Hydrolysis of Haloalkanes

Spec Mapping — OCR H432 Module 4.2.2(c)-(d) — Hydrolysis of haloalkanes, covering hydrolysis as a nucleophilic substitution in which water (or hydroxide ion) is the nucleophile; the use of aqueous silver nitrate in ethanol to (i) accelerate the hydrolysis by removing the halide as silver-halide precipitate and (ii) identify the halogen present by the colour of the precipitate; the order of rate of hydrolysis (R-I > R-Br > R-Cl ≫ R-F) and its relationship to C-X bond enthalpy; comparison with the trends in C-X bond polarity which would predict the opposite (refer to the official OCR H432 specification document for exact wording).

Hydrolysis of haloalkanes is one of the most important kinetics experiments at A-Level. The synthetic outcome — converting R-X into R-OH — is unremarkable: an alcohol that could have been made more cleanly from another route. What earns this reaction its place in the OCR specification is what its rate reveals about bond strength versus bond polarity. By comparing four haloalkanes that differ only in the halogen (R = same carbon skeleton, X = F, Cl, Br, I) under identical conditions, you can watch the periodic trend in carbon-halogen bond enthalpy unfold as a wall-clock measurement: yellow silver-iodide precipitate appears in seconds, cream silver-bromide in a minute, white silver-chloride in five minutes, and no detectable silver-fluoride at all. The result is the opposite of what a naive electronegativity argument would predict (C-F is the most polar bond, so R-F "ought" to be the most reactive towards a nucleophile), and that mismatch is the central teaching point: for a polar covalent bond, rate is set by bond enthalpy, not bond polarity. This lesson develops the standard PAG 7 experimental protocol, the role of silver nitrate as both rate accelerator and halide-identifying reagent, the C-X bond-enthalpy explanation of the rate order, the silver-halide colour palette (white AgCl, cream AgBr, yellow AgI) and its ammonia-solubility extension test, two worked Arrhenius-style calculations, and the synoptic crossover with Lesson 6 (where changing the carbon class — primary vs tertiary — switches the mechanism from SN2 to SN1 and a different argument applies).

Key Definition: hydrolysis is a substitution reaction in which water (or aqueous hydroxide) acts as the nucleophile, breaking a bond in the substrate and incorporating an -OH group into the organic product. For a haloalkane, hydrolysis converts R-X into R-OH with elimination of HX (or, when AgNO₃ is present, with precipitation of AgX).

What hydrolysis actually does

Hydrolysis literally means "splitting with water". For a haloalkane the water molecule attacks the δ+ carbon carrying the halogen, the C-X bond breaks heterolytically, and an alcohol forms alongside the hydrogen halide:

$R\text{-}X + H_2O \rightarrow R\text{-}OH + HX$

Compared with the aqueous-NaOH substitution of Lesson 4, hydrolysis with neutral water is much slower. The lone pair on a water oxygen is a less effective nucleophile than the lone pair on a hydroxide ion because (i) it carries no negative charge, so electrostatic attraction to the δ+ carbon is weaker, and (ii) the resulting intermediate is positively charged (R-OH₂⁺) and has to lose a proton before the neutral alcohol is recovered. Despite the kinetic penalty, the reaction does occur at A-Level temperatures (~50 °C), and adding silver nitrate dramatically accelerates it.

R-X + H2O → R-OH + HX

In the presence of Ag+ (aqueous AgNO3 in ethanol, 50 C):

R-X + H2O + Ag+ → R-OH + AgX(s) + H+ The AgX precipitate is the observable evidence of hydrolysis; the colour identifies X.

Why silver nitrate?

Silver cations have two simultaneous effects in the hydrolysis flask, both of which speed the reaction up and make it visible:

Lewis-acid pull on the halide. Ag⁺ has a strong affinity for halide ions (the AgX precipitates have extremely low solubility products: Ksp(AgCl) ≈ 1.8 × 10⁻¹⁰, Ksp(AgBr) ≈ 5 × 10⁻¹³, Ksp(AgI) ≈ 8 × 10⁻¹⁷). As soon as the C-X bond begins to ionise, the halide is captured into a solid lattice and removed from solution. By Le Chatelier's principle, this depletes the product side of the equilibrium and pulls the forward reaction along.
Diagnostic colour. Each silver halide precipitate has a characteristic shade and a characteristic solubility in ammonia. The colour identifies the halogen that was in the original haloalkane.

Silver halide	Colour	Solubility in dilute NH₃(aq)	Solubility in conc. NH₃(aq)
AgF	(none — fully soluble in water)	n/a	n/a
AgCl	White	Soluble (gives [Ag(NH₃)₂]⁺)	Soluble
AgBr	Cream	Insoluble	Soluble
AgI	Yellow	Insoluble	Insoluble

The ammonia-solubility extension test resolves any borderline AgCl / AgBr colour ambiguity: if the precipitate dissolves in dilute NH₃ it must be AgCl; if it needs concentrated NH₃, it is AgBr; if it dissolves in neither, it is AgI.

The standard PAG 7 experiment — comparing rates

The OCR PAG 7 protocol uses four haloalkanes that share the same butyl carbon skeleton (so steric, inductive and class effects are matched) and differ only in the halogen.

Method

Set up a water bath at 50 °C and let it stabilise.
Place four clean test tubes in a rack inside the water bath.
To each tube add 1 cm³ ethanol (the co-solvent — the haloalkane is not soluble in pure water but is miscible with ethanol, and silver nitrate is soluble in the aqueous-ethanol mixed solvent).
Add 1 cm³ aqueous silver nitrate (0.1 mol dm⁻³) to each tube and allow to reach 50 °C (5 minutes).
To tube 1 add three drops of 1-fluorobutane; to tube 2 three drops of 1-chlorobutane; to tube 3 three drops of 1-bromobutane; to tube 4 three drops of 1-iodobutane — simultaneously, using calibrated dropping pipettes.
Start the stopwatch. Record the time at which a precipitate first becomes visible in each tube, and note its colour.
After 10 minutes, terminate and record final colour and appearance.

Expected results

Haloalkane	C-X bond enthalpy / kJ mol⁻¹	Precipitate colour	Time to first precipitate
1-Fluorobutane	484	None observed	(no reaction in 10 min)
1-Chlorobutane	338	White (AgCl)	~5 minutes
1-Bromobutane	276	Cream (AgBr)	~1 minute
1-Iodobutane	238	Yellow (AgI)	seconds

Order of rate of hydrolysis: R-I > R-Br > R-Cl ≫ R-F.

This is the reverse of the order of C-X bond polarity (which would put R-F at the top because fluorine is the most electronegative halogen) and instead matches the order of C-X bond enthalpy (weakest → fastest). The mark-scheme phrase to commit to memory is "rate of hydrolysis depends on C-X bond enthalpy, not on C-X polarity".

flowchart TD
  A["Test tube at 50 C: 1 cm3 ethanol + 1 cm3 aq AgNO3"] --> B[Add 3 drops haloalkane]
  B --> C{Which halogen?}
  C -->|R-I, C-I 238 kJ/mol| D["Yellow AgI precipitate<br/>in seconds - fastest"]
  C -->|R-Br, C-Br 276 kJ/mol| E["Cream AgBr precipitate<br/>in about 1 min"]
  C -->|R-Cl, C-Cl 338 kJ/mol| F["White AgCl precipitate<br/>in about 5 min"]
  C -->|R-F, C-F 484 kJ/mol| G["No visible precipitate<br/>in lesson timescale"]

Why bond enthalpy beats polarity

The hydrolysis is rate-limited by the step that breaks the C-X bond. For a primary haloalkane, this is the SN2 transition state where the C-X bond is partially broken as the C-Nu bond is partially formed; for a tertiary haloalkane it is the SN1 ionisation where the C-X bond is fully broken before water arrives. Either way, the activation energy is dominated by the cost of disrupting C-X.

Bond	Mean bond enthalpy / kJ mol⁻¹	C-X bond length / pm
C-F	484	138
C-Cl	338	178
C-Br	276	193
C-I	238	214

Two trends accumulate down the group. First, the bond length increases (larger halogen valence orbitals), which reduces orbital overlap and weakens the bond. Second, orbital matching worsens — the carbon 2p orbital overlaps best with the fluorine 2p (similar size) and progressively worse with chlorine 3p, bromine 4p, iodine 5p. Together these effects nearly halve the bond enthalpy from C-F to C-I.

Why polarity is the wrong predictor

Pauling electronegativities give C = 2.55, F = 3.98, Cl = 3.16, Br = 2.96, I = 2.66. The C-X dipole therefore decreases F > Cl > Br > I. If you imagined a nucleophile being pulled in electrostatically to the δ+ carbon, you would expect R-F to react fastest because its carbon is most δ+. Experiment refutes this — R-F is essentially unreactive at 50 °C. The lesson is that transition-state energy (a bond-breaking cost) trumps ground-state electrostatic attraction (a docking energy of a few kJ mol⁻¹) when the two effects pull in opposite directions.

flowchart LR
  A[Two competing trends down Group 17] --> B[C-X bond enthalpy decreases<br/>F > Cl > Br > I<br/>484 > 338 > 276 > 238 kJ/mol]
  A --> C[C-X bond polarity decreases<br/>F > Cl > Br > I<br/>delta+ on C shrinks]
  B --> D[Predicts: R-I fastest hydrolysis]
  C --> E[Predicts: R-F fastest hydrolysis]
  D --> F[Bond enthalpy wins by experiment]
  E --> F

Arrhenius reasoning

The Arrhenius equation $k = A e^{-E_a / RT}$ tells you that the rate constant is exponentially sensitive to activation energy. If we assume the activation energy of hydrolysis is approximately the C-X bond enthalpy (a crude but useful approximation — in reality it is somewhat less because the new C-O bond is partially formed in the transition state), then at $T = 323$ K (= 50 °C) and $R = 8.314$ J K⁻¹ mol⁻¹ the ratio of rate constants for C-I vs C-F is approximately:

$\frac{k_{\text{C-I}}}{k_{\text{C-F}}} \approx \exp\left(-\frac{238{,}000 - 484{,}000}{8.314 \times 323}\right) = \exp(91.6) \approx 10^{40}$

A 40-order-of-magnitude rate advantage is the reason R-F appears completely inert on the bench. Real transition states benefit from partial bond formation, so the true ratio is smaller, but it is still vast — easily 10⁵ or larger. This is also why CFCs are environmentally persistent: their C-F and C-Cl bonds resist hydrolysis at all atmospheric temperatures, so the molecules survive long enough to reach the stratosphere (Lesson 7).

Using hydrolysis as a halide-identification test

Because each silver halide has a distinctive colour and ammonia-solubility, the same hydrolysis experiment doubles as an identification test. Procedure for an unknown:

Warm a small sample of the unknown haloalkane with aqueous AgNO₃ and ethanol at 50 °C in a test tube.
Note the colour and onset time of any precipitate.
Test solubility: add dilute NH₃(aq) to a portion of the precipitate. If it dissolves, the halide is chloride.
If insoluble in dilute NH₃, add concentrated NH₃(aq). If it dissolves now, the halide is bromide.
If insoluble in concentrated NH₃, the halide is iodide.

Precipitate observed	Colour	Dilute NH₃	Concentrated NH₃	Halogen
AgCl	White	Dissolves	Dissolves	Cl
AgBr	Cream	Insoluble	Dissolves	Br
AgI	Yellow	Insoluble	Insoluble	I
(none)	—	—	—	F or non-halide

The ammonia-solubility step disambiguates the two pale precipitates (white AgCl vs cream AgBr), which can look very similar in dim laboratory lighting.

Worked example 1 — comparing 1-bromobutane and 2-bromo-2-methylpropane

Compare the rates of hydrolysis of 1-bromobutane and 2-bromo-2-methylpropane with aqueous silver nitrate in ethanol at 50 °C. The tertiary substrate gives a cream precipitate within seconds; the primary substrate takes more than a minute. Explain.

Both haloalkanes contain the same C-Br bond with the same bond enthalpy (≈ 276 kJ mol⁻¹). The trend "weaker C-X → faster" cannot explain the difference. Instead, the mechanism differs:

1-Bromobutane (primary) reacts by SN2. Water has to attack the back of the carbon bearing Br while Br is still partially bonded. The transition state is pentacoordinate and the reaction is concerted. The rate is limited by orbital overlap of the incoming water with the C-Br σ* and by steric access to the back face.
2-Bromo-2-methylpropane (tertiary) reacts by SN1. The C-Br bond ionises first to give a tertiary carbocation $(CH_3)_3C^+$ that is strongly stabilised by inductive donation and hyperconjugation from the three methyl groups. Water then attacks the planar cation in a fast second step. The rate-limiting ionisation is greatly accelerated relative to a primary substrate because of carbocation stability.

So when you change only the halogen (same R), bond enthalpy controls the rate (this lesson). When you change only the carbon class (same X), mechanism (SN2 vs SN1) controls the rate (Lesson 4 and Lesson 6). The two arguments are independent and complementary.

Worked example 2 — Arrhenius-style estimate

Calculate the percentage decrease in C-X bond enthalpy on going from C-F (484 kJ mol⁻¹) to C-I (238 kJ mol⁻¹). Estimate the ratio of rate constants at 50 °C, assuming the activation energy equals the C-X bond enthalpy. Comment.

Decrease = $484 - 238 = 246$ kJ mol⁻¹.

Lesson 5: Hydrolysis of Haloalkanes