Homologous series

Test 1 — one general formula. Every member of the series must fit a single algebraic formula in terms of the number of carbon atoms, $n$. For the three hydrocarbon series studied at this level:

where: $n$ is the number of carbon atoms (a positive integer, with the series starting at the smallest chemically valid $n$), and the exponents on hydrogen encode the bonding — alkanes are fully saturated (only single bonds), alkenes carry one C=C double bond, and alkynes carry one C≡C triple bond, each double or triple bond removing two hydrogens compared with the saturated case.

Test 2 — one functional group, similar chemistry. Alcohols ($C_nH_{2n+1}OH$, functional group $-OH$) and carboxylic acids ($C_nH_{2n+1}COOH$, functional group $-COOH$) are homologous series built around a functional group rather than a bond type. Because every member carries the same functional group, all members undergo the same characteristic reactions — this is why one set of reactions taught for "alcohols" applies to methanol, ethanol, propanol and every higher member alike.

Test 3 — constant $-CH_2-$ difference. Moving from one member to the next always adds exactly one carbon and two hydrogens, a $-CH_2-$ unit. Since carbon has atomic mass 12 and hydrogen has atomic mass 1, each step up the series raises the molecular mass by $12 + (2\times1) = 14$ units — a number worth memorising, because it is the fastest way to check whether two given compounds are genuine homologues.

Test 4 — gradation in physical properties. Melting point, boiling point, density and solubility all change gradually and predictably along a series, even though the chemical behaviour stays similar. The reason is intermolecular, not intramolecular: longer chains have greater molecular mass and greater surface area of contact with neighbouring molecules, so the weak intermolecular forces of attraction between molecules strengthen member by member, and more thermal energy is needed to separate the molecules into the gas phase — hence boiling point rises steadily up the series.

Question. Show that methane ($CH_4$), ethane ($C_2H_6$) and propane ($C_3H_8$) belong to the same homologous series, predict the molecular formula and molecular mass of the next member, and use the boiling points methane $-162\,^\circ\text{C}$, ethane $-89\,^\circ\text{C}$, propane $-42\,^\circ\text{C}$, butane $0\,^\circ\text{C}$ to justify the trend.

Step 1. Check the general formula $C_nH_{2n+2}$ for each: $n=1$ gives $C_1H_4 = CH_4$; $n=2$ gives $C_2H_6$; $n=3$ gives $C_3H_8$. All three fit, so they belong to the alkane series.

Step 2. Check the constant difference is $-CH_2-$: $C_2H_6 - CH_4 = CH_2$, and $C_3H_8 - C_2H_6 = CH_2$, matching the 14-unit mass rule derived above ($30-16=14$, $44-30=14$).

Step 3. Predict the next member using $n=4$: $C_4H_{(2\times4)+2} = C_4H_{10}$ (butane), with molecular mass $14(4)+2 = 58$, exactly 14 more than propane's 44.

Step 4. Explain the boiling-point trend: mass rises steadily (16, 30, 44, 58), so surface area of contact between neighbouring molecules grows, intermolecular forces of attraction strengthen, and more energy is needed to boil the liquid — matching the steady rise $-162 \to -89 \to -42 \to 0\,^\circ\text{C}$. The three compounds are confirmed homologues, the fourth member is butane ($C_4H_{10}$, mass 58), and its higher boiling point of $0\,^\circ\text{C}$ follows directly from its greater molecular mass and stronger intermolecular attraction.

• Reading "homologous" as "identical" — members have similar chemical properties but genuinely different, gradually changing physical properties; boiling point is not the same for every member.
• Forgetting the increment must be specifically $-CH_2-$ (14 mass units); any random formula difference does not create a homologous series.
• Confusing the general formula ($C_nH_{2n+2}$, describing the whole series) with a molecular formula (a specific compound once $n$ is fixed).
• Treating alkanes, alkenes and alkynes as one series — they are three separate series with three separate general formulae and bonding patterns; a compound cannot cross between them and remain a homologue.

Ask the tutor: Want to check whether a list of unfamiliar organic compounds actually forms one homologous series, or want more boiling-point-trend practice? Ask the AI tutor for guided questions.