Introduction to Organic Chemistry

Asked in 6 of the 23 papers. A hydrocarbon is a compound of carbon and hydrogen atoms, and it contains those two elements only. When the question asks what the term means, or why an alkene is an unsaturated hydrocarbon, those are two separate marks, and "only" (or "no other elements") scores only once you have named carbon and hydrogen; in the two-mark "why is it not a hydrocarbon" version the whole definition is one mark and "it contains oxygen" is the other. Say "atoms": "carbon and hydrogen molecules" is rejected, and so is calling a hydrocarbon an element.

To explain why a compound is not one, add the point that it contains oxygen as well (or whatever extra element its formula shows). For an unsaturated hydrocarbon the third mark is the carbon–carbon double bond; give the definition marks too, because the question is worth three.

Asked in 12 of the 23 papers. Count every atom for the molecular formula, then divide by the highest common factor for the empirical formula: C₄H₆ becomes C₂H₃, while C₃H₈ is the same in both because 3 and 8 share no factor. If asked what "empirical formula" means, say it is the simplest whole-number ratio of the atoms and show it on the compound given.

Keep the types apart: CH₃CH₂OH is a structural formula, and as a molecular formula ethanol is C₂H₆O (one scheme rejected C₂H₅OH there). Every alkene with one C=C (the CₙH₂ₙ series) has the empirical formula CH₂, because it has twice as many hydrogens as carbons, whereas CₙH₂ₙ₊₂ cannot be divided through by n, so each alkane has its own: C₂H₆ gives CH₃ and C₄H₁₀ gives C₂H₅.

Asked in 3 of the 23 papers, always as any two from the same list: same general formula, same functional group, neighbouring members differ by one CH₂ unit, similar chemical properties (they undergo the same reactions), and a trend in physical properties such as boiling point. Any two score, so pick the two you can word exactly.

The wording that fails is "same" or "similar" physical properties: physical properties change steadily along the series, and two schemes reject that phrase. "Similar reactivity" was not accepted for the chemical-properties mark either. One question gave "same general formula" in the stem and asked for two others, so read the stem before you write, and naming a particular series instead of describing the family is ignored.

Asked in 7 of the 23 papers, so learn the two marks word for word: isomers are compounds with the same molecular formula but different structural (or displayed) formulae. "The same number of atoms of each element" earns the first mark and "a different arrangement of atoms" the second, and one scheme marks the two independently, so a correct second half still scores if the first is wrong.

The first mark is where answers fail. "The same empirical formula" or "the same general formula" is not credited (every alkene shares CH₂ and CₙH₂ₙ without being an isomer of the others), "the same chemical formula" is rejected as too vague, and isomers are compounds, not elements.

Set in 5 of the 23 papers, one mark per correct isomer. Start from the straight chain, then shorten it and attach the spare carbons as branches: C₄H₁₀ is butane and methylpropane (three carbons in a row with a CH₃ on the middle one), and C₅H₁₂ adds methylbutane and dimethylpropane (a central carbon carrying four CH₃ groups) to pentane. Draw every bond and check that each carbon has four.

For alkene isomers, either move the double bond along the chain (but-1-ene to but-2-ene) or branch the chain (methylpropene). But-2-ene drawn with its two CH₃ groups on the same side of the C=C, and drawn with them on opposite sides, both scored, but a ring such as cyclobutane was rejected because the question asked for alkenes.

Asked in 7 of the 23 papers, for pentane, butane, propene, chloroethene (as a multiple-choice item), propan-1-ol and ethanoic acid so far; naming an ester is covered with the esters. Count the carbons in the longest chain for the stem, then read the ending from the functional group: no functional group gives -ane, a C=C gives -ene, an –OH gives -anol and a –COOH gives -anoic acid. Spelling must be correct: one scheme says so explicitly for pentane.

For an alcohol, number the carbon that carries the –OH from the nearer end. CH₃CH₂CH₂OH is propan-1-ol; "propanol" and "1-propanol" were both allowed, but propan-2-ol was rejected because it names a different compound.

Asked in 10 of the 23 papers, often as a multiple-choice item with combustion, decomposition, neutralisation, polymerisation or precipitation as the distractors. Decide from the products. An alkane reacting with a halogen in ultraviolet light gives a halogenoalkane and HBr or HCl: one atom has been swapped for another, so it is substitution. An alkene reacting with bromine, chlorine or hydrogen gives one product only, the double bond having opened, so it is addition.

The one-mark reason an alkane cannot undergo addition is that it is saturated: it has no double bond, only single bonds. One scheme also allowed "redox" for an alkane with bromine, but substitution is the name every scheme credits.

Set in 6 of the 23 papers, four complete and two incomplete. For complete combustion the first mark is for correct formulae throughout, with carbon dioxide and water as the products; the second, which depends on the first, is for balancing. Balance the carbons first, then the hydrogens, and count the oxygens last. An alcohol already contains one oxygen atom, so subtract it before halving: the products of pentanol hold 16 oxygen atoms, 16 − 1 = 15, so 7.5 O₂. Fractions are accepted (3½O₂ scores in full), and so is the doubled version. State symbols are ignored unless asked for; then they carry their own mark, and (g) is accepted for water.

For incomplete combustion the products are carbon monoxide (sometimes with carbon, as soot, or some carbon dioxide alongside it) and water; subtract the alcohol's oxygen there too (butanol burning to carbon dioxide, carbon monoxide and water needed 5 O₂). Balance the oxygen from those products, and expect fewer O₂ than the complete equation needs.

Asked in 6 of the 23 papers. For two marks, name the product first: carbon monoxide forms because the supply of oxygen is limited. The second mark depends on naming it and is for saying that carbon monoxide lowers the blood's ability to carry oxygen; a reference to it binding to haemoglobin (forming carboxyhaemoglobin) is accepted instead. Some schemes credit "toxic" or "poisonous" on its own, but others want the effect on the blood, so always give it. "Harmful" is ignored.

Soot (carbon) is the other product. Two schemes credit it as an alternative first mark, with the second then for breathing or respiratory problems, but the carbon monoxide route is the one every scheme takes.