Topic 10 · page 20 of 27

Core Practical 6: chlorination of 2-methylpropan-2-ol

Shake a tertiary alcohol with concentrated hydrochloric acid, separate the upper layer, wash it, dry it and distil it. Every technique in 10.19 is used once, and the boiling temperature shows whether the chloroalkane is pure.

14 min read10.1410.19

Read first: Separating, drying and checking a liquid, Replacing the OH group with a halogen

The reaction 10.14

Core Practical 6 is the chlorination of 2-methylpropan-2-ol with concentrated hydrochloric acid. The –OH group of a tertiary alcohol is replaced by –Cl. It sits in the alcohol pages because it starts from an alcohol, although the specification lists it with the halogenoalkanes.

(CHX3)X3COH+HCl→(CHX3)X3CCl+HX2O \ce{(CH3)3COH + HCl -> (CH3)3CCl + H2O}
Left: displayed formula of 2-methylpropan-2-ol, a central carbon bonded to three methyl groups and an OH group. An arrow labelled concentrated HCl, shake, room temperature. Right: 2-chloro-2-methylpropane, the same skeleton with Cl in place of OH. The OH and the Cl are drawn in orange.
The OH (orange) is replaced by Cl (orange). The product is 2-chloro-2-methylpropane, a tertiary halogenoalkane.

A tertiary alcohol reacts with concentrated hydrochloric acid at room temperature, with no need for heating or for PCl₅. The product is a tertiary halogenoalkane. It is a liquid that does not mix with water, and it boils at a much lower temperature than the alcohol.

Substance Boiling temperature / °C
2-methylpropan-2-ol (the alcohol) 82
2-chloro-2-methylpropane (the product) 51

The alcohol can form hydrogen bonds between its molecules. The chloroalkane cannot, so it boils at a lower temperature.

Method 10.1410.19

A five-step flow chart. 1 Shake: alcohol and concentrated HCl in a separating funnel, release the pressure. 2 Separate: run off and discard the lower aqueous acid layer, keep the upper layer. 3 Wash: shake with sodium hydrogencarbonate solution, release CO2, discard the lower layer. 4 Dry: anhydrous calcium chloride, swirl until clear. 5 Distil: collect the liquid that boils between 49 and 52 degrees Celsius.
The five steps of Core Practical 6. Each step removes a different impurity.

PracticalMethod

  1. Put 2-methylpropan-2-ol and concentrated hydrochloric acid in a separating funnel. Shake gently and release the pressure often. Leave the layers to separate.
  2. Run off and discard the lower aqueous layer (the acid). The product is the upper layer.
  3. Add sodium hydrogencarbonate solution to the upper layer and shake, releasing the pressure of the carbon dioxide often. This removes the remaining acid. Run off and discard the lower layer.
  4. Transfer the upper layer to a dry flask. Add anhydrous calcium chloride and swirl until the liquid is clear and some salt is free-flowing. Decant into a dry distillation flask.
  5. Distil and collect the liquid that boils between 49 and 52 °C.
Step Technique Removes or does
Shake with acid extraction in a separating funnel makes the product; the layers are then separated
Run off lower layer separating funnel removes the aqueous acid
Wash with NaHCO₃ separating funnel removes acid left in the product
Add CaCl₂ drying removes water
Distil, 49–52 °C distillation and boiling temperature removes unreacted alcohol (82 °C), shows purity

The hydrogencarbonate wash is an acid–base reaction:

HX+(aq)+HCOX3X−(aq)→HX2O(l)+COX2(g) \ce{H+(aq) + HCO3-(aq) -> H2O(l) + CO2(g)}

The fizzing shows that acid is being removed. The pressure of CO₂ builds up in the funnel, so it must be released.

Yield 10.14

Worked examplePercentage yield (illustrative)

10.0 g of 2-methylpropan-2-ol gives 8.1 g of purified 2-chloro-2-methylpropane. Calculate the percentage yield.

  1. MrM_\text{r} of (CH₃)₃COH = 4(12.0) + 10(1.0) + 16.0 = 74.0.
  2. MrM_\text{r} of (CH₃)₃CCl = 4(12.0) + 9(1.0) + 35.5 = 92.5.
  3. n(alcohol) = 10.0 ÷ 74.0 = 0.1351 mol.
  4. The equation is 1 : 1, so the maximum amount of product is 0.1351 mol, with mass 0.1351 × 92.5 = 12.50 g.
  5. Percentage yield = 8.1 ÷ 12.50 × 100 = 65% (2 s.f.).

Yield is lost because some alcohol does not react, because product stays in the aqueous layers or on the drying agent, and because some is lost in the transfers and the distillation.

Safety 10.14

Check yourself

  1. Write the equation for the reaction in Core Practical 6.
  2. Explain why the product is the upper layer in the separating funnel.
  3. Explain why the product is washed with sodium hydrogencarbonate, and write the ionic equation for what happens.
  4. Explain why the liquid is dried before distillation.
  5. A student collects the distillate at 49–52 °C. Explain what this shows, and why a distillate at 80 °C would be a problem.
  6. Calculate the percentage yield if 15.0 g of the alcohol gives 11.0 g of the product.

Answers cover these until you have tried the questions

  1. (CHX3)X3COH+HCl→(CHX3)X3CCl+HX2O\ce{(CH3)3COH + HCl -> (CH3)3CCl + H2O}
  2. The chloroalkane does not mix with water and is less dense than the aqueous acid, so it forms the upper layer.
  3. To remove unreacted hydrochloric acid that is still dissolved in the product. HX+(aq)+HCOX3X−(aq)→HX2O(l)+COX2(g)\ce{H+(aq) + HCO3-(aq) -> H2O(l) + CO2(g)}. It fizzes.
  4. To remove water. A wet liquid would distil with water and the boiling temperature would not be sharp.
  5. A narrow range around 51 °C shows that the product has distilled and is pure. A distillate at 80 °C is close to the boiling temperature of the alcohol (82 °C), so it would contain unreacted alcohol.
  6. n(alcohol) = 15.0 ÷ 74.0 = 0.2027 mol. Maximum mass = 0.2027 × 92.5 = 18.75 g. Yield = 11.0 ÷ 18.75 × 100 = 58.7% (59% to 2 s.f.).