Okay, let's pick up where we left off. 5 (a) iii) The reaction of compound E ($\text{CH}_3\text{CH}=\text{CH}_2$, propene) with $\text{Br}_2/\text{CCl}_4$ is an electrophilic addition reaction. Step 1: The $\pi$ electrons of the alkene attack one bromine atom, forming a cyclic bromonium ion and releasing a bromide ion. $$ \text{CH}_3\text{CH}=\text{CH}_2 + \text{Br}_2 \longrightarrow \begin{bmatrix} \text{CH}_3\text{CH} \\ \text{CH}_2 \end{bmatrix}^{\text{+}} \text{Br} + \text{Br}^- $$ Step 2: The bromide ion ($\text{Br}^-$) acts as a nucleophile and attacks one of the carbons of the cyclic bromonium ion from the opposite side, opening the ring. $$ \begin{bmatrix} \text{CH}_3\text{CH} \\ \text{CH}_2 \end{bmatrix}^{\text{+}} \text{Br} + \text{Br}^- \longrightarrow \text{CH}_3\text{CH}(\text{Br})\text{CH}_2\text{Br} $$ The product is 1,2-dibromopropane. 6 (a) i) Step 1: Calculate the moles of each element in 100 g of the compound. Given percentages: C = 35.0%, H = 6.6%, Br = 58.4%. R.A.M: C = 12, H = 1, Br = 79.9. Moles of C $= \frac{35.0 \text{ g}}{12 \text{ g/mol}} = 2.9167 \text{ mol}$ Moles of H $= \frac{6.6 \text{ g}}{1 \text{ g/mol}} = 6.6 \text{ mol}$ Moles of Br $= \frac{58.4 \text{ g}}{79.9 \text{ g/mol}} = 0.7309 \text{ mol}$ Step 2: Divide by the smallest number of moles to find the simplest mole ratio. Smallest number of moles is 0.7309 (for Br). Ratio of C $= \frac{2.9167}{0.7309} \approx 3.99 \approx 4$ Ratio of H $= \frac{6.6}{0.7309} \approx 9.03 \approx 9$ Ratio of Br $= \frac{0.7309}{0.7309} = 1$ Step 3: Write the empirical formula. The empirical formula is $\boxed{\text{C}_4\text{H}_9\text{Br}}$. ii) Step 1: Calculate the empirical formula mass. Empirical formula mass of $\text{C}_4\text{H}_9\text{Br} = (4 \times 12) + (9 \times 1) + (1 \times 79.9) = 48 + 9 + 79.9 = 136.9 \text{ g/mol}$. Step 2: Compare the empirical formula mass with the given molecular weight. Given molecular weight = 136.9. Since the empirical formula mass (136.9) is equal to the molecular weight (136.9), the molecular formula is the same as the empirical formula. The molecular formula is $\boxed{\text{C}_4\text{H}_9\text{Br}}$. 6 (b) i) To distinguish between $\text{CH}_3\text{CH}_2\text{CHO}$ (propanal, an aldehyde) and $\text{CH}_3\text{COCH}_3$ (propanone, a ketone): Use Tollens' reagent. • Propanal (aldehyde) will react with Tollens' reagent (ammoniacal silver nitrate) upon warming to form a silver mirror on the inside of the test tube. • Propanone (ketone) will show no reaction. ii) To distinguish between $\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$ (but-1-yne, a terminal alkyne) and $\text{CH}_3\text{C}\equiv\text{CCH}_3$ (but-2-yne, an internal alkyne): Use ammoniacal silver nitrate solution. • But-1-yne (terminal alkyne) will react to form a white precipitate of silver acetylide. • But-2-yne (internal alkyne) will show no reaction. 6 (c) Organic chemistry is the branch of chemistry that studies the structure, properties, composition, reactions, and preparation of carbon-containing compounds. Two unique properties of carbon: 1. Catenation: Carbon atoms can form strong covalent bonds with other carbon atoms, leading to the formation of long chains, branched chains, and rings. 2. Tetravalency: Carbon atoms can form four covalent bonds, allowing for a vast diversity of stable molecular structures. 6 (d) i) A functional group is a specific group of atoms within a molecule that is responsible for the characteristic chemical reactions of that molecule. ii) The systematic name of the compound $\text{CH}_3\text{CH}(\text{CH}_3)\text{CONH}_2$ is 2-Methylpropanamide. iii) The compound $\text{CH}_3\text{CH}(\text{CH}_3)\text{CONH}_2$ exhibits optical isomerism because it contains a chiral carbon atom (the carbon bonded to the methyl group, hydrogen, and the amide group). 6 (e) i) The asymmetric carbon atom in $\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}$ is the central