Reactions of alcohols

An alcohol molecule has two reactive sites: the weakly acidic O–H hydrogen and the polar C–O bond. Everything in this article follows from attacking one site or the other. Attack the O–H bond and you get salts (sodium alkoxide) or esters — the oxygen stays put, only the hydrogen leaves. Attack the C–O bond and the whole –OH group is swapped out, usually for a halogen, using $HX$ , $PCl_3$ , $PCl_5$ or $SOCl_2$ . A third, separate pathway — oxidation — removes hydrogen from the carbinol carbon itself, and here the class of alcohol (1°, 2°, 3°) decides everything: it fixes whether an aldehyde, a ketone, or nothing at all forms. That same class-dependence resurfaces in Lucas' test, the standard qualitative tool for telling 1°, 2° and 3° alcohols apart by how fast they turn cloudy. Master these five reaction families and you can write the equation for almost anything an exam throws at an alcohol.

The core insight

Alcohols react at O–H (acidity, ester formation) or at C–O (halogenation, oxidation), and the class of alcohol — 1°, 2° or 3° — governs both the rate of C–O substitution and the outcome of oxidation.

The core idea

Acidic nature. The O–H bond is weakly polarised, so alcohols are very weak acids — weaker than water, strong enough to react with reactive metals but too weak to react with a base as mild as $NaHCO_3$ :

2R\text{–OH} + 2Na \rightarrow 2R\text{–ONa} + H_2\uparrow

where: $R\text{–ONa}$ is a sodium alkoxide and the evolved $H_2$ gas is the usual test for an –OH group.

Reaction with hydrogen halides. $HX$ replaces –OH with –X in one step. Reactivity of the halide follows $HI > HBr > HCl$ (iodide is the best leaving-group-forming acid), and reactivity of the alcohol follows $3° > 2° > 1°$ (a more substituted carbon stabilises the intermediate carbocation better):

R\text{–OH} + HX \rightarrow R\text{–X} + H_2O

Reaction with $PCl_3$ , $PCl_5$ , $SOCl_2$ . All three chlorinate the alcohol, but $SOCl_2$ (thionyl chloride) is the synthetic favourite because both by-products are gases that simply leave the flask, giving a clean alkyl chloride with no residue to separate:

R\text{–OH} + SOCl_2 \rightarrow R\text{–Cl} + SO_2\uparrow + HCl\uparrow

3R\text{–OH} + PCl_3 \rightarrow 3R\text{–Cl} + H_3PO_3

R\text{–OH} + PCl_5 \rightarrow R\text{–Cl} + POCl_3 + HCl

Reaction with acid chlorides/anhydrides. This gives esters directly and fast, an alternative to slow, reversible Fischer esterification with a carboxylic acid:

R\text{–OH} + CH_3COCl \rightarrow CH_3COOR + HCl

R\text{–OH} + (CH_3CO)_2O \rightarrow CH_3COOR + CH_3COOH

Oxidation depends entirely on class. Primary alcohols go to aldehydes under mild, controlled oxidation and on to carboxylic acids under excess/strong oxidant; secondary alcohols go to ketones, which resist further oxidation of the same skeleton; tertiary alcohols resist oxidation altogether under normal conditions, because there is no hydrogen on the carbinol carbon for the oxidant to remove.

Lucas' test. Conc. $HCl$ + anhydrous $ZnCl_2$ (Lucas' reagent) reacts with alcohols at very different rates: 3° alcohols turn turbid immediately (fast $S_N1$ via a stable carbocation), 2° alcohols turn turbid within a few minutes, and 1° alcohols show no turbidity at room temperature (slow $S_N2$ on a primary carbon).

Figure

A hub diagram showing an alcohol at the centre with five arrows branching out to sodium alkoxide, alkyl halide, ester, aldehyde or ketone, and alkyl chloride, each arrow labelled with its reagent

Why it works

The two pathways — O–H attack and C–O attack — trace back to a single structural fact: oxygen is far more electronegative than carbon or hydrogen. That pulls electron density away from both the O–H bond (making the hydrogen slightly acidic and removable by a strong-enough base like sodium metal) and the C–O bond (making the carbon slightly electrophilic and open to attack by a good nucleophile like $Cl^-$ or $Br^-$ , once the –OH has been "activated" into a better leaving group by protonation or by $ZnCl_2$ ). Substitution at carbon explains the reactivity order $3° > 2° > 1°$ for $HX$ and Lucas' test together. A tertiary carbon can lose its leaving group first to form a stable, more-substituted carbocation, and only then get attacked by halide — this is the $S_N1$ pathway, and it is fast precisely because the intermediate is stabilised by three alkyl groups pushing electron density toward the positive carbon. A primary carbon cannot form a stable carbocation this way, so it must instead be attacked by halide at the same moment the leaving group departs ( $S_N2$ ), a slower, more crowded, one-step process — hence no visible reaction with Lucas' reagent at room temperature. Oxidation runs on a different requirement: the oxidant needs a hydrogen sitting directly on the carbinol carbon (the one bonded to –OH) to pull away, together with the O–H hydrogen, forming a carbon–oxygen double bond. A primary carbon has two such hydrogens, so oxidation can go twice — first to an aldehyde, then, if the oxidant persists, on to a carboxylic acid. A secondary carbon has only one such hydrogen, so oxidation stops cleanly at the ketone stage; there is no further hydrogen to remove without breaking a carbon–carbon bond. A tertiary carbon has none at all, so mild oxidants simply cannot act on it.

Worked example

Convert propan-1-ol to 1-chloropropane using three different reagents, and identify which gives the cleanest product.

Step 1. Using conc. $HCl$ with heat (a slow route for a primary alcohol, since it must go through the less-favoured $S_N2$ pathway):

CH_3CH_2CH_2OH + HCl \xrightarrow{\Delta} CH_3CH_2CH_2Cl + H_2O

Step 2. Using $PCl_5$ :

CH_3CH_2CH_2OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl + POCl_3 + HCl

Step 3. Using $SOCl_2$ (thionyl chloride):

CH_3CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2Cl + SO_2\uparrow + HCl\uparrow

Step 4. Compare cleanliness. The $HCl$ route is an equilibrium and is intrinsically slow for a primary substrate. The $PCl_5$ route works but leaves $POCl_3$ , a liquid, mixed with the product. The $SOCl_2$ route is preferred: both by-products, $SO_2$ and $HCl$ , are gases that escape the reaction mixture on their own, so 1-chloropropane is obtained essentially free of contaminating by-products. Answer: $SOCl_2$ gives the cleanest 1-chloropropane, because its only by-products are gases.

Competitive stretch

If asked to also distinguish propan-1-ol from an unknown 2° or 3° isomer at the same time, run Lucas' test first (fast, no heating needed) before committing to a chlorinating agent — a 1° alcohol showing no turbidity at room temperature confirms the substrate class before you choose $SOCl_2$ over $PCl_5$ on cost or scale grounds.

Common traps

Assuming every alcohol oxidises the same way — only 1° and 2° alcohols are readily oxidised under mild conditions; 3° alcohols are not oxidised without breaking a C–C bond.
Forgetting that oxidation of a 1° alcohol can run all the way to the carboxylic acid with excess acidified $K_2Cr_2O_7$ / $KMnO_4$ and heating, unless the aldehyde is deliberately distilled off as it forms.
Reversing the Lucas' test timing — the correct order is 3° immediate, 2° in minutes, 1° no reaction without heat; students very commonly swap 1° and 3°.
Confusing the fast, essentially irreversible ester route via acid chlorides/anhydrides with slow, reversible, acid-catalysed Fischer esterification from a carboxylic acid — both give esters, by different mechanisms and rates.

Ask the tutor: Still mixing up when an alcohol reacts by SN1 versus SN2, or unsure how far oxidation of your specific alcohol goes under given conditions — ask the AI tutor to walk through your exact reagent and substrate.

Why does a tertiary alcohol resist oxidation but react fastest in Lucas test?

These are two different bonds reacting by two different mechanisms, so the same structural feature can make one reaction fast and the other impossible. Oxidation needs a hydrogen on the carbinol carbon to remove; a tertiary carbon, bonded to three other carbons, has none, so mild oxidants cannot touch it. Lucas' test instead needs the leaving group (–OH, activated by $ZnCl_2$ ) to depart and form a carbocation; a tertiary carbon does this exceptionally well because three alkyl groups stabilise the resulting positive charge, so substitution there is the fastest of the three classes.

Why is SOCl2 preferred over PCl5 for making an alkyl chloride if both work

Both give the alkyl chloride in good yield, but the by-products differ. $SOCl_2$ produces $SO_2$ and $HCl$ , both gases that bubble out of the reaction mixture on their own, leaving behind essentially pure alkyl chloride. $PCl_5$ produces $POCl_3$ , a liquid that stays mixed with the product and must be separated by further work-up. For a clean, low-effort synthesis, $SOCl_2$ is therefore the standard laboratory choice.

Does excess oxidant always take a primary alcohol all the way to the carboxylic acid?

Under the exam-standard conditions — acidified $K_2Cr_2O_7$ or $KMnO_4$ with heating and excess oxidant — yes, the default expected product is the carboxylic acid, because the intermediate aldehyde remains in the reaction mixture and keeps getting oxidised further. The only way to stop cleanly at the aldehyde is to remove it as it forms, for example by distilling it off before it can react again, or by switching to a milder, more selective oxidant that is beyond the ISC syllabus. If a question gives no such condition and just says excess oxidant plus heat, assume the acid is the final product.