Here are the solutions to the organic chemistry questions: SECTION C: ORGANIC CHEMISTRY 5. (a) Compound A of molecular mass 103 g/mol contains carbon, hydrogen, nitrogen and oxygen. On analysis it is found to contain 46.6% carbon, 8.74% hydrogen, 13.6% nitrogen and 31.1% oxygen. (i) The empirical formula of A: Step 1: Assume 100 g of the compound. Mass of C = 46.6 g Mass of H = 8.74 g Mass of N = 13.6 g Mass of O = 31.1 g Step 2: Convert mass to moles using atomic masses (C=12.01, H=1.008, N=14.01, O=16.00). Moles of C $= \frac{46.6 \, \text{g}}{12.01 \, \text{g/mol}} \approx 3.880 \, \text{mol}$ Moles of H $= \frac{8.74 \, \text{g}}{1.008 \, \text{g/mol}} \approx 8.671 \, \text{mol}$ Moles of N $= \frac{13.6 \, \text{g}}{14.01 \, \text{g/mol}} \approx 0.9707 \, \text{mol}$ Moles of O $= \frac{31.1 \, \text{g}}{16.00 \, \text{g/mol}} \approx 1.944 \, \text{mol}$ Step 3: Divide by the smallest number of moles (0.9707 mol for N) to get the mole ratio. C: $\frac{3.880}{0.9707} \approx 4.00$ H: $\frac{8.671}{0.9707} \approx 8.93 \approx 9$ (rounding to nearest whole number) N: $\frac{0.9707}{0.9707} = 1.00$ O: $\frac{1.944}{0.9707} \approx 2.00$ The empirical formula is $\boxed{\text{C}_4\text{H}_9\text{NO}_2}$. (ii) The molecular formula of A: Step 1: Calculate the empirical formula mass (EFM) of $\text{C}_4\text{H}_9\text{NO}_2$. EFM $= (4 \times 12.01) + (9 \times 1.008) + (1 \times 14.01) + (2 \times 16.00)$ EFM $= 48.04 + 9.072 + 14.01 + 32.00 = 103.122 \, \text{g/mol}$ Step 2: Determine the ratio $n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}$. Given Molecular Mass = 103 g/mol. $n = \frac{103 \, \text{g/mol}}{103.122 \, \text{g/mol}} \approx 1$ Step 3: Multiply the subscripts in the empirical formula by $n$. Since $n=1$, the molecular formula is the same as the empirical formula. The molecular formula is $\boxed{\text{C}_4\text{H}_9\text{NO}_2}$. (iii) How can you identify the presence of nitrogen in compound A? Give the reagents and expected results only. Reagents: Sodium metal, distilled water, iron(II) sulfate solution, concentrated sulfuric acid (or dilute $\text{HCl}$ followed by $\text{FeCl}_3$ solution). Expected results: Formation of a Prussian blue* precipitate or coloration. (This is the Lassaigne's test, where nitrogen is converted to cyanide, which then reacts to form ferrocyanide and finally Prussian blue.) 5. (b) The organic compound P, named 2-hydroxypropanenitrile exhibits isomerism. (i) Draw two isomeric structures of P. 2-hydroxypropanenitrile has the formula $\text{CH}_3\text{CH(OH)CN}$. The central carbon atom (C2) is chiral, as it is bonded to four different groups ($\text{CH}_3$, $\text{H}$, $\text{OH}$, $\text{CN}$). Therefore, it exists as a pair of enantiomers. Here are the two enantiomeric structures (Fischer projections): Isomer 1: ` CN | H-C-OH | CH3 ` Isomer 2: ` CN | HO-C-H | CH3 ` (ii) State the type of isomerism exhibited by the two structural forms of P. The type of isomerism exhibited is $\boxed{\text{Optical isomerism}}$ (a type of stereoisomerism). (iii) Give a test you would use to distinguish the isomers of P in b(i) above. Test: Use a $\boxed{\text{polarimeter}}$. Expected results: One isomer will rotate the plane of plane-polarized light in one direction (e.g., clockwise, dextrorotatory), while the other isomer will rotate the plane of plane-polarized light by an equal amount in the opposite direction (e.g., counter-clockwise, levorotatory).