Functional groups and homologous series: Exercises — Chemistry deep dive

Before we start, one picture fixes every idea below: a molecule is a spectator chain plus a reactive button (the functional group). In the figure, the black zig-zag is the inert carbon stick and the single red dot is the button — notice how everything reactive is concentrated in that one red spot while the long black chain just sits there. Keep this image in mind — in every problem below we point to "the button" (the group) and ignore "the stick" (the chain).

The five levels chain together like a staircase: first you spot the button (L1), then apply a formula to it (L2), then compare a trend across a family (L3), then combine that with isomerism (L4), and finally run a full deduction (L5). Each level below reuses everything learned in the one before it.

Level 1 — Recognition

"Can you name the group when you see it?"

Q1.1

Identify the functional group and the class name of each: (a) $C H_{3} C H_{2} O H$ (b) $C H_{3} C O O H$ (c) $C H_{3} C H O$ (d) $C H_{3} C O C H_{3}$ (e) $C H_{3} N H_{2}$ .

Recall Solution Q1.1

On our picture, the $C H_{3}$ / $C H_{2}$ pieces are the stick; find the button (the atoms that carry the electron imbalance — a loose $π$ -bond cloud, a lone pair, or a polar bond).

(a) $- O H$ → hydroxyl → alcohol (ethanol).
(b) $- C O O H$ → carboxyl → carboxylic acid (ethanoic/acetic acid).
(c) $- C H O$ → aldehyde group → aldehyde (ethanal/acetaldehyde).
(d) the shorthand $> C = O$ means a carbon that is double-bonded to oxygen and single-bonded to two other carbons — the two lines " $>$ " on the left are just those two C–C bonds drawn sideways. That carbon-in-the-middle carbonyl → ketone (propanone/acetone).
(e) $- N H_{2}$ → amino → amine (methanamine).

Why the stick is ignored: the C–H and C–C bonds are strong and nonpolar — chemically lazy. The reaction always happens at the button.

Q1.2

Which of these is NOT a member of a homologous series with the others? $C H_{4}, C_{2} H_{6}, C_{3} H_{8}, C_{3} H_{6}, C_{4} H_{10}$ .

Recall Solution Q1.2

Test the alkane general formula $C_{n} H_{2 n + 2}$ .

$C H_{4}$ : $2 (1) + 2 = 4$ ✓ · $C_{2} H_{6}$ : $2 (2) + 2 = 6$ ✓ · $C_{3} H_{8}$ : $8$ ✓ · $C_{4} H_{10}$ : $10$ ✓
$C_{3} H_{6}$ : $2 (3) + 2 = 8 \neq = 6$ ✗ — it has two H fewer than a full alkane.

What "two H fewer" means (degree of unsaturation). A fully "saturated" chain carries the maximum possible hydrogens — every carbon bond that isn't C–C is filled with an H. Each time you remove one pair of H (a " $- 2 H$ "), the two freed bonds must join each other, forming one extra bond — either a ring or, here, a $C = C$ double bond (the extra sideways $π$ -bond cloud we defined above). That one missing H-pair is called one degree of unsaturation (one extra ring or $π$ -bond). So $C_{3} H_{6}$ has exactly one: it is an alkene $C_{n} H_{2 n}$ (here the button is the $C = C$ ).

Answer: $C_{3} H_{6}$ — it belongs to the alkene family, not the alkanes.

Level 2 — Application

"Use a formula or a rule to compute."

Q2.1

The 5th member of the alcohol series has how many carbons, and what is its molecular formula? (Count methanol as the 1st.)

Recall Solution Q2.1

Alcohol general formula: $C_{n} H_{2 n + 1} O H$ , i.e. $C_{n} H_{2 n + 2} O$ .

Reading the shorthand $C_{n} H_{2 n + 1} O H$ . The piece $C_{n} H_{2 n + 1}$ is just an alkyl group — an alkane $C_{n} H_{2 n + 2}$ with one H removed so it has a free bond to hang onto something. That free bond is joined to the $- O H$ button. So $C_{n} H_{2 n + 1} O H$ literally reads "an alkyl stick bonded to a hydroxyl button." (button $= - O H$ ; stick $= C_{n} H_{2 n + 1}$ grows with $n$ .)

Methanol is $n = 1$ , so the 5th member is $n = 5$ (pentanol).
$H = 2 (5) + 2 = 12$ , one $O$ . $C_{5} H_{12} O (pentan-1-ol)$ Why $n = 5$ and not $6$ : the counting starts at $n = 1$ , so member number = carbon number here.

Q2.2

A compound in a homologous series has molar mass $58 g/mol$ . The member just below it has molar mass $44$ . Confirm the $C H_{2}$ gap and identify both if the series is the aldehyde series $C_{n} H_{2 n} O$ .

Recall Solution Q2.2

Gap: $58 - 44 = 14 =$ mass of one $== C H_{2} ==$ ✓ (that is the definition of consecutive members — the stick grew by one $C H_{2}$ , the $- C H O$ button unchanged).
Aldehyde mass $= 12 n + 1 (2 n) + 16 = 14 n + 16$ .
Set $14 n + 16 = 44 \Rightarrow n = 2$ → ethanal $C_{2} H_{4} O$ ( $C H_{3} C H O$ ).
Set $14 n + 16 = 58 \Rightarrow n = 3$ → propanal $C_{3} H_{6} O$ ( $C H_{3} C H_{2} C H O$ ).

Why the linear form $14 n + 16$ : every extra $C H_{2}$ adds exactly $14$ , so molar mass is a straight line in $n$ — solving is one subtraction.

Level 3 — Analysis

"Compare, order, and justify with the right physical reason."

Q3.1

Order the boiling points (lowest → highest): $C H_{3} C H_{2} C H_{3}$ (propane), $C H_{3} C H_{2} C H_{2} O H$ (propan-1-ol), $C H_{3} C H_{2} C H_{2} C H_{3}$ (butane).

Recall Solution Q3.1

Two forces compete (see Intermolecular Forces):

London/van der Waals — grows with chain length/surface (this is the stick's contribution).
Hydrogen bonding — only when the button is $- O - H$ .

Reasoning:

Propane vs butane: same button-less family (alkanes), butane is longer → butane > propane.
Propan-1-ol has an $- O H$ button → hydrogen bonds, a far stronger sticking force than London forces at this size → it beats both alkanes. $propane < butane < propan-1-ol$ Boiling points ( $^{\circ} C$ ): $- 42 < - 1 < 97$ . ✓

Why H-bonding dominates: separating molecules to boil requires breaking the $- O H \dots O$ network — that costs much more energy than pulling apart weakly polarizable alkanes.

Q3.2

$C H_{3} F$ and $C H_{3} I$ are both haloalkanes. Which C–X bond is more polar, and which molecule would you expect to boil higher? Are these the same answer? Explain.

Recall Solution Q3.2

Polarity comes from electronegativity difference. F is far more electronegative than I, so C–F is the more polar bond (larger dipole) — an inductive argument.
Boiling point is dominated by London forces here: iodine is huge and highly polarizable, so $C H_{3} I$ (bp $4 2^{\circ} C$ ) boils far above $C H_{3} F$ (bp $- 7 8^{\circ} C$ ).
They are NOT the same answer. Bond polarity ≠ intermolecular stickiness.

Why the mismatch: polarity is about electrons inside one bond; boiling point is about forces between whole molecules, where sheer size/polarizability wins for heavy halogens.

Level 4 — Synthesis

"Combine functional-group logic with isomerism and reactions."

Q4.1

$C_{3} H_{6} O$ can be propanal ( $C H_{3} C H_{2} C H O$ ) or propanone ( $C H_{3} C O C H_{3}$ ). (a) What kind of isomers are these? (b) Which reduces Tollens' reagent (silver mirror)? (c) Give the oxidation ladder position of each.

Recall Solution Q4.1

(a) Same molecular formula, different button — $- C H O$ (aldehyde, button at a chain end) vs $> C = O$ (ketone, button in the middle, carbon flanked by two carbons) → functional isomers (Isomerism).
(b) Propanal (aldehyde) reduces Tollens' — aldehydes are easily oxidised further to acids, giving the silver mirror. Ketones are not (their carbonyl carbon has no H to lose).
(c) Explicit ladder positions (see the figure below and Oxidation of Alcohols): the ladder runs alcohol → aldehyde → carboxylic acid.
- Propanal is the first oxidation product — the aldehyde rung (middle): it comes from oxidising a primary alcohol and can still climb one more step to propanoic acid.
- Propanone is the ketone rung — an endpoint: it is the oxidation product of a secondary alcohol and climbs no further under mild oxidation.

In the figure below, the three black dots are the three rungs (alcohol → aldehyde → acid) and the arrows are the " $+ O / - H$ " oxidation steps. The red circle marks the aldehyde rung — the only rung that gives a silver mirror with Tollens'. Propanal sits on that red rung; propanone would sit off this ladder entirely, as a ketone endpoint.

Q4.2

Both $C_{2} H_{5} O H$ and $C H_{3} O C H_{3}$ are $C_{2} H_{6} O$ . Balance the reaction of the reactive one with sodium and state how many moles of $H_{2}$ come from $2$ mol of it.

Recall Solution Q4.2

Ethanol ( $- O H$ button) reacts; dimethyl ether ( $- O -$ , no O–H) does not. $2 C_{2} H_{5} O H + 2 N a \to 2 C_{2} H_{5} O N a + H_{2}$ Ratio: $2$ mol ethanol → $1$ mol $H_{2}$ . (Each $- O H$ gives half an $H_{2}$ ; two of them make one.)

Why ether is inert here: there is no acidic O–H hydrogen for Na to displace — the button is simply absent.

Level 5 — Mastery

"Full exam-style deduction."

Q5.1

Compound X, $C_{n} H_{2 n + 2} O$ , is the 4th member of the alcohol series. On mild oxidation it gives Y, which turns orange $K_{2} C r_{2} O_{7}$ green and gives a silver mirror with Tollens'. Further oxidation of Y gives Z, which turns blue litmus red. Identify $n$ , X, Y, Z and write the class of each.

Recall Solution Q5.1

Step 1 — find $n$ . Alcohol series counted from methanol ( $n = 1$ ): 4th member ⇒ $n = 4$ . Recall $C_{n} H_{2 n + 1} O H$ means an alkyl stick $C_{n} H_{2 n + 1}$ bonded to the $- O H$ button: $X = C_{4} H_{10} O = C_{4} H_{9} O H (butan-1-ol, a primary alcohol; here C_{4} H_{9} - is the butyl alkyl group)$ (Primary, because it must oxidise to an aldehyde in Step 2 — the button sits at a chain end.)

Step 2 — identify Y. Mild oxidation of a primary alcohol gives an aldehyde; the silver mirror confirms an aldehyde (only $- C H O$ reduces Tollens'), and $C r_{2} O_{7}^{2 -}$ green confirms oxidation occurred. $Y = C_{3} H_{7} C H O = C_{4} H_{8} O (butanal, an aldehyde)$

Step 3 — identify Z. Further oxidation of an aldehyde gives a carboxylic acid, which is acidic (blue litmus → red). $Z = C_{3} H_{7} C O O H = C_{4} H_{8} O_{2} (butanoic acid)$

Ladder (same picture as Q4.1): alcohol (X) → aldehyde (Y) → carboxylic acid (Z). Each step removes H / adds O — the oxygen ladder from Oxidation of Alcohols.

Answer: $n = 4$ ; $X = C_{4} H_{10} O$ (butan-1-ol, alcohol); $Y = C_{4} H_{8} O$ (butanal, aldehyde); $Z = C_{4} H_{8} O_{2}$ (butanoic acid, carboxylic acid).

Recall Master self-quiz

Constant mass gap between consecutive homologues ::: 14 (one $C H_{2}$ ). General formula of an alcohol ::: $C_{n} H_{2 n + 1} O H$ i.e. $C_{n} H_{2 n + 2} O$ . What does the $C_{n} H_{2 n + 1}$ part of an alcohol formula mean ::: an alkyl group (an alkane minus one H) bonded to the $- O H$ button. What is a $π$ -bond ::: the extra sideways electron cloud of a second shared pair, present in every double bond; loosely held, so reactive. Which oxidises to a ketone: primary or secondary alcohol ::: secondary. Does dimethyl ether react with Na ::: No — it has no O–H. Molar mass form of aldehyde series ::: $14 n + 16$ . One missing H-pair (one ring or one $π$ -bond) is called ::: one degree of unsaturation.

Connections

Functional groups and homologous series — the parent note this drills.
Oxidation of Alcohols — the ladder used in Q4.1 and Q5.1.
Isomerism — functional isomers in Q1.2, Q4.1, Q4.2.
Intermolecular Forces — boiling-point reasoning in Q3.1.
Inductive and Mesomeric Effects — bond polarity in Q3.2.
IUPAC Nomenclature — naming every compound above.
Hydrocarbons — the inert skeleton (stick) behind the buttons.