Topic 5 · page 10 of 15

The mechanism for ethene and hydrogen bromide

Hydrogen bromide adds to ethene by the same two-step pattern as bromine. The difference is that H–Br is permanently polar, so no dipole has to be induced.

11 min read5.65.6(i)

Read first: The mechanism for ethene and bromine

The mechanism 5.65.6(i)

The overall reaction is:

CHX2=CHX2+HBr→CHX3CHX2Br \ce{CH2=CH2 + HBr -> CH3CH2Br}

Hydrogen bromide is polar: Hδ+–Brδ−. So there is no need for an induced dipole: the Hδ+ atom is the electrophile from the start.

Step 1 of the electrophilic addition of hydrogen bromide to ethene. The H-Br molecule lies above the C=C, with a delta plus on hydrogen and a delta minus on bromine. One curly arrow starts in the middle of the C=C and points to the hydrogen atom. A second curly arrow starts in the H-Br bond and points to the bromine atom. The products are the carbocation CH3CH2 with a plus charge, and a bromide ion.
Step 1. Arrow 1 from the C=C to the δ+ hydrogen. Arrow 2 from the H–Br bond to the bromine.

Step 1. The first arrow starts in the middle of the C=C and points to the Hδ+: the π electrons form a new C–H bond. The second arrow starts in the H–Br bond and points to Br: the bond breaks heterolytically and Br leaves with both electrons as Br⁻. The other carbon is left with a + charge: the carbocation CH₃CH₂⁺.

Step 2 of the electrophilic addition of hydrogen bromide 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 bromoethane.
Step 2. A lone pair on Br⁻ forms a bond to the positive carbon. The product is bromoethane.

Step 2. A lone pair on Br⁻ forms a bond to the positive carbon: the third arrow starts at the lone pair and ends on the C⁺. The product is bromoethane, CH₃CH₂Br.

Ethene + Br₂ Ethene + HBr
Electrophile Brδ+ (induced dipole) Hδ+ (permanent dipole)
Carbocation BrCH₂CH₂⁺ CH₃CH₂⁺
Negative ion Br⁻ Br⁻
Product 1,2-dibromoethane bromoethane

Worked exampleHydrogen chloride and ethene

Draw the mechanism for ethene and hydrogen chloride.

  1. H–Cl is polar: Hδ+–Clδ−.
  2. Arrow 1: from the middle of the C=C to the Hδ+.
  3. Arrow 2: from the H–Cl bond to the Cl, giving CH₃CH₂⁺ and Cl⁻ with a lone pair.
  4. Arrow 3: from a lone pair on Cl⁻ to the C⁺, giving CH₃CH₂Cl (chloroethane).
The mechanism of the electrophilic addition of hydrogen chloride to ethene in two steps. Step 1: a curly arrow from the middle of the C=C to the delta plus hydrogen, and a curly arrow from the H-Cl bond to chlorine, giving the carbocation CH3CH2 with a plus charge and a chloride ion. Step 2: a curly arrow from a lone pair on the chloride ion to the positive carbon, giving chloroethane.
Ethene + HCl: the same mechanism with chlorine in place of bromine.

Check yourself

  1. Explain why hydrogen bromide can act as an electrophile without any induced dipole.
  2. Draw the mechanism for the reaction of ethene with hydrogen bromide.
  3. Name the intermediate in the reaction and state the type of bond fission in the H–Br bond.
  4. Give one similarity and one difference between the mechanisms for ethene with HBr and ethene with Br₂.
  5. Ethene is passed into a solution of hydrogen iodide. Name the product and say what type of reaction occurs.
  6. A student writes "H⁺ attacks the C=C, then Br⁻ attacks". Explain why this is not the best description of the first step.

Answers cover these until you have tried the questions

  1. Bromine is more electronegative than hydrogen, so the H–Br bond is permanently polar: H is δ+ and Br is δ−. The Hδ+ atom is therefore already an electrophile.
  2. Ethene with four H; Hδ+–Brδ− above it. Arrow 1 from the middle of the C=C to Hδ+. Arrow 2 from the H–Br bond to Br. Draw CH₃CH₂⁺ and Br⁻ with a lone pair. Arrow 3 from a lone pair on Br⁻ to the C⁺. Product CH₃CH₂Br. (See the figures on this page; the HCl figure has the same shape.)
  3. The intermediate is the carbocation CH₃CH₂⁺. The H–Br bond breaks by heterolytic fission: both electrons go to Br.
  4. Similar: both are electrophilic addition, with a carbocation intermediate and a second step in which Br⁻ attacks it. Different: with Br₂ the dipole must be induced by the π electrons and the carbocation is BrCH₂CH₂⁺; with HBr the dipole is permanent and the carbocation is CH₃CH₂⁺.
  5. Iodoethane, CH₃CH₂I. It is an electrophilic addition.
  6. In the first step the π electrons attack the δ+ hydrogen of a polar H–Br molecule; there is no free H⁺ ion. The arrow must start in the C=C and end at H, and a second arrow must show the H–Br bond breaking. "H⁺ attacks the C=C" does not show where the electrons come from.