peak at 122 is the M + 2 peak. It occurs because chlorine has two isotopes 35C and 37C in a 3:1 ratio. The m/z = 43 occurs due to the alpha cleavage. The acylium ion has an m/z of 43. This fragment is particularly stable due to resonance.
Mass Spectroscopy Part-I
Mass Spectroscopy Part-II
Mass Spectroscopy Problems
Mass Spectroscopy Multiple choice questions
Proton magnetic resonance spectroscopy
