Khadijah Hanim Abdul Rahman PTT 102: Organic Chemistry PPK Bioproses, UniMAP Week 4: 6/10/2011

 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

 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.

 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 sp 2 carbons, it is easy to predict the product of the reaction  The electrophile (H + ) adds to 1 of the sp 2 carbons, and the nucleophile (X - ) adds to the other sp 2 carbon- doesn’t matter to which C it will attach to- same product.

- 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 sp 2 C that did not form a bond with H has lost a share in an electron pair. in 2 nd step of reaction: a lone pair on the –vely charged bromide ion forms a bond with the +vely charged C of the cabocation.

 1 st step of reaction: the addition of H + to a sp 2 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 1 st 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.

 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!

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 sp 2 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?

 When alkene has different substituents on its sp 2 carbons, undergoes electrophilic addition reaction- the electrophile can add to 2 different sp 2 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. CH 3 CH=CHCH 2 CH 3 + HBr  CH 3 CHCH 2 CH 2 CH 3 + CH 3 CH 2 CHCH 2 CH 3 Br 2-bromopentene 50% 3-bromopentene 50%

 Markovnikov’s rule: the electrophile adds to the sp 2 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.

 What alkene should be used to synthesize 3- bromohexane? ? + H-Br  CH 3 CH 2 CHCH 2 CH 2 CH 3 Solution: 1. 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

- 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 sp 2 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.

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

 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 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 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.

 Which of the following carbocations would be expected to rearrange?

 The halogens Br 2 and Cl 2 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.

As the electrons of the alkene approach a molecule of Br 2, 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 sp 2 carbon. Thus, the cyclic brominium ion reacts with a nucleophile, the bromide ion product is vicinal dibromide. Vicinus: near

The product for 1 st step: cyclic bromonium ion. NOT carbocation- Br elecron cloud is close to the other sp 2 carbon- form a bond. bromonium ion more stable- its atom have complete octets. positively charged carbon of carbocation – does not have complete octet.

 Cl 2 adds to an alkene- a cyclic chloronium ion is formed.  Final product- vicinal dichloride  If H 2 O rather than CH 2 Cl 2 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- 3 steps. 1 st step: a cyclic bromonium ion/chloronium is formed in the 1 st step because Br+/Cl+ the only electrophile in the reaction mixture 2 nd 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.

 Notice that in the preceding reaction, the electrophile (Br + ) end up on the sp 2 carbon bonded to the greater no of H. why?  In the 2 nd 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 sp 2 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 sp 2 carbon that is bonded to the greater no of H.

 When nucleophiles other than H 2 O 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 (Br 2 /Cl 2 )- the added nucleophile most likely to participate in the 2 nd step reaction.

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