Which bromobutane is which: timing the silver nitrate test
Three colourless liquids, A, B and C, are known to be 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane, but the labels have been lost. A student puts five drops of each into separate test tubes containing 5.0 cm3 of ethanol and 5.0 cm3 of 0.020 mol dm-3 silver nitrate solution, warms them together in a 60 °C water bath and times the first appearance of a pale yellow precipitate: A, 8 s; B, 95 s; C, 12 min. The student claims this identifies all three. A classmate objects that all three compounds release the same bromide ion, so the only thing being timed is how quickly the tubes were shaken, and that a fair test would need equal concentrations, which were not measured. The teacher allows a second experiment: each liquid is boiled with ethanolic potassium hydroxide and the gas produced is passed through bromine water.
1. Assign A, B and C to the three isomers and justify the assignment in terms of mechanism and the rate-determining step.
Read the worked answer
A (8 s) = 2-bromo-2-methylpropane, tertiary. B (95 s) = 2-bromobutane, secondary. C (12 min) = 1-bromobutane, primary.
In ethanol containing silver ions, the rate-determining step is heterolysis of the C-Br bond to give a carbocation, with Ag+ helping by removing Br- as insoluble AgBr. The rate therefore follows carbocation stability: tertiary > secondary > primary. Three alkyl groups on the tertiary cation donate electron density inductively and stabilise it by hyperconjugation; a primary cation is barely stabilised, so its activation energy is much higher. This is why the times differ by two orders of magnitude rather than by a small margin.
The primary halide reacts mainly by an SN2 route with ethanol or water as nucleophile, which is also slow. Precipitation of AgBr is effectively instantaneous once bromide is released, so the timed event reports the ionisation step, not the silver nitrate.
2. Evaluate the classmate's objection, and state what would happen to the time for each tube if the concentration of the halogenoalkane were doubled.
Read the worked answer
The objection is partly right and partly wrong. It is true that the identity of the precipitate cannot distinguish the isomers, and that quantities should have been equalised; five drops is not a measured amount, so concentrations were not matched.
But the argument that only shaking is being timed does not survive the data. All tubes were handled the same way, and the times differ by a factor of ninety. Shaking affects mixing and diffusion, not the rate at which a C-Br bond breaks, and mixing cannot produce a ninety-fold difference.
Effect of doubling [halogenoalkane]:
- A and B ionise by a unimolecular SN1 route, rate = k[RX], first order. Doubling concentration roughly halves the time.
- C reacts mainly by SN2, rate = k[RX][nucleophile], first order in the halide. Doubling the halide again roughly halves the time, but raising the water or ethanol concentration would also speed it up.
So the ranking would survive, which is why the test works even with an untidy method.
3. The elimination results show that one of the three liquids gives two alkenes and the other two each give a single alkene. Identify which liquid gives two, name the products, and explain how this supports the assignment made from the rate data.
Read the worked answer
The liquid that gives two alkenes is B, 2-bromobutane.
Boiling with ethanolic KOH removes H+ from a carbon adjacent to the C-Br carbon. In 2-bromobutane there are two different adjacent carbons: C1 (giving but-1-ene) and C3 (giving but-2-ene). But-2-ene predominates because the more highly substituted alkene is more stable (Zaitsev's rule).
1-Bromobutane (C) has only C2 adjacent to the halogen, so it can give only but-1-ene. 2-Bromo-2-methylpropane (A) has three equivalent methyl groups and gives only 2-methylpropene.
The elimination pattern therefore identifies B uniquely as the secondary isomer, independently of any rate measurement. Combined with the silver nitrate timings, which separate A from C, all three labels are fixed. Two independent pieces of evidence agreeing is much stronger than either test alone, and it answers the objection about shaking: the alkene pattern cannot be produced by mixing.