(a)(i) List two characteristics of homologous series.
(ii) Consider the compound represented by the following formula: CH\(_3\)(CH\(_2\))\(_2\)CH\(_3\). I. Which homologous series does the compound belong? II. Write the structures of three possible isomers of the coumpound. III Name the three possible isomers in (a)(ii)II.
(b) Write the structure of the major product formed in each of the following reactions: (i) ethanol with excess acidified potassium tetraoxomanganate (VII);
(iii) ethanol with propanoic acid in the presence of few drops of concentrated tetraoxosulphate(VI) acid.
(c) Name the major product in each reaction in (b).
(i) Give the IUPAC name of each compound.
(ii) State a chemical test for the functional group in each compound.
(e) An organic compound with relative molecular mass 136 contains 70.57% carbon, 5.90% hydrogen and 23.53% oxygen. Determine its: (i) empopirical formula; (ii) molecular formula. [H= 1.00, C= 12.0, H = 16.0]
(a)(i) Two characteristics of a homologous series
- All members are represented by the same general molecular formula.
- Successive members differ from one another by a
CH2 unit (relative mass 14).
(They also share the same functional group, show a steady gradation in physical properties, and have similar chemical properties.)
(a)(ii) The compound CH3(CH2)2CH3 is butane, molecular formula C4H10.
I. Homologous series: the alkanes (saturated aliphatic hydrocarbons).
II. Possible isomers. The formula C4H10 has only two possible structural (chain) isomers; a third isomer of this formula cannot be drawn. The two are:
CH3-CH2-CH2-CH3CH3-CH(CH3)-CH3 i.e. a central carbon bearing three methyl groups and one hydrogen.
III. Names: (1) butane (n-butane); (2) 2-methylpropane (isobutane). No third isomer exists for C4H10.
(b) & (c) Major products and their names
| Reaction | Structure of major product | Name of product |
|---|
| (i) ethanol + excess acidified KMnO4 (full oxidation) | CH3COOH | ethanoic acid |
| (ii) excess ethane + chlorine in sunlight (monosubstitution favoured) | CH3CH2Cl | chloroethane |
| (iii) ethanol + propanoic acid + few drops conc. H2SO4 (esterification) | CH3CH2COOCH2CH3 | ethyl propanoate |
(d) The two compounds shown
X is (CH3)3COH and Y is a benzene ring bearing -COOH, i.e. C6H5COOH.
(i) IUPAC names: X = 2-methylpropan-2-ol (2-methyl-2-propanol); Y = benzoic acid (benzenecarboxylic acid).
(ii) Chemical test for the functional group:
- X (hydroxyl, -OH): add solid phosphorus(V) chloride, PCl5; misty white fumes of hydrogen chloride are evolved, confirming the -OH group. (Alternatively a piece of sodium gives effervescence of hydrogen.)
- Y (carboxyl, -COOH): add aqueous sodium hydrogentrioxocarbonate(IV), NaHCO3; brisk effervescence of carbon dioxide occurs, and the gas turns lime water milky, confirming the -COOH group.
(e) Empirical and molecular formula (using C = 12.0, H = 1.00, O = 16.0; the third figure in the data is O, not H).
Divide each percentage by the atomic mass:
\[C:\ \frac{70.57}{12.0}=5.88,\qquad H:\ \frac{5.90}{1.00}=5.90,\qquad O:\ \frac{23.53}{16.0}=1.47\]
Divide through by the smallest ratio (1.47):
\[C:\ \frac{5.88}{1.47}=4,\qquad H:\ \frac{5.90}{1.47}=4,\qquad O:\ \frac{1.47}{1.47}=1\]
(i) Empirical formula = C4H4O.
Empirical formula mass \(=4(12.0)+4(1.00)+16.0 = 68\).
\[n=\frac{\text{molar mass}}{\text{empirical mass}}=\frac{136}{68}=2\]
(ii) Molecular formula = (C4H4O)2 = C8H8O2.
