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3.2.2Unit 3. Practical Skills in Chemistry I

Chlorination of 2-Methylpropan-2-ol — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 8 of 10 Unit quiz

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Part 1 of 6The Reaction

3.2.2 Chlorination of 2-Methylpropan-2-ol

A tertiary alcohol reacts with concentrated hydrochloric acid at room temperature.

Equation
  • (CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O converts the alcohol into the chloroalkane and water.
Why tertiary
  • Tertiary alcohols react readily with concentrated HCl, whereas primary and secondary alcohols need other reagents.
Layers
  • The chloroalkane is insoluble in water and less dense, so it forms the upper organic layer in the separating funnel.
Fume cupboard
  • Concentrated hydrochloric acid releases hydrogen chloride gas, so the work is done in a fume cupboard.

Part 2 of 6Preparation Steps

1
Measure 8 cm³ of 2-methylpropan-2-ol into a separating funnel in the fume cupboard.
2
Add 20 cm³ of concentrated hydrochloric acid in small portions, releasing the pressure by opening the stopper after each addition.
3
Leave the mixture for about 20 minutes, gently swirling every few minutes so that the two liquids mix well.
4
Allow the layers to separate and run off the lower aqueous layer through the tap.
5
Wash the organic layer with sodium hydrogencarbonate solution, venting often, then run off the aqueous layer again.

Part 3 of 6Purification

The crude product contains acid and water.

Purification
Neutralising
  • Sodium hydrogencarbonate solution reacts with leftover acid and releases CO₂, so the funnel must be vented often.
Drying
  • Transfer the organic layer to a dry flask and add anhydrous magnesium sulfate, which absorbs water until the liquid is clear.
Decanting
  • Pour the dried liquid off the solid drying agent into the distillation flask, leaving the solid behind.
Distillation
  • Distil and collect the fraction that boils within a narrow range around 51 °C, the boiling point of the pure product.

Part 4 of 6Safety and Notes

Pressure
  • Gas forms during mixing and washing, so frequent venting prevents the stopper from being forced out.
Flammable
  • The chloroalkane and the alcohol are flammable, so any heating must use an electric heater or water bath.
Fume hood
  • The sash must stay below the safety line so that the extraction can draw fumes away properly.
Yield
  • Product is lost in transfers and by incomplete separation, so the percentage yield is always below 100%.

Part 5 of 6Worked Example

Question

  • 8.0 cm³ of 2-methylpropan-2-ol (density 0.78 g cm⁻³) gave 4.68 g of 2-chloro-2-methylpropane.
  • Calculate the percentage yield (Ar: C = 12.0, H = 1.0, O = 16.0, Cl = 35.5).

  1. Mass of alcohol = 8.0 × 0.78 = 6.24 g; Mr of (CH₃)₃COH = 74.0.
  2. n(alcohol) = 6.24 ÷ 74.0 = 0.0843 mol, and the ratio to the chloroalkane is 1 : 1.
  3. Mr of (CH₃)₃CCl = 92.5, so theoretical mass = 0.0843 × 92.5 = 7.80 g.
  4. Percentage yield = 4.68 ÷ 7.80 × 100 = 60.0 %.

Answer Percentage yield = 60.0 %

Part 6 of 6Exam Checklist & Tips

3.2.2 – Key Points for the Exam

  • Separating funnel Glassware used to separate two immiscible liquids by running off the denser lower layer.
  • Drying agent An anhydrous salt, such as magnesium sulfate, that removes traces of water from an organic liquid.
  • Distillation Separation of liquids by boiling point, collecting the vapour that condenses.
  • Percentage yield Actual mass obtained divided by theoretical mass, multiplied by 100.

Common Mistakes

  • Closing the stopper without venting the separating funnel, which allows pressure to build up.
  • Collecting the lower layer as the product when the organic chloroalkane is the upper layer.
  • Adding drying agent to a wet mixture of two layers, instead of separating the layers first.
  • Using a wide boiling range in the distillation, which allows impurities into the product.

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