ndition as using an excess of ammonia ($\\text{NH}_3$) B1","marks":2,"order":1,"markingRubric":null,"explanation":{"methods":[{"id":"direct_recall","label":"Direct Recall of Reaction Mechanism and Conditions","steps":[{"type":"text","order":0,"title":"Identify the mechanism","content":"In **Reaction 2**, the reactant 1-bromobutane ($\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{Br}$) is treated with ammonia ($\\text{NH}_3$) to yield the primary amine butylamine ($\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{NH}_2$). Because the ammonia molecule acts as a nucleophile by donating its lone pair of electrons to the electron-deficient carbon atom and substituting the bromine atom, the reaction mechanism is **nucleophilic substitution**."},{"type":"text","order":1,"title":"Determine the condition","content":"The primary amine product (butylamine) has a lone pair of electrons on the nitrogen atom, making it nucleophilic as well. If it reacts with unreacted 1-bromobutane, further substitution will occur, leading to secondary and tertiary amines. To prevent this and ensure the primary amine is the major product, the reaction must be carried out using an **excess of ammonia**."}]}],"generatedAt":"2026-06-11T08:42:31.683Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"cf79fc14-e9cc-4cf5-9048-4d0427d9d273","label":"3","content":"Calculate the mass, in grams, of \$\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{NH}_2\$ produced from \$16.4\\text{ g}\$ of \$\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{Br}\$ in **Reaction 2** assuming a \$65.0\\%\$ yield. Give your answer to 3 significant figures. (Relative atomic masses: \$\\text{H} = 1.0\$, \$\\text{C} = 12.0\$, \$\\text{N} = 14.0\$, \$\\text{Br} = 79.9\$)","markscheme":"- Calculates the molar mass of 1-bromobutane ($136.9\\text{ g mol}^{-1}$) and butylamine ($73.0\\text{ g mol}^{-1}$) M1\n- Calculates the moles of 1-bromobutane: $\\text{moles} = \\frac{16.4}{136.9} \\approx 0.1198\\text{ mol}$ M1\n- Calculates the moles of butylamine produced taking into account the $65.0\\%$ yield: $0.1198 \\times 0.650 = 0.07787\\text{ mol}$ M1\n- Calculates the mass of butylamine: $0.07787 \\times 73.0 = 5.68\\text{ g}$ (or $5.69\\text{ g}$ if using $0.120\\text{ mol}$) A1","marks":4,"order":2,"markingRubric":null,"explanation":{"methods":[{"id":"mole_yield_first","label":"Applying Yield to Moles","steps":[{"type":"text","order":0,"title":"Calculate molar masses","content":"To start, we need to calculate the relative formula mass ($M_r$) of the starting material, 1-bromobutane ($\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{Br}$), and the product, butylamine ($\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{NH}_2$).\n\nUsing the given relative atomic masses:\n- **$M_r$ of 1-bromobutane ($\\text{C}_4\\text{H}_9\\text{Br}$):**\n $$M_r = (4 \\times 12.0) + (9 \\times 1.0) + 79.9 = 136.9\\text{ g mol}^{-1}$$\n- **$M_r$ of butylamine ($\\text{C}_4\\text{H}_{11}\\text{N}$):**\n $$M_r = (4 \\times 12.0) + (11 \\times 1.0) + 14.0 = 73.0\\text{ g mol}^{-1}$$\n\n*Awarding 1 mark ($M1$) for calculating both molar masses correctly.*","highlights":[{"text":"Relative atomic masses: \\text{H} = 1.0, \\text{C} = 12.0, \\text{N} = 14.0, \\text{Br} = 79.9","location":"specification"}]},{"type":"text","order":1,"title":"Calculate starting moles","content":"Next, we calculate the number of moles of 1-bromobutane we started with using the formula:\n\n$$\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}$$\n\nSubstitute the given mass of 1-bromobutane ($16.4\\text{ g}$):\n\n$$\\text{moles of 1-bromobutane} = \\frac{16.4}{136.9} \\approx 0.1198\\text{ mol}$$\n\n*Awarding 1 mark ($M1$) for calculating the initial moles of reactant.*","calculator":[{"gdc":{"expression":"16.4 / 136.9","instructions":"Type 16.4 / 136.9 and press ENTER."},"ti84":{"expression":"16.4 / 136.9","instructions":"Type 16.4 / 136.9 and press ENTER."},"desmos":{"expr

Halogenoalkanes

Question 17

Halogenoalkanes Questions

Halogenoalkanes

Question 17

Halogenoalkanes Questions