Topic 10 · page 17 of 27

Replacing the OH group with a halogen

Three reagent systems turn an alcohol into a chloro-, bromo- or iodoalkane. PCl₅ also gives a test for –OH, and the bromine and iodine reagents make the active compound in the flask.

12 min read10.1710.17(ii)

Read first: Burning alcohols and losing water

Three ways to swap OH for a halogen 10.1710.17(ii)

Each of these reactions converts an alcohol, R–OH, into a halogenoalkane, R–X. This is the reverse of the hydrolysis of a halogenoalkane. The mechanisms are not required.

Ethanol in the centre with three arrows to products. PCl5, dry, room temperature, gives chloroethane with POCl3 and HCl, and steamy white fumes. KBr and 50 percent concentrated sulfuric acid, heated under reflux, gives bromoethane. Red phosphorus and iodine, heated under reflux, gives iodoethane.
Three ways to turn ethanol into a halogenoalkane. The reagent decides which halogen is added.
Reagent Conditions Product from ethanol
PCl₅ room temperature, dry chloroethane
KBr and 50% concentrated H₂SO₄ heat under reflux bromoethane
red phosphorus and iodine heat under reflux iodoethane

Chloroalkanes with PCl₅ 10.1710.17(ii)

Phosphorus(V) chloride reacts with an alcohol at room temperature. It gives off steamy white fumes of hydrogen chloride.

CX2HX5OH+PClX5→CX2HX5Cl+POClX3+HCl \ce{C2H5OH + PCl5 -> C2H5Cl + POCl3 + HCl}

The reaction is a test for the –OH group. Add PCl₅ to a dry sample. If steamy white fumes of HCl form, the sample contains –OH. The test is positive for alcohols, but also for water and for carboxylic acids, because they too contain –OH. A positive result shows that an –OH group is present, not which compound it is.

Bromoalkanes with KBr and sulfuric acid 10.1710.17(ii)

Sulfuric acid reacts with potassium bromide to make hydrogen bromide in the flask. The HBr then reacts with the alcohol.

KBr+HX2SOX4→KHSOX4+HBr \ce{KBr + H2SO4 -> KHSO4 + HBr}
CX2HX5OH+HBr→CX2HX5Br+HX2O \ce{C2H5OH + HBr -> C2H5Br + H2O}

The acid is 50% concentrated because fully concentrated sulfuric acid is a strong oxidising agent. It would oxidise the bromide ions to bromine (Topic 8) and the yield of HBr would fall.

Iodoalkanes with red phosphorus and iodine 10.1710.17(ii)

Red phosphorus and iodine react together to make phosphorus(III) iodide, PI₃, in the flask. The PI₃ then reacts with the alcohol.

2 P+3 IX2→2 PIX3 \ce{2P + 3I2 -> 2PI3}
3 CX2HX5OH+PIX3→3 CX2HX5I+HX3POX3 \ce{3C2H5OH + PI3 -> 3C2H5I + H3PO3}

The reagent is made in the flask because PI₃ is unstable and reacts with moisture, so it is better to make it and use it at once.

Worked exampleWrite the equations for propan-1-ol

Write the equation for propan-1-ol with (a) PCl₅ and (b) HBr, made from KBr and sulfuric acid.

  1. (a) The OH is replaced by Cl, and POCl₃ and HCl form: CHX3CHX2CHX2OH+PClX5→CHX3CHX2CHX2Cl+POClX3+HCl\ce{CH3CH2CH2OH + PCl5 -> CH3CH2CH2Cl + POCl3 + HCl}. Product: 1-chloropropane. Check chlorine: 5 on the left; 1 + 3 + 1 = 5 on the right.
  2. (b) CHX3CHX2CHX2OH+HBr→CHX3CHX2CHX2Br+HX2O\ce{CH3CH2CH2OH + HBr -> CH3CH2CH2Br + H2O}. Product: 1-bromopropane.

Check yourself

  1. Write the equation for ethanol with phosphorus(V) chloride, and say what you would see.
  2. Name the reagents and the condition for turning ethanol into bromoethane.
  3. Write the two equations that explain how iodoethane forms from ethanol, red phosphorus and iodine.
  4. Explain why 50% concentrated sulfuric acid is used rather than concentrated acid.
  5. A student adds PCl₅ to a liquid and sees steamy white fumes. State what this shows and one limitation.
  6. Write the equation for butan-1-ol with PCl₅ and name the organic product.

Answers cover these until you have tried the questions

  1. CX2HX5OH+PClX5→CX2HX5Cl+POClX3+HCl\ce{C2H5OH + PCl5 -> C2H5Cl + POCl3 + HCl}. Steamy white fumes of hydrogen chloride.
  2. Potassium bromide and 50% concentrated sulfuric acid, heated under reflux.
  3. 2 P+3 IX2→2 PIX3\ce{2P + 3I2 -> 2PI3}, then 3 CX2HX5OH+PIX3→3 CX2HX5I+HX3POX3\ce{3C2H5OH + PI3 -> 3C2H5I + H3PO3}.
  4. Concentrated sulfuric acid would oxidise bromide ions to bromine, so less HBr is made and the product is contaminated. The diluted acid does not oxidise bromide.
  5. The liquid contains an –OH group. The limitation: water and carboxylic acids also give fumes, so the test does not show which compound it is.
  6. CHX3CHX2CHX2CHX2OH+PClX5→CHX3CHX2CHX2CHX2Cl+POClX3+HCl\ce{CH3CH2CH2CH2OH + PCl5 -> CH3CH2CH2CH2Cl + POCl3 + HCl}. The product is 1-chlorobutane.