2013年5月IB化学SL真题下载-Paper3
A1. Compound P contains a carbonyl group (C=O) and has the molecular formula C3H6O.
(a) Draw the two possible structures of compound P.
(b) Explain why the infrared spectra of the structures in (a) are very similar.
(c) Explain how the mass spectra of the structures in (a) can be used to distinguish between them.
(d) Pentan-2-one has the following mass spectrum.
(i) Deduce the formulas of the species with the m/z values at 86, 71 and 43.
m/z = 86:
m/z = 71:
m/z = 43:
(ii) Suggest a reason for the peak at m/z = 43 having an exceptionally high relative abundance.
A2. The diagram shows the apparatus used in column chromatography.
(a) A mixture is run through a chromatography column. Explain how the components of the mixture are separated.
(b) Identify one advantage of using column chromatography rather than thin-layer chromatography.
A3. The low resolution 1H NMR spectrum of compound Q is shown.
(a) Identify what information from the spectrum allows the determination of the relative numbers of hydrogen atoms producing each peak.
(b) Deduce which of the following compounds is Q.
CH3CH2CH3 CH3CH2COCH2CH3 CH3CH2OH
(c) Identify the wavenumbers of two peaks in the infrared spectrum of compound Q,using Table 17 of the Data Booklet.
A4. The electromagnetic spectrum is given in Table 3 of the Data Booklet.
Different types of electromagnetic radiation are used to excite atoms and molecules.
(a) Identify the type of radiation
- (i) whose photons have the lowest frequency.
- (ii) that causes molecules to rotate faster.
- (iii) that causes a change in bond polarity.
(b) Visible radiation causes electronic transitions in atoms. These transitions are responsible for absorption and emission spectra. Identify one similarity and one difference in the appearance of absorption and emission spectra.
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