Topic 5 · page 5 of 15

Adding hydrogen and halogens

Alkenes react by addition. The π bond breaks and two atoms add across the C=C. Hydrogen with a nickel catalyst gives an alkane, and a halogen gives a di-substituted halogenoalkane.

14 min read5.45.4(i)5.4(ii)

Read first: Alkenes and the double bond

Alkenes react by addition 5.4

Alkenes are reactive because of the π bond. In their reactions, the π bond breaks and two atoms or groups add across the C=C, one to each carbon. The specification limits you to five reactions.

Map of the five addition reactions of ethene. Ethene on the left gives: ethane with hydrogen and a nickel catalyst at 150 degrees Celsius; 1,2-dibromoethane with bromine at room temperature; bromoethane with hydrogen bromide at room temperature; ethanol with steam and a phosphoric acid catalyst at 300 degrees Celsius and 60 atmospheres; ethane-1,2-diol with acidified potassium manganate(VII) at room temperature.
The five reactions of ethene in the specification. Learn the reagent and condition on each arrow.

This page covers the first two. The next two pages cover the others.

The general pattern of addition. A C=C double bond and a molecule X-Y react. The pi bond, drawn in orange, breaks. X adds to one carbon and Y to the other, giving a single saturated product with a C-C single bond.
The π bond breaks. X adds to one carbon and Y to the other.

Hydrogen: hydrogenation 5.45.4(i)

Hydrogen adds across the C=C, and the product is the alkane with the same carbon skeleton.

CHX2=CHX2+HX2→Ni catalyst,150°CCHX3CHX3 \ce{CH2=CH2 + H2 ->[Ni catalyst, 150 °C] CH3CH3}

The same reaction with a ring alkene gives the cycloalkane:

CX6HX10+HX2→Ni catalyst,150°CCX6HX12 \ce{C6H10 + H2 ->[Ni catalyst, 150 °C] C6H12}

The reaction is also called hydrogenation. It adds hydrogen, so it is a reduction as well as an addition. It is used to harden vegetable oils into margarine: some of the C=C bonds in the oil are turned into C–C bonds, which raises the melting temperature.

Worked exampleVolume of hydrogen needed

What volume of hydrogen, at room temperature and pressure, is needed to hydrogenate 0.280 g of ethene completely? (Molar volume 24.0 dm³ mol⁻¹.)

  1. C₂H₄ + H₂ → C₂H₆, so the ratio is 1 : 1.
  2. MrM_\text{r} (C₂H₄) = 2 × 12.0 + 4 × 1.0 = 28.0.
  3. n(C₂H₄) = 0.280 ÷ 28.0 = 0.0100 mol, so n(H₂) = 0.0100 mol.
  4. V = 0.0100 × 24.0 = 0.240 dm³ = 240 cm³.

Halogens: a di-substituted halogenoalkane 5.45.4(ii)

Bromine and chlorine add across the C=C at room temperature, with no catalyst.

CHX2=CHX2+BrX2→CHX2BrCHX2Br \ce{CH2=CH2 + Br2 -> CH2BrCH2Br}

The product is a di-substituted halogenoalkane: it has two halogen atoms, one on each of the two carbons that were joined by the C=C. For ethene and bromine it is 1,2-dibromoethane. Chlorine gives 1,2-dichloroethane in the same way.

Skeletal formulae. Propene reacts with bromine to give 1,2-dibromopropane: one bromine atom, in orange, is attached to each of the two carbons that were joined by the double bond.
Propene + bromine → 1,2-dibromopropane. The orange atoms are the added Br.

The numbers in the name show where the halogens are. In 1,2-dibromopropane, C1 and C2 were the two carbons of the C=C of propene, so the numbers must be adjacent.

Worked exampleMass of product

Ethene (2.80 g) reacts completely with excess bromine. What mass of 1,2-dibromoethane forms?

  1. C₂H₄ + Br₂ → C₂H₄Br₂, ratio 1 : 1.
  2. n(C₂H₄) = 2.80 ÷ 28.0 = 0.100 mol, so n(C₂H₄Br₂) = 0.100 mol.
  3. MrM_\text{r} (C₂H₄Br₂) = 24.0 + 4.0 + 2 × 79.9 = 187.8.
  4. Mass = 0.100 × 187.8 = 18.78 g = 18.8 g.

Check yourself

  1. Give the reagent and conditions needed to convert propene into propane, and write the equation.
  2. Name the product of the reaction between ethene and chlorine.
  3. Write an equation for the reaction of but-2-ene with bromine and name the product.
  4. State the type of reaction in questions 1 and 3, and explain what happens to the π bond.
  5. What volume of hydrogen, measured at room temperature and pressure, is needed to hydrogenate 4.20 g of propene? (MrM_\text{r} of propene = 42.0; molar volume 24.0 dm³ mol⁻¹.)
  6. Cyclohexene (8.20 g) reacts completely with excess bromine. Calculate the mass of 1,2-dibromocyclohexane formed. (ArA_\text{r}: C 12.0, H 1.0, Br 79.9.)

Answers cover these until you have tried the questions

  1. Hydrogen with a nickel catalyst, at about 150 °C. CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃.
  2. 1,2-dichloroethane, CH₂ClCH₂Cl: CH₂=CH₂ + Cl₂ → CH₂ClCH₂Cl.
  3. CH₃CH=CHCH₃ + Br₂ → CH₃CHBrCHBrCH₃, named 2,3-dibromobutane (the Br atoms are on the two carbons of the old C=C, which were C2 and C3).
  4. Both are addition reactions. (The hydrogen one is also a reduction.) The π bond breaks and each carbon of the old C=C forms a new σ bond to an added atom, leaving a C–C single bond.
  5. n(C₃H₆) = 4.20 ÷ 42.0 = 0.100 mol. The ratio C₃H₆ : H₂ is 1 : 1, so n(H₂) = 0.100 mol. V = 0.100 × 24.0 = 2.40 dm³ = 2400 cm³.
  6. MrM_\text{r} (C₆H₁₀) = 6 × 12.0 + 10 × 1.0 = 82.0. n = 8.20 ÷ 82.0 = 0.100 mol. The ratio is 1 : 1, so n(C₆H₁₀Br₂) = 0.100 mol. MrM_\text{r} (C₆H₁₀Br₂) = 72.0 + 10.0 + 159.8 = 241.8. Mass = 0.100 × 241.8 = 24.18 g = 24.2 g.