Functional groups and homologous series: Worked examples

Before we count anything, two ideas we will use constantly:

The scenario matrix

Every question about this topic lands in one of these cells. Each example below is tagged with the cell it covers. The figure below is a map of the whole page, laid out on real axes: read the horizontal axis as "what is varying in the question" (left→right: the formula/atoms, then the group, then a physical trend) and the vertical axis as "how hard the twist is" (bottom→top: routine → edge case → exam trap). Every tile sits at the (variation, difficulty) coordinate that matches its case class, and carries the example number that solves it. Glance back at it whenever you meet a new question — first decide which column (what varies) and which row (how tricky), and the tile tells you the example to imitate.

Figure — Functional groups and homologous series

Figure 1 — the case-class map. X-axis = what varies (atoms → group → physical trend); Y-axis = difficulty (routine → edge case → exam trap). Each tile (A–I) is placed at its coordinate; colour only groups tiles by row. The label names the trap and points to its example.

Cell A — Plain series identification

Worked example Ex 1 — Are

$C_{2} H_{6}, C_{3} H_{8}, C_{4} H_{10}$ one series? Forecast: Guess first — same family or not? What single number would prove it?

Step 1 — Subtract consecutive formulas (atom-by-atom). Using the atom-tally rule above: $C_{3} H_{8} - C_{2} H_{6} = (3 - 2, 8 - 6) = (1, 2) = C H_{2}$ . Then $C_{4} H_{10} - C_{3} H_{8} = (1, 2) = C H_{2}$ again. Why this step? The defining test of a homologous series is a constant $C H_{2}$ gap (mass $14$ ). A constant gap means we are just clipping identical building blocks onto the same skeleton.

Step 2 — Check a general formula fits. Try $C_{n} H_{2 n + 2}$ (the alkane formula). For $n = 2$ : $H = 2 (2) + 2 = 6$ ✓. For $n = 4$ : $2 (4) + 2 = 10$ ✓. Why this step? A shared general formula confirms the same type of skeleton with no extra functional group — these are pure hydrocarbons.

Verify: Masses: $C_{2} H_{6} = 30$ , $C_{3} H_{8} = 44$ , $C_{4} H_{10} = 58$ . Differences $= 14, 14$ ✓. Answer: yes — consecutive alkanes.

Cell B — Same formula, different group (functional isomers)

Worked example Ex 2 — Both are

$C_{2} H_{6} O$ . Same series? Compare ethanol $C_{2} H_{5} O H$ and dimethyl ether $C H_{3} O C H_{3}$ . Forecast: They weigh the same ( $46$ ). Does that make them the same family? Guess before reading.

Step 1 — Locate the functional group in each. Ethanol: $- O H$ (hydroxyl) → alcohol. Ether: $- O -$ between two carbons (alkoxy) → ether. The two structures are drawn side-by-side in the figure below — same atoms, different wiring. Why this step? A homologous series is defined by the group, not the molecular formula. Two molecules with identical formula but different groups are functional isomers.

Figure 2 — same formula $C_{2} H_{6} O$ , two different structures. Left: ethanol, oxygen carries a free $O - H$ (red) — sodium can grab it. Right: dimethyl ether, oxygen is trapped between two carbons with no $O - H$ — sodium finds nothing to attack.

Step 2 — Predict a distinguishing reaction. Add sodium metal. The $- O H$ has an $O - H$ bond sodium can attack: $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2} ↑$ The ether has no $O - H$ — nothing happens. Why this step? Reactivity comes from the polar/labile bond in the group. The ether's oxygen is locked between two carbons, so no $H_{2}$ is released.

Verify (atom balance): Left of the arrow $2 C_{2} H_{5} O H = C_{4} H_{12} O_{2}$ plus $2 N a$ ; right side $2 C_{2} H_{5} O N a = C_{4} H_{10} O_{2} N a_{2}$ plus $H_{2}$ . Carbons $4 = 4$ , oxygens $2 = 2$ , sodiums $2 = 2$ , hydrogens $12 = 10 + 2$ ✓. Answer: different series — the formula lied.

Cell C — Zero / degenerate input (the very first member)

Worked example Ex 3 — Does the aldehyde formula survive at

$n = 1$ ? Aldehydes are written two ways: $C_{n} H_{2 n} O$ and $C_{n - 1} H_{2 n - 1} - C H O$ . First let's see why these are the same thing, then test the $n = 1$ member where the carbon chain shrinks to nothing. Forecast: Name the $n = 1$ aldehyde. Does anything "break" when the skeleton shrinks to nothing?

Step 0 — Derive one form from the other. Peel the functional group $- C H O$ off the molecule. That group contains 1 C, 1 H, 1 O. The leftover chain (an "alkyl" piece $C_{k} H_{2 k + 1}$ ) attaches to it. If the whole molecule has $n$ carbons, the leftover chain has $k = n - 1$ carbons, i.e. $C_{n - 1} H_{2 (n - 1) + 1} = C_{n - 1} H_{2 n - 1}$ . Now glue the group back on, adding its atoms:

= C_{(n-1)+1}\,H_{(2n-1)+1}\,O = C_nH_{2n}O.$$ *Why this step?* It shows the two formulas are **the same tally written from two viewpoints** — "chain + group" versus "total". This is the atom-vector idea from the intro used in reverse: we *add* the group's atom counts back in. **Step 1 — Plug $n=1$ into $C_nH_{2n}O$.** $C_1H_2O = CH_2O$. *Why this step?* Testing the smallest ($n=1$) member is the "zero-input" edge case — general formulas are only trustworthy if they survive the extremes. **Step 1½ — What does the "$C_0$ chain" actually mean?** At $n=1$ the chain piece $C_{n-1}H_{2n-1}$ becomes $C_{0}H_{1}$, i.e. **zero carbons and one hydrogen**. "Zero carbons" does *not* mean "nothing" — it means the $-CHO$ group's carbon has **no alkyl chain attached at all**, and the empty attachment slot is simply filled by that **single leftover H**. So instead of $\text{alkyl}\text{–}CHO$ we get $\mathbf{H}\text{–}CHO$: a hydrogen sitting where a carbon chain would normally be. In practice "$C_0H_1$" is just "one H, no carbons", which is exactly the extra $H$ on the front of $H\text{–}CHO$. *Why this step?* The symbol $C_0$ can look like a typo; spelling out that it stands for "no chain, plug the slot with H" removes the confusion before we name the molecule. **Step 2 — Draw it.** $CH_2O$ is **methanal (formaldehyde)**, $H\text{–}CHO$: the $-CHO$ group with just an $H$ instead of an alkyl chain. *Why this step?* We must confirm a *real* molecule exists — here the "chain" is degenerate (a single H), yet the $-CHO$ functional group is fully intact, so it still counts as the first aldehyde. **Verify:** Mass of $CH_2O = 12 + 2 + 16 = 30$. Next member ethanal $C_2H_4O = 44$; difference $= 14$ ✓. The formula holds even at the degenerate first member. **Answer: methanal, $CH_2O$.**

Cell D — Two groups at once (which suffix wins?)

Worked example Ex 4 — A molecule has both

$- O H$ and $- C O O H$ . What class is it? Consider $H O C H_{2} C O O H$ (glycolic acid). Forecast: Two groups, two possible suffixes ( $- o l$ and $- o i c a c i d$ ). Which one names the family?

Step 1 — Rank the groups by priority. IUPAC priority (high → low): carboxylic acid $>$ ester $>$ aldehyde $>$ ketone $>$ alcohol $>$ amine. $- C O O H$ outranks $- O H$ . Why this step? When groups compete, exactly one becomes the principal characteristic group (gets the suffix); the loser becomes a prefix (here $- O H \to$ "hydroxy-").

Step 2 — Assign class and behaviour. Principal group $= - C O O H \Rightarrow$ it is a carboxylic acid (a hydroxy-acid). Its dominant chemistry is acidic. Why this step? The higher-priority group also dominates the reactivity we predict — this molecule will react as an acid first.

Verify: Formula $H O C H_{2} C O O H = C_{2} H_{4} O_{3}$ , mass $= 2 (12) + 4 (1) + 3 (16) = 76$ . It sits in the hydroxy-acid family, suffix from $- C O O H$ . Answer: carboxylic acid (2-hydroxyethanoic acid).

Cell E — Sign / direction of a trend

Worked example Ex 5 — Order the boiling points of

$C H_{3} O H, C H_{3} C H_{2} O H, C H_{3} C H_{2} C H_{2} O H$ . Forecast: Do boiling points go up or down as the chain grows? Predict the direction first.

Step 1 — Same group, so H-bonding is common to all. All are alcohols ( $- O H$ ) → all form the same kind of hydrogen bonds. Why this step? To compare cleanly we hold the group (and its force type) constant and vary only $n$ .

Step 2 — Add the size-dependent force. Bigger chain ⇒ more electrons ⇒ larger polarizable surface ⇒ stronger London (van der Waals) forces ⇒ more energy needed to separate ⇒ higher boiling point. Why this step? This is the direction-setting term: it always increases with $n$ , so the trend rises. The plotted curve in the next figure shows this rise (and previews Cell F's flattening).

Figure 3 — boiling point of the straight-chain alcohols versus chain length $n$ . The curve climbs steadily (Cell E: direction is up). The mint labels show the size of each $C H_{2}$ step: the increment first grows ( $+ 13$ then $+ 19$ ) and only later begins to shrink, so the slope tends to zero only for large $n$ — it is not decreasing from the very start.

Verify (real data, °C): $C H_{3} O H \approx 65$ , $C_{2} H_{5} O H \approx 78$ , $C_{3} H_{7} O H \approx 97$ . Strictly increasing ✓. Answer: $C H_{3} O H < C_{2} H_{5} O H < C_{3} H_{7} O H$ .

Cell F — Limiting behaviour ( $n \to \infty$ and branching)

Worked example Ex 6 — Does the

$+ 14$ mass step ever "run out", and why can branching flip a boiling-point? Forecast: As chains get huge, does adding one more $C H_{2}$ raise the boiling point by the same amount forever? And which boils higher: n-pentane or neopentane (both $C_{5} H_{12}$ )?

Step 1 — The mass gap is exactly constant. Each added $C H_{2}$ adds mass $14$ regardless of $n$ — even at $n \to \infty$ . So the formula step never runs out. Why this step? Separates a strict algebraic fact (constant $14$ ) from a physical trend that merely "gradually" changes.

Step 2 — The boiling-point increment per $C H_{2}$ is NOT monotonic; it only tapers for large $n$ . Look at Figure 3: the first step $C H_{3} O H \to C_{2} H_{5} O H$ lifts the boiling point by $\approx 13 ° C$ , but the next step $C_{2} H_{5} O H \to C_{3} H_{7} O H$ lifts it by more, $\approx 19 ° C$ . So the increment does not shrink from $n = 1$ onward — it actually rises near the start. Only much later, for large $n$ , does each extra $C H_{2}$ add a smaller and smaller amount, so the increment finally decreases toward zero (the curve keeps rising but ever more gently, never levelling off at a hard ceiling for any finite $n$ ). Why this step? This is the true "limiting behaviour": the boiling point always increases with $n$ , but its slope tends to $0$ only as $n \to \infty$ , not steadily for every step. (Correcting a common misstatement: the slope dies away eventually; only the mass step $14$ stays fixed for all $n$ .)

Step 3 — Branching = degenerate shape. Neopentane is a compact sphere; n-pentane is a long rod. Rods pack with more surface contact ⇒ stronger London forces ⇒ higher boiling point. Why this step? Same formula, same group, but shape changes the contact area, so branching can lower the boiling point relative to the straight isomer.

Verify (real data, °C): n-pentane $\approx 36$ , neopentane $\approx 9.5$ . So n-pentane $>$ neopentane ✓, exactly as the surface-area argument predicts.

Cell G — Reaction prediction from group only

Worked example Ex 7 — Predict what happens to

$C H_{3} C H_{2} C H_{2} O H$ on mild oxidation. Forecast: The chain is 3 carbons — does that matter? Guess the product class first.

Step 1 — Identify the group and ignore the chain. Group $= - O H$ on a carbon bearing one alkyl group and one H → a primary alcohol. Why this step? The oxidation outcome is set entirely by the type of alcohol carbon, not by chain length (the extra $C H_{2}$ 's are spectators).

Step 2 — Apply the oxidation ladder. Primary alcohol $[O]$ aldehyde $[O]$ carboxylic acid (see Oxidation of Alcohols): $C H_{3} C H_{2} C H_{2} O H [O] C H_{3} C H_{2} C H O [O] C H_{3} C H_{2} C O O H$ Why this step? Each $[O]$ removes two H / adds an O at the functional carbon — the "Al-Co-Hol Carries Acid" ladder from the parent note.

Verify (formulas): $C_{3} H_{8} O (60) \to C_{3} H_{6} O (58) \to C_{3} H_{6} O_{2} (74)$ . Each aldehyde step loses $2$ H (mass $- 2$ ), the acid step gains one O (mass $+ 16$ ) — consistent ✓. Answer: propanal, then propanoic acid.

Cell H — Real-world word problem

Cell I — Exam twist (general-formula algebra)

Worked example Ex 9 — A compound of formula

$C_{n} H_{2 n}$ has molar mass $70$ . Find $n$ and name it. Forecast: Which general formula do we set equal to $70$ , and how do we solve for $n$ ?

Step 1 — Write the mass of $C_{n} H_{2 n}$ . Mass $M = 12 n + 1 \cdot (2 n) = 14 n$ . Why this step? Every carbon costs $12$ , every hydrogen $1$ ; with $2 n$ hydrogens the whole thing collapses to a clean $14 n$ — notice $14$ appears again (it's the mass of one $C H_{2}$ -like unit).

Step 2 — Solve $14 n = 70$ for $n$ . Divide both sides by $14$ : $n = 70/14 = 5$ . So the molecular formula is $C_{5} H_{10}$ . Why this step? This is the whole point of the exam twist — turning a given molar mass into the carbon count is just a one-line division once we have $M = 14 n$ .

Step 3 — Beware the twist: $C_{n} H_{2 n}$ names TWO families. $C_{5} H_{10}$ (indeed any $C_{n} H_{2 n}$ ) fits both an open-chain alkene (one $C = C$ double bond, e.g. pent-1-ene) and a cycloalkane (a ring of single bonds, e.g. cyclopentane). Both have exactly $2 n$ hydrogens because closing a ring removes the same 2 H that forming a double bond does — one "degree of unsaturation" either way. They are ring/chain isomers. Why this step? The molecular formula alone cannot decide between them; the classic exam trap is to accept "alkene" and forget the ring. To choose, you need a reactivity clue — e.g. only the alkene decolourises bromine water, because only it has a $C = C$ for bromine to add across.

Verify: $C_{5} H_{10}$ mass $= 5 (12) + 10 (1) = 70$ ✓ for both structures. Answer: $n = 5$ ; formula $C_{5} H_{10}$ — pent-1-ene or cyclopentane (a bromine-water test decides).

Recall Quick self-check across the whole matrix

Which cell? "Butanal vs butan-2-one, both $C_{4} H_{8} O$ " ::: Cell B — same formula, different group (aldehyde vs ketone), functional isomers. Which cell? "Boiling point of the 20th alkane vs the 21st" ::: Cell F — limiting behaviour; increment per $C H_{2}$ is small and tapering for large n but still positive. Which cell? "Name the $n = 1$ member of the alcohols" ::: Cell C — degenerate first member, $C H_{3} O H$ (methanol). Which cell? " $H O C H_{2} C H_{2} C O O H$ — which suffix?" ::: Cell D — two groups, $- C O O H$ wins by priority (a hydroxy-acid). Which cell? " $C_{4} H_{8}$ — alkene or ring?" ::: Cell I — $C_{n} H_{2 n}$ fits both an alkene and a cycloalkane; need a test to decide.

Connections

Parent: Functional groups & homologous series
Isomerism — Cells B, D & I live here (functional isomers, competing groups, ring/chain)
IUPAC Nomenclature — suffix priority used in Cells D, I
Oxidation of Alcohols — the ladder used in Cell G
Intermolecular Forces — force reasoning behind Cells E, F
Hydrocarbons — the alkane/alkene skeletons of Cells A, I
Inductive and Mesomeric Effects — deeper "why the group reacts"