Question

what are possible structures for each spectra

100 MS 80 - 60 Relative Intensity 40 20- 0 10 20 30 40 60 70 80 50 m/z 100- 80 Relative Intensity 40 20 0 10 15 20 25 30 35 40 50 55 60 65 70 75 80 45 m/z

Answer 1

What are possible structure for each spectra! Solution: 1. Spectra is due to 2-chloropropane 100- 80 60 Relative Intensity 40 20- o torte 10 20 30 40 60 70 80 50 m/z The Mass Spectrum of 2-Chloropropane 100 43 CH3CHCH: CI Relative abundance 50 41 63 78 65 80 0 0 10 20 30 40 50 60 70 80 90 100 m/z

CH3 CH3CH + CI m/2 = 43 CH3 CH: CHỊCH + CHỊCH m/2 = 78 m/2 = 80 CH3 CHỊCH-CI 2-chloropropane CH3 CH3 CHẶCHVOS + CHICAGO m/z= 78 m/2 = 80 + "CH3 CH2CH261 + CH3CH2261 m/z=63 m/z= 65 ΕΙ - H2 ܘܬ Leo m/z = 43 m/z = 41 + H2C: 4 m/z = 27 Peak at m/z = 80 is because of isotopic peak due to M+2 shown by chlorine.

2. Spectra is due to benzene 100 80 60 Relative Intensity 40 20- 10 15 20 25 30 35 60 65 70 75 80 40 45 50 55 m/z Mass spectrum of benzene: 100- 78 80- 60- 40- 771 20- 51, 39 L 15 30 45 60 75 90 m/z

Molecular ion fragmentation can produce characteristic ions: 1. m/z 78 (M+H)* 2. m/z 77 (M-H)", 3. m/z 51 (C4H3+), 4. m/z 26 (HC=CH) and 5. m/z 39 (C3H3+) aromatic ions. Peak at m/z = 79 is because of isotopic peak due to M+1 shown by Carbon and hydrogen. Benzene contains only carbon and hydrogen. Relative intensity (%) of the M+1 signal at m/z 79 is 6.55 % the intensity of signal at m/z 78. Useful information: Isotope abundance factors (%) of M+ 1 relative to M for CnHm: C 1.08n, H 0.012m. Verify the relative intensity of M+1 peak is 6.55 % the intensity of signal at m/z 78: Molecular formula is C6H6 Relative intensity (%) of the M+1 signal = 6 x 1.08 + 6 * 0.012 = 6.55 %