Conversion and distinction

Ask an exam question "convert ethanol into propan-1-ol" and a student who only remembers isolated reactions freezes — there is no bottle labelled "make the chain one carbon longer." Yet every reagent needed is already in the alcohols chapter: halogenating agents, magnesium metal, oxidising and reducing agents, dehydrating conditions. Conversion of one alcohol to another is the skill of chaining these into a short synthetic route, always through an intermediate functional group, never in one step. The companion skill is distinguishing alcohol classes — telling a 1°, 2°, or 3° alcohol apart using nothing but a test tube and an observation. Two tests do this: Lucas' test, which reads the speed of a substitution reaction, and oxidation with acidified dichromate or permanganate, which reads a colour change (or its absence). ISC examiners routinely combine both skills: plan a conversion and justify it, or identify unlabelled alcohols using more than one line of chemical evidence.

The core insight

Alcohol-to-alcohol conversions always route through an intermediate (halide, Grignard reagent, carbonyl, or alkene) since no single reagent edits a carbon skeleton directly, while Lucas' test (reaction speed: 3° instant, 2° minutes, 1° none) and dichromate oxidation (colour change for 1°/2°, none for 3°) independently and jointly distinguish alcohol classes.

The core idea

Why an intermediate is unavoidable. An alcohol's identity is fixed by its carbon skeleton and the position of –OH. No reagent bonds directly to a C–OH carbon and rearranges the skeleton in one step — so every conversion strips the transformation into pieces the chapter's known reactions can each handle, then reassembles the target.

Strategy 1 — change the chain length. Convert the alcohol to an alkyl halide using $HX$ , $PCl_5$ , or $SOCl_2$ ; turn the halide into a Grignard reagent with magnesium in dry ether; then react the Grignard reagent with a chosen carbonyl compound. The identity of the carbonyl compound (formaldehyde, another aldehyde, or a ketone) controls exactly how many carbons are added and whether the new alcohol is 1°, 2°, or 3°.

Strategy 2 — change the position or class of –OH. Oxidise the alcohol to its carbonyl compound, then either reduce it back with a different reagent, or dehydrate the alcohol to an alkene and re-hydrate it under conditions that place –OH on a different carbon (Markovnikov acid-catalysed hydration versus anti-Markovnikov hydroboration–oxidation). Both strategies obey the same logic: alcohol → different functional group → alcohol, never alcohol → alcohol directly.

Reading the two class-tests. Lucas' test exploits the fact that carbocation stability, and hence $S_N1$ reaction rate with $HCl/ZnCl_2$ , falls sharply from 3° to 2° to 1°:

\text{Rate}(S_N1) \propto \text{carbocation stability}: \quad 3^\circ > 2^\circ > 1^\circ

where: rate is the speed at which the alcohol reacts with conc. $HCl$ /anhydrous $ZnCl_2$ to give a turbid alkyl chloride layer, and stability ranks the carbocation formed once $ZnCl_2$ activates –OH as a leaving group. Oxidation is diagnosed by how far the reaction can go: a 1° alcohol has two oxidisable C–H bonds on the carbinol carbon (alcohol → aldehyde → acid), a 2° alcohol has one (alcohol → ketone, no further), a 3° alcohol has none, so it resists mild oxidation entirely.

Figure

Decision map: a Lucas test track split into instant minutes and no reaction, alongside an oxidation track split into acid ketone and no reaction, converging on 3° 2° and 1° alcohol labels

Derivation AWhy the chain-extension route works: ethanol to propan-1-ol adds exactly one carbon

Show that treating an alcohol with $PCl_5$ , converting the halide to a Grignard reagent, and reacting it with formaldehyde is guaranteed to lengthen the carbon chain by exactly one carbon while placing the new –OH on a terminal (primary) carbon, regardless of which primary alcohol is the starting material.

Step 1. Start from a generic primary alcohol $R\text{–}CH_2\text{OH}$ . Treating it with $PCl_5$ replaces –OH with –Cl without touching any C–C bond, since $PCl_5$ attacks only the oxygen-bearing carbon:

R\text{–}CH_2OH + PCl_5 \rightarrow R\text{–}CH_2Cl + POCl_3 + HCl

The carbon count of $R\text{–}CH_2Cl$ is identical to the starting alcohol — this step changes the leaving group only, not the skeleton.

Step 2. Magnesium inserts into the C–Cl bond in dry ether, again without altering the carbon skeleton, to give the Grignard reagent $R\text{–}CH_2MgCl$ . The carbanion-like carbon of this reagent is the same carbon that originally carried –OH.

Step 3. Formaldehyde, $HCHO$ , has exactly one carbon and two hydrogens on it — no other alkyl substituent. When the Grignard reagent's nucleophilic carbon attacks formaldehyde's carbonyl carbon and the product is hydrolysed, one new carbon is bonded on, and because formaldehyde offers no second alkyl group, the new carbinol carbon carries two hydrogens: it is necessarily a primary alcohol.

R\text{–}CH_2MgCl + HCHO \rightarrow R\text{–}CH_2\text{–}CH_2OMgCl \xrightarrow{H_2O/H^+} R\text{–}CH_2\text{–}CH_2OH

Step 4. Comparing the product $R\text{–}CH_2\text{–}CH_2OH$ to the starting material $R\text{–}CH_2OH$ , the chain is exactly one carbon longer and the new terminal carbon is still primary, for any choice of $R$ — which is why this three-reagent sequence ( $PCl_5$ , then $Mg$ /dry ether, then $HCHO$ /hydrolysis) is the general, reliable method for one-carbon chain extension examined as "conversion of one alcohol to another," and why swapping formaldehyde for a different aldehyde or ketone instead gives a secondary or tertiary product.

Worked example

Devise a synthetic route converting ethanol ( $CH_3CH_2OH$ ) into propan-1-ol ( $CH_3CH_2CH_2OH$ ), then use Lucas' test and the oxidation test to confirm that the product is indeed primary and distinguish it from propan-2-ol and 2-methylpropan-2-ol in unlabelled tubes.

Step 1 — Apply the chain-extension route from the derivation. With $R = CH_3$ , treat ethanol with $PCl_5$ to get chloroethane, then form the Grignard reagent with $Mg$ /dry ether, then react with formaldehyde and hydrolyse:

CH_3CH_2OH \xrightarrow{PCl_5} CH_3CH_2Cl \xrightarrow{Mg,\ \text{dry ether}} CH_3CH_2MgCl \xrightarrow{HCHO,\ \text{then}\ H_2O/H^+} CH_3CH_2CH_2OH

This yields propan-1-ol, one carbon longer than ethanol, exactly as the derivation predicts, since formaldehyde supplies the extra carbon with no second substituent.

Step 2 — Set up the distinguishing test. Suppose three unlabelled tubes contain propan-1-ol, propan-2-ol, and 2-methylpropan-2-ol in unknown order. Add conc. $HCl$ /anhydrous $ZnCl_2$ (Lucas reagent) to each and time the appearance of turbidity.

Step 3 — Read the Lucas results. The tube showing turbidity immediately is 2-methylpropan-2-ol (3°), because its tertiary carbocation forms fastest via $S_N1$ . The tube turning turbid within a few minutes is propan-2-ol (2°). The tube showing no turbidity at room temperature is propan-1-ol (1°) — the product from Step 1.

Step 4 — Confirm with the oxidation test. Warm the 1°-assigned and 2°-assigned tubes with acidified $K_2Cr_2O_7$ . Both show the orange dichromate turning green: the 1° tube (propan-1-ol) oxidises on to propanoic acid via propanal, while the 2° tube (propan-2-ol) stops at propan-2-one (acetone). The 3°-assigned tube (2-methylpropan-2-ol) shows no colour change, confirming it resists oxidation.

Both tests agree, so the propan-1-ol synthesised in Step 1 is confirmed primary: no turbidity with Lucas' reagent, and orange-to-green with acidified dichromate, oxidising all the way to the acid.

Competitive stretch

If asked to convert propan-1-ol into propan-2-ol (relocating –OH without changing carbon count), the route is: dehydrate propan-1-ol to propene ( $\text{conc. } H_2SO_4$ , heat), then re-hydrate under Markovnikov conditions (dil. $H_2SO_4$ , $H_2O$ ) so –OH lands on the more substituted (middle) carbon, giving propan-2-ol — Strategy 2 from the core idea, using dehydration–rehydration rather than oxidation–reduction.

Common traps

Trying to convert one alcohol into another with a single reagent — no such reagent exists; every valid route passes through at least one intermediate (halide, Grignard reagent, carbonyl, or alkene), shown explicitly in exam answers.
Assuming oxidation of a 2° alcohol to a ketone reverses to the same alcohol with the same oxidising reagent — reduction (e.g. $NaBH_4$ or $H_2/Ni$ ) is a distinct step, not a reversal of the oxidant.
Misordering Lucas' test results, for instance assuming 1° reacts fastest — the correct order is 3° (instant) greater than 2° (minutes) greater than 1° (no reaction).
Treating "no colour change" in the oxidation test as inconclusive rather than the positive signature of a tertiary alcohol — a null result is itself the diagnostic answer.

Ask the tutor: stuck on planning a specific alcohol-to-alcohol conversion, or want to check your Lucas'-test/oxidation-test reasoning for a set of unlabelled samples?

Why can't I just use one strong reagent to jump straight from one alcohol to another?

Because a reagent that attacks the C–OH carbon directly would have to simultaneously break a C–C bond somewhere else in the skeleton and form a new one — no single reagent in this syllabus does both at once. Every real route breaks the problem into steps a known reaction can each handle: leave the –OH group (via a halide), rebuild the skeleton (via a Grignard reagent and a chosen carbonyl compound), then restore the –OH group at hydrolysis. That is why "one-step conversion" answers lose marks even if the final product is correct.

If Lucas' test already tells me the class, why do I also need the oxidation test?

Lucas' test and the oxidation test probe different chemistry — carbocation stability for the first, oxidisability of the carbinol carbon for the second — so using both gives independent confirmation rather than relying on one observation that could be misread (for example, a faint turbidity that is hard to time precisely). Examiners often ask for "two chemical tests" precisely so you demonstrate both mechanisms, not because either test alone is unreliable in principle.

Does the chain-extension route only add one carbon, or can I add more at once?

The number of carbons added equals the number of carbons in the carbonyl compound you choose for the Grignard step: formaldehyde ( $HCHO$ ) adds exactly one carbon and gives a primary alcohol; any other aldehyde adds its own carbon count and gives a secondary alcohol (since it has an alkyl group already on the carbonyl carbon); a ketone adds its carbon count and gives a tertiary alcohol (two alkyl groups already present). Choosing the carbonyl compound is therefore how you control both how many carbons are added and what class the new alcohol will be.

Conversion and distinction

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads