Physical and Chemical Properties of Alcohols

Why boiling points and solubility behave the way they do. Every alcohol molecule carries an O–H bond in which oxygen, being far more electronegative than hydrogen, pulls the bonding electrons toward itself. The hydrogen ends up carrying a partial positive charge and can be attracted electrostatically to a lone pair on the oxygen of a neighbouring molecule — a hydrogen bond. Breaking a liquid alcohol into vapour means breaking large numbers of these intermolecular hydrogen bonds one by one, which costs energy that an alkane or an ether of the same molar mass simply does not have to pay (ethers have the polar C–O–C linkage but no O–H hydrogen to donate, so they cannot hydrogen-bond to each other, only accept one from water). That is why boiling points climb steeply within the alcohol family compared to hydrocarbons of similar size, and why the lower alcohols — methanol, ethanol, propan-1-ol — are completely miscible with water: each can hydrogen-bond into the water's own three-dimensional network. As the hydrocarbon tail grows past about four or five carbons, that non-polar tail dominates the molecule's overall behaviour and water solubility falls away steadily, even though the O–H group itself is unchanged.

Why the O–H bond is the site of every reaction. The same electron pull that creates the hydrogen bond also makes the O–H hydrogen weakly acidic and leaves the carbon–oxygen bond polarised the other way, so the molecule offers two reactive sites: the O–H hydrogen can be removed by a strong enough base (or an active metal), while the whole –OH group can be protonated and expelled as neutral water, turning the carbon into an electrophilic centre. The reaction with sodium metal shows the acidic O–H hydrogen being removed:

where: $\text{R}$ is any alkyl group, $\text{R–OH}$ the alcohol, $\text{R–ONa}$ the sodium alkoxide salt formed, and $\text{H}_2\uparrow$ the hydrogen gas evolved and used as the positive test for an O–H group. Esterification uses the same acidic hydrogen from the opposite side, with an acid catalyst: $\text{R–OH} + \text{R}'\text{COOH} \xrightleftharpoons[]{\text{H}^+} \text{R}'\text{COOR} + \text{H}_2\text{O}$. Reaction with hydrogen halides, $\text{R–OH} + \text{HX} \rightarrow \text{R–X} + \text{H}_2\text{O}$, instead breaks the C–O bond after protonation of oxygen, and its rate follows tertiary > secondary > primary because a tertiary carbocation is far more stable than a primary one — the basis of the Lucas test described below.

Chemical properties, gathered as a set. Oxidation is stepwise and depends on substitution: a primary alcohol is oxidised by acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$ first to an aldehyde and then, under the same conditions, on to a carboxylic acid, $\text{RCH}_2\text{OH} \xrightarrow{[O]} \text{RCHO} \xrightarrow{[O]} \text{RCOOH}$; only a milder, selective oxidant such as PCC (pyridinium chlorochromate) stops cleanly at the aldehyde. A secondary alcohol oxidises only as far as a ketone, since the ketone has no remaining C–H on the carbinol carbon to remove further; a tertiary alcohol resists oxidation by these reagents altogether, for the same structural reason. Dehydration removes water entirely, using concentrated $\text{H}_2\text{SO}_4$ (or $\text{Al}_2\text{O}_3$) and heat, to give an alkene, with the more substituted (Zaitsev) product favoured whenever a choice of double-bond position exists.

The goal here is to see why the reactivity order tertiary greater than secondary greater than primary, which governs both the Lucas test and dehydration, follows directly from carbocation stability, and why a primary alcohol still dehydrates cleanly even though a primary carbocation is far too unstable to form as a genuine intermediate.

Step 1. In every acid-mediated reaction of an alcohol — dehydration, or substitution with HX — the first event is the same: the acid protonates the oxygen, turning the poor leaving group –OH into the good leaving group –OH₂⁺, at comparable rates for all three classes, since it only involves oxygen's lone pair.

Step 2. What happens next depends on how well the developing positive charge on carbon is stabilised as water leaves. A tertiary carbon has three alkyl groups feeding in electron density by the inductive effect and hyperconjugation, so a full, discrete tertiary carbocation forms rapidly and survives long enough to react — why Lucas reagent turns turbid almost instantly with tertiary alcohols.

Step 3. A secondary carbon has only two stabilising alkyl groups, so its carbocation forms more slowly — the Lucas test needs several minutes and gentle warming. A primary carbon has essentially none; a discrete primary carbocation is too high in energy to form at all, so primary alcohols show no reaction with Lucas reagent even on warming.

Step 4. Yet primary alcohols do dehydrate. For a primary substrate, loss of water and removal of a β-hydrogen happen together in one concerted, E2-like step, never passing through the unstabilised cation — the base ($\text{HSO}_4^-$) removes the β-hydrogen at the same moment the C–O bond breaks. This is why full carbocation-rearrangement chemistry belongs to secondary and tertiary substrates, not primary ones.

Write the mechanism for the acid-catalysed dehydration of ethanol to ethene, state the conditions, and predict the major product if propan-1-ol is replaced by butan-2-ol under the same conditions.

Step 1. Protonation: concentrated $\text{H}_2\text{SO}_4$ protonates the oxygen of ethanol, converting the poor leaving group –OH into the good leaving group –OH₂⁺: $\text{CH}_3\text{CH}_2\text{OH} + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{OH}_2^+$.

Step 2. Loss of water: because ethanol is primary, this step is concerted with removal of a β-hydrogen (E2-like), never passing through a discrete primary carbocation (see Why it works, Step 4).

Step 3. Formation of the alkene: removal of a β-hydrogen by a base (here $\text{HSO}_4^-$), simultaneous with C=C π-bond formation, gives ethene: $\text{CH}_3\text{CH}_2\text{OH}_2^+ \rightarrow \text{CH}_2=\text{CH}_2 + \text{H}_3\text{O}^+$. Conditions: excess concentrated $\text{H}_2\text{SO}_4$, heated to about 443 K (170 °C); the acid is regenerated as water is removed, driving the equilibrium $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[443\text{ K}]{\text{conc. }\text{H}_2\text{SO}_4} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$ toward the alkene.

Step 4. For butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$, the carbinol carbon is secondary, so a genuine secondary carbocation now forms, and there is a real choice of which β-hydrogen leaves: removing one from C1 gives but-1-ene, while removing one from C3 gives but-2-ene, the more substituted alkene. Answer: ethanol gives ethene under concentrated $\text{H}_2\text{SO}_4$ at 443 K by a concerted, primary-alcohol pathway; butan-2-ol dehydrates through a discrete secondary carbocation and gives but-2-ene as the major product by Zaitsev's rule, with but-1-ene as the minor product.

• Writing a discrete carbocation-rearrangement mechanism for dehydration of a primary alcohol — even the accepted picture avoids a genuine primary cation; that level of rearrangement chemistry belongs to secondary/tertiary substrates.
• Confusing the reactivity order with HX/Lucas reagent (tertiary greater than secondary greater than primary, from carbocation stability) with the order of ease of oxidation, which runs the other way in outcome: primary alcohols oxidise all the way to acids, secondary alcohols stop at ketones, tertiary alcohols do not oxidise at all.
• Assuming all alcohols dissolve freely in water because the lower members do — solubility drops off sharply beyond about four or five carbons once the hydrophobic chain dominates the molecule's behaviour.
• Assuming oxidation of a primary alcohol with $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$ stops automatically at the aldehyde — under normal aqueous acidic conditions it proceeds through to the carboxylic acid; only a mild, selective oxidant like PCC halts it at the aldehyde stage.

Ask the tutor: want to walk through the concerted dehydration mechanism for a different primary alcohol, or practise predicting Lucas-test timings and oxidation products for a mixed set of alcohols?