Alkenes & Markownikoff’s Rule
Addition of Unsymmetrical Electrophiles to Unsymmetrical Alkenes
Unsymmetrical electrophiles
eg
HBr compared to Br2 (Br – Br).
Unsymmetrical Alkenes
eg
CH2 = CH – CH2 – CH3 compared to CH3 – CH = CH – CH3
When HBr attacks propene, there are two possible reaction pathways: -
Route ‘A’ results in the formation of a secondary carbocation intermediate and is preferred to route ‘B’, which
involves the formation of a primary carbocation intermediate. Secondary carbocations are more stable than
primary because secondary ions have two alkyl groups bonded to the carbon atom carrying the positive charge.
Alkyl groups have a +I inductive effect and push electrons from the C – C bond towards the + carbon atom,
reducing the + charge on the carbon atom, so tending to spread the + charge throughout the molecule. Two alkyl
groups directly bonded to the + carbon atom are more effective than one alkyl group even though it may be a
larger alkyl group, as it is in the case of the primary carbocation. This “spreading out” of the positive charge
stabilises the carbocation. Concentrated positive charge on an atom that is not particularly electropositive
makes the ion very susceptible to nucleophilic attack or very unstable and unlikely to be formed.
When 2 – methylpropene reacts with HBr, only one product results.
Route ‘A’ involves the formation of a tertiary carbocation, which is far more stable than the primary carbocation
involved in route ‘B’. Because there is a large difference in the relative stabilities of the carbocation
intermediates, only one product is formed.
Product mixtures result when carbocations of comparative stability are formed. 1 – methylcyclohexene forms
only one product with HBr and that is 1–bromo–1–methylcyclohexane. Equal amounts of
2 –
bromopentane and 3 – bromopentane are formed when HBr reacts with pent – 2 – ene. You should verify these
last two statements for yourself.
A rule that will help to remember which way an electrophile adds across a double bond is Markownikoff’s rule. It
must be borne in mind that the rule is only an aid to memory and does not serve as an explanation of why the
addition product is one molecule compared to another possible molecule.
Markownikoff’s Rule
When HX adds across a double bond of an alkene, the + H atom bonds onto the
least substituted carbon atom (the one with the most H atoms bonded to it) whilst the  X atom bonds to the
most substituted carbon atom (the one with the least number of H atoms bonded to it).
Practical Details
The addition of a hydrogen halide to an alkene is carried out by bubbling the dry gaseous hydrogen halide directly
into the alkene. A moderately polar solvent such as ethanoic acid can be used which will dissolve both the
hydrogen halide and the alkene.
The DRY gaseous hydrogen halide has to be used because the addition of an AQUEOUS solution of the hydrogen
halide results in the formation of the alcohol as well as the desired haloalkane. This is because after very slow
electrophilic attack by the H+ ion, very rapid nucleophilic attack on the resulting carbocation by water molecules
follows. There are very many more water molecules present in aqueous solution than there are Br  ions present.
Reaction of Alkenes with Cold Concentrated Sulphuric Acid
If the gaseous alkene is bubbled through cold, concentrated sulphuric acid, (liquid alkenes are added slowly,
dropwise and mixed) alkyl hydrogensulphates are formed. The mechanism is exactly the same as in the case of
attack by HBr. The carbocation intermediate formed by the initial electrophilic attack by + H is followed by
nucleophilic attack.
If this reaction is then followed by dilution of the product with water, or if dilute sulphuric acid is used in the
first instance, propan–2–ol is formed. Either, the alkyl hydrogensulphate undergoes nucleophilic substitution by
attack by water, the HSO4 nucleophile being replaced by H2O:, which then loses a proton. Or, in the case of
reaction of the alkene with dilute H2SO4, water acts as a nucleophile after the initial protonation of the alkene.
The reaction is a catalytic electrophilic hydration, as the mechanism shows.
The mechanism shows the catalytic hydration of 2–methylpropene to produce 2–methylpropan–2–ol, which is of
industrial importance.
Remember, ethanol is made by the direct hydrolysis of ethene by mixing ethene with steam at 300 oC and 65
atmospheres pressure and passing the mixture over a phosphoric (V) acid on Celite catalyst. (See Page 2)
Epoxyethane
Epoxyethane is important in that it is used to make important derivatives, such as ethane–1,2–diol (antifreeze),
non-ionic surfactants (detergents) and polyesters such as Terylene.
Ethene is mixed with air and pressurised to about 10 atmospheres (1-2 Mpa). It is then passed over a catalyst of
silver present as a finely divided layer on an alumina base at about 300°C.
O
Ag
2 CH2 = CH2
+
O2
catalyst
2 CH2 – CH2
H =  210 kJ mol1
About 80% conversion takes place because some ethene is lost by oxidation to CO 2 and H2O. Great care must be
taken during manufacture because epoxyethane is both flammable and explosive and the reaction must be
carefully monitored because it is exothermic and the danger of explosion increases with increased temperature.
An efficient system of heat exchangers is used and problems with static electricity and “hot spots” on the
catalyst surface have to be avoided. It is also very toxic and affects the respiratory system and the brain. The
gas is an excellent sterilising agent. The epoxyethane is a gas with a b.pt. of 10 oC and is purified by fractional
distillation.
Epoxyethane is very reactive towards nucleophiles because of its very strained ring system. With bond angles of
60 o, it doesn’t take much to open up the ring and bring about an exothermic reaction.
Water
80% of all epoxyethane produced is converted into ethane – 1,2 – diol or antifreeze. Epoxyethane is mixed with
water in a ratio of 1:10 in molar quantities, at 60 oC in the presence of sulphuric acid, which acts as a catalyst.
The resulting solution is concentrated to 70% ethane–1,2– diol by evaporation and then further purified by
fractional distillation.
O
CH2 – CH2
+ H 2O
HOCH2CH2OH
H =  80 kJ mol1
First, the oxygen atom of the three-membered ring is protonated by the acid catalyst.
The large excess of water is to try to ensure as high a yield as possible, but still only 90% conversion is achieved.
The main side reaction or bi – product is HOCH2CH2OCH2CH2OH. This is formed by some of the ethane–1,2– diol
product reacting with epoxyethane reactant.
If the proportion of water is lowered in the hydration of epoxyethane, more complex polymeric products are
formed in a stepwise series of reactions.
These products are called “polyethylene glycols” and are used in the manufacture of plasticisers, polyurethanes
and polyester resins, and which use they are put to depends on the value of ‘n’ in the equation. They are also used
as solvents and lubricants.
Ethane–1,2–diol, or glycol itself is used as an antifreeze in internal combustion engines. It has a low m.pt. of
12oC and a high b.pt. of 198 oC and it is completely miscible with water. A 60% mixture with water freezes 
40oC.
Ethane–1,2–diol is also used to make Terylene, a polymer fibre used in the clothing industry.