Topic 5 · page 9 of 15

The mechanism for ethene and bromine

Bromine adds to ethene in two steps. The π electrons attack an induced Br⁺ end, forming a carbocation, and then a bromide ion attacks it. The products formed in bromine water are the evidence for two steps.

14 min read5.65.6(i)

Read first: Electrophiles and curly arrows

Step 1: the electrophile attacks the π bond 5.65.6(i)

The overall reaction is the one from Adding hydrogen and halogens:

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

The mechanism has two steps. In the first step a bromine molecule approaches the C=C. The π electrons push the bonding pair of the Br–Br bond away, which gives the nearer Br atom a δ+ charge (an induced dipole).

Step 1 of the electrophilic addition of bromine to ethene. A bromine molecule lies above the C=C, with a delta plus on the nearer bromine atom and a delta minus on the other. One curly arrow starts in the middle of the C=C and points to the delta plus bromine atom. A second curly arrow starts in the Br-Br bond and points to the other bromine atom. The products are a carbocation, BrCH2CH2 with a plus charge, and a bromide ion with a lone pair.
Step 1. Arrow 1 starts at the π bond. Arrow 2 starts at the Br–Br bond. The result is a carbocation and a bromide ion.

Two curly arrows describe what happens.

  1. The first arrow starts in the middle of the C=C and points to the Brδ+ atom. The π electrons form a new C–Br bond.
  2. The second arrow starts in the Br–Br bond and points to the other Br atom. The Br–Br bond breaks heterolytically, and that Br atom leaves with both electrons as a bromide ion, Br⁻, with a lone pair.

The carbon that did not form a bond with bromine has lost its share of the π electrons, so it is a carbocation: BrCH₂CH₂⁺.

Step 2: the bromide ion attacks the carbocation 5.65.6(i)

The carbocation is a positive ion, and the bromide ion has a lone pair of electrons. They are attracted to each other.

Step 2 of the electrophilic addition of bromine to ethene. A curly arrow starts at a lone pair on the bromide ion and points to the positively charged carbon of the carbocation. A new C-Br bond forms and the product is 1,2-dibromoethane.
Step 2. The arrow starts at a lone pair on Br⁻ and ends on the C⁺, forming the second C–Br bond.

The third curly arrow starts at a lone pair on Br⁻ and points to the positive carbon. The lone pair becomes the bond between C and Br. The product is 1,2-dibromoethane, and each carbon now has one Br atom.

MethodDrawing the mechanism for ethene and bromine

  1. Draw ethene with its C=C and four H atoms.
  2. Draw Br–Br above the C=C and label the nearer Br δ+ and the other δ−.
  3. Arrow 1: from the middle of the C=C to the δ+ Br.
  4. Arrow 2: from the Br–Br bond to the δ− Br.
  5. Draw the carbocation BrCH₂CH₂⁺ (with the + on the carbon) and Br⁻ (with a lone pair).
  6. Arrow 3: from the lone pair on Br⁻ to the C⁺.
  7. Draw the product, 1,2-dibromoethane.

Evidence for the two-step mechanism 5.6

How do we know there are two steps, with an intermediate in between? The evidence comes from a change to the reaction mixture.

Ethene is bubbled into bromine water that also contains sodium chloride. Three species with lone pairs are present that could attack a carbocation: Br⁻, Cl⁻ and H₂O. A species that attacks a positive centre by donating a lone pair is called a nucleophile (an electron-pair donor). Chemists look at the organic products.

Displayed formulae of the compounds looked for when ethene is bubbled into bromine water containing sodium chloride. 1,2-dibromoethane, 1-bromo-2-chloroethane and 2-bromoethanol are found. 1,2-dichloroethane, which has no bromine atom, is never found.
Three products are found. The fourth, 1,2-dichloroethane, is never found.

The products are 1,2-dibromoethane, 1-bromo-2-chloroethane and 2-bromoethanol. 1,2-dichloroethane is never found.

With ordinary bromine water, water is by far the most plentiful of these species, so the main product is 2-bromoethanol. The bromine colour still disappears, so the test for C=C still works.

Check yourself

  1. Explain how a non-polar bromine molecule acts as an electrophile towards ethene.
  2. Draw the mechanism for the reaction of ethene with bromine, showing all curly arrows, charges, dipoles and lone pairs.
  3. Name the intermediate in this reaction and give its formula.
  4. Chlorine reacts with ethene in the same way as bromine. Write the equation and name the product.
  5. Ethene reacts with bromine water containing sodium chloride. Name three organic products and explain how they support a two-step mechanism.
  6. State why 1,2-dichloroethane is not found in this experiment.

Answers cover these until you have tried the questions

  1. As the Br₂ molecule approaches the C=C, the π electrons repel the Br–Br bonding pair, which induces a dipole: the nearer Br becomes δ+. The Brδ+ accepts a pair of electrons from the π bond, so Br₂ acts as an electrophile.
  2. Draw ethene with four H, and Br–Br above it with δ+ on the near Br and δ− on the far Br. Arrow 1: from the middle of the C=C to Brδ+. Arrow 2: from the Br–Br bond to the Brδ−. This gives BrCH₂CH₂⁺ (with the + on carbon) and Br⁻ (with a lone pair). Arrow 3: from a lone pair on Br⁻ to the C⁺. Product: BrCH₂CH₂Br. (The figure in this page shows the answer.)
  3. A carbocation, BrCH₂CH₂⁺.
  4. CH₂=CH₂ + Cl₂ → CH₂ClCH₂Cl, 1,2-dichloroethane.
  5. 1,2-dibromoethane, 1-bromo-2-chloroethane, 2-bromoethanol. Br⁺ (from Br₂) always adds first, forming a carbocation intermediate; the intermediate is then attacked by whichever of Br⁻, Cl⁻ or H₂O is nearby. If both atoms added together, the second atom could not vary.
  6. The first atom to add must be the electrophile Brδ+. Cl⁻ is a nucleophile, not an electrophile, and Cl₂ is not present, so there is no way to start the reaction with Cl.