5. Mass spectrum
Mass spectrum
A graph of the relative abundance of each
fragment plotted against its m/z value.
The x-axis, in units of m/z, represents the
formula weight of the detected ions.
The y-axis represents the relative abundance
of each detected ion.
The most abundant ion (tallest peak) is called
the base peak.
The peak at high m/z, is known as molecular
ion peak
.
Fragmentation
peaks
Base
peak
6. 13 RULE
The Rule of Thirteen is so named because, to generate a base formula
(containing only carbon and hydrogen); the molar mass of the substance
is divided by 13. The numerator from division gives the number of carbon
atoms in the base formula. The remainder, when added to the numerator,
gives the number of hydrogen’s in the base formula.
M+/13 H=n+r CnHn+r
If you have heteroatoms, you adjust the formula. For example:
For O, add O and subtract CH4.
For N, add N and subtract CH2.
For 35Cl, add Cl and subtract C2H11,etc.
7. Nitrogen rule
If a compound contain odd number of nitrogen then, its molecular ion peak
will appear at an odd mass value.
If a compound contain even number of nitrogen or no nitrogen then, its
molecular ion peak will appear at an even mass value.
This rule is very useful determining the nitrogen content of an unknown
compound.
8. Hydrogen Deficient Index (HDI)
OR
Index Of Hydrogen Deficiency
Or
Degree Of Unsaturation
Its used to determine how many rings, double bonds, and triple bonds are
present in the compound to be drawn.
Formula for HDI=(x+1)-1/2(y)+1/2(z)
x=tetravalent y=monovalent z=trivalent
10. Molecularion
peakat86
By applying above three rules on
mass spectrum
For carbon
13 rule=molecular ion peak/13
13)86(6 numerator (n)
-78
8 reminder(r)
It means over compound have 6
carbon.
For hydrogen = n+r
=6+8=14
So base formula is C6H14
HDI=(x+1)-1/2(y)+1/2(z)
=(6+1)-1/2(14)+1/2(0)
=(7)-(7)+(0)
HDI=0
HDI is zero so there will be no
unsaturation .
Nitrogen rule as its molecular ion peak
does not appear at an odd mass
value, so presence of single nitrogen
is not possible
Finally we can say that it’s a compound that doesn’t have any
ring or any double bond
Also we knew that no single nitrogen is present and our
comound have C6H14 no any other hint is given for
heteroatom.
M-15
M-29
M-43
M-57
11. • Molecular ion peak observed,
their intensity is inversely
proportional to the No: of carbon
atoms.
• Their fragments are at M-15,
M-29, M-43 etc.
• Base peak is observed at 43, 57
etc. Depend upon quantity of
carbon in the chain.
• Bond cleavage takes place at
the site of branches.
• Due to stable carbocation the
further fragmentation is
favored as a result weak M+
peak occurs.
• No: of branches of alkanes is
inversely proportional to the
intensity of molecular ion peak.
13. • Base peak is at 43 due to
higher abundance of propyl
ion
• CH3-CH2-CH2-CH2-CH3
• Possibility of base peak is:
• CH3-CH2-CH2-CH2-CH3
• CH3-CH2-CH2
+ & .CH2-CH3
pentane
14. Mass spectrum of pentane vs. 2-methylbutane: Both mass spectra are similar to each other with two
notable exception, one the peak in 2-methylbutane at 57 is much more intense than normal pentane and
other is the molecular ion peak is less intense in 2-methylbutane.
.
15. • Fairly prominent M+ peak.
• Terminal alkenes lose allylic cation if possible.
• Most common fragmentation is the cleavage
of allylic bond.
• Positive charge usually remains with the fragments containing double bond in order to give
a resonance stabilized allylic cation.
McLafferty rearrangement may occur when there is a gamma carbon with hydrogen.
• 1-butene and 2-butene Mass spectra are identical –not a good method for alkene
isomer.
17. M+2 (isotopic) peak is observed.
• For bromine isotopic abundance is 1:1 and for chlorine it is of 1:3, due
to natural abundance of halogen (bromine & chlorine).
• Elimination reaction takes place.
• α- cleavage takes place.
• Molecular ion peak is observed.
20. • Intensity of molecular ion peak is depend upon substitute.
• Substitute is inversely proportional to the abundance of molecular ion
peak.
• Primary M+ peak is weak.
• Secondary M+ peak is weakest.
• Tertiary M+ peak is usually absent.
• M-17 peak is observed due to removal of –OH group.
• M-18 peak is observed due to formation of water.
(Elimination reaction)
21. • Base peak is observed at:
• m/z = 31 for primary.
• m/z = 45 for secondary.
• m/z = 59 for tertiary.
Possible Fragmentation
• Elimination.
• α- cleavage.
• β-cleavage.
24. • Base peak is usually lose of alkyl radical.
• The peak formed due to acylium ion is most abundant peak at m/z = 43
• Loss of large alkyl radical favors the ketone in α- cleavage.
• McLafferty rearrangement also occurs.
• α- cleavage takes place.
• Molecular ion peak is observed.
26. M-17 (-OH group) indicates the presence of carboxylic acid.
• Peaks observed at:
• Molecular ion is observed but low abundance.
M-45 (-COOH group) is also observed.
Possible Fragmentation
• α- cleavage.
• β-cleavage.
• McLafferty rearrangement
28. Identify molecule.
Identify the functional group.
Identify molecular ion peak.
Construct the molecule by using base peak, fragmentation & functional
group.
29. • Introduction to Spectroscopy by Pavia.
• Elementary Organic Spectroscopy by Y.R.Sharma.
• Organic chemistry 4th edition by paula yurkanis bruice chapter no#13
mass spectrometry & infrared spectroscopy
• Organic Chemistry by Robert Thornton Morrison, 6th edition.
• Web Sources.
30. I am extremely grateful to almighty ALLAH for His blessings over me
during up and downs of my life.