IB Docs (2) Team MC test: Spectroscopic identification of organic compounds
Multiple choice test on 11.3 Spectroscopic identification of organic compounds
Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).
If you get an answer wrong, read through the explanation carefully to learn from your mistakes.
Cholesterol has the molecular formula C27H46O.
What is the index of hydrogen deficiency of cholesterol?
Oxygen does not affect the IHD. If it had been fully saturated it would be C27H56. A difference of ten H atoms is equivalent to an IHD of 5. You can also use the formula IHD = (2x + 2 − y) ÷ 2 (where x is 27 and y is 46).
Which compound has an IHD different to the other three?
C2H7N is fully saturated and has an IHD of 0. The other three compounds all have an IHD of 1.
Retinal has the molecular formula C20H28O. It contains five C=C double bonds and one C=O bond. How many rings does retinal possess?
The IHD of retinal is 7. Since it contains 6 double bonds it must contain one ring.
Which row gives the correct information about 1H NMR and IR spectroscopy?
| 1H NMR | spectroscopy | Infrared | spectroscopy | ||
| Row | Type of transition | Frequency | Type of transition | Frequency | |
| 1 | Nuclear spin | Higher than IR | Molecular vibrations | Lower than 1H NMR | |
| 2 | Molecular vibrations | Higher than IR | Nuclear spin | Lower than 1H NMR | |
| 3 | Nuclear spin | Lower than IR | Molecular vibrations | Higher than 1H NMR | |
| 4 | Molecular vibrations | Lower than IR | Nuclear spin | Higher than 1H NMR |
1H NMR absorbs in the radio region of the electromagnetic spectrum which has a much lower energy (and frequency) than the infrared region.
A compound shows a molecular ion peak in its mass spectrum at 58 m/z, an IHD of 1 and a sharp absorption at about 1725 cm−1 in its infrared spectrum.
Which compounds are consistent with this data?
I. 2-propen-1-ol, CH2CHCH2OH
II. propanone, CH3COCH3
III. propanal, CH3CH2CHO
All three compounds have a molar mass of 58 g mol−1 and an IHD of 1 but only the aldehyde and ketone have an absorption at about 1725 cm−1 in their infrared spectrum due to the C=O bond.
Which compound gives the 1H NMR spectrum shown?
There are two signals in the ratio of 3:1 so the protons must be in two different chemical environments in the ratio of 3:1.
Butan-2-ol and 2-methylpropan-1-ol are isomers. Their mass spectra show that both have a molecular ion peak at 74 m/z and their IR spectra both show a broad absorption at about 3300 cm−1 due to the −OH bond. Which statement about their 1H NMR spectra can be used to distinguish between the two isomers.
The protons in butan-2-ol are in 5 different chemical environments 
in the ratio of 3:3:2:1:1
The protons in 2-methylpropan-1-ol are in 4 different chemical environments 
in the ratio 6:2:1:1.
What is the simplest ratio of the areas under the signals in the 1H NMR spectrum of pentan-3-one?
Pentan-3-one, CH3CH2COCH2CH3 is symmetrical around the carbonyl group so the the two CH3− groups are in the same chemical environment as are the two CH2− groups giving an actual ratio of 6:4 with the simplest ratio of 3:2.
Which will give a peak at (M−29)+ (where M is the mass of the molecular ion peak of the compound) in their mass spectrum?
I. ethanol, CH3CH2OH
II. ethanal, CH3CHO
III. ethyl ethanoate, CH3COOCH2CH3
A peak at (M−29)+ can be due to the fragment caused by the loss of either C2H5 or CHO as both have a relative mass of 29.
A hydrocarbon contains 85.7% carbon by mass. Its 1H NMR spectrum shows two signals in the ratio of 3:1.
Which compound is consistent with this data?
From the percentage composition the empirical formula is CH2 (so it cannot be propane). The protons in 2-methylprop-1-ene, 
are in two different chemical environments in the simplest ratio of 3:1. But-1-ene will give four signals in the ratio of 3:2:2:1 and propene three signals in the ratio 3:2:1.
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