The mass spectrum of an unknown compound had the following relative intensities
ID: 520624 • Letter: T
Question
The mass spectrum of an unknown compound had the following relative intensities for the M (m/z=86), M+1, and M+2 peaks, respectively: 18.5, 1.15, and 0.074 (percentage of base peak). From the partial list of isotopic abundance ratios below, determine the molecular formula of the unknown.
Isotopic-Abundance Ratios (M=100%)
Formula
M+1
M+2
C4H6O2
4.50
0.48
C4H8NO
4.87
0.30
C4H10N2
5.25
0.11
C5H10O
5.60
0.33
C5H12N
5.98
0.15
C6H14
6.71
0.19
Isotopic-Abundance Ratios (M=100%)
Formula
M+1
M+2
C4H6O2
4.50
0.48
C4H8NO
4.87
0.30
C4H10N2
5.25
0.11
C5H10O
5.60
0.33
C5H12N
5.98
0.15
C6H14
6.71
0.19
Explanation / Answer
the m/z value of Molecular ion peak is: 86. This is even number. so according to nitrigen rule, the compound must be having zero or even no. of nitrogens.
We consider the intensity of the M ion peak is 100%, then relative intensities for M+1 and M+2 peaks will be 6.21% and 0.4% respectively. By using the relative intensity of the M+1 peak, we can find the number of carbon atoms present in the unknown compound. (since relative natural abundance of 13C isotope is 1.1%) here,
(M+1)%/1.1% = 6.21/1.1 = 5.64. so this compound containing 5 or 6 carbons.
By using the relative intensity of the M+2 peak, we can check the presence of S, Cl, or Br. since relative intensity is <4%, here are no S, Cl, or Br atoms.
From these data, we expect the molecular formula of the compound will be C5H10O or C6H14.
The Hydrogen Rule states that the maximum number of hydrogens in the molecular formula is (2C+N+2).where, C: no. of carbons, N: no. of nitrogens.
for C5H10O: 2(5)+0+2 = 12, Double bond equivalents (DBE) = C – (H/2) + (N/2) + 1 = 5--(10/2)+(0)+1 = 1
Based on Isotopic-Abundance Ratios (given in table), molecular formula of unknown compound will be C5H10O.
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