Topic 10 · page 6 of 27

Hydroxide: substitution or elimination

The hydroxide ion does two different jobs with a halogenoalkane. In aqueous potassium hydroxide it is a nucleophile and gives an alcohol. In ethanolic potassium hydroxide it is a base and gives an alkene.

12 min read10.810.8(i)10.8(ii)

Read first: Primary, secondary and tertiary halogenoalkanes

Aqueous alkali: substitution 10.810.8(i)

Warm a halogenoalkane with aqueous potassium hydroxide, under reflux, and the halogen is replaced by –OH. An alcohol forms.

CHX3CHX2Br+OHX−→CHX3CHX2OH+BrX− \ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}

The full equation, with the spectator ion K⁺ included:

CHX3CHX2Br+KOH→CHX3CHX2OH+KBr \ce{CH3CH2Br + KOH -> CH3CH2OH + KBr}

The hydroxide ion has lone pairs on oxygen. It is a nucleophile: a lone pair is donated to the δ+ carbon, and the bromide ion leaves with the bonding pair. The reaction is nucleophilic substitution, and because aqueous alkali breaks the bond it is also hydrolysis.

Reagent KOH(aq), aqueous potassium hydroxide
Condition warm; heat under reflux
Product an alcohol (ethanol from bromoethane)
Role of OH⁻ nucleophile
Type nucleophilic substitution (also hydrolysis)

Ethanolic alkali: elimination 10.810.8(ii)

Change the solvent to ethanol, use hot, concentrated potassium hydroxide, and a different reaction takes over. The same ion removes H⁺ from the carbon next to the C–Br carbon, the C–Br bond breaks, and a C=C forms. An alkene forms, and a small molecule, HBr, is lost.

CHX3CHX2Br+KOH→CHX2=CHX2+KBr+HX2O \ce{CH3CH2Br + KOH -> CH2=CH2 + KBr + H2O}
Left: bromoethane with aqueous KOH gives ethanol. The hydroxide ion bonds to the carbon that carried bromine and Br minus leaves. Right: bromoethane with ethanolic KOH gives ethene. The hydroxide ion takes the hydrogen from the neighbouring carbon, forming water, and Br minus leaves. The hydrogen that is removed and the OH group that is added are drawn in orange.
One halogenoalkane, one ion, two jobs. In water OH⁻ attacks carbon. In ethanol OH⁻ removes an H from the next carbon.

In elimination the hydroxide ion is a base: a proton acceptor. It uses a lone pair to take an H⁺, not to attack carbon.

Reagent KOH in ethanol, ethanolic potassium hydroxide
Condition heat under reflux
Product an alkene (ethene from bromoethane)
Role of OH⁻ base
Type elimination

Worked example2-bromopropane with each reagent

Name the organic product of 2-bromopropane with (a) aqueous KOH and (b) ethanolic KOH, and give the role of OH⁻ in each.

  1. (a) The Br is replaced by OH: CHX3CHBrCHX3+OHX−→CHX3CH(OH)CHX3+BrX−\ce{CH3CHBrCH3 + OH- -> CH3CH(OH)CH3 + Br-}. The product is propan-2-ol. OH⁻ is a nucleophile.
  2. (b) H and Br are removed from neighbouring carbons: CHX3CHBrCHX3+KOH→CHX3CH=CHX2+KBr+HX2O\ce{CH3CHBrCH3 + KOH -> CH3CH=CH2 + KBr + H2O}. The product is propene. OH⁻ is a base.

Elimination needs a hydrogen on a carbon next to the C–X carbon. Chloromethane, CH₃Cl, has no neighbouring carbon, so it cannot eliminate.

Check yourself

  1. Write the equation for bromoethane with aqueous potassium hydroxide, and name the type of reaction and the role of the hydroxide ion.
  2. Write the equation for bromoethane with ethanolic potassium hydroxide, and name the type of reaction and the role of the hydroxide ion.
  3. 1-bromopropane is heated with ethanolic KOH. Name the organic product.
  4. State the changes to the conditions that make a halogenoalkane give an alkene rather than an alcohol.
  5. Explain why chloromethane cannot give an alkene when heated with ethanolic KOH.
  6. A student writes: "Both reactions form a product containing OH, because OH⁻ is a nucleophile." Correct this.

Answers cover these until you have tried the questions

  1. CHX3CHX2Br+OHX−→CHX3CHX2OH+BrX−\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-} (or with KOH and KBr). Nucleophilic substitution (also hydrolysis). OH⁻ is a nucleophile.
  2. CHX3CHX2Br+KOH→CHX2=CHX2+KBr+HX2O\ce{CH3CH2Br + KOH -> CH2=CH2 + KBr + H2O}. Elimination. OH⁻ is a base.
  3. Propene, CH₂=CHCH₃: the H on carbon 2 and the Br on carbon 1 are removed.
  4. Use potassium hydroxide dissolved in ethanol (not aqueous), and heat under reflux.
  5. Chloromethane has only one carbon, so there is no neighbouring carbon to lose a hydrogen, and no C=C can form.
  6. With aqueous KOH, OH⁻ acts as a nucleophile and an alcohol forms. With ethanolic KOH, OH⁻ acts as a base: it removes H⁺, water forms, and the product is an alkene, which does not contain OH.