Addition reactions of alkenes

What an addition reaction is

• In an addition reaction the C=C double bond opens and one atom or group adds to each of the two carbons.
• Nothing is lost — there is only one product, and it contains every atom from both reactants.
• The alkene becomes saturated.

This is why alkenes are the starting point of organic chemistry: four different families can be made from one.

The four additions you need

Reagent	Conditions	Product family	Example product from ethene
$H_2$	Pt catalyst	alkane	ethane
$Cl_2$ or $Br_2$	room temperature	dihaloalkane	1,2-dibromoethane
$H_2O$	conc. $H_2SO_4$, then water ($H^+$ catalyst)	alcohol	ethanol
$HX$ (HCl, HBr, HI)	room temperature	haloalkane	bromoethane

Adding hydrogen — hydrogenation

• Hydrogen adds across the double bond over a platinum catalyst, turning the alkene into an alkane.

$$CH_2{=}CH_2 + H_2 \xrightarrow{Pt} CH_3CH_3$$

• One H goes to each carbon.

Adding a halogen — halogenation

• Chlorine or bromine adds one halogen atom to each carbon, giving a dihaloalkane with the halogens on adjacent carbons.

$$CH_2{=}CH_2 + Br_2 \rightarrow CH_2BrCH_2Br$$

• The product is named 1,2-dibromoethane — note both locants and the di-.
• This is the reaction behind the bromine water test on the next page.

Adding water — hydration

• Water adds across the double bond with an acid catalyst, giving an alcohol.
• Conditions: concentrated sulfuric acid, then water.

$$CH_2{=}CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$$

• The H goes to one carbon and the –OH to the other.
• This is industrially how much ethanol is made.

Adding a hydrogen halide

• HX adds with the H on one carbon and the X on the other, giving a haloalkane.

$$CH_2{=}CH_2 + HBr \rightarrow CH_3CH_2Br$$

• With ethene it does not matter which carbon gets which, because both are the same.
• With an asymmetric alkene like propene, the two carbons are different, and you get a major and a minor product — that is the next page but one.

How to work out any addition product

• Find the C=C.
• Break the double bond down to a single bond.
• Split the reagent into its two parts ($H|H$, $Br|Br$, $H|OH$, $H|Br$).
• Attach one part to each of the two carbons.
• Check every carbon still has four bonds, then name the product.