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A-Level Representing organic molecules and their shapes
What the A-Level syllabus expects for Representing organic molecules and their shapes, and how to practise it.
What the syllabus expects
- Read and apply the naming, general formulae and displayed formulae of these compound families:
- the hydrocarbons, namely alkanes, alkenes and arenes
- the halogen derivatives, namely halogenoalkanes and halogenoarenes
- the hydroxyl compounds, namely alcohols and phenols
- the carbonyl compounds, namely aldehydes and ketones
- the carboxylic acids together with acyl chlorides and esters
- the nitrogen compounds, namely amines, amides, amino acids and nitriles
- Describe sp3 hybridisation with ethane, sp2 hybridisation with ethene and benzene, and sp hybridisation with ethyne.
- Relate the sigma and pi carbon-carbon bonds in ethane, ethene, benzene and ethyne to the shape and bond angles of each molecule.
- Forecast the geometry and bond angles of molecules resembling those in (c).
How it's examined
About 2% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (5 marks)
Because cocaine is a controlled substance, related compounds are employed to give comparable effects. Compound Y (molecular formula C9H11NO2) is such a derivative and serves as a local anaesthetic. Work out the structure of Y from the following 1H NMR measurements. 7.849 ppm, integration 2, doublet. 6.626 ppm, integration 2, doublet. 4.305 ppm, integration 2, quartet. 4.100 ppm, integration 2, singlet. 1.354 ppm, integration 3, triplet.
Show the worked answer
Degrees of unsaturation for C9H11NO2 = (2x9+2+1-11)/2 = 5, consistent with a benzene ring (4) plus one C=O (1). Signal assignment: - 7.849 ppm (2H, doublet) and 6.626 ppm (2H, doublet): an AA'BB' pattern of two pairs of equivalent aromatic H that couple to each other = a 1,4- (para-) disubstituted benzene ring. The upfield pair (6.63) is shielded, indicating an electron-donating group (NH2) on that side; the downfield pair (7.85) is deshielded, next to an electron-withdrawing C=O. - 4.305 ppm (2H, quartet) and 1.354 ppm (3H, triplet): a -CH2CH3 ethyl group. The CH2 quartet is strongly deshielded (~4.3 ppm), so it is O-CH2, i.e. an ethyl ester -COO-CH2CH3. - 4.100 ppm (2H, singlet): two exchangeable, uncoupled protons = -NH2. Putting the pieces on a para-ring: an ester group -COOCH2CH3 on one side and -NH2 on the other. This is ethyl 4-aminobenzoate (benzocaine): H2N-C6H4-COO-CH2CH3. Check formula: ring C6H4 + NH2 + COO + CH2CH3 = C9H11NO2. Correct, and it is a known local anaesthetic.
More A-Level H2 Chemistry topics
The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry