Question

1. For the ethylbenzene molecule, find the intensity ratios of the peaks at m/z = 106, m/z = 107 and m/z = 108 that you would expect due to natural carbon isotope distribution.

2. Write down the fragmentation pathway(s) that give(s) rise to the structure with m/z = 77 and the structure with m/z = 51 in the EI-MS of ethylbenzene.

3. In a mass spectrometry experiment, what peaks does the ion CH3Cl+ give rise to? - List m/z and relative intensities.

Answer 1

1.

The molecular formula of ethyl benzene is C8H10. The carbon atom has two isomers with distribution ratio of 98.9% : 1.1% for C12 and C13 isotopes. The C11 and C14 isotopes are radioactive and present in nature in trace amount.

Thus, ethyl benzene will give peaks at M (m/z=106) and M+1 (m/z=107). No peak will be observed at M+2 (m/z=108)

The intensity ratio of (M+1 / M ) is calculated as follows:

M+1 / M = Number of carbon atoms x (percentage of C13/percentage of C12)

M+1 / M = 8 x (1.1 / 98.9)

M+1 / M = 0.0889 / 1

Intensity ration in terms of percentage is given as follows:

M+1 / M = 8.89 % / 100%

Therefore, peak at m/z=107 will have intensity of 8.89% of the peak at m/z=106.

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2.

The molecular ion of ethyl benzene loses ethyl fragment to give fragment with m/z = 77. This fragment further loses a molecule of acetylene to give a fragment C₄H₃⁺ having m/z = 51

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3.

The molecule of CH3Cl contians a chlorine atom. The natural abundance of isomers of chlorine is 3:1 (Cl³⁵:Cl³⁷).

Therefore MS spectrum of CH3Cl+ will show two peaks at M and M+2 with a intensity ratio of 3:1

M of CH3Cl has m/z of 50 and M+2 has m/z of 52.