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

2. 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

3. 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.

4. 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.

5. 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

6. 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.

7. 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.

8. Since the only product formed is tert-butyl chloride- tert-butyl cation is formed faster than isobutyl cation.

9. 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

10. 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.

11. 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.

12. 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.

13. Exercise
Which of the following is the most stable carbocation?

14. 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.

15. 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

16. 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%

17. 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.

18. 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

19. 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.

20. Exercise
What alkene should be used to synthesize 2- bromopentane?

21. 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

22. 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.

23. 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.

24. 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.

25. 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.

26. 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.

27. exERCISE
Give the major product obtained from the reaction of HBr with:
1-methylcyclohexene
3-methylcyclohexene

28. 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.

29. 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

30. The product for 1st step: cyclic bromonium ion
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.

31. 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:

32. 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.

33. 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.

34. 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.

35. 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.

36. Example
Complete the following reaction and provide a detailed, step-by-step mechanism for the process.
Answer:

37. 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 botanical insecticide

38. Extraction of limonene from orange peels using petroleum ether- essential oil
Purify the essential oils using distillation
Characterize the limonene using FTIR.