Now that you have been introduced to the terminology and fundamental ideas of reaction mechanisms, it is time to focus on the reactions of alkenes.
Alkenes react by electrophilic addition which means that the first step in the mechanism is the addition of an electrophile to the C=C double bond.
Electrophile: a positively charged ion or neutral molecule that accepts a pair of electrons from an electron donor to form a covalent bond such as carbocations (e.g. CH3CH2+) and molecules with a 𝛅+ atom (e.g. HBr, H2O)
The C=C double bond consists of a sigma bond and a weaker pi bond – the electrons from the pi bond can be deployed in an attack on an electrophile and be used to make a new covalent bond (sigma).
There reactivity of an alkene depends on its stability …
Let’s look at the reaction mechanism for electrophilic addition, starting with the reaction between propene and HBr.
Markownikov’s rule states that on the addition of HX (electrophile) to an alkene, the hydrogen bonds to the carbon atom with the fewest alkyl groups and the X bonds to the carbon with the most alkyl groups.
In the example above, this means we would always make 2-bromopropane, never 1-bromopropane.
The bad news is that you need to learn the reaction conditions for all these organic reactions, but the good news is you have the perfect excuse to make lots of flashcards 😆.
Practice questions
- When 2-methylbuta-1,3-diene reacts with excess hydrogen bromide, four isomeric products are formed. Draw the skeletal structures of these four molecules.
2. (a) Name the homologous series to which the following molecule belongs, and give the general formula for this series.
(b) What reagents and conditions are needed to saturate this molecule?
(c) Complete the flowchart shown below, filling in either the reagents and conditions for reaction (i) and the missing products A and B:
3. (a) State the reagents and conditions needed to complete this reaction:
(b) Would you describe this reaction as hydrolysis, oxidation, reduction or substitution?
(c) Name another commercially important process that involves this type of reaction.
4. Consider the reactions of 1,1-dichloroethene as shown below:
- State the reagents and conditions needed for reaction 2. Name the product of this reaction.
- Draw out a full mechanism for reaction 1 given that the electrophile is an oxonium ion, H3O+. Include relevant charges, dipoles and curly arrows.
5. 3-methylpent-2-ene reacts with hydrogen bromide to form a mixture of 3-bromo-3-methylpentane and 2-bromo-3-methylpentane. Name and outline the mechanism for the formation of 3-bromo-3-methylpentane. Explain why 3-bromo-3-methylpentane is the major product of this reaction.
6. (a) Explain how Markownikoff’s rule can be used to predict the products of a reaction between an alkene such as propene and an electrophile such as hydrogen bromide.
(b) Write an equation using displayed formulae to show the reaction between propene and an aqueous bromine solution.
(c) Propene also reacts with steam to form an alcohol. State a suitable catalyst and draw the structure of the main product.
Answers
2. (a) alkene; CnH2n
(b) hydrogen gas; finely divided nickel catalyst at 150° and 5 atm / platinum catalyst at 1 atm.
(c) i = concentrated sulphuric acid;
3. (a) hydrogen gas; finely divided nickel catalyst at 150° and 5 atm / platinum catalyst at 1 atm
(b) this is a reduction reaction
4. (a) bromine / organic solvent (CCl4) ; 1,2-dibromo-1,1-dichloroethane (must name alphabetically, not numerically)
(b)
5. 3-bromo-3-methylpentane is the preferred product because of Markownikoff’s rule. The carbocation intermediate is more stable as the positive charge is stabilised by the
electron donating (positive induction) effect of three alkyl groups. In the case of 2-bromo-3-methylpentane, the intermediate carbocation would have the positive charge on the second carbon atom in the chain, and stabilised by only two ally groups.
6. (a) Markownikoff’s rule states that in the case of HBr adding to propene, the H will bond to the C of the C=C double bond which has two H atoms already bonded to it and the Br will bond to the C which has a H atom and a methyl group already bonded to it.
You can see the full mechanism for the reaction between aqueous bromine with an alkene here 👈
