质谱同位素

e.g. k=370;j=13;n=12,13
R(370,13)=C(357,13)/[C(357,13)+C(358,12)] =(2.416798e-3)*(1.107e-2)/[(2.675386e5)+(3.287701e-1)] Q(370,13)=2*R(370,13)
源自文库
• The interesting features of the 13C composition: 1.The highest-mass isotopic peak has a 13C composition of 2 ------the highest-mass peak is made of all-heavy isotopes,so both carbon atoms must be 13C 2.The odd-mass peaks all have a 13C compositon of 1 ----the Br3Cl3 has a strictly even-mass isotopic distribution 3.The even-mass peaks have a very small contribution from 13C ---- 13C forms a small percentage of the total carbon ,the probability of finding two 13C atoms in amolecule is very low.
• 2.Traditional methods for isotopic distribution 2.1Binomial Approach---two isotope forms e.g.Cl2,Br2
2.2Polynomial Approach ---several nonmonoisotopic elements (a1+a2+a3+…)m(b1+b2+b3)n(c1+c2+c3)o+…
Step3:Add the mass of the isotope of interest to the mass---13C,distribution in Step2 ,shifted by 13mass units
Step4:Divide the intensities of the peaks in Figure 1c by the intensities of the corresponding peaks in Figure 1a
Isotopic Composition and Accurate Mass
1. Carbon:An X+1 Element e.g. A total of 1000CH4 molecules ,there will be 11 molecules containing 13C (0.011070) instead of 12C，therefore，the ratio of relative intensities of the peaks at m/z 16 and m/z 17 is defined by the ratio 989/11 or by usual normalization 100/1.1.
Step5:Multiply this result by the elemental isotopic abundance for the isotope of interest for the atom removed in step 2--13C(0.01107)
Step6:Multiply this result by the number of atoms of the element under consideration in the formula for the parent compound---2
Graphical form ：PbCl2
e.g. C19H35O2Cl3 (A12C+A13C)19(A1H+A2H)35(A16O+A17O+A18O)2(A35Cl+A3 )3 7Cl *The number of terms would be 219*235*32*23 =1.297*1018
When applied to isotopically complex molecules, polynomial based methods have very unfavorable scaling properties because the number of terms to be calculated undergoes a combinatorial explosion as molecular weight increases and calculation time and memory requirements rapidly grow to levels that make the method impractical.
PM+1 / PM = w[c/（100-c）] PM+2 / PM =[ w（w-1）c2 ]/[2（100-c）2] e.g. C60 PM+1 / PM = 60*1.1/98.9=0.667 If the monoisotopic peak at m/z 720 due to 12C60 is regard as 100%，the M+1 peak due to 12C 13C will have 66.7% relative intensity. 59 In a simplified manner ，60*1.1%=66%. *It is helpful to read out the PM+1 / PM from a mass spectrum to calculate the approximate number of carbon atoms .
r=c/（100-c） PM =[(100-c)/100]w ） PM+1 = w*[c/（100-c）] [(100-c)/100]w
r ： is the ratio of 13C and 12C PM ：is the probability to have only 12C in a molecular ion M consisting of w carbons PM+1 ： is the the probability to have one 13C
• 3 An algorithm to calculate the isotopic composition of individual isotopic peaks----performs a direct calculation of the probabilities (steps 1-5)and then folds the result with the chemical formula to yield the isotopic composition of the isotopic peaks(step 6) • The compound under consideration ----parent compound (C2Br3Cl3) • Product compound -----CBr3Cl3
• Proof probability matrix
370(C2Br3Cl3)is the combination of (i,j)=(357,13) with (358,12)
• K---molecular weight of the selected isotopic peak of the parent compound • j---the atom weight of the isotope of the atom • n---is the dummy variable used to generate the sum(12,13) • R(k,j)---- is the probability that the atom will be of isotopic mass j if it is removed at random from a parent ion of isotopic mass k.
The Algorithm
Step1:Compute the isotopic distribution of the parent compound---- C2Br3Cl3
Step2:Remove the atom and compute the isotope distribution of the product compound ---CBr3Cl3