Worked examples — Alcohols — preparation, acidity (pKa ~16), oxidation (PCC, Jones, K₂Cr₂O₇), Lucas test

This page is the drill floor for the Alcohols topic note. We enumerate every kind of situation the chapter can throw at you, then work a labelled example for each. Before you read a solution, cover it and forecast — guessing first is where the learning happens.

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The scenario matrix

Each row is a case class the topic can test. The worked examples that follow are tagged with the cell they cover, so together they touch every box.

#	Case class	What it stresses	Covered by
A	1° alcohol, mild+dry oxidant	PCC stops at aldehyde	Ex 1
B	1° alcohol, strong+wet oxidant	over-oxidation to acid	Ex 2
C	2° alcohol, any oxidant	stops at ketone (no more C–H)	Ex 3
D	3° alcohol, oxidant	degenerate: no reaction	Ex 3
E	Lucas test across classes	carbocation-stability ranking	Ex 4
F	Zero-carbon limit (methanol / formaldehyde)	edge of the 1°/2°/3° scale	Ex 5
G	Acidity ranking (sign of substituent effect)	anion stability, resonance	Ex 6
H	Grignard retrosynthesis (build a target)	choosing carbonyl class	Ex 7
I	Real-world word problem	pick reagent to hit a stopping point	Ex 8
J	Exam twist — molecule with two OH of different class	apply rules locally	Ex 9

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Example 1 — Cell A: 1° alcohol + PCC (mild, dry)

Forecast: 1° alcohol going up the ladder — but which rung do we stop on, aldehyde or acid?

1. Identify the class. The carbinol carbon ($C{H}_{2}OH$) touches one other carbon → 1°. Why this step? Class decides how far oxidation can go and whether it happens at all.
2. Read the reagent's personality. PCC is mild and anhydrous (no water). Why this step? Over-oxidation of an aldehyde to an acid needs water to first make a gem-diol (a carbon with two OH). No water → that pathway is blocked.
3. Remove two H's, make C=O. Strip one H from O–H and one from the carbinol C–H, form the double bond: $C{H}_{3}C{H}_{2}OH\stackrel{\text{PCC}}{\to }C{H}_{3}CHO$ Why this step? Oxidation here is "lose 2 H, gain a C=O."
4. Stop. Product = ethanal (acetaldehyde), $C{H}_{3}CHO$.

Verify: Count carbons — start with 2 C, end with 2 C (no C–C broken ✔). The aldehyde carbon still carries one H, so in principle it could go further, but dry PCC never lets it. Molecular formula went $C_2{H}_{6}O\to C_2{H}_{4}O$ (lost $H_2$) ✔.

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Example 2 — Cell B: same 1° alcohol, strong wet oxidant

Forecast: Same starting material as Ex 1 — will the answer be the same aldehyde, or further?

1. Class: still 1° (as in Ex 1).
2. Reagent personality: $K_{2}C{r}_2{O}_7/H^{+}$ is strong and aqueous. Why this step? Now water is present, so the aldehyde formed en route gets hydrated to a gem-diol $C{H}_{3}CH(OH{)}_2$, which has a fresh C–H to lose.
3. First oxidation gives the aldehyde: $C{H}_{3}C{H}_{2}OH\to C{H}_{3}CHO$.
4. Second oxidation (via the gem-diol) gives the acid: $C{H}_{3}CHO\stackrel{[O],H_{2}O}{\to }C{H}_{3}COOH$ Why this step? The hydrate's C–H is exactly what an oxidant removes to build the second C=O of $–COOH$.
5. Colour cue: dichromate goes orange → green ($C{r}^{6+}\to C{r}^{3+}$), confirming reaction.

Verify: Final product ethanoic acid (acetic acid) $C{H}_{3}COOH$, formula $C_2{H}_4{O}_2$. Contrast with Ex 1: same substrate, different reagent → different rung. That contrast is the whole point of cells A vs B.

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Example 3 — Cells C & D: 2° stops, 3° refuses

Forecast: One gives a ketone, one gives nothing. Which is which, and why does one refuse?

1. Classify both.
   • Propan-2-ol: carbinol C touches 2 carbons → 2°.
   • 2-methylpropan-2-ol: carbinol C touches 3 carbons → 3°. Why this step? The count of C–H bonds on the carbinol carbon is the deciding factor.
2. Count carbinol C–H bonds.
   • 2° has one C–H there.
   • 3° has zero C–H there (all four bonds to C or O). Why this step? Oxidation must remove a C–H from that carbon. No C–H → no oxidation, full stop.
3. Oxidise the 2°: $C{H}_{3}CH(OH)C{H}_3\stackrel{[O]}{\to }C{H}_{3}COC{H}_3(\text{propanone / acetone})$ It stops here — the carbonyl carbon now has no H left (Cell C).
4. The 3° refuses: $(C{H}_3{)}_{3}COH\stackrel{[O]}{\to }\text{no reaction}$ (Cell D — the degenerate input.)

Verify: Propanone $C_3{H}_{6}O$ — start $C_3{H}_{8}O$, lost $H_2$ ✔. For the 3°: dichromate stays orange (no colour change), a lab-observable confirmation that nothing happened ✔.

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Example 4 — Cell E: Lucas test ranking

Forecast: All three have an OH. Do they cloud at the same rate? If not, which is fastest?

1. Recall the mechanism. Lucas is SN1-like: $ZnC{l}_2$ makes OH a good leaving group, the C–O breaks to a carbocation $R^{+}$, then $C{l}^{-}$ caps it to an insoluble $R-Cl$ (the cloudiness). Why this step? If rate is set by cation formation, then rate follows cation stability.
2. Rank carbocation stability. More alkyl groups donate electron density / offer hyperconjugation: $3°>2°>1°$ Why this step? This is the ordering that turns into a timing difference you can watch.
3. Translate to observation.
   • 3° (2-methylpropan-2-ol): turbidity immediately.
   • 2° (butan-2-ol): turbidity in ~5 minutes, faster on warming.
   • 1° (butan-1-ol): no turbidity at room temperature (1° cation too unstable to form).

Verify: The order matches carbocation stability exactly, and gives a clean three-way discrimination — which is why the Lucas test works as an identifier. Sanity check on the limiting case: a 1° alcohol showing nothing at RT is the diagnostic negative result, not a failed experiment.

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Example 5 — Cell F: the zero-carbon edge (methanol & formaldehyde)

Forecast: Methanol's carbinol carbon touches zero other carbons — does it break the rules?

1. Place methanol on the scale. Its carbinol carbon has no attached carbons. By convention it is treated as the limit below 1° — it still has C–H bonds on the carbinol carbon, so it can be oxidised. Why this step? The oxidation rule cares about C–H count on the carbinol carbon, not the alkyl-carbon count; methanol has three C–H there.
2. Oxidise methanol with PCC: removes 2 H → methanal (formaldehyde) $HCHO$. Why this step? Dry + mild stops at the aldehyde, just like any 1°.
3. Grignard on formaldehyde. $HCHO$ has no carbon substituents on its carbonyl carbon; adding one R group gives a carbinol carbon touching exactly one carbon: $C{H}_{3}MgBr+HCHO\to C{H}_{3}C{H}_{2}OH(\text{after }{H}^{+}\text{ workup})$ Why this step? Formaldehyde is the unique carbonyl that yields a 1° alcohol.

Verify: Product is ethanol, a genuine 1° alcohol (carbinol C touches 1 carbon ✔). This confirms the edge rule: HCHO → 1°, any other aldehyde → 2°, ketone → 3°.

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Example 6 — Cell G: acidity ranking (the sign of the substituent effect)

Forecast: They all have an O–H. Guess the order before reading — where do the two alcohols land relative to water?

1. Acidity = stability of the anion $A^{-}$ left behind. More stable conjugate base → stronger acid → lower pKa. Why this step? It reframes "which is more acidic" into "whose anion is happier."
2. Acetic acid ($p{K}_a\approx 4.8$): its $–CO{O}^{-}$ spreads charge over two equivalent oxygens by resonance → very stable → strongest here.
3. Phenol ($p{K}_a\approx 10$): its $O^{-}$ delocalises into the benzene ring → stabilised → next. (See Phenols — Acidity and Resonance and Carboxylic Acids — Acidity.)
4. Water ($p{K}_a\approx 15.7$) vs ethanol ($p{K}_a\approx 16$): almost tied; ethanol's single alkyl group is weakly electron-donating, pushing charge onto $O^{-}$ and destabilising it slightly → ethanol a hair weaker than water.
5. 3° alcohol: more alkyl donation + poorer solvation of the bulky anion → weakest acid of the set. Why this step? This is the anti-intuition: 3° cations are stable, but 3° anions are destabilised — acidity is about the anion.

Verify: Order = acetic acid > phenol > water > ethanol > 3° alcohol. Numerically $4.8<10<15.7<16<\sim 18$, and "lower pKa = stronger acid" places them in exactly this sequence ✔.

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Example 7 — Cell H: Grignard retrosynthesis

Forecast: A 3° alcohol has three carbons on the carbinol carbon. Two come from the carbonyl, one from R — which carbonyl?

1. Confirm it's 3°. The carbinol carbon carries $C{H}_3$, $C_6{H}_5$, and $C{H}_3$ — three carbons. Why this step? A 3° target demands a ketone partner (a ketone already supplies two carbons on its C=O).
2. Disconnect one C–C bond at the carbinol carbon. Break the bond to one $C{H}_3$: that $C{H}_3$ becomes the Grignard $C{H}_{3}MgBr$; the rest becomes the ketone. Why this step? The Grignard R group is the piece you "add"; the carbonyl is the piece already holding the other two carbons.
3. Assemble forward: $C_6{H}_5-CO-C{H}_3(\text{acetophenone})+C{H}_{3}MgBr\to (C{H}_3)(C_6{H}_5)C(OH)(C{H}_3)$ after aqueous $H^{+}$ workup. (See Grignard Reagents and Carbonyl Compounds — Aldehydes and Ketones.)

Verify: Carbon carbinol count = 3 ✔ (3° achieved). Rule check: ketone + Grignard → 3° alcohol, matching Cell H. Alternative valid disconnection: $C_6{H}_{5}MgBr+(C{H}_3{)}_{2}CO$ (acetone) gives the same target — both routes obey "ketone → 3°."

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Example 8 — Cell I: real-world word problem

Forecast: Both benzaldehyde and its acid smell very different. Will the strong wet oxidant deliver the aldehyde?

1. Classify the substrate: $C_6{H}_{5}C{H}_{2}OH$ is 1° (carbinol C touches one carbon, the ring carbon).
2. Predict the junior's outcome. Strong + aqueous $K_{2}C{r}_2{O}_7/H^{+}$ over-oxidises the 1° all the way: $C_6{H}_{5}C{H}_{2}OH\stackrel{K_{2}C{r}_2{O}_7/H^{+},\Delta }{\to }{C}_6{H}_{5}COOH(\text{benzoic acid})$ Why this step? Water lets the intermediate aldehyde hydrate and oxidise again — wrong product, no almond smell.
3. Fix it — pick a stopping-point reagent. Use PCC in dry $C{H}_{2}C{l}_2$: $C_6{H}_{5}C{H}_{2}OH\stackrel{\text{PCC}}{\to }{C}_6{H}_{5}CHO$ Why this step? Mild + anhydrous halts at the aldehyde — exactly the target.

Verify: Desired product $C_6{H}_{5}CHO$ ($C_7{H}_{6}O$) is reached only with PCC; the junior's route lands on $C_7{H}_6{O}_2$ (benzoic acid, a solid, odourless-ish). The decision hinges purely on the water/strength switch, confirming cell I's lesson.

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Example 9 — Cell J: exam twist, two OH of different class

Forecast: Two hydroxyls in one molecule — do both react, one, or neither? Treat each locally.

1. Classify each OH separately.
   • The $C2$ hydroxyl: that carbon bears two $C{H}_3$ and the chain → 3°.
   • The $C4$ hydroxyl: that carbon bears one $C{H}_3$ and the chain (two carbons total) → 2°. Why this step? Oxidation is a local event at each carbinol carbon; you judge each independently.
2. Apply the rule at C2 (3°): no C–H on that carbon → no reaction there.
3. Apply the rule at C4 (2°): one C–H → oxidises to a ketone, and stops (carbonyl carbon then has no H): $\text{C4: }-CH(OH)-C{H}_3\to -CO-C{H}_3$
4. Combine. Product = $(C{H}_3{)}_{2}C(OH)-C{H}_2-CO-C{H}_3$ (a hydroxy-ketone): the 3° OH survives, the 2° OH becomes a ketone.

Verify: Only one degree of oxidation occurred → net loss of $H_2$ once. Start $C_5{H}_{12}{O}_2$, product $C_5{H}_{10}{O}_2$ (lost $H_2$, kept both oxygens: one as OH, one as C=O) ✔. This is the classic twist: apply class rules per-carbon, not per-molecule.

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Recall Self-test (cover the answers)

Ethanol + PCC gives what? ::: Ethanal (acetaldehyde) $C{H}_{3}CHO$ — stops at the aldehyde. Ethanol + hot $K_{2}C{r}_2{O}_7/H^{+}$ gives what? ::: Acetic acid $C{H}_{3}COOH$ — water lets it over-oxidise. Fastest Lucas turbidity: 1°, 2° or 3°? ::: 3° — most stable carbocation, SN1 is fastest. Which is the stronger acid, ethanol or a 3° alcohol? ::: Ethanol — alkyl groups on the 3° destabilise the anion. Grignard + formaldehyde gives which alcohol class? ::: Primary (1°). Grignard + ketone gives which class? ::: Tertiary (3°). Why does a 3° alcohol resist oxidation? ::: Its carbinol carbon has no C–H to remove.

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Connections used: Haloalkanes — SN1 and SN2 (Lucas mechanism), Carbonyl Compounds — Aldehydes and Ketones, Grignard Reagents, Phenols — Acidity and Resonance, Carboxylic Acids — Acidity, Markovnikov and Anti-Markovnikov Addition.