Reactions of alkenes with bromine

Alkenes contain a C=C double bond. This double bond is the reactive part of the molecule (the 'functional group'). It can open up so that new atoms join onto the two carbon atoms. This type of reaction is called an ADDITION reaction.

In an addition reaction:

• The C=C double bond becomes a C–C single bond.
• A small molecule (such as Br₂, H₂ or H₂O) splits and adds across the double bond — one part joins one carbon, the other part joins the other carbon.
• No atoms are lost — every atom of both reactants ends up in a single product.

Why ADDITION and not substitution?

• The C=C double bond is electron-rich — there is a lot of negative charge between the two carbons.
• It is also weaker and more exposed than the single bonds in an alkane.
• This means small molecules are attracted to the C=C and can open it easily, so they ADD across it.
• Alkanes have only single bonds (no C=C), so they cannot do addition reactions — they are much less reactive at room temperature.

This is the key idea behind all three addition reactions you need: with bromine, with hydrogen, and with steam.

The most important addition reaction in Double Award Chemistry is the reaction of an alkene with bromine (Br₂).

Conditions: none needed — the reaction happens rapidly at room temperature with no catalyst.

What happens: the C=C double bond opens, and one bromine atom adds to each carbon.

Equation (ethene + bromine):

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br

or, using molecular formulae:

C₂H₄ + Br₂ → C₂H₄Br₂

Product name: 1,2-dibromoethane — the '1,2' tells you a bromine atom is on carbon-1 and another on carbon-2.

The colour change you see. Bromine (and bromine water) is orange-brown. When you add it to an alkene, the Br₂ is used up in the addition reaction, so the orange-brown colour disappears. We say the bromine is DECOLOURISED (it turns from orange-brown to colourless).

This is the test for a C=C double bond (unsaturation):

• Alkene → bromine water is DECOLOURISED (orange-brown → colourless).
• Alkane → bromine water stays orange-brown (no reaction, no C=C to add across).

Mark-scheme keyword: write 'decolourised' or 'orange-brown to colourless' — NOT 'turns clear'. Both orange bromine water and plain water are clear (see-through), so 'clear' does not score the colour-change mark.

As well as bromine, you must recall two more addition reactions: with hydrogen and with steam. Both open the C=C in exactly the same way.

(1) Addition of hydrogen (hydrogenation) → an alkane.

• Reagent: hydrogen gas, H₂.
• Conditions: nickel (Ni) catalyst, temperature ~150 °C.
• Product: the corresponding alkane (saturated — no more C=C).

Equation (ethene + hydrogen):

CH₂=CH₂ + H₂ → CH₃CH₃ (Ni, ~150 °C)

So ethene → ethane. One H atom adds to each carbon of the C=C.

Use: hydrogenation turns liquid vegetable oils (which contain C=C bonds) into solid margarine by removing the double bonds.

(2) Addition of steam (hydration) → an alcohol.

• Reagent: steam (water as a gas, H₂O).
• Conditions: phosphoric acid (H₃PO₄) catalyst, temperature ~300 °C, pressure 60 atm.
• Product: an alcohol.

Equation (ethene + steam):

CH₂=CH₂ + H₂O → CH₃CH₂OH (H₃PO₄, ~300 °C, 60 atm)

So ethene + steam → ethanol. One H atom adds to one carbon and an –OH group adds to the other carbon. This is the industrial way to make ethanol from ethene.

Quick summary table.

Reagent	Conditions	Product type	Example product
Bromine (Br₂)	Room temperature	Dibromoalkane	1,2-dibromoethane
Hydrogen (H₂)	Nickel catalyst, ~150 °C	Alkane	Ethane
Steam (H₂O)	H₃PO₄ catalyst, ~300 °C, 60 atm	Alcohol	Ethanol

Notice the pattern: in every case the double bond opens and atoms add across it. Learn the reagent + conditions + product for each row.

Spec 4.25 asks you to write balanced equations for addition reactions. The good news: addition equations are easy to balance, because no atoms are lost.

The simple rule: the product contains every atom of the alkene PLUS every atom of the small molecule added.

Worked check — ethene + bromine.

• Left side: C₂H₄ has 2 C and 4 H; Br₂ has 2 Br.
• Right side: C₂H₄Br₂ has 2 C, 4 H and 2 Br. ✓ Balanced.

So: C₂H₄ + Br₂ → C₂H₄Br₂.

Using displayed (structural) formulae. Examiners often want you to show the C=C opening:

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br

This makes it clear that the double bond became a single bond and a Br atom joined each carbon.

The same works for any alkene. For example, with propene (C₃H₆):

• C₃H₆ + Br₂ → C₃H₆Br₂ (1,2-dibromopropane).
• C₃H₆ + H₂ → C₃H₈ (propane).
• C₃H₆ + H₂O → C₃H₇OH (propanol).

Top tips for full marks.

• State symbols are not usually required at this level, but write them if asked.
• For hydrogen and steam, add the conditions over the arrow (Ni 150 °C, or H₃PO₄ 300 °C 60 atm) if the question asks for conditions.
• Double-check by counting each type of atom on both sides — they must match.
• The product is one molecule (addition has a single product) — do not write a second product.