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

False. Acidity is set by the stability of the ANION $R{O}^{-}$, not any cation. Electron-donating alkyl groups pile electron density onto $O^{-}$ and destabilise it, so tertiary is the WEAKEST alcohol acid in solution.

True. Ethanol $p{K}_a\approx 16$, water $p{K}_a\approx 15.7$ — nearly identical because both stabilise the negative charge only on a bare, non-resonating oxygen.

True. In phenoxide the $O^{-}$ charge delocalises into the aromatic ring — figure s02 shows the resonance structures putting the negative charge on the ortho and para carbons — so the charge is spread out. An alkoxide $R{O}^{-}$ has no such resonance and traps the charge on one oxygen, so it is less stable and the alcohol is a weaker acid. See Phenols — Acidity and Resonance.

False. Since $p{K}_a(ROH)\approx 16>p{K}_a(H_{2}O)\approx 15.7$, the equilibrium barely favours the alkoxide. You must use reactive Na metal, which drives the reaction by releasing $H_2$ gas.

False. Tertiary alcohols resist oxidation because the carbinol carbon carries no C–H bond to remove — oxidation always strips one H from O–H and one from C–H to build a C=O.

False. PCC (mild, anhydrous) stops at the aldehyde; Jones (strong, aqueous) overshoots to the carboxylic acid. The presence or absence of water decides the stopping point.

False. It stops at the ketone. The carbonyl carbon of a ketone has no remaining C–H bond, so there is nothing left to oxidise.

False. Only tertiary alcohols cloud immediately; secondary take ~5 min (warming helps); primary give no turbidity at room temperature because a 1° carbocation is too unstable to form.

False. Acid-catalysed hydration is Markovnikov (OH on the more substituted carbon); hydroboration–oxidation is anti-Markovnikov (OH on the less substituted carbon) and is syn — figure s03 shows both new groups arriving on one face. See Markovnikov and Anti-Markovnikov Addition.

PCC is mild and dry, so it stops at the aldehyde (propanal). Reaching the acid needs water plus a strong oxidant like acidified dichromate, which hydrates the aldehyde to a gem-diol that gets oxidised again.

Lucas needs a carbon that can ionise to a carbocation. Phenol's C–O bond is part of the aromatic ring and will not ionise, so no alkyl chloride and no turbidity form.

The stability statement is right for the CATION, but acidity depends on the alkoxide anion, which alkyl groups destabilise. The person is fusing two unrelated pieces of chemistry.

A ketone $R_{2}CO$ has two carbons on the carbonyl carbon, so reduction gives a secondary alcohol $R_{2}CHOH$. Only an aldehyde reduces to a primary alcohol. See Carbonyl Compounds — Aldehydes and Ketones.

Formaldehyde $HCHO$ contributes zero carbons besides the carbonyl one, so after $R$ adds you get only one carbon on the carbinol carbon → a primary alcohol. See Grignard Reagents.

A 3° substrate with the strong base/nucleophile $O{H}^{-}$ favours elimination (E2 — base pulls off a neighbouring H as the halide leaves), giving an alkene, not substitution. Use a route like hydration or Grignard instead. See Haloalkanes — SN1 and SN2.

Butan-2-ol is secondary, so it oxidises to butan-2-one (a ketone) and stops — the carbonyl carbon has no C–H left. Only primary alcohols climb all the way to acids.

The oxidant chromium goes from $C{r}^{6+}$ (orange dichromate) to $C{r}^{3+}$ (green) as it accepts electrons from the alcohol. The colour is a visible readout that the alcohol was successfully oxidised.

$ZnC{l}_2$ is a Lewis acid that bonds to the alcohol oxygen, forming a zinc–alkoxide complex (roughly $R-O-ZnC{l}_2^{-}$, protonated to leave as neutral $HOZnC{l}_2$ / with chloride as $[ZnC{l}_3{]}^{-}$). That turns a hopeless leaving group (bare $O{H}^{-}$) into an excellent one, so the C–O bond can break to give the carbocation the SN1 path needs — figure s01.

Oxidation must remove a hydrogen from the carbinol carbon to form the C=O. In a 3° alcohol that carbon is bonded to three other carbons and zero hydrogens, so there is simply no C–H to remove.

More alkyl groups donate more electron density onto the $O^{-}$, destabilising the negative charge, and their bulk also blocks solvent from stabilising it. Both effects make the more-substituted alkoxide less stable and its alcohol less acidic.

Without water the aldehyde cannot be hydrated to a gem-diol. Since the gem-diol is the species carrying the extra C–H needed for further oxidation, no water means no over-oxidation.

Protonation of the alkene forms the more stable (more substituted) carbocation, and water attacks that carbon. The OH therefore ends up on the more substituted carbon.

Boron is the electrophile and is bulky, so it adds to the less hindered (less substituted) carbon; H goes to the other carbon, and later oxidation replaces the B with OH in place — so OH ends up on the less substituted carbon. Because B and H add through one four-centre transition state on the same face, the addition is syn — figure s03.

With no electron-donating alkyl group to destabilise the anion, methoxide is relatively well-behaved, making methanol slightly MORE acidic than typical larger alcohols and comparable to water.

Nothing visible — the solution stays clear. A primary carbocation is far too unstable to form, so the SN1 pathway does not proceed and no cloudy alkyl chloride appears.

It is over-oxidised to a carboxylic acid, because water hydrates the aldehyde to a gem-diol whose fresh C–H is oxidised again. This is exactly why anhydrous PCC is needed to isolate an aldehyde.

It stops at the ketone. The ketone's carbonyl carbon has no C–H bond left, and without a C–H there is nothing for the oxidant to remove.

A fizz of $H_2$ gas is released as $2ROH+2Na\to 2R{O}^{-}N{a}^{+}+H_2\uparrow$. The bubbling is a quick qualitative test for an O–H group.

A carboxylate ($p{K}_a\approx 4.8$) spreads its negative charge over two equivalent oxygens by resonance, whereas an alkoxide traps the charge on one oxygen with no resonance. Less delocalisation means a less stable anion and a weaker acid. See Carboxylic Acids — Acidity.