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

Level 1 — Recognition

Recall Solution L1.1

What we do: find the carbinol carbon and count other carbons attached to it.

(a) The $O H$ sits on the end carbon $C H_{2}$ . That carbon touches exactly 1 other carbon → 1° (primary).
(b) The $O H$ sits on the middle carbon, which touches 2 carbons (a $C H_{3}$ on each side) → 2° (secondary).
(c) The $O H$ carbon touches 3 methyl carbons → 3° (tertiary).

Recall Solution L1.2

1° + PCC → stops at the aldehyde ( $R C H O$ ). Why: PCC is mild and anhydrous — no water means no gem-diol to over-oxidise.
1° + Jones or $K_{2} C r_{2} O_{7} / H^{+}$ → goes all the way to the carboxylic acid ( $R C O O H$ ). Why: water hydrates the aldehyde to a gem-diol, which has a fresh $C - H$ that gets oxidised again.
3° + any of them → no reaction. Why: the carbinol carbon has no $C - H$ to remove; oxidation here needs one $C - H$ on the carbinol carbon.

Level 2 — Application

Recall Solution L2.1

(a) $C H_{3} C H_{2} C H_{2} O H PCC C H_{3} C H_{2} C H O$ (propanal). Dry + mild ⇒ stop at aldehyde.
(b) $C H_{3} C H_{2} C H_{2} O H K_{2} C r_{2} O_{7} / H^{+} C H_{3} C H_{2} C O O H$ (propanoic acid). Aqueous + strong ⇒ overshoot to acid; solution turns orange → green ( $C r^{6 +} \to C r^{3 +}$ ).
(c) $C H_{3} C H (O H) C H_{2} C H_{3} Jones C H_{3} C O C H_{2} C H_{3}$ (butan-2-one). A 2° alcohol gives a ketone and stops — the new carbonyl carbon has no $C - H$ left.
(d) $(C H_{3})_{3} C O H \to$ no reaction (3°, no carbinol $C - H$ ).

Recall Solution L2.2

Count the carbons ending up on the new carbinol carbon.

(a) Methanal's carbon starts with 0 other carbons; add $R$ (1 carbon) ⇒ the carbinol carbon holds 1 carbon ⇒ 1° alcohol.
(b) An aldehyde carbon already holds 1 carbon ( $R^{'}$ ); add $R$ ⇒ 2 carbons ⇒ 2° alcohol.
(c) A ketone carbon already holds 2 carbons; add $R$ ⇒ 3 carbons ⇒ 3° alcohol.

See Grignard Reagents for the carbanion mechanism and Carbonyl Compounds — Aldehydes and Ketones for the $C = O$ target.

What the reagent does: $Z n C l_{2}$ is a Lewis acid that grabs the oxygen lone pair, turning the poor leaving group $- O H$ into a great one ( $- O Z n C l_{2}^{-}$ ). The $C - O$ bond then breaks first, releasing a carbocation $R^{+}$ ; only afterwards does $C l^{-}$ cap it. This "leave first, attack later" order is the SN1 pattern (see Haloalkanes — SN1 and SN2). Rate-limiting step: forming $R^{+}$ . So the whole speed is set by how stable that cation is. Stability order: $3° > 2° > 1°$ , because each alkyl group donates electron density (hyperconjugation + induction) into the empty carbon, cushioning the positive charge.

3°: most stable cation ⇒ forms fast ⇒ turbidity immediately.
2°: moderately stable ⇒ ~5 min (gentle warming helps).
1°: cation too unstable to form ⇒ no turbidity at room temperature.

Recall Solution L3.2

Lower $p K_{a}$ = stronger acid, and $p K_{a} = - lo g K_{a}$ , so acid strength is set by how stable the conjugate anion $A^{-}$ is. Order: acetic acid $>$ phenol $>$ water $>$ ethanol.

Acetic acid (4.8): the negative charge is spread over two equivalent oxygens (carboxylate resonance) — very stable anion. See Carboxylic Acids — Acidity.
Phenol (10): the $O^{-}$ charge delocalises into the benzene ring (ortho/para positions). Real but weaker spreading than carboxylate. See Phenols — Acidity and Resonance.
Water (15.7) vs ethanol (16): the alkoxide $C H_{3} C H_{2} O^{-}$ has no resonance — just a bare, localised charge on oxygen. Worse still, ethyl is weakly electron-donating, pushing extra density onto the already-negative oxygen and destabilising it. So ethanol is slightly weaker than water.

Level 4 — Synthesis

Recall Solution L4.1

(a) $C H_{3} C H_{2} C H_{2} O H PCC, C H_{2} C l_{2} C H_{3} C H_{2} C H O$ . Dry + mild traps the aldehyde.
(b) $C H_{3} C H_{2} C H_{2} O H K_{2} C r_{2} O_{7} / H_{2} S O_{4}, Δ C H_{3} C H_{2} C O O H$ . Aqueous + strong drives it to the acid.
Reverse (acid → 1° alcohol): $C H_{3} C H_{2} C O O H L i A l H_{4} C H_{3} C H_{2} C H_{2} O H$ . We need the strong hydride $L i A l H_{4}$ here because $N a B H_{4}$ is too mild to reduce a carboxylic acid (it only touches aldehydes/ketones). $L i A l H_{4}$ delivers $H^{-}$ to the carbonyl carbon and drops the acid back to the 1° alcohol.

Recall Solution L4.2

Target class: the carbinol carbon carries three carbons (two $C H_{3}$ and one $C H_{2} C H_{3}$ ) ⇒ 3° alcohol ⇒ we need Grignard + ketone. One route: disconnect the new $C - C$ bond as ethyl + acetone. $C H_{3} C H_{2} M g B r + (C H_{3})_{2} C O ⟶ (C H_{3})_{2} C (O M g B r) C H_{2} C H_{3} H_{3} O^{+} (C H_{3})_{2} C (O H) C H_{2} C H_{3}$ Ethylmagnesium bromide's carbanion $C H_{3} C H_{2}^{-}$ attacks the ketone carbon, forming the $C - C$ bond; aqueous work-up protonates the alkoxide to the alcohol. Alternative disconnection: $C H_{3} M g B r + C H_{3} C O C H_{2} C H_{3}$ (butan-2-one) — the methyl carbanion adds to that ketone to give the same 3° alcohol.

Level 5 — Mastery

Recall Solution L5.1

(a) $C H_{3} C H_{2} O H + H_{2} O \to C H_{3} C O O H + 4 H^{+} + 4 e^{-}$ . The chromium goes orange $C r^{6 +}$ → green $C r^{3 +}$ as it is reduced. (b) Electrons lost per alcohol: ethanol → acetic acid removes 4 electrons (2 for alcohol→aldehyde, 2 for aldehyde/gem-diol→acid). One $C r_{2} O_{7}^{2 -}$ accepts $6 e^{-}$ . $electrons needed = 0.10 \times 4 = 0.40 mol e^{-}$ $n (C r_{2} O_{7}^{2 -}) = \frac{0.40}{6} = 0.0667 mol$ So about $0.067 mol$ of dichromate is required.

Recall Solution L5.2

(a) For an acid–base reaction, $lo g K = p K_{a} (acid consumed) - p K_{a} (acid formed)$ . Here water is formed (its conjugate $O H^{-}$ is consumed) and ethanol is consumed: $lo g K = p K_{a} (H_{2} O) - p K_{a} (E tO H) = 15.7 - 16.0 = - 0.3$ $K = 1 0^{- 0.3} \approx 0.50$ So the equilibrium barely reaches half-way — only a small fraction of alkoxide at any instant. (b) With $K \approx 0.5$ , $N a O H$ cannot deprotonate ethanol appreciably. Sodium metal solves it by an irreversible redox: $2 C H_{3} C H_{2} O H + 2 N a \to 2 C H_{3} C H_{2} O^{-} N a^{+} + H_{2} ↑$ The escaping $H_{2}$ gas removes product permanently, dragging the reaction fully to alkoxide (and the fizz doubles as a quick test for $- O H$ ).

Recall Solution L5.3

Step 1 — Lucas test (separates by cation stability / class):

Immediate turbidity ⇒ 3° ⇒ 2-methylpropan-2-ol.
Turbidity in ~5 min ⇒ 2° ⇒ butan-2-ol.
No turbidity at room temperature ⇒ 1° ⇒ butan-1-ol.

That already distinguishes all three. Confirm with $K_{2} C r_{2} O_{7} / H^{+}$ :

butan-1-ol (1°): orange → green, oxidises to butanoic acid. ✔
butan-2-ol (2°): orange → green, oxidises to butan-2-one. ✔
2-methylpropan-2-ol (3°): stays orange (no reaction). ✔

The two tests cross-check: the bottle that fails Lucas fastest is the same one that fails to change the dichromate colour — both flag the 3° alcohol from opposite directions.

Recall Quick self-check (close the book first)

Lucas order fastest→slowest ::: 3° > 2° > 1° (carbocation stability) Acidity order strongest→weakest among alcohols ::: 1° > 2° > 3° (alkyls destabilise the anion) PCC on a 1° alcohol gives ::: the aldehyde (dry, mild — stops there) $K_{2} C r_{2} O_{7} / H^{+}$ on a 1° alcohol gives ::: the carboxylic acid (aqueous, strong) Grignard + ketone gives which class ::: 3° alcohol Why sodium metal, not NaOH, for alkoxides ::: $K \approx 0.5$ , so NaOH barely works; Na releases $H_{2}$ and drives it fully

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