Question

1. Identify the number of compounds in the image.

Compound 15: 2,3-dibromobutane a. Draw out the structure. b. Label the molecular ion peak and base peak. C. Assign structures to the fragment ion peaks at m/e 135 & m/e 137, m/e 107 & m/e 109, and m/e 55. Mass Spectrum 100 80 135 137 Relative Intensity 107 109 214 216 20 40 60 80 100 120 140 me 160 180 200 220 240 260

2. Find the RT of each compound in the image.

3. Find the percent of each compound in the image.

Answer 1

Br 2,3-dibromobutane CH3 Hąc Br M+ion peak (i.e. molecular ion peak) is 214 216 is the M+2 peak. Due to presence of Bromine the M+ and M+2 ion peaks have almost the equal intensity. Base peak is the peak with 100% intensity. Base peak is 55. Br нс сн H2CH CH3 m/z = 135 and 137 This peak is formed due to loss of one Br. As there is one more Br present it show m+2 peak with equal intensity. m/z = 107 and 109 This peak is formed due to loss of CH3-CHBr radical. As there is one more Br present it show m+2 peak with equal intensity. BV CH2 HỌC - HCVCHz - H m/z = 55

2) 1) As there are two main peaks with sheded area there are two compounds. 2) RT for first compound is 47 s and for second compound RT is 60 s. 3) Intiaration value for first peak area under the curve is = 5.187 e-V and that for the second curve is = 7.782 e-V. Thus the total area under curve is = 7.782+5.187 = 12.969 e-V Now the % of first Compound with RT:47s = (5.187/12.969)*100=40% the % of second Compound with RT:60s = (7.782/12.969)*100=60%