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This question is about alcohols.<br>a) The alcohols are members of a homologous series.<br>i) Write down the general formula for this homologous series.<br>ii) Write down the names and draw the structures of the two monohydric alcohols (alcohols that have only one $$-\mathrm{OH}$$ group) containing three carbon atoms per molecule.<br>b) On analysis, a 10.00 g sample of an alcohol was found to contain 3.75 g of carbon, 1.25 g of hydrogen, and 5.00 g of oxygen. Use this data to find the empirical formula of this alcohol. Suggest its molecular formula.<br>c) The dissociation energy of ethanol is shown below<br>$$\mathrm{C_2H_5OH(g) \rightarrow 2C(g) + 6H(g) + O(g)}$$, $$\Delta H = +3234\ \text{kJ/mol}$$ of ethanol<br>Using bond energy values (given below) calculate the bond energy of C-O per mole.<br>[Bond energies, H-H = 436 kJ/mol, C-H = 435 kJ/mol, O-H = 464 kJ/mol, C-C = 347 kJ/mol]
Correct Answer: 248
(a)(i) $$\mathrm{C_nH_{2n+1}OH}$$ (that is, $$\mathrm{C_nH_{2n+2}O}$$).<br>(ii) Propan-1-ol, $$\mathrm{CH_3CH_2CH_2OH}$$ (the OH on an end carbon), and propan-2-ol, $$\mathrm{CH_3CH(OH)CH_3}$$ (the OH on the middle carbon).<br>(b) Moles of C $$= \frac{3.75}{12} = 0.3125$$, of H $$= \frac{1.25}{1} = 1.25$$ and of O $$= \frac{5.00}{16} = 0.3125$$, a ratio of $$1 \colon 4 \colon 1$$. The empirical formula is $$\mathrm{CH_4O}$$, and since this already fits $$\mathrm{C_nH_{2n+1}OH}$$ with $$n = 1$$, the molecular formula is $$\mathrm{CH_3OH}$$ (methanol).<br>(c) Ethanol contains one C-C, five C-H, one C-O and one O-H bond, so $$347 + 5(435) + E + 464 = 3234$$. This gives $$2986 + E = 3234$$, so the C-O bond energy is $$E = 248\ \text{kJ/mol}$$.
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