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Organic Chemistry I Exam 2 Notes

by: Denice Arnold

Organic Chemistry I Exam 2 Notes Chem 241

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Denice Arnold

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These notes cover chapters 6, 7, 11, and half of 8 in Wade's Organic Chemistry textbook. This is all the information covered over Exam 2.
Organic Chemistry I
Dr. Winkler
orgo, Organic Chemistry, Chemistry, Chem241
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This 13 page Bundle was uploaded by Denice Arnold on Sunday January 24, 2016. The Bundle belongs to Chem 241 at University of Pennsylvania taught by Dr. Winkler in Fall 2015. Since its upload, it has received 42 views. For similar materials see Organic Chemistry I in Chemistry at University of Pennsylvania.


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Date Created: 01/24/16
Orgo Chapter 6 Saturday, October 24, 2015 5:54 PM Chapter 6 First-OrderElimination:E1 Alkyl Halides: Nucleophilic Substitution and Elimination (p. 218-272) Eliminationinvolves the loss of two atoms from the substrate, usually with the formation of a pi bond Alkyl halide- halogen atom bonded to one of the sp3 hybrid carbon atoms of an alkyl group Vinyl halide- halogen atom bonded to one of the sp2 hybrid carbon atoms of an alkene Aryl halide- halogen atom bonded to one of the sp2 hybrid atoms of anaromatic ring Serve as intermediates in the synthesis of many other functional groups Halogen - group VII atoms H with one electron - hydrogen H. H with two electrons - hydride H:- H with no electrons - proton H+ Naming Alkyl Halides Haloalkane name (IUPAC) - treats an alkyl halide as an alkane with a halo-substituent Common or "trivial" names - name the alkyl group and then the halide E1 (elimination, unimolecular) Unimolecular because the rate limitingtransition state involves a single molecule rather than a collisionbetween two molecules The slow step of an E1 reaction is the same as the Sn1 reaction: unimolecular ionization to form a carbocation The second step in E1 involves a base taking a proton from the carbon atom adjacent to the C+ The electrons that once formed the C--H bond now form a pi bond between two carbon atoms Methylene halide - CH2X2 (methylene group with two halogens) Haloform - CHX3 Carbon tetrahalide - CX4 Methyl halide - CH3X Primary halide - RCH2X Secondary halide - R2CHX Tertiary halide - R3CX Geminal dihalide - two halogen atoms bonded to the same carbon atom Vicinal dihalide - two halogen atoms bonded to adjacent carbon atoms Structureof Alkyl Halides Halogen atom bonded to an sp3 hybrid carbon atom Halogen is more electronegative than carbon and the C--X bond is polarized C--halogen bond lengths increases as the halogen atoms become bigger The greater the bond length, the greater the dipole moment Electronegativities of halogens increase from I to Br to Cl to F More electronegative, the greater the charge separation, the greater the dipole moment Larger halogens have longer bonds, but weaker electronegativities, so overall dipole moments: dipole moment (u) = 4.8 * charge separation (delta) * bond length (d) Difficultto predict molecular dipole moments, which are the vector sum of the individual bond dipole moments Depend on bond angles and other factors Symmetrical oriented polar bonds (CX4) cancel to give a molecular dipole of zero Competition with the Sn1 reaction Physical Propertiesof Alkyl Halides Whenever a carbocation is formed, it can undergo either substitution or elimination Dehydrohalogenation- an elimination of hydrogen and a halogen atom Boiling points Larger the surface area, larger the boiling point Occurs in the absence of a strong base Ionization of the alkyl halide gives a carbocation intermediate which loses a proton to give the alkene Larger london attractions For this reason, Sn1 and E1 reactions of alkyl halides are not often used for organic synthesis Higher molecular weights, higher the boiling point Slower moving and greater surface area Allylic Bromination Free radical halogenation tends to be plagued by mixtures of products, so not usually used in lab Allylic Bromination- free radical bromination of alkenes can be carried out in a highly selective manner Allylic- carbon next to a carbon-carbon double bond Allylic intermediates are stabilized by resonance with the double bond - delocalized charge Less energy required to form a primary allylic radical than a typical secondary radical even though typically primary radicals are less stable than secondary radicals Bromination is highly selective - only the most stable radical is formed If there is an allylic hydrogen, allylic radical is usually formed Potential energy diagram for E1 is similar to Sn1 Ionization step is strongly endothermic Second step is a fast exothermic deprotonation by a base Base is not involved in the reaction until after the rate limiting step, so rate depends only on the concentration of the alkyl halide Weak bases are common in E1 Rate-limiting step involves unimolecular ionization of the alkyl halide, so the rate equation is first order Rearrangements in E1 Solvent acts as a base in the E1 reaction (and a nucleophile in the Sn1 reaction) Basic solvent can remove either adjacent proton in step 3 Either of the resonances can react with Br to give products Second compound is said to be the product of anallylic shift N-Bromosuccinimide(NBS) often used as the bromine source because it combines with the HBr side product to regenerate a constant low concentration of bromine If there was too high a concentration of bromine, it can unfavorably add to the double bond NBS also works well for brominating benzylic positions, next to an aromatic ring Orgo Page 1 NBS also works well for brominating benzylic positions, next to an aromatic ring Preparing alkyl halides from alkanes, use free radical halogenation From other starting points, different reactions are used different cases Reactions of Alkyl Halides Halide is a good leaving group- halogen leaves with its bonding pair of electrons to form a stable halide ion When another atom replaces the halide ion, the reaction is asubstitution When the halide ion leaves with another atom or ion (often H+) and forms a new pi bond, the reaction is anelimination A molecule of H--X is often lost from the alkyl halide to give an alkene Called dehydrohalogenationsbecause a hydrogen halide was removed from the alkyl halide Substitution and elimination often compete with each other In a nucleophilicsubstitution,a nucleophilereplaces a leaving group using its long pair of electrons to form a new bond to the carbon atom CarbocationReactions A carbocation can: (React with its own leaving group to return to the reactant) R+ + :X- --> R--X 1. React with a nucleophile to form a substitution product In the elimination, the reagent B:- reacts as a base, taking a proton from the alkyl halide Sn1 Most nucleophiles are also basic and can engage in either substitution or elimination 2. Lose a proton to form an elimination product (an alkene) Other compounds can undergo substitution and elimination aside from alkyl halides E1 Both substitution and elimination have leaving groups; only substitutions have nucleophiles 3. Rearrange to form a more stable carbocation, then react further Order of stability of carbocations is: Second Order NucleophilicSubstitution:SN2 Resonance stabilized, tertiary > secondary > primary Hydroxide ion is a strong nucleophile(donor of an electron pair) because the oxygen atom has unshared pairs of electrons and a negative charge PositionOrientationof Elimination:Zaitsev's Rule Substrate- attacked by the reagent - electrophilic because it is bonded to an electronegative atom, which will act as the leaving group We can predict which elimination product will predominate of a mixture of alkenes Zaitsev's Rule- in elimination reactions, the most substituted alkene usually predominates Substrate gains a partial positive charge that draws in the nucleophile Whenever a carbocation is formed next to a more highly substituted carbon, consider whether rearrangement might occur Second-OrderElimination:E2 Nucleophile attacks the backside of the electrophilic carbon atom, donating a pair of electrons to form bond One-step mechanism supported by kinetic information Rate doubles when the concentration of either reactant is doubled (substrate or nucleophile) - first order with respect to E2 (elimination, bimolecular) each reactant and second order overall Eliminations take place under second order conditions in the presence of a strong base Most strong nucleophiles are also strong bases Elimination commonly occurs when a strong base/nucleophile is used with a poor Sn2 substrate such as a tertiary or hindered secondary alkyl halide In general, strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group One-step nucleophilic substitution - SN2 mechanism (substitution, nucleophilic, bimolecular) Transition state of the rate-limiting step involves the collisionof two molecules Transitionstate - point of maximum energy (not to be confused with an intermediate) Exists only for an instant Reactivity of order reflects the greater stability of highly substituted double bonds Elimination of a tertiary halide gives a more substituted alkene than a secondary halide Lower activation energies and higher rates for eliminations that lead to highly substituted alkenes If there are two or more possible products, Zaitsev's rule predicts which will be the major product: the most substituted alkene (usually applies in E2 reactions unless the base and/or LG is unusually bulky) Stereochemistryof the E2 reaction Halogen exchange reactions- one halide displaces another Concertedmechanism- bond breaking and bond formation take place at the same time Many alkyl chlorides react with sodium iodide to give alkyl iodides Partial formation of new bonds lowers the energy of the transitions state Require specific geometric arrangements so that the orbitals of the bonds being broken can overlap with those being Strength of the Nucleophile formed and the electrons can flow smoothly A "stronger" nucleophile is an ion or molecule that reacts faster in the SN2 reaction Species with a negative charge are stronger than a similar, neutral species For Sn2, you need a back side attack For E2, you need a coplanar arrangement A base is always a stronger nucleophile than its conjugate acid Formation of the new pi bond implies that the two sp3 orbitals must be parallel Basicity and nucleophilicity are different properties Anti-coplanar- dihedral angle of 180 degrees and orbitals aligned Basicity is defined by the equilibrium constant for abstracting a proton Syn-coplanar- hydrogen and halogen eclipsed with a dihedral of 0 degrees Nucleophilicity is defined by the rate of attack on an electrophilic carbon atom If molecules rigidly held in eclipsed conformations, they are likely to undergo E2 by a concerted syn-coplanar mechanism If the new bond is to a proton, it has reacted as a base Syn-coplanar eliminations are unusual If the new bond is to a carbon, it has reacted as a nucleophile E2 is stereospecificbecause different stereoisomers of the starting material react to give different stereoisomers of the product Most good nucleophiles are also strong base Results from the anti-coplanar transition state Electrons that are more loosely held are morepolarizable, meaningthat the electrons can move more freely toward a positive charge, resulting in a stronger bonding in the transition state Comparisonof E1 and E2 Elimination Mechanisms 1. A species with a negative charge is stronger than a similar neutral species A base is always a stronger nucleophile than its conjugate acid E1 E2 2. Nucleophilicity decreases from left to right and up on the periodic table Base strength unimportant (usually weak) Requires strong bases The more electronegative, the less reactive they'll be as a nucleophile Requires a good ionizing solvent Solvent polarity is not so important The bigger or the more polarizable, the more nucleophilic Rate = k [RX] Rate = k [RX] [B:-] Steric effects on nucleophilicty Usually Zaitsev orientation Usually Zaitsev orientation To serve as a nucleophile, bulky groups hinder the close approach needed to attack the carbon atom No particular geometry required for the slow step Coplanar arrangement (usuallyanti) required for transition state Tert-butoxide is a stronger base, yet weaker nucleophile due to its three bulky methyl groups Ethoxide is a weaker base, yet it is unhindered, so it acts as a stronger nucleophile Rearrangements common No rearrangements Steric hindrance (bulkiness) hinders nucleophilicty more than it hinders basicity Tertiary > secondary Tertiary > secondary > primary Solvent effects on nucleophilicty A protic solventis one that has acidic protons, usually in the form of O--H or N--H groups Form hydrogen bonds to negatively charged nucleophiles Convenient solvents for nucleophilic substitutions because the reagents tend to be quite soluble Small ions are solvated more strongly than large anions in a protic solvent because the solvent approaches the small anion more closely and forms stronger hydrogen bonds When an anion reacts as a nucleophile, energy is required to "strip off" some of the solvent molecules, thus breaking some of the hydrogen bonds More energy is required to strip off solventmolecules from a smaller solvated ion like fluoride as opposed to iodide Smaller anions are less nucleophilic (reinforces the trend in polarizability) Nucleophilicity in protic solvents decreases with increasing atomic number Orgo Page 2 Smaller anions are less nucleophilic (reinforces the trend in polarizability) Nucleophilicity in protic solvents decreases with increasing atomic number Aprotic solvents(solvents without O--H or N--H) enhance the nucleophilicity of anions An anion is more reactive in an aprotic solvent because it is not so strongly solvated No hydrogen bonds to be broken But, most polar, ionic reagents are insolublein simple aprotic solvents such as alkanes Polar aproticsolventshave strong dipole moments to enhance solubility,yet they have no O--H or N--H groups to form hydrogen bonds with anions Examples of polar aprotic solvents: The strength of the base or Nu determines the order of the reaction If a strong Nu or base is present, it will force second order kinetics (Sn2 or E2) Primary halides usually undergo the Sn2 reaction(occasionally E2) Exception is if they are stabilized by resonance Tertiary halides usually undergo the E2 reaction (strong base) or a mixture of Sn1 and E1(weak base) Some nucleophiles and bases favor substitution or elimination A fluoride ion, normally a poor nucleophile in hydroxylic (protic) solvents, can be a good nucleophile in an aprotic solvent A bulky strong base (t-butoxide) enhances elimination A good nucleophile with limited basicity (highly polarizable that is the conj base of a strong acid) enhances substitution Reactivity of the Substrate in SN2 Reactions Examples are Br- and I- To be a good substrate, a molecule must have an electrophilic carbon atom with a good leaving group, and that carbon atom must not be too sterically hindered for a nucleophile to attack A good leaving group should be 1. Electron withdrawing, to polarize the carbon atom C--X where X = halogen C--O, C--N, C--S 2. Stable (not a strong base) once it has left Conjugate bases of strong acids 3. Polarizable, to stable the transition state To maintain partial bonding with the carbon atom in the transition state Lowers activation energy Steric effects on the substrate Rapidly with methyl halides and most primary substrates Tertiary halides fail to react at all by the SN2 mechanism Steric hindrance- Backside of the electrophilic carbon atom is crowded by the presence of bulky groups Stereochemistryof the SN2 Reaction Back-side attack turns the tetrahedron of the carbon atom inside out Inversionof configuration- nucleophile assumes a stereochemical position opposite the position the leaving group originally occupied SN2 displacement is a good example of a stereospecificreaction Different stereoisomers react to give different stereoisomers of the product In inversion of configuration, an R starting material does not always produce an S product and vice versa Don't confuse absolute configuration with designation of configuration First-OrderNucleophilicSubstitution:SN1 Solvolysis- a reaction that takes place with the solvent acting as the nucleophile Tert-butyl bromide placed in boiling methanol Methoxide replaces bromide, but methanol is a weak nucleophile and tert-butyl bromide is a hindered tertiary halide, so this reaction does not go through the SN2 mechanism Reaction rate is first order with respect to the alkyl halide and zeroth order with respect to the concentration of the nucleophile Therefore, the nucleophile is not present in the transition state of the rate-limiting step Nucleophile reacts after the slow step SN1 reaction- (substitution, nucleophilic,unimolecular) Unimolecular refers to the fact that there is only one molecule involved in the transition state of the rate-limiting step First step is a slowionization to form a carbocation Second step is a fast attack on the carbocation by a nucleophile Carbocation is a strong electrophile and reacts fast even with weak nucleophiles Nucleophiles in SN1 are usually weak because a strong nucleophile would be more likely to attack the substrate and force some kind of second order reaction Orgo Page 3 Step 3 would be deprotonation to form the product if Nu: is water or an alcohol Ionization step is highly endothermic with a large activation energy Nucleophilic attack is strongly exothermic - it effectively reacts with carbocation almost as soon as it forms Substituent effects Rate limiting step is the ionization to form a carbocation (strongly endothermic) so the first transition state resembles the carbocation (Hammond postulate) Rates of Sn1 reactions depend strongly on carbocation stability Carbocations are most stable as tertiary carbons due to the inductive effect- alkyl groups donate electron density through sigma bonds Sn1 reactivity is exactly opposite form Sn2: tertiary > secondary > primary > CH3X Alkyl groups hinder Sn2 by blocking the attack of the Nu, but alkyl groups enhance Sn1 by stabilizingthe carbocation intermediate Resonance stabilization of the carbocation can also promote Sn1 Vinyl and aryl halides generally do not undergo Sn1 or Sn2 reactions Sn1 would require ionization to form a vinyl or aryl cation, which is less stable than alkyl carbocations Sn2 would require back-side attack by the nucleophile, which is made impossibleby the repulsion of electrons in the double bond or aromatic ring Leaving group effects Leaving group breaks its bond to carbon in the rate-limiting ionization step in Sn1 A highly polarizable leaving group helps stabilize the rate-limiting transition state Leaving group should be a weak base, very stable after it leaves with the pair of electrons Solvent effects Sn1 occurs much more readily in polar solvents that stabilize ions Protic solvents such as alcohols and water are even more effective because anions form hydrogen bonds with the-- OH hydrogen atom and cations complex with the nonbonding electrons of the--OH oxygen Stereochemistryof the Sn1 Reaction Sn1 is not stereospecific Carbocation intermediate issp2 hybridized and planar A nucleophile can attack the carbocation from either face Planar + chiral givesracemization Both enantiomers of the product are formed, one with retention of configuration (from the front) and one with inversion of configuration (from the back) Deuterium has the same size and shape as the hydrogen and it undergoes the same reactions It is used to label and track a specific atom Distinguishesbetween two faces of the ring Racemic mixtures can slightly favor a configuration if the leaving group hinders approach of the nucleophilic solvent Rearrangementsin the Sn1 Reactions Rearrangements are structural changes that form more stable I'''ons May occur after a carbocation has formed or as the leaving group is leaving (Sn2 occurs in one step, so there is no opportunity for rearrangement) Hydrideshift - the movement of a hydrogen atom with its bonding pair of electrons Orgo Page 4 Methyl shift - migration of a methyl group together with its pair of electrons Primary halides and methyl halides rarely ionizeto carbocations in solution If a primary hallide ionizes, it will likely ionize with rearrangement Formation of a primary carbocation is very unstable In this case, rearrangement will usually occur at the same time the leaving group leaves Comparisonof Sn1 and Sn2 Sn1 Sn2 Nu strength unimportant (usually weak) Strong Nu required Tertiary > secondary (primary and CH3X unlikely) CH3X > primary > secondary Good ionizing solvent required, polar May go faster in a less polar solvent, aprotic First order rate equation Second order rate equation Rate = k [R--X] Rate = k [R--X] [Nu:-] Racemization (inversion and retention) Complete inversion Rearrangements common Rearrangements impossible Strong nucleophiles encourage second order reactions and weak nucleophiles more commonly react by first order mechanisms Sn2 unlikely with tertiary halides and Sn1 unlikely with primary halides, unless they are resonance stabilized For Sn1, AgNO3 forces ionization A reluctant first-order substrate can be forced to ionize by adding silvernitrate (one of the few soluble silver salts) Orgo Page 5 Orgo Chapter 7 and 11 Sunday, November 8, 2015 7:02 PM Chapter 7 Chapter 11 Structure and Synthesis of Alkenes Alcohols as Nucleophiles and Electrophiles; Formation of Tosylates (p. 476) E-Z system for cis-trans isomers is used to label molecules with double bonds that are O--H bond is broken when alcohols react as nucleophiles, both weak or strong capable of geometric isomerism When an alcohol reacts as an electrophile, the C--O bond is broken Zusammen "together" if two first-priority items are cis An alcohol is a weak electrophile because the hydroxyl group is a weak leaving group Entgegen "opposite" if two first-priority items are trans It becomes a good leaving group when it is protonated But a protonated alcohol requires a strongly acidic solution In addition, few good nucleophiles are stable in strongly acidic solutions Most strong nucleophiles are also basic and will protonate again To convert an alcohol to an electrophile that is compatible with basic nucleophiles, we can convert it to an alkyl halide or we can make its tosylate ester (ROTs) Use the same method as R or S to determine relative priorities Product of condensation of an alcohol with p-toluenesulfonic acid (TsOH) Stability of Alkenes Heat of hydrogenation - the heat given off (delta H) during catalytic hydrogenation Substitution effects 11 kJ/mol difference between the stabilities of a monosubstituted and a trans- disubstituted alkene Tosylate group is an excellent leaving group Often times, more reactive than the equivalent alkyl halide Trisubstituted alkene is more stable than the di by 14 kJ/mol These values are the energy it takes to remove the pi bond Tosylates are made from alcohols usingtosyl chloride (TsCl) in pyridine Tosylate leaving group is displaced by a wide variety of nucleophiles More substituted double bonds are usually more stable R must be an unhindered primary or secondary alkyl group is substitution is to predominate over elimination Energy differences of cis-trans isomers Trans isomers are generally more stable than the corresponding cis isomers Alkyl substituents are separated further in trans Chapter 10 Formation of sodium and potassium alkoxides (p. 436) Alkoxide ions are strong nucleophiles and strong bases Oxidation-reduction when the metal being oxidized and the hydrogen ion reduced to form hydrogen gas Stability of Cycloalkenes Cycloalkenes generally react like acyclic alkenes The presence of a ring only makes a major difference if there is ring strain Ring strain can arise from a trans double bond or just a small ring Rings that are five-membered or larger can easily accommodate double bonds The more acidic alcohols, like methanol and ethanol, react rapidly with sodium to form sodium Three or four-membered rings show evidence of ring strain methoxide and sodium ethoxide Orgo Page 1 The presence of a ring only makes a major difference if there is ring strain Ring strain can arise from a trans double bond or just a small ring Rings that are five-membered or larger can easily accommodate double bonds The more acidic alcohols, like methanol and ethanol, react rapidly with sodium to form sodium Three or four-membered rings show evidence of ring strain methoxide and sodium ethoxide In cyclobutane, normal ring strain arises from forcing the sp3 hybrid angles from 109.5 degrees to 90 degrees The double bond forces the sp2 hybrid angles from 120 degrees down to 90 degrees Makes double bond more reactive than a typical double bond In cyclopropene, the double bond forces the angle from 120 degrees all the way down to 60 degrees Highly strained In acyclic alkenes, the trans isomers are ususally more stable This is true for cycloalkenes as well, but only if the rings are large enough For cyclodecane and larger cycloalkenes, the trans isomer is nearly as stable as the cis isomer Bredt's Rule - a bridged bicyclic compound cannot have a double bond at a bridgehead position unless one of the rings contains at least eight carbon atoms If there is a double bond at the bridgehead carbon of a bridged bicyclic system, one of the two rings contains a cis double bond and the other must contain a trans double bond If the larger ring contains at least eight carbons, then it can contain a trans double bond and the bridgehead double bond is stable Orgo Page 2 Orgo Chapter 8 Sunday, November 8, 2015 8:25 PM Chapter 8 Reaction of Alkenes (p. 328 - 349) Reactivity of the Carbon-CarbonDoubleBond Hydrogenation of an alkene is an example of an addition Addition, elimination, and substitution are the three major reaction types we have studied Addition is the most common reaction of alkenes ElectrophilicAdditionto Alkenes Many different reagents could add to a double bond and form stable products (energetically favorable) In the second step of addition, the nucleophileattacks the carbocation, forming a stable product Electrophile is E+ while the nucleophile is Nuc:- Requires strong electrophile to attract the electrons of the pi bond and generate a carbocation in the rate limiting step Called the electrophilicadditionto alkenes The strong electrophile attracts the electrons out of the pi bond to form a new sigmabond and generate a carbocation The nucleophile will then react with the carbocation to form another sigma bond Regiochemistry- the orientation of addition Which part of the reagent adds to which end of the double bond Stereochemistry- if the reaction is stereospecific Orgo Page 1 Additionof Hydrogen Halidesto Alkenes Orientation of Addition: Markovnikov's Rule When the proton adds to the secondary carbon, a tertiary carbocation results, which is more stable Protonation of one carbon atom of a double bond gives the carbocation on the carbon that was not protonated Proton adds to the end of the doublebond that is less substituted to give the more substituted carbocation (more stable) Addition of hydrogen halides is said to be regioselective becauseone of the two possibleorientation of addition is preferred over the other Markovnikov's rule - the addition of a proton acid to the double bond of an alkene results in a product with the acid proton bonded to the carbon atom that already holds the greater number of hydrogen atoms Proton adds to the carbon with the most hydrogens In addition, in an electrophilic addition to an alkene, the electrophile adds in such a way as to generate the most stable intermediate Free-Radical Addition of HBr, Anti-Markovnikov Addition Anti-markovnikovreactions are most likely when the reagents or solvents were exposed to peroxide Peroxides give rise to free radicals that initiate the addition, causing it to occur by a radical mechanism Oxygen-oxygen bond in peroxides are weak, so it can break to give two alkoxy radicals Orgo Page 2 Oxygen-oxygen bond in peroxides are weak, so it can break to give two alkoxy radicals In the initiation step, free radicals generated from the peroxide react with HBr to form bromine radicals The bromine, as an electrophile, reacts with the double bond to form the most stable radical on the carbon atom The free radical then reacts with the HBr to form another bromine radical, which will then react with another molecule of the alkene, continuing the chain reaction The other product is the anti-Markovnikov product, in which H has added to the more substituted end of the double bond, the end with fewer hydrogens The Markovnikov orientation does occur in the presence of peroxides, but the peroxide-catalyzed reaction is faster If just a little bit of peroxide is present, a mixture of Markovnikov and anti-Markovnikov products result Reverse orientation in the presence of peroxides is called the peroxideeffect It only occurs with the addition of HBr to alkenes HCl reacting with an alkyl radical is strongly endothermic, so peroxide effect is not seen HI is also not observed because iodinereacting with an alkene is strongly endothermic Additionof Water: Hydration of Alkenes An alkene reacting with water in the presence of a strongly acidic catalyst to from an alcohol is called a hydration A hydrogen atom adds to one carbon and a hydroxyl group adds to the other Hydration is accomplished by adding excess water to drive equilibrium toward the alcohol Hydration and dehydration reactions are complementary Follow the same reaction pathway, just in reverse Orgo Page 3 Step 1 involves adding a proton to the less substituted end of the double bond to form the more substituted carbocation (similarto the addition of HBr) Hydration is regioselective and it follows Markovnikov's rule Hydration may take place with rearrangement because it is involved with a carbocation intermediate Hydrationby Oxymercuration-Demercuration Many alkenes do not easily undergo hydration in aqueous acid They could be insolubleor undergo sidereactions such as rearrangement, polymerization, or charring Oxymercuration-demercuration works with many alkenes that do not easily undergo direct hydration No free carbocation is formed and it works with Markovnikov orientation The reagent for mercuration is mercuric acetate, Hg(OCOCH3)2, or Hg(OAc)2 This reagent acts as an electrophile by dissociating slightly to form a positively charged species, -OAc Oxymercuration involves an electrophilic attack on the double bond by the positively charged mercury species Product is a mercurinium ion, an organometallic cation containing a three-membered ring Water from the solvent then attacks the mercurinium ion to give (after deprotonating) an organomercurial alcohol A subsequent reaction is demercuration, to remove the mercury Sodium borohydride (NaBH4, a reducing agent) replaces the mercuric acetate fragment with a hydrogen atom Orgo Page 4 In Step 2, water attacks the most substituted carbon (Markovnikov orientation) Oxymercuration-demercuration of an unsymmetrical alkene generally gives Markovnikov orientation of addition The electrophile, +Hg(OAc), remains bonded to the less substituted end of the double bond Oxymercuration-demercuration reliably adds water across the double bond of an alkene with Markovnikov orientation and without rearrangement Gives better yields than direct acid-catalyzed hydration Avoids rearrangements and does not involve harsh conditions Alkoxymercuration-demercuration Alkoxymercuration-demercuration convertsalkenes to ethers by adding an alcohol across the double bond of the alkene When mercuration takes place in an alcohol solvent, the alcohol serves as a Nu to attack the mercurinium ion The solvent attacks the mercurinium ion at the more substituted end of the double bond, giving the Markovnikov orientation of the addition The Hg(OAc) group appears at the less substituted end of the double bond Reduction gives the Markovnikov product, with hydrogen at the less substituted end of the double bond Nu attacks the carbon from the opposite side, leading to an anti-addition,where the reagents (HgOAc) and the nucleophile are added to opposite faces of the double bond Hydroborationof Alkenes Hydroboration used to convert an alkene to the anti-Markovnikov alcohol Diborane (B2H6) adds to alkenes with anti-Markovnikov orientation to form alkylboranes, which can be oxidized to give anti-Markovnikov alcohols Alone, diborane is an inconvenient reagent, but with THF, it can be more easily used as a complex BH3-THF is the form of borane commonly used in organic reactions Orgo Page 5 Borane is an electron deficient compound with only six valence electrons BH3 forms "banana" bonds to acquire an octet Hydroborationof the double bond is thought to occur in one step, with the boron atom adding to the less substituted end of the double bond In the transition state, the electrophilic boron atom withdraws electrons from the pi bond and the carbon at the other end of the double bond acquires a partial positive charge (more stable on the more substituted carbon atom) Boron bonded to the less substituted end of the double bond and hydrogen bonded to the more substituted end due to steric hindrance The boron is removed by oxidation, using aqueous sodium hydroxide and hydrogen peroxide (HOOH or H2O2) to replace the boron atom with a hydroxyl group Does not affect orientation (anti-Markovnikov orientation established in the first step when BH3 was added) Hydroboration - syn addition - H and BH2 add to the same side Orgo Page 6


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