Question

1. Consider the Mass Spectrum below to answer the following questions: MS-NU-S495 Relative Intensity 0- 10 prote 20 30 40 t 50 to 60 70 m/z 80 90 100 110 a. (1 point) What is the m/z of the molecular ion peak: b. (1 point) What is the m/z value of the base peak: c. (2 point) Explain why do we observe the peak at m/z = 114 d. (2 points) The molecular formula of this compound is C.HSX. Identify X and explain how you identified it based on the mass spectrum. e. (2 points) Draw the chemical structure of the fragment represented by m/z = 77

Answer 1

a. Molecular ion peak is at m/z = 112 since it corresponds to highest m/z value.

b. Base peak is the peak with highest intensity, so base peak is at m/z = 112

c. The peak at M⁺ and (M+2) correspond to peak at 112 and 114 respectively which indicates the presence of isotope of chlorine i.e. Cl ³⁷ along with Cl ³⁵. The ratio of these Cl³⁷ to Cl ³⁵ is 3:1.

d. The X in formula C₆H₅X is Cl. This can be identified based on presence of M and M+2 molecular ion peaks and their relative abundance in the ratio 3:1 which is possible for only Chlorine atom.

e. The m/z at 77 corresponds to C₆H₅⁺ i.e. benzene ring with one less proton.