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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

  • 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.

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

  • 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-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.


  • 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.

  • 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 the more stable tertiary 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.








  • Markovnikov’s the more stable tertiary 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 the more stable tertiary

  • What alkene should be used to synthesize 3-bromohexane?

    ? + H-Br  CH3CH2CHCH2CH2CH3


  • 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 the more stable tertiary + 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.

Exercise the more stable tertiary

  • What alkene should be used to synthesize 2-bromopentane?

A carbocation will rearrange if it can form a more stable carbocation
A the more stable tertiary 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

  • F.C. Whitmore- 1 the more stable tertiary st 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 the more stable tertiary 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.

  • In this reaction, after 3,3-dimethyl-1-butene acquires an the more stable tertiary 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 the more stable tertiary 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 the more stable tertiary 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.

exERCISE the more stable tertiary

  • 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 the more stable tertiary 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 the more stable tertiary 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

  • The product for 1 the more stable tertiary st 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 the more stable tertiary 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 the more stable tertiary 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 the more stable tertiary 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.

  • 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 achieved by adding the 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.

Example achieved by adding the

  • Complete the following reaction and provide a detailed, step-by-step mechanism for the process.

  • Answer:

Biological alkene limonene
biological achieved by adding the 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