Alkanes

## Identify an alkane from bonding and composition - Alkanes are saturated hydrocarbons with only carbon–carbon single bonds. The usual acyclic series has general formula CₙH₂ₙ₊₂. - The displayed butane model has four carbons and ten hydrogens. A branched acyclic alkane follows the same formula pattern; branching changes connectivity rather than saturation. - The formula constraint matters:a saturated monocyclic hydrocarbon has CₙH₂ₙ, so not every saturated hydrocarbon belongs to the acyclic alkane series. ## Treat crude oil and its fractions as mixtures - Crude oil contains many hydrocarbons, mainly alkanes in the course description, with other hydrocarbon families and sulfur-containing compounds also present. - A fraction groups substances with a useful boiling range. It is usually still a mixture rather than one pure molecular compound. - Fractional distillation changes which molecules are collected together. It does not convert every original molecule into one new fuel molecule. ## Explain the hot-to-cool fractionating column - Heated crude enters the lower region of a fractionating column. Vapours move into progressively cooler regions as they rise. - Higher-boiling components condense lower down; more volatile components travel farther before condensing, and some gases leave at the top. Heavy residue remains low in the system. - The refinery illustration gives a schematic order of cuts. Its printed temperatures are illustrative conditions, not universal boundaries or exact boiling points for every fraction. ## Separate intermolecular attraction from covalent bonding - In a comparable unbranched alkane series, larger molecules generally have more electrons and stronger London attractions. More energy is needed to separate molecules during boiling. - The carbon skeleton remains intact in ordinary vaporisation. Breaking C–C covalent bonds belongs to chemical conversion such as cracking. - Compare like structures:branching and shape affect contact between molecules. Carbon count alone is not a universal predictor across all hydrocarbon structures. ## Use a specified condensation model - A simplified question may assign condensation ranges or collection bands. Use those stated inputs rather than memorising a rigid universal fraction list. - The supplied single-component model cools a vapour through three regions. Its stated boiling threshold determines where the first liquid appears. - Real columns involve mixtures, pressure and repeated vapour–liquid contact. This model isolates the temperature comparison rather than predicting every real tray composition. ## Relate cracking to a stated supply-and-demand pattern - A refinery can use conversion processes to turn less-demanded heavy material into more useful lighter fuel and chemical feedstock streams. - The supplied fictional inventory has a surplus heavy fraction and a shortfall lighter fraction. It illustrates the economic reason to consider cracking, not a current market forecast. - Conversion has energy, catalyst and equipment costs. A product being useful does not prove every operating choice is economically or environmentally preferable. ## Conserve atoms when cracking a molecule - Cracking changes covalent connectivity by breaking larger hydrocarbon molecules into smaller products. The course contrasts this chemical process with distillation. - One balanced simplified route is C₁₀H₂₂→C₈H₁₈+C₂H₄. Its atom balance does not imply these are the only possible real products or a 100% selective yield. - The reaction system receives heat, and the cracking feed is kept from simply burning with oxygen. Industrial product distributions depend on conditions and catalyst. ## Compare thermal and catalytic cracking - Thermal cracking uses high temperature and high pressure and gives a high proportion of alkenes in the AQA comparison. - Catalytic cracking uses a zeolite catalyst, high temperature and slight pressure. Its useful products include motor-fuel components and aromatic hydrocarbons. - These are process families rather than one compulsory temperature for every unit. Do not claim catalytic cracking always gives exactly one smaller alkane and one alkene. ## Keep catalyst action distinct from heat supply - A catalyst provides an alternative pathway with lower activation energy. It can increase the reaction rate at suitable conditions without being consumed overall. - Lower activation energy does not mean the reaction enthalpy changes or the equilibrium constant is automatically increased. Heat supply and catalyst action answer different questions. - The illustrative energy figures are model inputs. They describe a barrier comparison rather than a measured commercial cracking unit. ## Write complete combustion from atom conservation - Complete combustion of an acyclic alkane forms carbon dioxide and water under suitable oxygen supply and reaction conditions. - For CₙH₂ₙ₊₂, the balanced amount ratio is one alkane to(3n+1)/2 oxygen, n carbon dioxide and(n+1) water. - This is an ideal complete-combustion model. Real flames and engines can have mixing or kinetic limitations even when an overall oxygen supply is large. ## Use a balanced octane equation quantitatively - The balanced integer equation is 2C₈H₁₈+25O₂→16CO₂+18H₂O. Coefficients give mole ratios as well as molecular counts. - One mole of octane therefore uses 12.5 mol O₂ and produces 8 mol CO₂ and 9 mol H₂O in this model. - A combustion equation must conserve oxygen atoms as well as carbon and hydrogen. An oxygen coefficient cannot be chosen solely from the carbon dioxide term. ## Identify the limiting reactant in an ideal combustion model - The supplied octane amount would need 0.250 mol O₂ for full consumption, but only 0.200 mol is available. Oxygen limits the extent of the specified model. - If the problem explicitly allows only the complete-combustion route for the reacting fraction,0.0160 mol octane reacts and 0.00400 mol remains. - This calculation is not a prediction that an oxygen-starved real flame produces only CO₂. Incomplete products may form unless the ideal product model is imposed. ## Allow more than one incomplete-combustion product - With limited or poorly mixed oxygen, carbon monoxide and/or carbon soot may form alongside water and some carbon dioxide. There is no single compulsory incomplete-product ratio. - Two ideal limiting equations for octane are 2C₈H₁₈+17O₂→16CO+18H₂O and 2C₈H₁₈+9O₂→16C+18H₂O. - These equations isolate different carbon-containing products. Real exhaust can contain mixtures and unburned fuel, so a product analysis is needed for exact emissions. ## Check oxidation states rather than saying every atom is oxidised - In these hydrocarbon combustion models, hydrogen is normally+1 in the fuel and remains +1 in water. Formation of water does not mean hydrogen must have increased its oxidation state. - Carbon undergoes net oxidation to+4 in CO₂ or to+2 in CO. Oxygen changes from 0 in O₂ to−2 in these products. - Bond breaking requires energy; energy released on forming product bonds can exceed that input. An exothermic overall reaction is not evidence that breaking any bond releases energy. ## Trace nitrogen oxides to hot air reactions - At high engine temperatures, nitrogen and oxygen from air can form NO. Nitrogen monoxide can then oxidise further to NO₂ in air. - Use N₂+O₂→2NO and 2NO+O₂→2NO₂ as balanced course equations. Fuel does not need to contain nitrogen for this thermal-NO route. - NOx contributes to atmospheric chemistry including acid formation and photochemical pollution. Avoid claiming that every pollutant has one identical environmental effect. ## Explain the catalytic converter’s reacting surface - The cutaway shows gas flow through many catalyst-coated channels. A large accessible surface allows gases to contact the catalytic material. - The course examples oxidise CO to CO₂ and use CO to reduce NO to N₂. Oxidation of unburned hydrocarbons can also produce CO₂ and water. - The illustration selects a few gas species and includes possible unused O₂; it is not the full composition of an exhaust stream or proof of 100% removal. The catalyst needs suitable operating conditions. ## Connect sulfur impurities with sulfur dioxide - Sulfur-containing material in a fuel can form SO₂ during combustion. A basic oxide or carbonate can be used to remove this acidic gas from flue gas. - The simplified sulfur balance S+O₂→SO₂ gives one mole of SO₂ per mole of sulfur oxidised by that route. - Removing SO₂ reduces one acidifying emission. It does not automatically remove the CO₂ generated by the fuel or all other exhaust components. ## Use countercurrent contact and forced oxidation - The scrubber cutaway shows rising flue gas meeting descending limestone slurry. Distributed spray creates contact between gas and liquid. - Air enters the sump in this forced-oxidation model, allowing absorbed sulfur(IV) products to be oxidised toward sulfate. The bleed removes slurry for further processing. - Treated gas is not necessarily pollutant-free. The drawing gives no fixed removal percentage and does not imply that collected solids are automatically pure gypsum. ## Distinguish sulfite capture from sulfate formation - A simplified capture equation is CaO+SO₂→CaSO₃. Calcium sulfite can then be oxidised:2CaSO₃+O₂→2CaSO₄. - In aqueous operation, hydration and dissolved species add detail. The net dry-form overall equation is CaO+SO₂+½O₂→CaSO₄. - Gypsum is CaSO₄·2H₂O rather than anhydrous CaSO₄. Include two waters per formula unit if the requested product is the dihydrate. ## Balance removal benefits against reagent and carbon costs - For the stated sulfate model, CaCO₃+SO₂+½O₂→CaSO₄+CO₂. Carbonate consumption and captured SO₂ have a 1:1mole ratio. - Calcium carbonate also releases CO₂ in this reaction. An SO₂control process is not automatically a carbon-capture process. - For gypsum output, include two waters in the overall balance. Actual reagent demand can exceed stoichiometry when purity and conversion are below 100%. ## Describe methane chlorination as substitution - The overall single-substitution equation is CH₄+Cl₂→CH₃Cl+HCl. A hydrogen atom is replaced by chlorine rather than a double bond being opened. - The course initiates the chain by UV light causing Cl–Cl homolysis. Artificial UV can supply this input; sunlight is not the only possible light source. - Keep an overall equation separate from the mechanism. It gives net atom balance without showing the radical intermediates. ## Mark the unpaired electron in initiation - UV light can split Cl₂ homolytically to give 2Cl•. Each radical receives one electron from the original shared pair. - The dot records an unpaired electron rather than an ionic charge. This is different from assigning both bond electrons to one atom. - The illustration shows the initiation event. A radical chain can subsequently undergo many propagation cycles before termination; UV does not have to supply a new photon for every individual propagation step. ## Form the carbon radical in the first propagation step - The first methane step is Cl•+CH₄→HCl+CH₃•. It transfers the radical character from chlorine to the methyl species while conserving atoms. - One radical appears on each side. The dot must remain onCH₃• in the product:writing ordinary CH₃ without explaining its radical state loses important electron bookkeeping. - The step is not the final net reaction. The methyl radical still needs a subsequent reaction to form chloromethane and regenerate chlorine radical. ## Regenerate the chlorine radical - The second step is CH₃•+Cl₂→CH₃Cl+Cl•. It forms chloromethane and regenerates the chlorine radical. - Adding the two propagation equations cancelsCH₃• and Cl• to giveCH₄+Cl₂→CH₃Cl+HCl. A radical can carry the chain without appearing in the net equation. - Cancellation does not mean intermediates were never formed. It means the two listed steps consume and regenerate them in matching amounts. ## Remove radicals by combination - Possible methane-system termination equations includeCl•+Cl•→Cl₂, CH₃•+Cl•→CH₃Cl and CH₃•+CH₃•→C₂H₆. - Two radicals combine without a radical product in these models. Forming ethane is a possible side-product pathway, not the intended single-substitution net equation. - Answer the requested radical combination. Chlorine recombination is chemically a termination event, but it does not answer a question asking specifically for an organic termination product. ## Explain why radical substitution gives mixtures - Chloromethane can undergo further substitution to CH₂Cl₂, CHCl₃ and CCl₄ under suitable conditions. One reaction mixture need not stop cleanly at the first substitution. - Using a large excess of methane relative to chlorine can favour fewer multiple substitutions, but it does not prove a pure isolated product or eliminate termination pathways. - Product yield must use the specified target and limiting reagent. A fully substituted formula is a possible endpoint model, not a guarantee that every molecule reaches it. ## Use symmetry when counting monochloro products - All hydrogen positions in ethane are equivalent for constitutional monochlorination, so only one chloroethane connectivity results. - Propane has equivalent terminal positions and a different middle position. Its distinct constitutional monochloro products are 1-chloropropane and 2-chloropropane. - This is a connectivity count, not a prediction of equal product amounts. Equivalent drawings and repeated hydrogens do not each become new structural isomers. ## Evaluate a stated emissions reduction - The supplied before-and-after masses quantify removal of one pollutant. Calculate the percentage using the original inlet mass as the denominator. - An 85% reduction leaves 15% of that pollutant in this data model. It is not proof that all exhaust species are removed or that treatment creates no other emissions. - Compare flow, measurement basis and conditions before combining concentration records. A lower concentration alone may reflect dilution rather than a lower emitted mass.