Hydrolysis of alkyl halides

Pour aqueous sodium hydroxide onto an alkyl halide and the halogen is quietly swapped for a hydroxyl group — the simplest laboratory route from R–XR\text{–X} to an alcohol. It looks like one clean reaction, but underneath it two different mechanisms compete for the job, and which one wins depends entirely on whether the halide carbon is primary, secondary or tertiary. That competition is the real content of this topic. A primary halide is forced through a single concerted step where hydroxide attacks from behind and flips the carbon inside out. A tertiary halide instead falls apart on its own into a carbocation, which water then mops up from either face. The two routes give opposite stereochemical outcomes and opposite rate laws, and examiners lean hard on both. This article rebuilds the hydrolysis equation, walks both mechanisms with worked numbers, and flags the traps — aryl halides, NaOH versus plain water, and mixing up which mechanism goes with which halide class.

The core insight

Alkyl halides are hydrolysed to alcohols by OH−OH^- or H2OH_2O displacing the halide, with primary substrates going by concerted SN2S_N2 (inversion, second-order) and tertiary substrates going by stepwise SN1S_N1 (racemisation, first-order, faster because the intermediate carbocation is stabilised).

The core idea

The overall transformation is a nucleophilic substitution: the electronegative halogen is a good leaving group, and OH−OH^- (a strong nucleophile) or H2OH_2O (a weaker one) takes its place at the same carbon.

R–X+NaOH(aq)→ΔR–OH+NaXR\text{–X} + NaOH_{(aq)} \xrightarrow{\Delta} R\text{–OH} + NaX

where: R–XR\text{–X} is the alkyl halide, NaOH(aq)NaOH_{(aq)} supplies the nucleophile OH−OH^-, R–OHR\text{–OH} is the alcohol product, and NaXNaX is the by-product salt (e.g. NaBrNaBr). Heating under reflux (Δ\Delta) is usually needed because hydrolysis, especially by the SN2S_N2 route, is slow at room temperature. Which mechanism operates is fixed by the class of the halide:

  • Primary halides react by SN2S_N2 — a single concerted step. OH−OH^- approaches the carbon from the side directly opposite the C–XC\text{–X} bond (backside attack) exactly as X−X^- leaves, so bond-breaking and bond-making happen together. Because both the substrate and the nucleophile appear in the rate-determining (only) step, the reaction is second order overall: rate=k[R–X][OH−]\text{rate} = k[R\text{–X}][OH^-]. If the carbon under attack is a stereocentre, the product shows inversion of configuration (Walden inversion), because the nucleophile can only approach from the open backside.
  • Tertiary halides react by SN1S_N1 — two steps. R–XR\text{–X} first ionises on its own, slowly, into a planar carbocation and X−X^-; then water or OH−OH^- attacks that carbocation fast, from either face. Since only R–XR\text{–X} appears in the slow, rate-determining step, the reaction is first order: rate=k[R–X]\text{rate} = k[R\text{–X}]. A stereocentre gives racemisation — a roughly equal mix of both configurations — because the flat carbocation offers no preferred face for attack.
  • Secondary halides sit in between and can react by either pathway, or a blend of both, depending on the nucleophile, solvent and temperature.
Figure
Two reaction pathways side by side, primary halide undergoing one concerted backside attack step labelled S N 2 giving inversion, and tertiary halide first ionising to a flat carbocation then being attacked from either face labelled S N 1 giving racemisation
Two hydrolysis pathways — SN2 versus SN1 Primary halide — SN2 (one step) HO⁻ C Br⁻ backside attack — concerted result: inversion at C Tertiary halide — SN1 (two steps) C-Br slow C+ flat carbocation H2O top H2O bottom result: racemisation rate = k[R–X][OH⁻] — second order rate = k[R–X] — first order Steric crowding blocks SN2 at a tertiary carbon; a tertiary carbocation is stable enough to form for SN1.
Derivation AWhy the SN1 rate law contains only the halide

The goal is to show, from the two-step SN1S_N1 mechanism alone, why the observed rate of hydrolysis for a tertiary halide depends on [R–X][R\text{–X}] and not at all on [OH−][OH^-] or [H2O][H_2O], in contrast to the single-step SN2S_N2 case where both concentrations appear.

Step 1. Write the two steps separately. Step (i): R–X→k1, slowR++X−R\text{–X} \xrightarrow{k_1,\ \text{slow}} R^+ + X^-. Step (ii): R++H2O→k2, fastR–OH2+→−H+R–OHR^+ + H_2O \xrightarrow{k_2,\ \text{fast}} R\text{–OH}_2^+ \xrightarrow{-H^+} R\text{–OH}. Two distinct events occur, and they occur at very different speeds.

Step 2. Identify the rate-determining step. The overall rate of a multi-step reaction is set by its slowest step, because every later step simply has to wait for the intermediate the slow step produces. Here step (i), the ionisation, is slow (breaking a C–XC\text{–X} bond with no help from the nucleophile), while step (ii), the carbocation being captured by water, is fast almost regardless of how dilute the water is, since water is the solvent and present in huge excess.

Step 3. Read off the rate law. Because only species present in the rate-determining step can appear in the rate expression, and step (i) involves only R–XR\text{–X} falling apart, the observed rate is rate=k1[R–X]\text{rate} = k_1[R\text{–X}] — first order overall, with no dependence on [OH−][OH^-]. This is the experimental signature that distinguishes SN1S_N1 from SN2S_N2.

Step 4. Contrast with SN2S_N2. There, OH−OH^- and R–XR\text{–X} collide in the same, single, rate-determining step, so both concentrations enter directly: rate=k[R–X][OH−]\text{rate} = k[R\text{–X}][OH^-]. Doubling [OH−][OH^-] doubles an SN2S_N2 rate but leaves an SN1S_N1 rate unchanged — a quick experimental test for which mechanism is operating.

Worked example

Write the hydrolysis equation for 1-bromopropane with aqueous NaOH, classify the mechanism, and state the stereochemical consequence.

Step 1. Identify the halide class. 1-bromopropane, CH3CH2CH2BrCH_3CH_2CH_2Br, has the C–BrC\text{–Br} bond on a carbon attached to only one other carbon, so it is primary — the mechanism will be SN2S_N2.

Step 2. Write the overall equation. CH_3CH_2CH_2Br + NaOH_{(aq)} \xrightarrow{\Delta} CH_3CH_2CH_2OH + NaBr Product: propan-1-ol.

Step 3. Show the concerted step. Hydroxide attacks the carbon bearing BrBr from the side directly opposite the leaving group, so C–BrC\text{–Br} breaking and C–OHC\text{–OH} forming happen at once: HO^- + CH_3CH_2CH_2\text{–Br} \rightarrow [HO\cdots CH_2CH_2CH_3\cdots Br]^{\ddagger} \rightarrow HO\text{–}CH_2CH_2CH_3 + Br^-

Step 4. State rate and stereochemistry. rate=k[R–Br][OH−]\text{rate} = k[R\text{–Br}][OH^-] (second order); since this primary carbon is not a stereocentre here, no optical consequence is observable in this particular case, but if it were, the product would show inversion. Final answer: propan-1-ol, via a single-step SN2S_N2 mechanism with rate =k[R–Br][OH−]= k[R\text{–Br}][OH^-].

Competitive stretch

Now compare with tert-butyl bromide, (CH3)3CBr(CH_3)_3CBr, hydrolysing under the same conditions. It is tertiary, so it ionises first: (CH3)3C–Br→slow(CH3)3C++Br−(CH_3)_3C\text{–Br} \xrightarrow{\text{slow}} (CH_3)_3C^+ + Br^-, stabilised by three methyl groups donating electron density. Water then attacks the flat cation from either face: (CH3)3C++H2O→fast(CH3)3C–OH2+→−H+(CH3)3C–OH(CH_3)_3C^+ + H_2O \xrightarrow{\text{fast}} (CH_3)_3C\text{–OH}_2^+ \xrightarrow{-H^+} (CH_3)_3C\text{–OH}, giving 2-methylpropan-2-ol. Because a stable tertiary carbocation forms readily while a bulky SN2S_N2 transition state is comparatively hard to reach, tert-butyl bromide hydrolyses faster than 1-bromopropane despite having a "harder" leaving-group carbon to attack directly.

Common traps

  • Assuming every alkyl halide hydrolyses by the same mechanism — the class (1°/2°/3°) decides whether SN1S_N1 or SN2S_N2 dominates; secondary halides can show both.
  • Mixing up the stereochemical outcomes: SN2S_N2 gives clean inversion, SN1S_N1 gives racemisation because the intermediate carbocation is planar and attackable from either side.
  • Writing chlorobenzene or other aryl halides as if they hydrolyse the same way — they do not undergo this simple substitution under these conditions, because partial double-bond character in the C–XC\text{–X} bond blocks backside attack; phenol needs harsher methods (Dow's process).
  • Treating hydrolysis by plain water and by aqueous NaOH as equally fast — OH−OH^- is a far stronger nucleophile than H2OH_2O, so alkaline hydrolysis is noticeably faster, especially for SN2S_N2-prone primary halides.

Ask the tutor: not sure whether your specific alkyl halide will go by SN1S_N1 or SN2S_N2 — ask the AI tutor to classify it and predict the stereochemical outcome.

Why doesn't chlorobenzene hydrolyse the same way as an alkyl chloride?

In chlorobenzene the chlorine's lone pair overlaps with the aromatic ring, giving the C–ClC\text{–Cl} bond partial double-bond character and pulling the carbon into the plane of the ring. That makes the bond shorter and stronger than a normal C–ClC\text{–Cl} single bond, and it leaves no accessible backside for a nucleophile to attack, so ordinary SN1S_N1/SN2S_N2 hydrolysis simply does not happen. Converting an aryl halide to phenol needs a completely different, much harsher route.

If a secondary halide can go by either mechanism, how do I know which one to write in an answer?

Look at what the question emphasises. If it stresses a strong nucleophile, high [OH−][OH^-], or a polar aprotic-type solvent, lean SN2S_N2; if it stresses a polar protic solvent, weak nucleophile (plain water), or heat with no added base, lean SN1S_N1. Unless the question specifies conditions that clearly favour one, it is safe to say secondary halides can show a mixture of both pathways rather than forcing a single answer.

Does the alcohol product depend on whether we use NaOH or just water?

No — the same alcohol forms either way, since OH−OH^- and H2OH_2O both deliver an –OH\text{–OH} group to the same carbon. What changes is the rate: OH−OH^- is a stronger nucleophile than neutral H2OH_2O, so hydrolysis with aqueous NaOH is faster, particularly for primary halides going by SN2S_N2 where the nucleophile actively participates in the rate-determining step. For a tertiary halide going by SN1S_N1, the rate barely changes with [OH−][OH^-] at all.

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Where this leads