Reactions of alkenes - Organic Chemistry PDF

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Reactions of alkenes 

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Steps for HBr alkene addition 1. First step is the rate determining step. Slow addition of electrophilic proton to nucleophilic alkene to form carbocation intermediate 2. 2nd step: positively charged carbocation intermediate (electrophile) reacts rapidly with negatively charged bromide ion (nucleophile) What alkenes have in common? Loosely held pi electrons are attracted to electrophile. Each rxn starts with addition of electrophile to one of the sp2 carbons of alkene and ends with addition of nucleophile to other sp2 carbon. If alkene has different substituents, the one that is formed faster is preferred. 1. All alkene electrophilic addition reactions: the electrophile adds to the sp2 carbon bonded to the greater number of hydrogens. o What happens when no stereocentre is formed? Only one product is formed since stereochemistry is irrelevant. o What happens when one stereocenter is formed? Stereochemistry is still irrelevant. In this case, 2 products are formed instead of 4. o When is stereochemistry relevant? When 2 stereocenters are formed. o What is stereospecificity? relates to whether the reaction is anti or syn. If neither, then the reaction is not stereospecific.

Carbocation Stability 

How is the stability of the carbocation increased? Stability of carbocation increases as the alkyl substituents bonded to the positively charged carbon increases because the alkyl substituents decrease the concentration of the positive charge on the carbon. Tertiary carbocation are therefore more stable than secondary and so on

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Why is there a positive charge on carbon in carbocation? Positive charge on carbon signifies an empty p orbital so movement of electrons from sigma bond orbital toward vacant p orbital of ethyl cation decreases charge on sp2 carbon and causes partial positive charge to develop on carbon bonded by sigma bond. This stabilizes carbocation since the charge is no longer localised on the carbon and spread out over a larger volume. This dispersion of charge stabilizes the carbocation.

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What is hyperconjugation? Delocalization of electrons by the overlap of the sigma bond with the vacant p orbital if in proper orientation (achieved since sigma bonds can freely rotate)

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What is resonance stabilization and what is its significance? Because of resonance stabilization, a primary allylic or benzylic carbocation is as stable as a secondary alkyl carbocation and secondary allylic or benzylic carbocation is as stable as a tertiary. (charge is spread over a larger volume which increases the stability of the carbocation) Tertiary allylic carbocation is more stable than tertiary carbocation.

Structure of transition state 

Is formation of carbocation intermediate endergonic or exergonic? o Formation of carbocation intermediate is an endergonic reaction, the structure of the transition state will more closely resemble the structure of the carbocation product. This means that the transition state will have a significant amount of positive charge on carbon so the same factors that stabilize carbocation intermediate stabilizes positively charged transition state.  In the case of an exergonic reaction, the structure of the transition state will more closely resemble the structure of the reactant. transition state more closely resembles the reactant than the product

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What determines rate of a reaction? o Rate of a reaction is determined by the free energy of activation which is difference between free energy of transition state and free energy of reactant so the more stable the transition state is, the smaller is the free energy of activation

Regioselectivity of electrophilic addition reactions

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What is regioselectivity? o Regioselectivity: preferential formation of one constitutional isomer over the other. o Degrees of regioselectivity: high, moderate or complete (one of the possible products may not be formed) o Addition of HBr to 2-pentene is not regioselective as addition of the proton to either sp2 carbon would form the same carbocation intermediate. o The 2 different products in electrophilic addition reactions are called constitutional isomers. A reaction in which 2 or more constitutional isomers are formed but one predominates is called a regioselective reaction. What determines which product will be the major product? o To determine, major product of an electrophilic addition rxn, we must know the relative stability of the carbocation intermediates. What is Markovnikov and Anti-Markovnikov rule? o Markovnikov’s Rule: Hydrogen adds to the sp2 carbon with greater number of hydrogens. o Anti-Markovnikov’s Rule: Hydrogen adds to the sp2 carbon with lesser number of hydrogens.

Addition of water and addition of alcohols 

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What happens when water is added to alkene by itself? o When water is added to alkene, no reaction takes place since there is no electrophile present. The O-H bonds in water are too strong and too weakly acidic to allow hydrogen to act as nucleophile. Reagents? o In presence of acid (H2SO4, HCl) provides an electrophile and the product will be an alcohol. (Hydration) o Protonated alcohols are very strong acids. Steps? o First 2 steps are the same as the electrophilic addition of hydrogen halide: acid-catalysed reaction in the presence of acid and water.

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Are both mechanisms the same? What about the products? o Mechanism of acid catalysed addition of alcohols is same as mechanism of acid catalysed addition of water. Addition of alcohols also requires an acid catalyst. Why is there an arrow pointing forward and backward (equilibrium)? By carefully controlling the amount of water present, we can favor one side of the eq over the other

If it is dilute acid, the forward reaction is favoured while if it's concentrated, then the opposite. 

What about the stereochemistry? The stereochemistry isnt in the syllabus but if one stereocenter was formed, then 2 enantiomers will be formed (racemic mixture)

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What do we do to favour the alcohol being produced? Use dilute acid.

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What do we do to favour the alkene being produced? Use concentrated acid.

Rearrangement of carbocations 

What are the types of rearrangements? Practice them. o Some electrophilic addition reactions don’t give expected product. Unexpected product results from rearrangement of carbocation intermediate to a more stable carbocation intermediate. o 3 types of rearrangements: 1,2-methyl shift 1,2-hydride shift and ring expansion

Addition of halogens (alkene halogenation) 

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What happens in the first step? o Cyclic bromonium ion forms in the first step, no carbocations. No carbocation rearrangements occur here. Reagents? o Reactions of alkenes with Br2 and Cl2 are carried out by mixing the alkene and the halogen in an inert solvent such as dichloromethane (CH2Cl2) that readily dissolves both reactants and does not participate in the reaction. Anti addition! What is the regiochemistry here? Markovnikov addition since it forms the more stable carbocation

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What is the stereochemistry here? Stereochemistry is irrelevant.

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What happens if a different solvent is used? o If water is used as solvent instead of dichloromethane, vicinal halohydrin would be the major product. What is a halohydrin?

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o Halohydrin: organic molecule that contains both OH and the halogen (Markovnikov and anti addition) Why is the stereochemistry anti? First, the alkene possesses a pi bond which is a region of high electron density so it acts as the nucleophile. The bond between the 2 bromine atoms is a covalent bond. When Br2 molecule approaches alkene, electron density of the pi bond repels electron density in Br2 molecule creating temporary dipole moment in Br2.

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Practice the mechanisms and post here for addition of halogen and halohydrin formation o Mechanism for halohydrin formation Mechanism? The alkene attacks Br2 to form bridged, bromonium intermediate. In case we are adding the halogen in presence of dichloromethane, the nucleophile would be the pi bond and the electrophile is the bromine. In case we are forming a halohydrin, there would be 2 nucleophiles. Water and bromide. Rather than bromide attacking the bromonium ion, water would instead since the bromonium ion is a high energy intermediate and is very reactive.

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Why does OH add to the more substituted carbon in formation of halohydrin? The more substituted carbon (tertiary carbon) can best handle the carbocationic character. Therefore, the more substituted carbon will possess more partially positive charge than the less substituted. This means that the tertiary carbon atom will be close in character to sp2 hybridised carbon atom. The geometry of the tertiary carbon atom will be somewhere between trigonal planar and tetrahedral. Since the geometry is close to trigonal planar, SN2 can occur for back side attack of water.

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Can dihalide and halohydrin be formed together? Yes but halohydrin would be the major product since water is in excess.

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What determines the regiochemistry? The regiochemistry will be determined by the preference for the reaction to proceed via the the more stable carbocation,

Oxymercuration-Reduction and Alkoxymercuration-Reduction 

How is this different from acid-catalysed addition of water and alcohol? o Does not require acidic conditions and does not form carbocation intermediates so no rearrangements are required.

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o Water adds to an alkene by acid catalysed reaction (industrially) but in lab conditions, water is added to an alkene by oxymercuration-reduction which yields the same product. Reagents? o Alkene is treated with mercuric acetate in aqueous tetrahydrofuran (THF) and then sodium borohydride (NaBH4) - converts C-Hg to C-H (reduction) o Addition of alcohol works better in presence of mercuric acetate (mercuric trifluoroacetate) which is alkoxymercuration-reduction and forms ether. What happens if a chiral centre is formed in the mercurinium ion? o If the mercurinium ion has a chiral center, then a pair of enantiomers can be obtained.

Addition of borane: Hydroboration-oxidation 

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Which is the electrophile? o Molecule does not have to have a positive charge to be an electrophile. BH3 is a neutral molecule but boron has only 6 shared electrons in its valence shell and readily accepts a pair of e- to complete octet. What is the regiochemistry? o Regiospecific with Br adding to the less substituted side of the double bond. o Regioselectivity: boron adds to sp2 carbon with higher number of hydrogens since it forms the more stable carbocation-like transition state. Why does the boron add to the less substituted carbon? BH2 is bigger and bulkier than H so it will have an easier time fitting over less substituted carbon (less sterically hindered position)

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What is the stereochemistry? o Stereoselective because addition of H and BH2 is concerted (syn addition) Reagents o Alkenes undergo electrophilic addition reactions with borane as electrophile. Nucleophile is hydrogen. When reaction is over, aqueous sodium hydroxide and hydrogen peroxide are added to form an alcohol. o Reagent used as source of BH3 in the 1st step of hydroboration-oxidation is borane-THF complex. o H2O2 and NaOH used as oxidizing agents. Does it take place in one step? o Addition of electrophilic boron and nucleophilic hydride ion takes place in one step so no intermediates. (concerted reaction and pericyclic) Does boron have an octet? No. It has an empty p orbital. Therefore, it is very reactive and acts as the electrophile. It reacts with itself to give dimeric structures called diborane.

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How is the empty p orbital stabilized?

Stabilized by a solvent called tetrahydrofuran (THF) donating electron density into the empty p orbital. 

Mechanism? A pi bond first attacks the empty p orbital of boron triggering simultaneous hydride shift.

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What is hydroboration? It is the formation of the trialkylborane which occurs when you mix an alkene with BH3 in THF.

Addition of radicals (relative stabilities of radicals)  

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What is the regiochemistry? o anti-Markovnikov addition of HBr (Br is the electrophile) How are radicals stabilized? o Radicals are stabilized by electron donating alkyl groups (tertiary is most stable) o An alkyl peroxide is a radical initiator since it creates radicals Reagents? o Addition of HBr in presence of peroxides forms radical intermediate rather than carbocation so does not rearrange. Why does presence of peroxides cause anti-Markovnikov addition? The reaction follows a mechanism that involves radical intermediates rather than ionic intermediates. Peroxides are used to generate bromine radicals.

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Mechanism? O-O bond of peroxide is broken in presence of light (hv) or heat. The bond is broken homolytically forming 2 radicals. Each of these radicals can then abstract a hydrogen atom from HBr to form reactive intermediate, bromine radical. Carbon radical then abstracts a hydrogen atom from HBr to give final product.

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What is the difference between this and deprotonation? Here, an entire hydrogen atom is abstracted with its electron and proton.

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Why is the regiochemistry the way it is? The same reason, alkyl groups stabilize the radical intermediate just like carbocation intermediates and so tertiary radical rather than secondary. This is why the Br is added to the less substituted carbon.

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Which is faster? ionic pathway or radical pathway? Radical reaction. Regiochemistry can be controlled by choosing the reagents.

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When a bond breaks such that both of its electrons stay with one atom, heterolysis.

Addition of hydrogen (relative stability of alkene) 

Reagents? o In the presence of metal catalysts such as platinum or palladium, hydrogen adds to the double bond of alkene to form an alkane. (.) without a catalyst, the energy barrier would be huge since H-H bond is strong

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What is hydrogenation? o Hydrogenation: require a catalyst so called catalytic hydrogenation and catalysts are insoluble in the rxn so they are called heterogeneous catalysts (catalysts decrease the activation energy by breaking H-H bond) syn addition. o The more alkyl substituents bonded to the sp2 carbon of an alkene, the more stable it is (stabilizing effect) What is the mechanism? o Hydrogen is adsorbed to the surface of the metal and alkene complexes with the metal by overlapping its p orbitals with vacant orbitals of metal. What is more stable? Cis or trans alkene? Why? o Both trans-2-butene and cis-2-butene have 2 alkyl groups bonded to their sp2 carbons but trans has a smaller heat of hydrogenation since alkyl groups are further away from each other. (In cis, electron clouds can interact with each other causing strain and making it less stable - steric strain) Explain regiochemistry Regiochemistry is irrelevant since we are adding both of the same group

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Explain stereochemistry Syn addition. The reaction is stereospecific.

Epoxidation (addition of 2 OH, anti addition)  

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What is stereochemistry? o To add 2 OH groups in anti-addition, we first make an epoxide and then open the epoxide with H2O under conditions of acid catalysis. What is the first step? o In the first step, a peroxyacid (strong oxidizing agent) reacts with the alkene to form an epoxide. What is the most common peroxyacid?

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o Common peroxyacid is meta-chloroperbenzoic acid (MCPBA) Reagents? peroxy acid such as MCPBA and H3O+

Syn hydroxylation (addition of 2 OH, syn addition)  

What is the first step? o In the first step, osmium tetroxide (OsO4) adds across alkene in concerted process (both oxygens add to the alkene at the same time) and then H2O2 (peroxide) Reagents? OsO4 and then H2O2 (osmium tetroxide in first step and then hydrogen peroxide)

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What other reagents can be used here? o The same process can be accomplished using cold potassium permanganate and hydroxide. mechanism? Osmium tetroxide adds across alkene in concerted process which adds two groups across the same face of the alkene.

Oxidative cleavage of an alkene 

What is this reaction called? Ozonolysis.

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Reagents? O3 and then DMS

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Mechanism? Ozone breaks the double bond and it forms a cleavage.

Reaction of oxiranes 

Substitution reaction Na+ N2-, H2O/dioxane and 30 degrees

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Grignard reagents used to form alcohol CH3CH2MgBr, 25 degrees and dioxane

Synthesis 

What do we do for changing position of functional group? Eliminate and then add.

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What do we do for changing position of pi bond? Add and then eliminate.

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What do we do when starting material has no LG or double bond? RADICAL BROMiNATION (NBS, hv) and the bromine ends up at most substituted carbon....