carbon carbon double bond as a substituents , markovnikov rule , anti markonikov rule ,

1. Alkenes as substituents
Prepared by
M.Maruthamuthu,M.pharm,
Assistant professor,
Department of pharmaceutical chemistry,
Svcp , Tiruchengode.

2.  In some cases, a group containing an
alkene may need to be treated as a
substituent.
 In these cases the substituent is named in
a similar fashion to simple alkyl
substituents.
 The method is required when the alkene
is not the priority group.
 The substituent is named in a similar way
to the parent alkene.
 It is named based on the number of
carbon atoms in the branch plus the suffix
- yl. i.e. alkenyl

3.  There are two common names that are widely used

4. CH3CH=CHCH(CH=CH2)2
 Functional group is a (tri) alkene, therefore suffix = -ene
 The longest continuous chain with 2 C=C is C6 therefore
root plus "a" = hexa
 There are two alkenes in the parent chain, so insert the
multiplier di
 Number from the right as drawn to give the C=C the
lowest locants : therefore 1- and 4-
 The substituent is a C2 alkenyl
group i.e. an ethenyl group
 The substituent is on C3
3-ethenylhexa-1,4-diene

5. Markovnikov rule
When an unsymmetrical reagent adds to an unsymmetrical
double bond, the positive part of the reagent becomes
attached to the double-bonded carbon atom which bears the
greatest number of hydrogen atoms.
Consider the addition of HBr to propene. The mechanism of
this reaction involves the following steps:
Step 1. Hydrogen bromide gives a proton (H) and a
bromide ion (Br)

6. Step 2. The proton (electrophile) attacks the 𝝅 bond of
propene to give a more stable carbonium
Remember: The tertiary (30) carbonium ion) carbonium ions
are more stable than Secondary (2) carbonium ions in turn are
more stable than primary (1) carbonium ions.

7. Step 3. The bromide ion (nucleophile) combines with the more
stable secondary carbonium ion to give the major product.
Morkovnikov rule may now be restated “ Addition of an
unsymmetrical reagent to an unsymmetrical double
bond proceeds in such a way as to involve the most
stable carbonium ion”

8. Halogenation of Alkenes
Reaction Overview: The alkene halogenation reaction,
specifically bromination or chlorination, is one in which a
dihalide such as Cl2 or Br2 is added to a molecule after breaking
the carbon to carbon double bond. The halides add to
neighboring carbons from opposite faces of the molecule. The
resulting product is a vicinal (neighboring) dihalide.

9. What Exactly Goes On In This Reaction?
the electrophile in this reaction is a neutral and nonpolar molecule. At first
glance the dihalide may not appear to be electrophilic. When the halogen
molecule is exposed to a nucleophile, its electrons are ‘scared away' in the
direction of the second attached halide. For this fraction of second, the
closest halogen has an induced partial positive charge and is considered to
be electrophilic.
Attack of the pi bond:
Nucleophilic electrons in a pi bond are strong enough to induce a temporary
polarity on a neutral halogen molecule such as Br2 or Cl2. The pi electrons will
then reach out to grab the temporarily partially positive halogen.

10. Alkene Halogenation Mechanism
 Nucleophilic pi bond attacks a halogen, pi bond breaks in
the process
 Halogen retaliates and ‘attacks back' resulting in a
halogen bridge, second halide breaks away
 Bridged halogen has a +1 charge due to 2 bonds and 2
lone pairs
 Second halides returns and attacks from the opposite side
of the molecule breaking the bridge
 Product formed is a vicinal dihalide

11. Key Reaction Notes
This reaction does NOT have a carbocation intermediate
Positive halogens are unhappy, unstable, and very reactive
Product is anti-addition with 1 halogen facing up, the other down
This reaction MUST take place in an inert solvent to avoid
unexpected products
This reaction is ideal for Cl2 and Br2. F2 and I2 are seldom used
The Cyclic Bromonium or Cyclic Chloronium:
Let's look back to our starting molecule. The carbon to carbon double
bond broke when the halogen got attacked. The 2 carbon atoms are
now each bound by a sigma (single) bond to the halogen. The
resulting intermediate is quite unstable with a 3-membered ring
composed of 2 carbon atoms and 1 halogen. To add insult to injury,
the halogen has a formal charge of +1 with 2 bonds, and just 2 lone
pairs.

12. The bridged intermediate is called a Bromonium when Br is
involved, and Chloronium when Cl is involved.
The Unhappy Positive Halide:
Since halogens are highly electronegative, they will be very
unstable or ‘unhappy' when stuck with a positive charge. The
positive bridged halogen will pull on both of the bonds that
connect it to carbon. This results in a slight concentration of
electrons around the halogen which helps balance that positive
charge. The attached carbon atoms each have a partial positive
charge as its electrons are pulled away.

13. Breaking the Bridge – Anti Addition
The second halide which broke away at the
start of this reaction is ready to attack. Despite
being happy with a complete octet, its
negative charge makes it highly nucleophilic.
As a nucleophile, the halogen is attracted to
partially positive carbon atoms in the bridged
molecule.
The resulting product has 2 halogens attached on opposite sides or ‘anti'
to each other. This happens when the incoming nucleophile is forced to
attack from the opposite face of the molecule when compared to the
bridge location.
If the halide attempts to attack from the same side of the molecule, the bridge
bonds will hinder or block its electrons from reaching the partially positive carbon
atom.

14. Instead, the halide must approach from the opposite side of the bridge
where it can use its lone pair of electrons to attack the carbon atom.
Since carbon can only have 4 bonds, this attack forces carbon to let go
of the bridge halogen, thus breaking the bridge. The unhappy halogen
was already pulling so hard on the bonding electrons that this bond is
considered easy-enough to break.
The resulting molecule is considered a vicinal dihalide given that the
2 halogen atoms are attached to neighboring carbons. The very
carbons that used to hold a pi bond.

15. A Note About The Solvent:
This reaction must be carried out in an inert solvent
such as CH2Cl2 and CCl4. An inert solvent serves a single
purpose – to dissolve the reactants. If this reaction is
carried out in a polar protic solvent such as water or
alcohol, the solvent molecules will ‘interfere' and take part
in the mechanism.

16. Addition of Radicals to Alkenes
in some early experiments in which peroxide contaminated reactants
were used, 1- bromobutane was the chief product. Further study
showed that an alternative radical chainreaction, initiated by
peroxides, was responsible for the anti-Markovnikov product. This
is shown by the following equations.

17. Introduction
This process was first explained by Morris Selig Karasch in his
paper: 'The Addition of Hydrogen Bromide to Allyl Bromide' in
1933.
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. HI and HCl cannot be used in radical reactions,
because in their radical reaction one of the radical reaction steps:
Initiation is Endothermic, this means the reaction is unfavorable.
To demonstrate the anti-Markovnikov regiochemistry

18. Initiation Steps
Hydrogen Peroxide is an unstable molecule, if we heat it,or shine it
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.

19. Propagation Steps
The Bromine Radical will go on and attack the LESS SUBSTITUTED carbon of
the alkene. This is because after the bromine radical attacked the alkene a carbon
radical will be formed. A carbon radical is more stable when it is at a more
substituted carbon due to induction and hyperconjugation. Thus, the radical will
be formed at the more substituted carbon, while the bromine is bonded to the less
substituted carbon. After a carbon radical is formed, it will go on and attack the
hydrogen of a HBr, which a bromine radical will be formed again.

20. Termination Steps
There are also Termination Steps, but we do not concern about
the termination steps as they are just the radicals combining to
create waste products. For example two bromine radical
combined to give bromine. This radical addition of bromine to
alkene by radical addition reaction will go on until all the
alkene turns into bromoalkane, and this process will take some
time to finish.

21. Sr.no Markovnikov Anti markovnikov
1. H of goes to double bonded carbon
atom which has greatest number of
hydrogen atom
Hof goes to double bonded carbon
atom which has least number of
hydrogen atom
2. Initially hydrogen goes to carbon to
carbocation
Initially halogen goes to carbon
atom which has greatest number of
hydrogen atom
3. Secondary carbocation intermediate
formed
free radical intermediate formed
4. Peroxide effect is not considered Peroxide effect is considered
5. Electrophilic addition reaction Free radical addition reaction

22. Substitution of Alkenes by Halogens (Allylic Substitution).
When alkenes are treated with Cl2 Br2, at high temperatures,
one of their allylic hydrogens is replaced by halogen atom. Allylic
position is the carbon adjacent to one of the unsaturated carbon
atoms.

23. MECHANISM
Allylic substitution takes place by a free-radical mechanism.
Following steps are involved
Step 1. Chain-Initiation Step. Chlorine molecule undergoes
homolytic fission to give chlorine
Step 2. Chain-Propagation Steps. (a) Chlorine free radical attacks
propene to give HCl and an ally free radical.

24. (b) The allyl free radical attacks chlorine molecule to produce allyl
chloride and a chlorine free radical
step(a)and (b) repeated over and over again
Step:3 chain termination steps:
The above Chain Reaction come to a halt when any two free
radicals combine

25. THE RESONANCE OF ALLYLIC SYSTEM
The resonance of allylic
systems results in the equal
lengths of the propenyl
radical's C-C bonds. Due to
the resonance, all four
terminal hydrogen atoms
display allylic character.
Only the hydrogen at the
central carbon is vinylic.

26. The specific characteristic of allylic systems is reflected in their
chemical reactions. Generally, at room temperature, chlorine and
bromine are added to the double bonds of alkenes. At higher
temperatures, however, the addition reaction is increasingly
superseded by the allylic substitution, especially when the halogen
concentration is low.
In such radical allylic substitutions, which are also used on a
large scale , a fair yield is obtained.
Eg. Shell process

27. 1. The allylic substitution is carried out at a high temperature in
the gas phase.
2. Under these reaction conditions, the chlorine molecule is
homolytically cleaved by thermal energy into two chlorine
radicals.
3. Principally speaking, the chlorine radicals may also be
attached to the double bond by a radical addition.
4. Nevertheless, due to the application of chlorine in a low
concentration, radical substitution exceeds the addition.
5. On a laboratory scale, selective halogenations - particularly
brominations - in allylic position are obtained by the
application of N-bromosuccinimide (NBS), which is a
continuous source of small amounts of bromine.

29. ORIENTATION AND REACTIVITY
All of the halides (HBr, HCl, HI, HF) can participate in this
reaction and add on in the same manner. Although different halides
do have different rates of reaction, due to the H-X bond getting
weaker as X (halogen size ) gets larger (poor overlap of orbitals)s.
VARIATION OF RATES WHEN YOU CHANGE THE
HALOGEN
Reaction rates increase in the order HF – HCl – HBr – HI.
Hydrogen fluoride reacts much more slowly than the other three,
and is normally ignored in talking about these reactions. When the
hydrogen halides react with alkenes, the hydrogen-halogen bond
has to be broken. The bond strength falls as you go from HF to HI,
and the hydrogen-fluorine bond is particularly strong. Because it is
difficult to break the bond between the hydrogen and the fluorine,
the addition of HF is bound to be slow.

30. Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to
symmetrical ones. For simplicity the examples given below are all
symmetrical ones- but they don’t have to be. Reaction rates increase
as the alkene gets more complicated – in the sense of the number of
alkyl groups (such as methyl groups) attached to the carbon atoms at
either end of the double bond.

31. There are two ways of looking at the reasons for
1. Alkenes react because the electrons in the pi bond attract things
with any degree of positive charge. Anything which increases
the electron density around the double bond will help this.
2. Alkyl groups have a tendency to “push” electrons away from
themselves towards the double bond.
3. The more alkyl groups you have, the more negative the area
around the double bonds becomes.
4. The more negatively charged that region becomes, the more it
will attract molecules like hydrogen chloride.
5. The more important reason, though, lies in the stability of the
intermediate ion formed during the reaction.

32. The three examples given above produce these carbocations
(carbonium ions) at the half-way stage of the reaction:
1. The stability of
the intermediate
ions governs the
activation energy
for the reaction.
2. As you go
towards the more
complicated
alkenes, the
activation energy
for the reaction
falls. That means
that the reactions
become faster

33. Markovnikov's rule allows us to predict the products of most
addition reactions, but constitutional isomers form in some reactions.
For example, the addition of HCl to 3-methyl-1-butene gives not
only the expected product , 2-chloro-3-methylbutane, but also 2-
chloro-2-methylbutane.
Allylic rearrangement

34. What is the origin of this second, apparently unexpected product?
The answer is:
the carbocation formed in the first step of the reaction
subsequently rearranges to a more stable species, which then
reacts with chloride.
The expected product forms from the reaction of the nucleophilic chloride ion
with the secondary cation that forms when a proton adds to the double bond
by Markovnikov addition.
1. The isomeric product forms when a nucleophilic chloride ion reacts with a
tertiary carbocation
2. This carbocation forms when the hydrogen atom at C-3 moves, with its
bonding pair of electrons, to the adjacent secondary carbocation center
3. This rearrangement is called a 1,2-hydride shift because a hydride ion
(H– ) moves between adjacent carbon atoms.

36. 1. Some of the secondary carbocation also reacts with chloride ion
without rearranging to give the expected product.
2. The relative amounts of rearranged and un rearranged products
depend on how efficiently the carbocation is captured by the
nucleophile versus the rate of the rearrangement process.
The reaction of 3,3-dimethyl-1-butene with HCl gives us an
example of this type of rearrangement.

37. Adding a proton to the less substituted carbon atom of the double
bond (Markovnikov’s rule again) gives a secondary carbocation.
This secondary carbocation reacts with the nucleophilic chloride ion
to give the expected product. The rearranged product forms by
reaction of chloride ion with a tertiary carbocation that forms by a
shift of a methyl group, with its bonding pair of electrons, from the
quaternary center to the adjacent secondary carbocation center. This
rearrangement is called a 1,2-methide shift because a CH3 unit
moves between adjacent carbon atoms.

38. Sample Solution
Write the structures of all of the possible addition products of
3,3-dimethylcyclohexene with HBr.
Adding a proton at C-2 gives a secondary carbocation. Capture of
this carbocation by bromide ion yields 1-bromo-3,3-
dimethylcyclohexane.

39. Adding a proton at C-1 gives a secondary carbocation at the original
C-2 atom. Capture of the carbocation by bromide ion can give 1-
bromo-2,2-dimethylcyclohexane.
However, the secondary carbocation can rearrange to two possible
tertiary carbocations. Migration of methyl followed by capture of
the carbocation gives 1-bromo-1,2-dimethylcyclohexane.

40. A 1,2-shift of a methylene group of the ring can also occur to give a
tertiary carbocation. Capture of the carbocation by bromide gives a
product containing a cyclopentane ring.
We can say, however, that when we consider carbocation
chemistry we should more or less "expect the unexpected."