IR SPECTRUM(1)
Spectra obtained from a liquid film of 1-bromobutane. The right-hand part of the of the infrared
spectrum of 1-bromobutane, wavenumbers ~1500 to 400 cm-1 is considered the fingerprint region for
the identification of 1-bromobutane and most organic compounds. It is due to a unique set of complex
overlapping vibrations of the atoms of the molecule of 1-bromobutane.
1-bromobutane, C4H9Br, CH3CH2CH2CH2Br, CH3-CH2-CH2-CH2-Br
Interpretation of the infrared spectrum of 1-bromobutane
The most prominent infrared absorption lines of 1-bromobutane
C-H stretching vibration absorptions peak at wavenumbers ~2845 to 2975 cm-1.
C-H bending/deformation vibrations absorption bands peaking at wavenumbers ~1270 to 1470 cm-1.
C-C-C skeletal vibrations from the CH2-CH2-CH2 grouping occur at wavenumbers ~1370 to 1250 cm-1
and 840 to 790 cm-1.
These absorption bands are typical of most alkyl structures you see in molecules like 1-bromobutane.
Characteristic groups of C-Br stretching vibration absorptions at wavenumbers ~550 to 750 cm-1 that
you expect in 1-bromobutane.
The absence of other specific functional group bands will show that a particular functional group is
absent from the 1-bromobutane molecular structure.
MASS SPECTRUM (2)
1-bromobutane, C4H9Br, CH3CH2CH2CH2Br, CH3-CH2-CH2-CH2-Br
Interpreting the fragmentation pattern of the mass spectrum of 1-bromobutane
[M]+ is the parent molecular ion peaks (M) with m/z values of 136 and 138 corresponding to [C4H9Br]+,
the original 1-bromobutane molecule minus an electron, [CH3CH2CH2CH2Br]+.
There are two possibilities because bromine consists of two isotopes, 79Br and 81Br in the ratio ~1 : 1.
Therefore the molecular ion can be [CH3CH2CH2CH279Br]+ and [CH3CH2CH2CH281Br]+, which should,
and does, show up as a double peak of ~equal heights (~equal abundance).
These are referred to as the M and M+2 peaks respectively, emphasising the two mass unit difference
due to the bromine isotopes in the two molecular ions of 1-bromobutane.
The two bromine isotopes also account for the 'twin peaks' of m/z ions 93 & 95 and 107 & 109.
Bromine consists of two isotopes, 79Br and 81Br in roughly equal proportions, therefore any molecular
ion or fragment containing a bromine atom will show up as a double peak of similar height (abundance)
two mass units apart e.g. m/z ions 93 and 95, and 107 and 109, plus the molecular ion peaks of m/z
values 136 and 138 (but still, all pairs of ~equal height!) in the mass spectrum of 1-bromobutane.
The very small small M+1 and M+3 peaks at m/z 137 and 139, corresponds to an ionised 1-bromobutane
molecule with one 13C atom in it i.e. an ionised 1-bromobutane molecule of formula [13C12C3H9Br]+
Carbon-13 only accounts for ~1% of all carbon atoms (12C ~99%), but the more carbon atoms in the
molecule, the greater the probability of observing this 13C M+1 peak.
1-bromobutane has 4 carbon atoms, so on average, ~1 in 25 molecules will contain a 13C atom.
The most abundant ion of the molecule under mass spectrometry investigation (1-bromobutane) is
usually given an arbitrary abundance value of 100, called the base ion peak, and all other abundances
('intensities') are measured against it.
Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of
1-bromobutane.
Unless otherwise indicated, assume the carbon atoms in 1-bromobutane are the 12C isotope.
Some of the possible positive ions, [molecular fragment]+, formed in the mass spectrometry of 1bromobutane.
The parent molecular ions of 1-bromobutane m/z 136: [C4H979Br]+ and m/z 138 [C4H981Br]+
m/z value of [fragment]+
57, all 12C atoms
138
[molecular fragment]+ [C4H981Br]+
[CH279Br]
[C4H9]+
136
109
[C4H979Br]+
[C4H9]+
107
95
[C2H481Br]+
93
58, with 13C atom
[C2H479Br]+
[CH281Br]
m/z value of [fragment]+
56
55
79
80
81
82
m/z ions of 79 to 82
have a very low abundance, just tiny peaks in the mass spectrum of 1-bromobutane.
[molecular fragment]+ [C4H8]+
[C4H7]+
[79Br]+ [H79Br]+
[81Br]+ [H81Br]+
m/z value of [fragment]+
15
42
40
28
43
[molecular fragment]+ [C3H7]+
[C2H5]+
[C2H4]+
41
[C3H6]+
[C2H3]+
39
[C3H5]+
[C2H2]+
29
[C3H2]+
[CH3]+
27
26
[C3H3]+
Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of 1bromobutane
Atomic masses: H = 1; C = 12; Br = 79 or 81 (1:1 isotope abundance ratio)
Bond enthalpies = kJ/mol: C-C = 348; C-H = 412; C-Br 276
Possible equations to explain some of the most abundant ion peaks of 1-bromobutane (tabulated above)
Formation of m/z 107 and 109 ions:
[CH3CH2CH2CH2Br]+ ===> [C2H4Br]+ + C2H5
C-C bond scission in the parent molecular ion, mass change 136/138 - 29 = 107/109.
Note the twin ~1:1 peaks due to the two bromine isotopes.
The C-Br bond is the weakest bond in the molecule, hence the most likely bond scission is C-Br with 1bromobutane (see below m/z 57 ion).
Formation of m/z 93 and 95 ions:
[CH3CH2CH2CH2Br]+ ===> [CH2Br]+ + C3H7
C-C bond scission in the parent molecular ion, mass change 136/138 - 43 = 93/95.
Note the twin ~1:1 peaks due to the two bromine isotopes.
The C-Br bond is the weakest bond in the molecule, hence the most likely bond scission is C-Br with 1bromobutane (see below m/z 57 ion).
Formation of m/z 57 ion:
[CH3CH2CH2CH2Br]+ ===> [C4H9]+ + Br
This alternative ionisation compared to above is much more likely, C-Br bond (weakest) scission in the
parent molecular ion, mass change 136/138 - 79/81 = 57.
The m/z 57 ion is the base peak ion, the most abundant and 'stable' ion fragment.
The m/z 58 ion is likely to be [13C12C3H9]+ i.e. as above but with a 13C atom in the hydrocarbon
fragment.
The m/z 57 ion can lose a hydrogen atom/molecule to give m/z ions 56 and 55.
There is a low probability that the bromine atom can also be ionised to give m/z 79 and 81 ions - you can
just about make out the tiny twin peaks.
Formation of m/z 56 ion:
[CH3CH2CH2CH2Br]+ ===> [C4H8]+ + HBr
Elimination of hydrogen bromide from the parent molecular ion can also give the m/z 56 ion.
Mass change 136/138 - 80/82 = 56.
There is a low probability that the hydrogen bromide molecule can also be ionised to give m/z 80 and 82
ions - you can just about make out the tiny twin peaks.
Formation of m/z 41 and 39 ions:
Possible reactions include:
m/z 41: [C4H8]+ ===> [C3H5]+ + CH3
m/z 39: [C3H5]+ ===> [C3H3]+ + H2
Formation of m/z 29, 28 and 27 ions:
Possible reactions include:
[CH3CH2CH2CH2Br]+ ===> [C2H5]+ + CH2CH2Br
From bond scission in the parent molecular ion.
m/z 27: [C4H8]+ ===> [C2H3]+ + C2H5
m/z 28: [C4H8]+ ===> [C2H4]+ + C2H4
m/z 29: [C4H8]+ ===> [C2H5]+ + C2H3
C13 NMR SPECTRUM (3)
TMS is the acronym for tetramethylsilane, formula Si(CH3)4, whose 13C atoms are arbitrarily given a
chemical shift of 0.0 ppm. This is the 'standard' in 13C NMR spectroscopy and all other 13C resonances,
called chemical shifts, are measured with respect to the TMS, and depend on the individual (electronic)
chemical environment of the 13C atoms in an organic molecule - 1-bromobutane here.
1-bromobutane, C4H9Br, CH3CH2CH2CH2Br, CH3-CH2-CH2-CH2-Br
Interpreting the C-13 NMR spectrum of 1-bromobutane
As you can see from the diagram above there are 4 different chemical shift lines in the C-13 NMR
spectrum of 1-bromobutane indicating 4 different chemical environments of the 4 carbon atoms of 1bromobutane.
CH3CH2CH2CH2Br
(Note the 4 different colours indicating the 4 different chemical environments of the 4 carbon atoms in
1-bromobutane).
13C chemical shifts (a) to (d) on the C-13 NMR spectrum diagram for 1-bromobutane.
(a) CH3-CH2-CH2-CH2-Br: The 13C NMR chemical shift of 13.2 ppm for the methyl group carbon atom 4.
(b) CH3-CH2-CH2-CH2-Br: The 13C NMR chemical shift of 21.4 ppm for carbon atom 3 of the CH2 group.
(a) CH3-CH2-CH2-CH2-Br: The 13C NMR chemical shift of 35.0 ppm for carbon atom 2 of the CH2 group.
(b) CH3-CH2-CH2-CH2-Br: The 13C NMR chemical shift of 33.4 ppm for carbon atom 1 joined to the
bromine atom.
The carbon-13 NMR spectra provides direct evidence of 4 different carbon atom environments for the 4
carbon atoms in the 1-bromobutane molecule, deduced from the presence of 4 different 13C NMR
chemical shifts (ppm).
H1 NMR SPECTRUM (4)
TMS is the acronym for tetramethylsilane, formula Si(CH3)4, whose protons are arbitrarily given a
chemical shift of 0.0 ppm. This is the 'standard' in 1H NMR spectroscopy and all other proton resonances,
called chemical shifts, are measured with respect to the TMS, and depend on the individual (electronic)
chemical environment of the hydrogen atoms in an organic molecule - 1-bromobutane here.
The chemical shifts quoted in ppm on the diagram of the H-1 NMR spectrum of 1-bromobutane
represent the peaks of the intensity of the chemical shifts of (which are often groups of split lines at high
resolution) AND the relative integrated areas under the peaks gives you the ratio of protons in the
different chemical environments of the 1-bromobutane molecule.
1-bromobutane, C4H9Br, CH3CH2CH2CH2Br, CH3-CH2-CH2-CH2-Br
Interpreting the H-1 NMR spectrum of 1-bromobutane
In terms of spin-spin coupling from the possible proton magnetic orientations, for 1-bromobutane I have
only considered the interactions of non-equivalent protons on adjacent carbon atoms e.g. -CH2-CH3, CH2-CH2-, protons etc.
For relatively simple molecules, the low resolution H-1 NMR spectrum of 1-bromobutane is a good
starting point (low resolution diagram above).
The 9 hydrogen atoms (protons) of 1-bromobutane occupy 4 different chemical environments so that
the low resolution NMR spectra should show 4 principal peaks of different H-1 NMR chemical shifts
(diagram above for 1-bromobutane).
CH3CH2CH2CH2Br
Note the proton ratio 3:2:2:2 of the 4 colours of the 9 protons of 1-bromobutane in the 4 chemically
different proton environments
Chemical shifts (a) to (d) on the H-1 NMR spectrum diagram for 1-bromobutane.
Although there are 9 hydrogen atoms in the molecule, the proton NMR spectrum shows there are only 4
possible different chemical environments for the hydrogen atoms in 1-bromobutane molecule.
The integrated signal proton ratio 3:2:2:2 observed in the high resolution H-1 NMR spectrum,
corresponds with the structural formula of 1-bromobutane.
The high resolution 1H NMR spectrum of 1-bromobutane
The high resolution spectra of 1-bromobutane also shows 4 groups of proton resonances and in the
3:2:2:2 ratio expected from the structural formula of 1-bromobutane, but we can now consider the
splitting of resonance lines from the spin-spin coupling in the molecule of 1-bromobutane.
The ppm quoted on the diagram represent the peak of resonance intensity for a particular proton group
in the molecule of 1-bromobutane - since the peak' is at the apex of a band of H-1 NMR resonances due
to spin - spin coupling field splitting effects - see high resolution notes on 1-bromobutane below.
So, using the chemical shifts and applying the n+1 rule to 1-bromobutane and make some predictions
using some colour coding! (In problem solving you work the other way round!)
1H NMR resonance (a) 1H Chemical shift 0.90: CH3-CH2-CH2-CH2-Br
The 1H resonance is split into a 1:2:1 triplet by the adjacent CH2 protons (n+1 = 3).
Evidence for the presence of a CH2 group in the molecule of 1-bromobutane
1H NMR resonance (b) 1H Chemical shift 1.31 ppm: CH3-CH2-CH2-CH2-Br
This resonance is split into 1:5:10:10:5:1 sextet by the CH3 and CH2 protons on either side (n+1 = 6).
Evidence for the presence of a CH3-CHx-CH2 grouping in the molecule of 1-bromobutane (x can be 1 or
2, as in this case).
1H NMR resonance (c) 1H Chemical shift 1.82 ppm: CH3-CH2-CH2-CH2-Br
This resonance is split into 1:4:6:4:1 quintet by the CH2 and CH2 protons on either side (n+1 = 6).
Evidence for the presence of a CH2-CHx-CH2 grouping in the molecule of 1-bromobutane (x can be 1 or
2, as in this case).
1H NMR resonance (d) 1H Chemical shift 3.51 ppm : CH3-CH2-CH2-CH2-Br
The 1H resonance is split into a 1:2:1 triplet by the adjacent CH2 protons (n+1 = 3).
Evidence for the presence of another CH2 group in the molecule of 1-bromobutane.
Note the decreasing effect on the 1H chemical shift as the proton is further from the more
electronegative bromine atom in 1-bromobutane.
(1) https://docbrown.info/page06/spectra2/1-bromobutane-ir.htm
(2) https://docbrown.info/page06/spectra2/1-bromobutane-ms.htm
(3) https://docbrown.info/page06/spectra2/1-bromobutane-nmr13c.htm
(4) https://www.docbrown.info/page06/spectra2/1-bromobutane-nmr1h.htm
(5) https://www.chemicalbook.com/SpectrumEN_109-65-9_IR2.htm
(6) https://www.chemicalbook.com/SpectrumEN_109-65-9_MS.htm
(7) https://www.chemicalbook.com/SpectrumEN_109-65-9_13CNMR.htm
(8) John C. Gilbert & Stephen F. Martin. Experimental Organic Chemistry: A Miniscale and Microscale
Approach. 6th E (2006). p.471
(9)
LASSAIGNE TEST (1)
A small amount of organic substance is fused with a small quantity of sodium metal in a fusion tube. The
red hot fusion tube is then plunged into distilled water and the contents are boiled for a few minutes,
then cooled and filtered. then cooled and filtered.
The filtrate obtained is called sodium fusion extract (SFE) or Lassaigne’s extract. It is usually alkaline. If it
is not alkaline, a few drops of NaOH solution may be added to make it alkaline.
Thus obtained SFE is used to detect the presence of elements like N,S,Cl, Br & I.
The elements in the organic compound react with sodium during fusion reaction as follows:
Na + C + N ------------> NaCN
(if N is present)
2Na + S ----------------> Na2S
(if S is present)
Na + S + C + N --------> NaSCN (if both N & S are present & insufficient amount of Na is used)
Na + X -------------------> NaX
(If halogens are present)
Where
X = Cl/Br/I
Hence SFE may contain any of or all of ionic forms of respective elements
The sodium fusion extract, SFE can be used to detect the presence of chlorine, bromine and iodine but
not fluorine. To detect their presence, the SFE is first acidified with HNO3 and then added with AgNO3
solution.
i) The formation of a curdy white precipitate that is soluble in NH4OH indicates the presence of chlorine
in the organic compound.
lassaigne's test for chlorine
ii) The formation of a pale yellow precipitate that is partially soluble in NH4OH confirms the presence of
bromine.
lassaignes test for bromine pale yellow precipitate
iii) Where as the formation of a yellow precipitate insoluble in NH4OH confirms the presence of iodine in
the organic compound.