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Reactions of alkenes Sem 1: 2012/2013. Khadijah Hanim Abdul Rahman PTT 102: Organic Chemistry PPK Bioproses , UniMAP Week 4 : 2/10/2012. Learning Outcomes. Reaction of alkenes: addition reactions

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reactions of alkenes sem 1 2012 2013

Reactions of alkenesSem 1: 2012/2013

KhadijahHanim Abdul Rahman

PTT 102: Organic Chemistry

PPK Bioproses, UniMAP

Week 4: 2/10/2012

learning outcomes
Learning Outcomes
  • 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
reactions of alkenes
Reactions of alkenes
  • 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.
addition of a hydrogen halide to an alkene
Addition of a hydrogen halide to an alkene
  • 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.
what happen if alkene does not have the same substituents
What happen- if alkene does not have the same substituents?

- 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
slide6
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.
slide7
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.
slide8
Since the only product formed is tert-butyl chloride- tert-butyl cation is formed faster than isobutyl cation.
why is the tert butyl cation formed faster
Why is the tert-butyl cation formed faster?
  • 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
slide10
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.
slide11
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-butycation.
how do alkyl groups decrease the concentration of ve charge on the carbon
How do alkyl groups decrease the concentration of +ve charge on the Carbon?
  • 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.
exercise
Exercise
  • Which of the following is the most stable carbocation?
electrophilic addition reactions are regioselective
Electrophilic addition reactions are regioselective
  • When alkene has different substituentson its sp2 carbons, undergoes electrophilic addition reaction- the electrophile can add to 2 different sp2 carbons- result in the formation of more stable carbocation.
slide15
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
slide16
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

Br

2-bromopentene

50%

3-bromopentene

50%

slide17
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.
example
Example
  • What alkene should be used to synthesize 3-bromohexane?

? + H-Br  CH3CH2CHCH2CH2CH3

Solution:

  • 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

Br

slide19
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 2-bromohexane.
  • 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 3-bromohexane.
  • Therefore, 3-hexene is the best alkene to use to prepare 3-bromohexane.
exercise1
Exercise
  • What alkene should be used to synthesize 2-bromopentane?
a carbocation will rearrange if it can form a more stable carbocation
A carbocation will rearrange if it can form a more stable carbocation
  • Sometimes, in the electrophilic addition reactions, the products obtained are not as expected.
  • For eg: the addition of HBr to 3-methyl-1-butene forms 2 products.
  • 2-bromo-3-methyl butane (minor product)-predicted
  • 2-bromo-2-methylbutane- unexpected product- major product
slide22
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.
slide23
Result of the carbocation rearrangement- 2 alkyl halides are formed
  • 1 from the addition of the nucleophile to the unrearrangecarbocation 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.
slide24
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.
slide25
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.
slide26
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.
exercise2
exERCISE
  • Give the major product obtained from the reaction of HBr with:
  • 1-methylcyclohexene
  • 3-methylcyclohexene
the addition of a halogen to an alkene
The addition of a halogen to an alkene
  • 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.
mechanism for the addition of bromine to an alkene
Mechanism for the addition of bromine to an alkene
  • 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
slide30
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.
halohydrin formation
Halohydrin formation
  • Cl2 adds to an alkene- a cyclic chloronium ion is formed.
  • Final product- vicinal dichloride
  • If H2O rather than CH2Cl2 is used as solvent- major product will be a vicinal halohydrin
  • Halohydrin- organic molecule that contains both halogen and an OH group.

The same reaction with chlorine affords a chloronium ion:

mechanism for halohydrin formation
Mechanism for halohydrin formation
  • 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 protonatedhalohydrin is strong acid- so it loses proton.
how to explain regioselectivity in the reaction
How to explain regioselectivity in the reaction?
  • 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.
slide34
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.
slide35
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.
example1
Example
  • Complete the following reaction and provide a detailed, step-by-step mechanism for the process.
  • Answer:
biological alkene limonene
biological alkene-Limonene
  • 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 botanicalinsecticide
slide38
Extraction of limonene from orange peels using petroleum ether- essential oil
  • Purify the essential oils using distillation
  • Characterize the limonene using FTIR.
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