Hydrocarbons and Homologous Series

Diamond, natural gas, candle wax, and the petrol in a scooter tank are all built from the same two elements — carbon and hydrogen — arranged differently. A compound made of only carbon and hydrogen is a hydrocarbon, and hydrocarbons turn out not to be a random jumble but a set of tidy families, each running from a tiny first member to giant chains, every member differing from its neighbour by exactly one −CH2−-\text{CH}_2- unit. That family pattern is called a homologous series, and once you see it, most of "organic chemistry" stops looking like a list of names to memorise and starts looking like arithmetic on a repeating unit. This article covers how hydrocarbons are classified by their carbon–carbon bonding (saturated versus unsaturated), the three general formulas for alkanes, alkenes and alkynes, and the four defining features of a homologous series — building directly on the covalent bonding and tetravalency of carbon covered earlier in this chapter.

The core insight

Hydrocarbons split into alkanes, alkenes and alkynes by their carbon–carbon bond type, and members of the same family form a homologous series that adds one −CH2−-\text{CH}_2- unit (14 u) per step while keeping the same general formula and similar chemical behaviour.

The core idea

Classifying by carbon–carbon bonding. Every hydrocarbon carbon is tetravalent — it forms exactly four bonds. Whether those bonds to neighbouring carbons are single, double or triple decides the family:

  • Saturated hydrocarbons (alkanes) have only single C–C bonds, so every remaining valency is used up by hydrogen. General formula:
CnH2n+2\text{C}_n\text{H}_{2n+2}

where: nn is the number of carbon atoms in the molecule. Examples: methane (CH4\text{CH}_4, n=1n=1), ethane (C2H6\text{C}_2\text{H}_6), propane (C3H8\text{C}_3\text{H}_8).

  • Unsaturated hydrocarbons contain at least one C=C or C≡C bond, which uses up two of a carbon's valencies at once and so leaves fewer positions for hydrogen. Alkenes (one C=C double bond) follow
CnH2n\text{C}_n\text{H}_{2n}

Examples: ethene (C2H4\text{C}_2\text{H}_4), propene (C3H6\text{C}_3\text{H}_6). Alkynes (one C≡C triple bond) follow

CnH2n−2\text{C}_n\text{H}_{2n-2}

Examples: ethyne (C2H2\text{C}_2\text{H}_2), propyne (C3H4\text{C}_3\text{H}_4).

Carbon skeletons. For a given family, the carbons themselves can be arranged straight-chain, branched-chain, or closed into a simple ring (cyclic) — e.g. cyclohexane, C6H12\text{C}_6\text{H}_{12}, at an introductory structural level. A ring closes two "loose ends," so a cyclic saturated compound loses two hydrogens compared with the open-chain alkane of the same carbon count — which is exactly why cyclopentane and pentene share the formula C5H10\text{C}_5\text{H}_{10} even though one is saturated and the other is not.

The homologous series pattern. A homologous series is a family of compounds sharing four features at once: the same general formula, a constant −CH2−-\text{CH}_2- difference (14 u) between successive members, similar chemical properties because they share the same bonding pattern, and a gradual rise in melting and boiling point as chain length (and molecular mass) increases, since bigger molecules feel stronger intermolecular forces of attraction.

Figure
A ladder of the first four alkanes showing methane ethane propane and butane each step adding one CH2 unit worth fourteen atomic mass units
The alkane homologous series — each rung adds one CH2, 14 u CH4 methane, 16 u C2H6 ethane, 30 u C3H8 propane, 44 u C4H10 butane, 58 u +CH2, +14 u +CH2, +14 u +CH2, +14 u Same general formula CnH2n+2, similar chemistry, boiling point rising steadily rung by rung
Derivation ADeriving the alkane general formula CnH2n+2 from carbon's fixed valency

The goal is to show, using nothing beyond carbon's tetravalency (four bonds per carbon atom) and hydrogen's single bond, why a saturated straight-chain hydrocarbon of nn carbons must carry exactly 2n+22n+2 hydrogen atoms — turning a memorised formula into a countable fact about bonds.

Step 1. Count the total bonding positions available. Each of the nn carbon atoms is tetravalent, so together they offer 4n4n bond positions in total, to be filled by bonds to other carbons or to hydrogens.

Step 2. Count how many of those positions get used up linking carbon to carbon. In a straight chain of nn carbons there are exactly n−1n-1 carbon–carbon single bonds (one between each adjacent pair), and each such bond uses one valency from each of the two carbons it joins, so it removes 22 positions from the pool for every C–C bond formed: 2(n−1)2(n-1) positions are used internally.

Step 3. Whatever bonding positions remain must be filled by hydrogen, since the molecule is saturated and has no other atoms to bond to. The number of hydrogens is therefore the total positions minus the ones used for C–C bonds:

H atoms=4n−2(n−1)=4n−2n+2=2n+2.\text{H atoms} = 4n - 2(n-1) = 4n - 2n + 2 = 2n + 2.

Step 4. Check against known members: for n=1n=1 (methane) there is no C–C bond to subtract, giving 4(1)=44(1) = 4 hydrogens, matching CH4\text{CH}_4; for n=2n=2 (ethane), 2(2)+2=62(2)+2 = 6, matching C2H6\text{C}_2\text{H}_6; for n=4n=4 (butane), 2(4)+2=102(4)+2 = 10, matching C4H10\text{C}_4\text{H}_{10}. The formula CnH2n+2\text{C}_n\text{H}_{2n+2} is not an arbitrary pattern to memorise — it falls directly out of counting carbon's four bonds per atom against the bonds used to link the chain together.

Worked example

Show that propane (C3H8\text{C}_3\text{H}_8) and butane (C4H10\text{C}_4\text{H}_{10}) belong to the same homologous series, find the mass difference between them, and predict the formula and mass of the next member.

Step 1. Check both fit the alkane general formula CnH2n+2\text{C}_n\text{H}_{2n+2}: for propane, n=3n=3, so 2n+2=82n+2=8 — matches C3H8\text{C}_3\text{H}_8. For butane, n=4n=4, so 2n+2=102n+2=10 — matches C4H10\text{C}_4\text{H}_{10}. Both fit, so both are alkanes in the same homologous series.

Step 2. Confirm the structural difference is exactly one −CH2−-\text{CH}_2- unit: butane is propane with one extra CH2\text{CH}_2 inserted into the chain — CH3−CH2−CH2−CH3\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3 versus CH3−CH2−CH3\text{CH}_3-\text{CH}_2-\text{CH}_3.

Step 3. Calculate the mass of one CH2\text{CH}_2 unit using atomic masses C = 12 u, H = 1 u:

Mass of CH2=12+2(1)=14 u.\text{Mass of CH}_2 = 12 + 2(1) = 14\ \text{u}.

This 14 u is exactly the mass difference expected between consecutive homologues.

Step 4. Predict the next member by adding one more CH2\text{CH}_2 to butane: n=5n = 5, so the formula is C5H12\text{C}_5\text{H}_{12} (pentane), and its mass is the mass of butane plus 14 u.

Result: propane and butane are consecutive members of the alkane homologous series, differing by one −CH2−-\text{CH}_2- unit (14 u); the next member is pentane, C5H12\text{C}_5\text{H}_{12}, obtained by the same +14 u step.

Competitive stretch

Given the molecular formula C5H10\text{C}_5\text{H}_{10}, identify the hydrocarbon family and name one possible structure. Compare against the three general formulas for n=5n=5: alkane needs 2n+2=122n+2=12, alkene needs 2n=102n=10, alkyne needs 2n−2=82n-2=8. Since C5H10\text{C}_5\text{H}_{10} has 10 hydrogens, it matches the alkene pattern exactly, so it is pentene, e.g. CH2=CH−CH2−CH2−CH3\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3 (pent-1-ene). It could equally be drawn as the saturated ring cyclopentane, since closing a chain into a ring also removes two hydrogens — a reminder that a molecular formula alone does not fix the structure.

Common traps

  • Mixing up the three general formulas — remember "alkane has the most H, alkyne has the least, for the same number of carbons": 2n+2>2n>2n−22n+2 > 2n > 2n-2.
  • Thinking a homologous series just means "compounds with similar names" — the real definition rests on the same general formula, a constant −CH2−-\text{CH}_2- (14 u) difference, and similar chemical properties, not naming alone.
  • Believing physical properties like boiling point jump erratically between homologues — the syllabus expects a gradual, steady rise with increasing chain length, not random variation.
  • Forgetting that a ring (cyclic) structure can share the same molecular formula as an open-chain unsaturated compound (e.g. cyclopentane and pentene both being C5H10\text{C}_5\text{H}_{10}) — the formula alone does not tell you whether the compound is a ring or a chain.

Ask the tutor: Want to practise classifying a hydrocarbon as an alkane, alkene, or alkyne straight from its formula, or work out the next member and mass jump in a homologous series? Ask the AI tutor for a few quick problems.

Why does adding one CH2 unit always add exactly 14 u, no matter which homologous series it is?

The 14 u comes purely from atomic masses, not from which family the compound belongs to: one carbon atom is 12 u and the two hydrogens that come with it in a −CH2−-\text{CH}_2- unit are 1 u each, so 12+1+1=1412 + 1 + 1 = 14 u every time. Whether you are moving from methane to ethane, or from ethene to propene, the step in the chain is always the same repeating unit, so the mass jump is always 14 u. This is exactly why a homologous series can be thought of as one basic formula with nn dialled up by one each time.

If cyclopentane and pentene share the formula C5H10, how do I tell them apart?

The molecular formula alone genuinely cannot distinguish them — you need the structure. Pentene is an open chain with one carbon-carbon double bond somewhere along it; cyclopentane is a fully saturated ring with no double bond, but the ring closure removes the same two hydrogens that the double bond would have removed in an open chain. In an exam, you are usually given either a structural formula or told explicitly whether the compound is cyclic or contains a double bond, so this ambiguity is a reason to read the question carefully, not a gap you need to resolve from the formula alone.

Do all four homologous series conditions have to hold for a set of compounds to count as one, or is the same general formula enough?

All four together are needed; the general formula alone is not sufficient proof. Two compounds could accidentally have formulas that look related without actually being consecutive members of a genuine series. The full definition — same general formula, a constant 14 u CH2 difference between successive members, similar chemical behaviour from sharing the same bonding pattern, and a steady gradation in physical properties like boiling point — is what makes "homologous series" a precise chemical idea rather than just compounds with similar-looking names.

Before this

Where this leads