Download Group B_reaction of alkenes

Khadijah Hanim Abdul Rahman
PTT 102: Organic Chemistry
PPK Bioproses, UniMAP
Week 4: 2/10/2012

-
-

-
Reaction of alkenes: addition reactions
DEFINE, REPEAT and APPLY electrophilic addition of
hydrogen halides which follows the Markonikov’s Rule
and the stability of carbocation according to
hyperconjugation theory.
DISCUSS regioselectivity of electrophilic addition
reaction
DISCUSS the carbocations rearrangements in hydrogen
halide addition to alkenes
Reaction of halogens to alkenes
DEFINE and REMEMBER the addition of halogen to
alkene and its mechanism
DISCUSS the concept of organic chemical process in
biotechnology industry



Alkenes are more reactive than alkanes due to the
presence of π bond.
The bond has high electron density or is electron rich site
and susceptible to be attacked by electrophiles (electron
deficient species/low electron density).
Alkenes undergo ADDITION reaction which means the C=C
are broken to form C-C bonds.

If the electrophilic reagent that adds to an alkene is a
hydrogen halide  the product of the reaction will be an
alkyl halide:

Alkenes in these reactions have the same substituents on
both sp2 carbons, it is easy to predict the product of the
reaction
The electrophile (H+) adds to 1 of the sp2 carbons, and the
nucleophile (X-) adds to the other sp2 carbon- doesn’t
matter to which C it will attach to- same product.

- Mechanism of reaction.
• Arrow shows- 2 electrons of the π bond of the alkene are attracted to
the partially charged H of HBr.
• π electrons of the alkene move toward the H, the H-Br bond breaks,
with Br keeping the bonding electrons
• notice that π electrons are pulled away from 1 C, but remain attached
to the other.
• thus, the 2 electrons that originally formed the π bond – form a new σ
bond between C and the H from HBr.
• the product is +vely charged since the sp2 C that did not form a bond
with H has lost a share in an electron pair.
• in 2nd step of reaction: a lone pair on the –vely charged bromide ion
forms a bond with the +vely charged C of the cabocation.
1st step of reaction: the addition of H+ to a sp2
carbon to form either tert-butyl cation or
isobutyl cation.
 Carbocation formation- rate-determining step
 If there is any difference in the formation of
these carbocations- the 1 that formed faster will
be the predominant product of the first step.
 Since carbocation formation-rate determining
step, carbocation that is formed in 1st step,
determines the final product of reaction.

Since the only product formed is tert-butyl chloride- tert-butyl cation is
formed faster than isobutyl cation.
 Factors
-
-
that affect the stability of
carbocations- depends on the no of alkyl
groups attached to the +vely charged carbon.
Carbocations are classified according to the
carbon that carries the +ve charge.
Primary carbocation- +ve charge on primary
C
Secondary carbocation- +ve charge on
secondary C
Tertiary carbocation- +ve charge on tertiary
C
 Thus,
the stability of carbocations increases
as the no of alkyl substituent attached to
+vely charged carbon increases.
 The
reason for decreasing stability: alkyl
groups bonded to the positively charged C
decrease the conc of +ve charge on the C.
 Decreasing conc of +ve charge makes the
carbocation more stable.
 Notice
that the blue (representing +ve
charge) in these electrostatic potential maps
is most intense for the least stable
carbocation (methyl cation).
 Least intense for the most stable tert-buty
cation.
• In ethyl cation: the orbital of an adjacent C-H σ bond can overlap the
empty p orbital (empty p orbital= positive charge on a C)
• Note: no such overlap is impossible for methyl cation.
• movement of electrons from the σ bond orbital toward the vacant p
orbital decreases the charge on the sp2 carbon- causes a partial positive
charge to develop on the atoms bonded by the σ bond
• Thus, the +ve charge is no longer concentrated on 1 atom but is
delocalized (spreading out).
• the dispersion of positive charge stabilizes the carbocation because a
charged species is more stable if its charge is spread out.
•Delocalization of electrons by overlap of a σ bond orbital with empty p
orbital on an adjacent carbon- hyperconjugation.
 Which
of the following is the most stable
carbocation?
 When
alkene has different substituents on
its sp2 carbons, undergoes electrophilic
addition reaction- the electrophile can add
to 2 different sp2 carbons- result in the
formation of more stable carbocation.
• in both cases, the major product- that results from forming the more
stable tertiary carbocation- it is formed more rapidly.
• the 2 products known as constitutional isomers – same molecular
formula, differ in how their atoms are connected.
• A reaction in which 2 or more constitutional isomers could be obtained
as products but 1 of them are predominates- regioselective reaction.
• 3 degrees of regioselective:
-Moderately regioselective
- highly regioselective
- completely regioselective

-
-

Completely regioselective:
1 of the possible products is not formed at all
For E.g: addition of a hydrogen halide to 2-methylpropane- 2
possible carbocations are tertiary and primary.
Addition of a hydrogen halide to 2-methyl-2-butene- 2 possible
carbocations are tertiary and secondary- closer in stability.
Addition of HBr to 2-pentene- not regioselective. Because the
addition oh H+ to either of the sp2 carbons produces a secondary
carbocation- same stability so both are formed with equal ease.
CH3CH=CHCH2CH3 + HBr  CH3CHCH2CH2CH3 + CH3CH2CHCH2CH3
Br
2-bromopentene
50%
Br
3-bromopentene
50%

Markovnikov’s rule: the electrophile adds to the sp2 carbon
that is bonded to the greater no of hydrogens.

In the above reaction, the electrophile (H+) adds
preferentially to C-1 because C-1- bonded to 2 H.

C-2 is not bonded to H.
Or we can say that: H+ adds to C-1 bacause it results
in the formation of secondary carbocation, which is
more stable than primary carbocation- would be
formed if H+ added to C-2.

 What
alkene should be used to synthesize 3bromohexane?
? + H-Br  CH3CH2CHCH2CH2CH3
Br
Solution:
1. List the potential alkenes that can be used to
produce 3-bromohexane.
Potential alkenes 2-hexene and 3-hexene
2. Since there are 2 possibilities- deciding whether
there is any advantage of using 1 over the
other
-
-
-
The addition of H+ to 2-hexene- form 2
different carbocations- both secondary, same
stability- equal amounts of each will be
formed. ½ 3-bromohexane and ½ 2bromohexane.
The addition of H+ to either of the sp2
carbons of 3-hexene- forms the same
carbocation because the alkene is
symmetrical. Thus, all product will be 3bromohexane.
Therefore, 3-hexene is the best alkene to use
to prepare 3-bromohexane.
 What
alkene should be used to synthesize 2bromopentane?
 Sometimes,
in the electrophilic addition
reactions, the products obtained are not as
expected.
 For eg: the addition of HBr to 3-methyl-1butene forms 2 products.
- 2-bromo-3-methyl butane (minor product)predicted
- 2-bromo-2-methylbutane- unexpected
product- major product
 F.C.
Whitmore- 1st to suggest that the
unexpected products results from a
rearrangement of the carbocation
intermediate.
 Carbocations rearrange if they become more
stable as a result of the rearrangement.
 Result
of the carbocation rearrangement- 2
alkyl halides are formed
 1 from the addition of the nucleophile to the
unrearrange carbocation and
 1 from the addition of the nucleophile to the
rearranged carbocation- major product.
 Because it entails the shifting of H with its
pair of electrons- the rearrangement is
called a hydride shift (1,2-hydride shift). The
hydride ion moves from 1 carbon to an
adjacent C.
• In this reaction, after 3,3-dimethyl-1-butene acquires an
electrophile to form a secondary carbocation, one of the methyl
groups, with its pair of electrons shifts to the adjacent +vely charged
C to form a stable tertiary carbocation.
• 1,2-methyl shift
•Major product- is the most stable carbocation.



If a rearrangement does not lead to a more stable
carbocation, then the rearrangement does not occur.
For eg: when a proton adds to 4-methyl-1-pentene, a
secondary carbocation is formed.
A 1,2-hydride shift would form a different secondary
carbocation- but since both carbocations are equally
stable-no advantage to the shift. Rearrangement does not
occur.
 Carbocation
rearrangements also can occur
by ring expansion- another type of 1,2-shift.
 Ring
expansion produces a carbocation that
is more stable because it is tertiary rather
than secondary- five-membered ring has less
angle strain than 4-membered ring.
 Give
the major product obtained from the
reaction of HBr with:
 1-methylcyclohexene
 3-methylcyclohexene
 The
halogens Br2 and Cl2 add to alkenes.
 It is not immediate apparent- electrophile
 Electrophile- necessary to start electrophilic
addition reaction
 Reaction is possible- the bond joining the 2
halogen atoms is relatively weak- easily
broken.
• As the electrons of the alkene approach a molecule of Br2, 1 of the Br
atoms accepts those electrons and releases the electrons of the Br-Br
bond to the other Br atom
• Br atom acts as nucleophile and electrophile- adds to the double
bond in a single step.
• the intermediate- unstable because there is considerable +ve charge
on the previously sp2 carbon.
•Thus, the cyclic brominium ion reacts with a nucleophile, the bromide
ion
• product is vicinal dibromide. Vicinus: near
• The product for 1st step: cyclic bromonium ion. NOT carbocation- Br
elecron cloud is close to the other sp2 carbon- form a bond.
• bromonium ion more stable- its atom have complete octets.
• positively charged carbon of carbocation – does not have complete
octet.
 Cl2
adds to an alkene- a cyclic chloronium ion
is formed.
 Final product- vicinal dichloride
The same reaction with chlorine
affords a chloronium ion:
If H2O rather than CH2Cl2 is used as solventmajor product will be a vicinal halohydrin
 Halohydrin- organic molecule that contains
both halogen and an OH group.

• Mechanism for halohydrin formation- 3 steps.
• 1st step: a cyclic bromonium ion/chloronium is formed in the 1st step
because Br+/Cl+ the only electrophile in the reaction mixture
• 2nd step: the unstable cyclic brominium ion rapidly reacts with any
nucleophile it bumps into. 2 nucleophiles present: H2O and Br-, but
because H2O is the solvent, its conc > Br-. Tendency to collide with H2O
is more.
• The protonated halohydrin is strong acid- so it loses proton.
Notice that in the preceding reaction, the
electrophile (Br+) end up on the sp2 carbon
bonded to the greater no of H. why?
 In the 2nd step of reaction: the C-Br bond has
broken to a greater extent than the C-O bond
has formed.
 As a result, there is a partial positive charge on
the carbon that is attacked by the nucleophile.

•Therefore, the more stable transition state is the 1 achieved by
adding the nucleophile to the more substituted sp2 carbon- carbon
bonded to fewer H.
• because, in this case the partial +ve charge is on a secondary carbon
rather than on a primary carbon.
• thus, this reaction too follows the general rule for electrophilic
addition reaction: the electrophile (Br+) adds to the sp2 carbon that is
bonded to the greater no of H.
 When
nucleophiles other than H2O are added
to the reaction mixture- change the product
of reaction, from vicinal dibromide to vicinal
bromohydrin
 Because, the concentration of the added
nucleophile will be greater than the conc of
halide ion (Br2/Cl2)- the added nucleophile
most likely to participate in the 2nd step
reaction.
 Complete
the following reaction and provide
a detailed, step-by-step mechanism for the
process.
 Answer:
Limonene is a colorless liquid hydrocarbon
Classified as cyclic terpenes.
 Strong smell of oranges.
 Use as a renewably based solvent in cleaning
products.
 Limonene takes its name from the lemon, as
the rind of the lemon, like other citrus fruits,
contains considerable amounts of this compound,
which contributes to their odor.
 Limonene is common in cosmetic products. As the
main odor constituent of citrus (plant
family Rutaceae), D-limonene is used in food
manufacturing and some medicines, e.g. as
a flavoring to mask the bitter taste of alkaloids, and
as a fragrant in perfumery; it is also used
as botanical insecticide

 Extraction
of limonene from orange peels
using petroleum ether- essential oil
 Purify the essential oils using distillation
 Characterize the limonene using FTIR.