Addition Reactions
The π bond of a double bond is weak, so it is easily broken. This allows alkenes to
undergo addition reactions. Because an alkene is a nucleophile, the first species it reacts with
is an electrophile.
When an alkene undergoes an electrophilic addition reaction with HBr, the first step is
a relatively slow addition of a proton (an electrophile) to the alkene (a nucleophile). A
carbocation intermediate (an electrophile) is formed, which then reacts rapidly with a
bromide ion (a nucleophile) to form an alkyl halide. Notice that each step involves the reaction
of an electrophile with a nucleophile. The overall reaction is the addition of an electrophile to
one of the sp2 carbons of the alkene and the addition of a nucleophile to the other sp2 carbon.
Examples:
The alkenes in the above reactions have the same substituents on both sp2 carbons, it is
easy to predict the product of the reaction: the electrophile (H+) adds to either one of the sp2
carbons, and the nucleophile adds to the other sp2 carbon. It does not matter which sp2 carbon
the electrophile adds to because the same product is obtained in either case.
But what happens if the alkene does not have the same substituents on both sp2 carbons?
Markovnikov’s rule
Markovnikov rule describes the regiochemistry where ‘The electrophile (H+) adds
preferentially to the sp2 carbon bonded to the most hydrogens’
In the above example, electrophile (in this case, H+) adds preferentially to C-1 because it is
the sp2 carbon bonded to the most hydrogens. Or we can say that H+ adds to C-1 to form a
secondary carbocation, which is more stable than the primary carbocation that would be
formed if H+ added to C-2.
When H+ adds to 1-pentene, a secondary and a primary carbocation could be formed. The
primary carbocation is unstable, hence not formed. Thus, 2-bromopentane is the only product
of the reaction which is formed via secondary carbocation.
Anti-Markovnikov rule
Anti-Markovnikov rule describes the regiochemistry where ‘The electrophile (H+)
adds preferentially to the sp2 carbon bonded to the least hydrogens’
Examples of Anti-Markovnikov includes Hydroboration-Oxidation and Radical Addition of
HBr.
Anti-Markovnikov radical addition of haloalkane can only happen to HBr and there
must be presence of Hydrogen Peroxide (H2O2). Hydrogen Peroxide is essential for this
process, as it is the chemical which starts off the chain reaction in the initiation step.
Initiation Steps
Hydrogen Peroxide is an unstable molecule, if we heat it or irradiate with sunlight, two free
radicals of OH will be formed. These OH radicals will go on and attack HBr, which will take
the Hydrogen and create a Bromine radical. Hydrogen radical do not form as they tend to be
extremely unstable with only one electron, thus bromine radical which is more stable will be
readily formed.
Propagation Steps
The Bromine Radical will go on and attack the LESS SUBSTITUTED carbon of the alkene.
More substituted carbon is more stable due to induction and hyperconjugation. After a carbon
radical is formed, it will go on and attack the hydrogen of a HBr.
Another example for anti-Markovnikov addition is hydroboration-oxidation,
The Addition of a Halogen to An Alkene
The halogens Br2 and Cl2 add to alkenes. The product of the reaction is a vicinal dihalide.
Mechanism for the addition of bromine to an alkene
■ As the π electrons of the alkene approach a molecule of Br2, one of the bromines accepts
those electrons and releases the shared electrons to the other bromine, which leaves as a
bromide ion. Because bromine’s electron cloud is close enough to the other sp2 carbon to form
a bond, a cyclic bromonium ion intermediate is formed rather than a carbocation intermediate.
■ The cyclic bromonium ion intermediate is unstable because of the strain in the three
membered ring and the positively charged bromine, which withdraws electrons strongly from
the ring carbons. Therefore, the cyclic bromonium ion reacts rapidly with a nucleophile (Br -).
Note: The addition of Br2 is anti because the two bromine atoms add to opposite sides of the
double bond.
Addition of a Hydrogen Halide to an Alkyne
The product of the electrophilic addition reaction of an alkyne with HCl is an alkene.
Therefore, a second addition reaction can occur if excess hydrogen halide is present. The
second addition—like other alkene addition reactions—is regioselective: the H+ adds to the
less substituted sp2 carbon.
The product of the second addition reaction is a geminal dihalide, a molecule with two
halogens on the same carbon. “Geminal” comes from geminus, which is Latin for “twin.”
The addition of a hydrogen halide to an alkyne can be stopped after the addition of one
equivalent of hydrogen halide because an alkyne is more reactive than the halo-substituted
alkene that is the reactant for the second addition reaction.
Mechanism for electrophilic addition of a hydrogen halide to an alkyne
■ The alkyne (a nucleophile) reacts with an electrophile to form a π-complex.
■ Chloride ion adds to the π-complex, forming a halo-substituted alkene.
Addition of a Halogen to an Alkyne
The halogens Cl2 and Br2 also add to alkynes. In the presence of excess halogen, a second
addition reaction occurs. The mechanism of the reaction is exactly the same as the mechanism
for the addition of Cl2 or Br2 to an alkene.
Addition of water
Alkene reacts with water to give alcohol.
Alkynes also undergo the acid-catalyzed addition of water. As expected, the electrophile (H+)
adds to the less substituted sp carbon.
The initial product of the reaction is an enol. An enol has a carbon–carbon double bond with
an OH group bonded to one of the sp2 carbons. (The suffix “ene” signifies the double bond,
and “ol” signifies the OH group.) The enol immediately rearranges to a ketone. A ketone and
its corresponding enol are called keto–enol tautomers. (Tautomers are constitutional isomers that
are in rapid equilibrium.)
Examples:
