Question

peak at 73. base peak 11. Consider the following MS spectrum and data: mt m/z % 15.0 2.9 26.0 2.5 27.0 8.7 28.0 11.6 29.0 13.4 32.0 3.1 39.0 41.0 1.3 42.0 5.3 43.0 100.0 44.0 2.5 57.0 9.5 71.0 1.4 72.0 34.2 73.0 1.7 2.4 Intensity lative 10 15 20 25 30 35 40 45 50 55 60 TTT 65 70 75 a. Calculate) how many carbons are in the molecule. mt=73 m=72 (1.7734.2) x100 =4.5 carbons b. Identify two molecule fragments using the spectra.

Answer 1

a. To calculate the number of C atoms, we first need to make the M+1 signal relative to the M signal. This is achieved as you've done it, by dividing them and multiplying by 100. This yields 4.99. Since the natural abundancy of the isotope ¹³C is 1.1%, we divide this by 1.1 and get 4.54, which means that the number of carbons is either 4 or 5. (This is the number you got). To exactly determine the number of carbons, we need to look at the fragmentation:

The most important fragmentations are the loss of 15 to yield the signal with m/z = 57 and a loss of 14 from that to yield the base peak. These are consistent with losses of -CH₃ and -CH₂, respectively. This means that there are no atoms different from C and H in this molecules (no losses of groups involving any other elements are there), so we can determine the number of C atoms considering the mass of the molecular ion, which is 72, consistent with C₅H₁₂, an alkane with 5 carbons.

b- Discussion about fragments was included in the previous item.