AS Chemistry 9701: The Silver Nitrate Test for Halogenoalkane Reactivity

Equal small volumes of 1-chlorobutane, 1-bromobutane and 1-iodobutane are each added to separate test tubes of aqueous silver nitrate dissolved in ethanol (ethanol is used because halogenoalkanes do not dissolve well in water alone), and the mixtures are warmed gently in a water bath. The time taken for a precipitate to appear is recorded for each. 1-Fluorobutane is not normally included in this experiment, because no precipitate is observed even after prolonged warming.

Approximate C–X bond enthalpies: C–F $\approx 484$ kJ mol⁻¹, C–Cl $\approx 338$ kJ mol⁻¹, C–Br $\approx 276$ kJ mol⁻¹, C–I $\approx 238$ kJ mol⁻¹.

(a) Give the ionic equation for the precipitation step common to all three halides once the halide ion, $\text{X}^-$ , has been released by hydrolysis. [1]

(b) State the colour of the precipitate formed from each of 1-chlorobutane, 1-bromobutane and 1-iodobutane. [3]

(c) Explain why 1-iodobutane produces its precipitate fastest, and 1-chlorobutane produces its precipitate slowest, of the three halides tested. [2]

(d) Explain why no precipitate is observed for 1-fluorobutane under these conditions. [2]

Part (a): The precipitation step

Once a halide ion has been released into solution by hydrolysis of the C–X bond, it reacts immediately with the silver ion already present: $\text{Ag}^+(\text{aq}) + \text{X}^-(\text{aq}) \rightarrow \text{AgX(s)}$

Part (b): Precipitate colours

1-Chlorobutane $\rightarrow$ white precipitate of $\text{AgCl}$ .
1-Bromobutane $\rightarrow$ cream (pale yellow) precipitate of $\text{AgBr}$ .
1-Iodobutane $\rightarrow$ yellow precipitate of $\text{AgI}$ .

Part (c): Why the timing differs

Before any $\text{Ag}^+$ can react, the C–X bond in the halogenoalkane must first break (hydrolyse) to release the free halide ion. This bond-breaking step is what actually limits how quickly each precipitate can appear. The weaker the C–X bond, the more readily it breaks, and the sooner $\text{X}^-$ is released to react with $\text{Ag}^+$ .

From the bond enthalpy data, C–I (238 kJ mol⁻¹) is the weakest of the three bonds tested, so 1-iodobutane hydrolyses fastest and its precipitate appears first. C–Cl (338 kJ mol⁻¹) is the strongest of the three, so 1-chlorobutane hydrolyses slowest and its precipitate appears last, with 1-bromobutane (276 kJ mol⁻¹) intermediate.

Part (d): Why 1-fluorobutane gives no precipitate

The C–F bond enthalpy (about 484 kJ mol⁻¹) is far higher than any of C–Cl, C–Br or C–I. It is the strongest carbon–halogen bond of the four. Under the mild conditions of this test (gentle warming), this bond is simply too strong to break at any measurable rate, so essentially no fluoride ion is released into solution, and consequently no $\text{AgF}$ precipitate can form, even after prolonged warming.

Final answers

(a) $\text{Ag}^+(\text{aq}) + \text{X}^-(\text{aq}) \rightarrow \text{AgX(s)}$
(b) White (AgCl, from 1-chlorobutane); cream (AgBr, from 1-bromobutane); yellow (AgI, from 1-iodobutane)
(c) Weakest bond (C–I) breaks fastest → 1-iodobutane precipitates first; strongest bond of the three (C–Cl) breaks slowest → 1-chlorobutane precipitates last
(d) The C–F bond is far stronger than C–Cl/Br/I, so it does not break at a measurable rate under these mild conditions, and no precipitate forms