1. No category

Analytical Chemistry 2 – Other Problems
1.
Infrared spectroscopy is commonly used as an analytical technique by inorganic, physical
and organic chemists.
(a)
Explain why hydrogen bromide is IR active whereas bromine is IR inactive.
(1)
(b)
- HBr
 polar molecule  bond stretches and vibrations involve changes in
bond dipole ∴ IR active
- Br2
 non-polar molecule (non-polar bond)  bond stretches and vibrations
do NOT involve changes in bond dipole ∴ IR inactive
The IR spectrum and mass spectrum of an unknown compound, X, of molecular
formula C5H10O2, are as follows.
don
(i)
In the IR spectrum, identify the bond responsible for each of the absorptions
labelled I, II and III.
I: O-H
II: C-H
III: C=O
(3)
(ii)
In the mass spectrum, deduce which fragments the m/z values at 102, 57 and 45
correspond to.
m/z = 102: C5H10O2+ (or molecular ion, M+)
m/z = 57: C4H9+
m/z = 45: COOH+
Remember that formulas for the fragments must have a + charge. Only ions are
detected in a mass spectrometer - the other non-charged particles are removed
by the vacuum.
(3
(iii)
Deduce the structure of X.
Given ONLY the IR and MS data, the structure CH3CH2CH2CH2COOH would
be acceptable.
NMR data (not given here) provides more information that shows the structure
is actually this structural isomer:
2.
(a)
Explain why the nitrogen molecule, N2, does not absorb infrared radiation. (2)
- N2 has a non-polar bond and any bond stretch/vibration would not change the bond
dipole
(b)
Describe two vibrations in the water molecule that absorb infrared radiation. (2)
- symmetrical stretching (v1 below)
- asymmetrical stretching (v3 below)
- bending (v2 below)
3.
Describe the operating principles of a double-beam IR spectrometer. (4)
- beam of monochromatic radiation (same wavelength) used
- splitter splits (IR) light into two beams (of same wavelength)
- one beam passes through sample and the other passes through the reference
- photomultiplier converts radiation into electrical signal
- detector compares the transmittance of reference and the sample
- IR spectrum produced by changing the wavelength of the radiation used
4.
Consider the IR spectra of the following three compounds.
A = CH3(CH2)3COOH
B = CH3COOC(CH3)3
C = (CH3CH2)3COH
Determine which IR spectrum corresponds to each compound A, B and C. Explain your
reasoning. IR data can be found in Table 17 of the Data Booklet.
Compound
Spectrum
A
I
Reason
-1
- broad peak at 2500-3300 cm  O-H in carboxylic acid
-1
- sharp peak at 1700 cm  C=O
-1
- sharp peak at 1700 cm  C=O (in ester)
B
III
- missing broad peak at either 2500-3300 or 3200-3600
-1
cm  no hydroxyl group
-1
- C-O bond in fingerprint region (<1500 cm )  cannot
use
-1
C
II
- only sample with abroad peak at 3200-3600 cm  O-H
in alcohol
-1
- no sharp peak at 1700 cm  no C=O