Hydration of alkenes

Alcohols can be built directly from alkenes by adding the elements of water, H and OH, across the carbon–carbon double bond. The ISC Class 12 syllabus names three distinct routes for doing this: direct acid-catalysed hydration, indirect hydration through an alkyl hydrogen sulphate intermediate, and hydroboration–oxidation. The first two follow Markovnikov's rule and go through a carbocation; the third deliberately reverses the regiochemistry by routing addition through boron instead. The same starting alkene can therefore give two genuinely different alcohols depending on which method is used — which is exactly why examiners like to ask "predict the product" questions that hinge on knowing which rule applies to which method.

The core insight

Acid-catalysed hydration (direct or indirect) adds water Markovnikov-fashion through a carbocation, while hydroboration–oxidation adds it anti-Markovnikov through electron-deficient boron, so the method chosen — not just the alkene — decides which alcohol you get.

The core idea

Direct hydration. An alkene reacts with water in the presence of a catalytic acid ( $H_2SO_4$ or $H_3PO_4$ ) at high temperature and pressure. The mechanism is stepwise: the acid protonates the double bond first, generating the more stable possible carbocation, and water then attacks that carbocation.

CH_2=CH_2 + H_2O \xrightarrow[\text{catalyst}]{H^+} CH_3CH_2OH

where: $H^+$ is a catalyst, regenerated at the end, so the equation is written with it over the arrow rather than consumed; the reaction is reversible, the exact reverse of acid-catalysed dehydration of the same alcohol.

Indirect hydration (via alkyl hydrogen sulphate). The alkene is first absorbed in cold, concentrated $H_2SO_4$ , adding $H$ and $HSO_4^-$ across the double bond in Markovnikov fashion; the resulting alkyl hydrogen sulphate is then hydrolysed by boiling with water.

CH_2=CH_2 + H_2SO_4 \xrightarrow{\text{cold}} CH_3CH_2OSO_3H \xrightarrow[\Delta]{H_2O} CH_3CH_2OH + H_2SO_4

where: the sulphuric acid consumed in step one is regenerated in step two, so overall it behaves catalytically even though it is a genuine two-step, two-intermediate route — distinct from single-step direct hydration.

Hydroboration–oxidation. Diborane ( $B_2H_6$ , effectively $BH_3$ ) adds across the double bond with boron going to the less substituted carbon, because boron is the electron-deficient atom that attacks the alkene and prefers the less hindered site; oxidation of the resulting trialkylborane with alkaline hydrogen peroxide replaces C–B with C–OH, retaining the same orientation.

3\,CH_3CH=CH_2 + (BH_3)_2 \rightarrow 2\,(CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2/OH^-} 3\,CH_3CH_2CH_2OH

where: the addition is anti-Markovnikov (boron, carrying H, ends up on the terminal carbon) and proceeds with syn stereochemistry — both atoms add to the same face of the double bond — a mechanistic detail that follows from a concerted four-centred transition state rather than a discrete carbocation.

Figure

Two divergent arrows from propene one via a carbocation to propan two ol and one via boron addition to propan one ol

Why the two rules differ

The contrast between Markovnikov and anti-Markovnikov selectivity comes down to which species is electron-deficient in the slow, rate-determining step. In direct and indirect hydration, the alkene attacks a proton first, so the intermediate is a genuine carbocation, and carbocation stability ( $3^\circ > 2^\circ > 1^\circ$ ) decides where the positive charge — and hence the incoming –OH — ends up: the more substituted carbon. In hydroboration, it is the electron-deficient boron atom of $BH_3$ that attacks the alkene, not a proton, and boron simply goes to whichever carbon is less crowded; there is no discrete carbocation at any stage, so there is nothing to rearrange, and the reaction is also stereospecifically syn, a point not separately tested at ISC level but useful for remembering why hydroboration behaves so differently from the acid-catalysed routes.

Derivation APredicting the major alcohol from carbocation stability

The goal is to show, for a given unsymmetrical alkene undergoing acid-catalysed (direct or indirect) hydration, exactly which carbon receives the proton and which receives the incoming water or hydrogen-sulphate group, by comparing the stability of the two carbocations that could form and showing why only the more stable one controls the product in practice.

Step 1. Write the general unsymmetrical alkene as $R_2C=CHR'$ , where one alkene carbon carries more alkyl substitution than the other. Protonation by $H^+$ can occur at either carbon, giving two possible carbocations: attack at the less-substituted carbon gives a cation on the more-substituted carbon, $R_2\overset{+}{C}\text{–}CH_2R'$ , while attack at the more-substituted carbon gives a cation on the less-substituted carbon, $R_2CH\text{–}\overset{+}{C}HR'$ .

Step 2. Compare the stability of these two possible cations using the established order $3^\circ > 2^\circ > 1^\circ$ (tertiary more stable than secondary more stable than primary), which holds because each additional alkyl group donates electron density toward the positive carbon through hyperconjugation and the inductive effect, spreading out and lowering the charge.

Step 3. Because carbocation formation is the slow, rate-determining step of the mechanism, the pathway through the more stable cation has the lower activation energy and is therefore overwhelmingly the faster, dominant pathway — the other pathway is not forbidden, only comparatively negligible. This is the chemical content of Markovnikov's rule stated in mechanistic terms: "H adds to the carbon already bearing more hydrogens" is simply the outcome of protonation choosing whichever site produces the more substituted, more stable carbocation.

Step 4. Water (or $HSO_4^-$ in the indirect route) then attacks this single dominant carbocation at the positively charged carbon, so the –OH (or –OSO_3H, hydrolysed later to –OH) ends up on the more substituted carbon — exactly where the more stable cation sat. Check: for propene, $CH_3CH=CH_2$ , protonation at C1 gives the secondary cation $CH_3\overset{+}{C}HCH_3$ at C2, more stable than the primary cation protonation at C2 would give at C1; water attacks C2, giving propan-2-ol as the major product — matching the worked example below.

Worked example

Problem. Predict the major product when propene ( $CH_3CH=CH_2$ ) undergoes (a) direct acid-catalysed hydration and (b) hydroboration–oxidation, and state which alcohol is primary and which is secondary.

Step 1 — Direct hydration. Using the Step 1–3 argument from the derivation above, $H^+$ attacks the terminal $=CH_2$ carbon (C1), placing the positive charge on C2 as the secondary carbocation $CH_3\overset{+}{C}HCH_3$ , which is more stable than the primary carbocation that the other orientation would give. Water then attacks C2:

CH_3CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH(OH)CH_3

Major product: propan-2-ol (isopropyl alcohol), a secondary alcohol, since the –OH-bearing carbon is attached to two other carbons.

Step 2 — Hydroboration–oxidation. Boron (carrying its attached H) adds to the less hindered terminal carbon, C1, in a concerted syn-addition step; oxidation with $H_2O_2/OH^-$ then replaces the C–B bond at C1 with C–OH, with retention of orientation:

CH_3CH=CH_2 \xrightarrow{B_2H_6} (CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2/OH^-} CH_3CH_2CH_2OH

Major product: propan-1-ol, a primary alcohol, since the –OH-bearing carbon (C1) is attached to only one other carbon.

Step 3 — Compare. The same starting alkene gives two chemically different alcohols purely because the two methods route the addition through different intermediates: propan-2-ol from carbocation-controlled (Markovnikov) direct hydration, and propan-1-ol from boron-controlled (anti-Markovnikov) hydroboration–oxidation.

Competitive stretch

Problem. Ethene is converted to ethanol by the indirect (sulphuric acid) route. Write both steps and explain, using the derivation above, why no regiochemistry question arises for this particular alkene.

Step 1. Ethene is symmetrical: $CH_2=CH_2$ has identical carbons on either side of the double bond, so protonation (or $H^+/HSO_4^-$ addition) at either carbon gives the same primary carbocation, $CH_3\overset{+}{C}H_2$ — there is no "more stable" option to choose between, since both possible cations are identical.

CH_2=CH_2 + H_2SO_4 \xrightarrow{\text{cold}} CH_3CH_2OSO_3H \xrightarrow[\Delta]{H_2O} CH_3CH_2OH + H_2SO_4

The product is ethanol, and because ethene has no unsymmetrical substitution, Markovnikov's rule gives only one possible product rather than distinguishing a major and minor one — this is exactly why ethene is the standard textbook example for introducing the mechanism before unsymmetrical alkenes like propene are tackled.

Common traps

Applying Markovnikov's rule to hydroboration–oxidation — the boron step is deliberately anti-Markovnikov, so the final alcohol sits on the opposite carbon from what direct or indirect hydration would give.
Treating direct hydration as a one-way, complete reaction — it is catalysed and reversible (the exact reverse of dehydration of the same alcohol), so equilibrium conditions matter, not just the forward equation.
Confusing indirect hydration (via alkyl hydrogen sulphate, two distinct steps) with direct hydration (single catalytic step) — they use different reagents, different intermediates, and different conditions even though both are Markovnikov-selective.
Writing hydroboration and oxidation as a single combined step — the intermediate trialkylborane must be shown explicitly; both the addition step ( $B_2H_6$ ) and the oxidation step ( $H_2O_2/OH^-$ ) are separately examinable.

Ask the tutor: if you keep mixing up which method gives the Markovnikov versus anti-Markovnikov product for a specific alkene, ask the AI tutor to work through your exact alkene with all three hydration routes side by side.

Why is direct hydration industrially less convenient than it sounds, if it is only one step?

Direct hydration needs high temperature and pressure with an acid catalyst to get a reasonable equilibrium yield, since the reaction is reversible and mildly exothermic in the forward direction — running it too hot favours the reverse dehydration instead. Historically, indirect hydration via cold concentrated sulphuric acid could absorb even dilute, low-pressure alkene streams under milder conditions, which is why both routes remain in the syllabus even though modern industrial ethanol production increasingly favours direct catalytic hydration.

Does hydroboration-oxidation ever go through a carbocation like the other two methods?

No, and that is precisely why it gives the opposite regiochemistry. In hydroboration, boron itself is the electron-deficient species that attacks the alkene in a concerted step, with both boron and hydrogen adding to the same face of the double bond simultaneously. Since no carbocation ever forms, there is nothing for carbocation stability to control, and boron simply ends up on the less hindered carbon by a steric argument rather than an electronic one.

If I am only given a product alcohol, how do I work backward to figure out which method made it?

Look at whether the –OH sits on the more substituted or less substituted carbon of the corresponding alkene skeleton. If –OH is on the more substituted (higher-priority Markovnikov) carbon, either direct or indirect acid-catalysed hydration was used; if –OH is on the less substituted, typically terminal, carbon of an otherwise unsymmetrical alkene, hydroboration-oxidation was used. Symmetrical alkenes like ethene will not distinguish the methods this way, since all three give the same single alcohol.