Topic 5 · page 8 of 15

Electrophiles and curly arrows

The π bond is a region of high electron density, so it attracts electrophiles. Before drawing any mechanism, you need the rules for curly arrows and for how a bromine molecule can act as an electrophile.

12 min read5.6

Read first: Alkenes and the double bond, Adding hydrogen and halogens

Why alkenes attract electrophiles 5.6

A mechanism shows, step by step, how the bonds in a reaction break and form. The mechanisms for the reactions of alkenes follow from one fact: the C=C contains a π bond, and the π electrons form a region of high electron density above and below the molecule.

So an electrophile is attracted to the π bond of an alkene. The alkene donates its π electrons, and the π bond breaks. The overall type of reaction is electrophilic addition: an electrophile starts the reaction, and two atoms are added across the C=C.

Electrophiles are usually positive ions or the δ+ end of a polar molecule. Two cases matter here.

Hydrogen bromide is polar, because bromine is more electronegative than hydrogen: Hδ+–Brδ−. The Hδ+ end is the electrophile.

Bromine, Br₂, is non-polar, because the two atoms are identical. So how can it be an electrophile? When a Br₂ molecule approaches the C=C, the π electrons repel the electrons in the Br–Br bond. The bonding pair is pushed towards the far Br atom. The nearer Br atom becomes δ+ and the far one δ−. The alkene has induced a dipole in the bromine molecule. The Brδ+ end is then the electrophile.

Once the electrophile has taken a pair of electrons, the other end of the X–Y bond is left with a pair of electrons of its own. The X–Y bond breaks heterolytically: both electrons of the bond go to Y, which becomes a negative ion (Br⁻ from HBr or Br₂). This is heterolytic fission, as in Topic 4.

Curly arrows 5.6

Curly arrows show the movement of a pair of electrons.

Rule What it means
The arrow shows a pair of electrons moving it has a full head (two barbs). Half-arrows, with one barb, are for single electrons in radical reactions.
It starts at a bond or a lone pair never at an atom symbol, never at a + charge
It ends at the atom that receives the pair, or in the space where a new bond forms the arrowhead touches the atom or the gap between atoms
Charges and bonds must balance afterwards the carbon that lost a pair of π electrons is left with a + charge
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.
A preview of the mechanism on the next page: two curly arrows, one from the C=C and one from the Br–Br bond.

In the first step of every mechanism in this topic, the first arrow starts from the middle of the C=C, because that is where the π electrons are.

Once the π pair has gone to the electrophile, one of the two carbons of the old C=C has lost its share of the π electrons. It has only three bonds and no electrons to spare, so it is left with a positive charge. A species with a positively charged carbon is a carbocation. It is an intermediate: it forms in the first step and is used up in the second.

Check yourself

  1. Define the term electrophile.
  2. Explain why alkenes are attacked by electrophiles.
  3. Hydrogen bromide and bromine are both electrophiles towards ethene. Explain how each one is able to act in this way.
  4. State where a curly arrow must start and where it must end.
  5. Explain what the + charge on the carbocation intermediate shows about the electrons in the C=C.
  6. A student draws a curly arrow from the minus sign on Br⁻ to the carbocation. Explain what is wrong and how to correct it.

Answers cover these until you have tried the questions

  1. An electrophile is an electron-pair acceptor.
  2. The C=C contains a π bond, a region of high electron density above and below the molecule. An electrophile is attracted to this region and accepts a pair of electrons from it.
  3. HBr has a permanent dipole: Br is more electronegative than H, so H is δ+ and is the electrophile. Br₂ is non-polar, but as it approaches the C=C the π electrons repel the electrons of the Br–Br bond, so a dipole is induced: the nearer Br becomes δ+ and acts as the electrophile.
  4. It must start at a bond or at a lone pair of electrons, and end at the atom that receives the pair of electrons (or at the point where the new bond is formed).
  5. The two π electrons were donated to the electrophile to make a new bond. The carbon that did not receive a new bond is left one electron pair short, so it has only three bonds and a positive charge.
  6. A curly arrow must start at a lone pair (or a bond), not at a charge. Draw the lone pair on the Br⁻ ion as a pair of dots and start the arrow at that lone pair, ending on the C⁺.