REACTIONS OF ALKENES

SEM 1: 2011/2012Khadijah Hanim Abdul Rahman

PTT 102: Organic Chemistry

PPK Bioproses, UniMAP

Week 4: 6/10/2011

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 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 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?

- 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? 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.

Alkyl groups decrease the concentration of positive Charge- makes the carbocation more stable!

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 carbon)• Note: no such overlap is imposible 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 Which of the following is the most stable

carbocation?

ELECTROPHILIC ADDITION REACTIONS ARE REGIOSELECTIVE 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 Br2-bromopentene 50%

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.

EXAMPLE What alkene should be used to

synthesize 3-bromohexane? ? + H-Br CH3CH2CHCH2CH2CH3

Solution:1. List the potential alkenes that can be used

to produce 3-bromohexane. - Potential alkenes 2-hexene and 3-hexene2. Since there are 2 possibilities- deciding

whether there is any advantage of using 1 over the other

Br

- 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.

EXERCISE What alkene should be used to

synthesize 2-bromopentane?

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

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.

EXAMPLE Which of the following carbocations

would be expected to rearrange?

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

• 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.

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- 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.

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.

•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.

EXAMPLE Complete the following reaction and

provide a detailed, step-by-step mechanism for the process.

Answer: