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Reactions of alkenes Sem 1: 2011/2012

Reactions of alkenes Sem 1: 2011/2012. Khadijah Hanim Abdul Rahman PTT 102: Organic Chemistry PPK Bioproses , UniMAP Week 4: 6/10/2011. Learning Outcomes. Reaction of alkenes: addition reactions

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Reactions of alkenes Sem 1: 2011/2012

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  1. Reactions of alkenesSem 1: 2011/2012 KhadijahHanim Abdul Rahman PTT 102: Organic Chemistry PPK Bioproses, UniMAP Week 4: 6/10/2011

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

  11. 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.

  12. Exercise • Which of the following is the most stable carbocation?

  13. 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.

  14. 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

  15. 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%

  16. 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.

  17. 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

  18. 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.

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

  20. 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

  21. 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.

  22. 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.

  23. 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.

  24. 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.

  25. 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.

  26. example • Which of the following carbocations would be expected to rearrange?

  27. 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.

  28. 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

  29. 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.

  30. 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:

  31. 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.

  32. 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.

  33. 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.

  34. 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.

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

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