Preparing 1-bromobutane: yield, washing steps and a change of alcohol
A student prepares 1-bromobutane by refluxing 7.40 g of butan-1-ol with 12.0 g of sodium bromide and 15 cm³ of concentrated sulfuric acid, then distilling. The crude distillate is orange-brown. It is washed with water, then with concentrated sulfuric acid, then with sodium hydrogencarbonate solution, dried over anhydrous calcium chloride and distilled, collecting the fraction boiling at 99–103 °C. The colourless product weighs 4.10 g and boils sharply. The student's report states: 'Yield 30 %. The low yield is because the reaction is an equilibrium that does not go to completion.' Data: butan-1-ol, bp 118 °C, M = 74.1; 1-bromobutane, bp 102 °C, M = 137.0; but-1-ene, bp −6 °C; dibutyl ether, bp 142 °C; NaBr, M = 102.9.
1. Calculate the theoretical and percentage yield, and evaluate the student's explanation for the low figure.
Read the worked answer
n(butan-1-ol) = 7.40/74.1 = 0.0999 ≈ 0.100 mol n(NaBr) = 12.0/102.9 = 0.117 mol, so the alcohol is limiting.
Theoretical m(1-bromobutane) = 0.100 × 137.0 = 13.7 g % yield = (4.10/13.7) × 100 = 29.9 % ≈ 30 %
The explanation is wrong. The substitution ROH + HBr → RBr + H₂O is not a seriously reversible equilibrium under these conditions; with excess HBr and hot concentrated acid the position lies essentially completely on the product side. The back reaction (hydrolysis of the bromide) is slow. A 70 % loss must therefore be accounted for practically:
- Elimination (E2) to but-1-ene, favoured by hot concentrated H₂SO₄; the alkene is volatile (bp −6 °C) and is lost.
- Ether formation, as butan-1-ol attacks protonated bromobutane in an S_N2 step, giving dibutyl ether (bp 142 °C).
- Oxidation of HBr by hot concentrated H₂SO₄, destroying reagent.
- Physical losses: 1-bromobutane is volatile (bp 102 °C), so material is lost during the washes, during drying, and as hold-up in the flask and condenser; some product also remains in the distillation residue.
So the 30 % figure reflects competing reactions and handling losses, not an equilibrium ceiling. A well-run preparation of this compound typically does considerably better, which is itself evidence against the equilibrium explanation.
2. Explain the purpose of each wash, and account for the orange-brown colour of the crude distillate with an equation.
Read the worked answer
Water wash: removes most of the sulfuric acid, sodium hydrogen sulfate and residual butan-1-ol (which is appreciably water-soluble), and stops the acid attacking the product during the later steps.
Concentrated sulfuric acid wash: protonates and dissolves the remaining butan-1-ol and any dibutyl ether, removes but-1-ene (alkenes are absorbed/protonated by concentrated acid), and dries the organic layer. It must be done carefully and quickly, because strong acid can also attack the bromide.
Sodium hydrogencarbonate wash: neutralises any acid carried through and removes sulfur dioxide; a little sodium sulfite or thiosulfate is often added to reduce bromine to bromide. Effervescence here is CO₂.
Colour: hot concentrated sulfuric acid oxidises hydrogen bromide:
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
Bromine is volatile and distils with the product, giving the orange-brown colour. It is removed by the sulfite/thiosulfate and hydrogencarbonate washes; the product is colourless once bromine has gone.
Optional check of purity: the sharp boiling point (102 °C, not 118 °C or 142 °C) and a refractive index or gas chromatogram would confirm that the alcohol and ether have been removed.
3. The student repeats the preparation with 2-methylpropan-2-ol instead of butan-1-ol, expecting a better yield, but obtains mainly 2-methylpropene and little 2-bromo-2-methylpropane. Explain this using mechanisms, and state what it reveals about the mechanism of the butan-1-ol reaction.
Read the worked answer
Butan-1-ol (primary): the mechanism is S_N2. HBr is generated in situ; Br⁻ attacks the protonated alcohol from the side opposite the leaving water molecule, in a single step with inversion of configuration. Because no carbocation is formed, the competing elimination is minor and substitution dominates.
2-Methylpropan-2-ol (tertiary): the mechanism is S_N1. Protonation and loss of water give the relatively stable tertiary carbocation, (CH₃)₃C⁺, which can either be captured by Br⁻ or lose a proton from a neighbouring methyl group. Loss of a proton gives 2-methylpropene (bp ≈ −7 °C), which escapes as a gas and drives the elimination forward. Under hot strongly acidic conditions, E1 therefore competes effectively with, and here outruns, substitution - hence little tertiary bromide.
What this shows: the first preparation works because a primary alcohol cannot form a stable carbocation, so the S_N1/E1 pathway is closed and the reaction is forced through S_N2. Going to a tertiary alcohol opens the carbocation route and with it the elimination route, so 'more substituted alcohol = better yield' is a false generalisation. To obtain the tertiary bromide one would use a less forcing method - brief reaction at low temperature, or a reagent such as PBr₃ which delivers bromide without a strongly acidic, hot medium.