Alkene Reaction Guide

Addition of Hydrochloric Acid (HCl)Description: Treatment of alkenes with hydrochloric acid (HCl) will result in the formation of alkyl chlorides. The chlorine always ends up at the more substituted carbon of the alkene, since protonation of the alkene will result in the more stable carbocation.

Notes: Note that this reaction is “Markovnikoff selective”. Since it goes through a carbocation, rearrangements are possible.Examples:

Notes: When a secondary carbocation is formed adjacent to a tertiary or quaternary carbon, rearrangements are possible. Note the third example – where Markovnikoff’s rule gives no clear preference a mixture will be obtained.Mechanism: Electrons from the C1-C2 π bond attack the hydrogen of HCl, expelling the Cl anion and leading to the formation of a carbocation (Step 1, arrows A and B). Note that the most stable carbocation is formed preferentially. The chloride anion then attacks the carbocation, leading to formation of the alkyl chloride (Step 2, arrow C).

Addition of HBr to Alkenes

Description: Treatment of alkenes with hydrobromic acid will result in the formation of alkyl bromides.Notes: This is an addition reaction. Note that the bromine always ends up at the more substituted carbon of the alkene (Markovnikoff-selectivity)Examples:

Notes: When a secondary carbocation is formed adjacent to a tertiary or quaternary carbon, rearrangements are possible. Note the third example – where Markovnikoff’s rule gives no clear preference, a mixture will be obtained. Note that the right hand molecule of the third example contains a stereocenter, which will be obtained as a racemic mixture (1:1 mixture of enantiomers).Mechanism: Electrons from the C1-C2 π bond attack the hydrogen of HBr, expelling the bromide anion and leading to the formation of a carbocation (Step 1, arrows A and B). Note here that the carbocation preferentially forms on C2 (secondary) and not C1 (primary) since secondary carbocations are more stable. The bromide anion then attacks the carbocation, leading to formation of the alkyl bromide (Step 2, arrow C)

Notes: Since the carbocation is planar (flat) there is no preferred direction of attack of the bromide ion. If it’s possible to form a stereocenter, a mixture of stereoisomers will be obtained.

Additional example (advanced) : If the molecule contains adjacent stereocenters, the two directions of attack on the carbocation will no longer be of equal energy and a mixture of diastereomers will be obtained. In this example, attack of the bromide ion on the less hindered “top face” of the carbocation is favored, and a mixture of products (diastereomers) will be obtained.

Addition of Hydroiodic Acid (HI)Description: Treatment of alkenes with hydroiodic acid (HI) leads to the formation of alkyl iodides. Note that the iodine always ends up on the more substituted carbon.

Notes: Note that this reaction is Markovnikoff selective. Since it goes through a carbocation, rearrangements are possible in some situations.Examples:

Notes: When a secondary carbocation is formed adjacent to a tertiary or quaternary carbon, rearrangements are possible. Note the third example – where Markovnikoff’s rule gives no clear preference a mixture will be obtained.Mechanism: Electrons from the C1-C2 π bond attack the hydrogen of HI, expelling the iodide ion and leading to the formation of a carbocation (Step 1, arrows A and B). Note that the most stable carbocation is formed preferentially. The iodide anion then attacks the carbocation, leading to formation of the alkyl iodide (Step 2, arrow C).

Alcohols from Alkenes (H3O+)Description: Addition of aqueous acid to alkenes leads to the formation of alcohols.

Notes: Note that the alcohol is formed at the most substituted carbon of the alkene (Markovnikoff selectivity!) Since this reaction proceeds through a carbocation, rearrangements can occur in some cases.Aqueous acid is often written as “H3O(+)” . An alternative way to depict aqueous acid is H2O/H2SO4. There’s no essential difference for our purposes.Examples:

Notes: Note that the last example is a rearrangement!Mechanism: Protonation of the alkene by acid (Step 1, arrows A and B) leads to the more substituted (i.e. stable) carbocation, which is then attacked by water (Step 2, arrow C). Deprotonation of the oxygen then gives the neutral alcohol (Step 3, arrows D and E). Note that the acid is regenerated in step 3, so it acts as a catalyst.

Notes: It’s probably more reasonable to show water as the base in step 3, but HSO4(-) was used here for simplicity’s sake. Note that the acid is catalytic.In this example a secondary carbocation is formed: if C-3 were a tertiary or quaternary carbon, a rearrangement would have occurred.Chlorination (Cl2)Description: Treatment of alkenes with chlorine (Cl2) leads to formation of vicinal dichlorides (1,2-dichlorides)

Notes: Note that the chlorines add to opposite faces of the double bond (“anti-addition”)Examples:

Notes: Again note that the stereochemistry is trans.Mechanism: Attack of the alkene on chlorine (Step 1, arrows A and B) gives the chloronium ion, which is then attacked by chloride at carbon to give the 1,2-dichloride (vicinal dichloride). Note that this should occur at the more substituted carbon of the chloronium ion where appropriate (“Markovnikoff” addition).

Notes: Note that only one enantiomer is shown here: chlorine has an equal chance of adding to either face of the alkene, and thus a 1:1 mixture of enantiomers will be formed.Bromination (Br2)

Description: Treatment of alkenes with bromine (Br2) gives vicinal dibromides (1,2-dibromides). Notes: The bromines add to opposite faces of the double bond (“anti addition”). Sometimes the solvent is mentioned in this reaction – a common solvent is carbon tetrachloride (CCl4). CCl4 actually has no effect on the reaction, it’s just to distinguish this from the reaction where the solvent is H2O, in which case a bromohydrin is formed (see also).

Notes: Again note that in the first example CCl4 is merely the solvent and in these cases has no effect on the reaction (unlike when water or alcohols are the solvent – see bromohydrin formation page).Mechanism:Attack of the alkene on bromine (Step 1, arrows A and B) gives the bromonium ion, which is attacked at the backside by bromide ion to give the trans-dibromo product. Note that the bromines are delivered to opposite sides of the alkene (“anti” addition).

Notes: Note that a 1:1 mixture of enantiomers will be formed in this reaction. The enantiomer formed will depend on which face of the alkene the bromine adds to.

Iodination (I2)Description: Treatment of alkenes with iodine (I2) leads to the formation of vicinal diiodides (1,2-diiodides).

Notes: The iodides add to opposite faces of the double bond (“anti addition”)Examples:

Notes: Note that the stereochemistry is trans in these products.Mechanism: Attack of the alkene on iodine gives the iodonium ion (Step 1, arrows A and B). The iodonium ion is then attacked by iodide ion at the most substituted carbon where applicable (Markovnikoff) to give the 1,2-diiodide.

Chlorohydrin Formation (Cl2/H2O)Description: Alkenes treated with chlorine (Cl2) in the presence of water will form chlorohydrins. The stereochemistry of the products is anti.

Notes: Note that this reaction is Markovnikoff selective (OH ends up on most substituted carbon of the alkene) and the OH and Cl groups have opposite stereochemistry [where possible]Examples:

Notes:Note how in examples 1 and 2 the oxygen is attached to the most substituted carbon (Markovnikoff). Note the trans stereochemistry in example 1. Note that the third example this is an example of an intramolecular reaction.Mechanism: The alkene attacks Cl2 to form a chloronium ion (Step 1, arrows A and B) which is attacked on the more substituted carbon by water (Step 2, arrows C and D). The positively charged oxygen is then deprotonated to give the neutral alcohol.

Notes: Note that it’s actually more reasonable to show water acting as the base in step 3 (it’s a stronger base). Also, although not shown here the oxygen and chlorine should be trans to each other since the bromonium ion undergoes backside attack by the water.Bromohydrin formation (Br2/H2O)Description: Alkenes treated with bromine (Br2) in the presence of water will form bromohydrins.

Notes: Note that this reaction is Markovnikoff selective and delivers the trans-product. This is essentially the same reaction as with NBS.Examples:

Notes: Note how in examples 1 and 2 the oxygen is attached to the most substituted carbon (Markovnikoff). The trans-stereochemistry is evident in example 1. Finally in the third example this is an example of an intramolecular reaction – often gives students trouble!Mechanism: The alkene attacks Br2 to form a bromonium ion (Step 1, arrows A and B) which is then attacked on the more substituted carbon by water (Step 2, arrows C and D). The positively charged oxygen is then deprotonated to give the neutral alcohol.Acid-Catalyzed Addition of AlcoholsDescription: Alkenes treated with acid in the presence of an alcohol (usually as solvent) form ethers.

Notes: The acid used typically has a poorly nucleophilic counter-ion, like H2SO4 or p-TsOH. Often, H+ is just written as the acid. [If an acid with a relatively nucleophilic counter ion like HCl were written, then there might be some competition with Cl- adding to the carbocation that is formed].The alcohol is generally used as solvent. Note that the addition is “Markovnikov” (H adds to least substituted carbon of the alkene, oxygen adds to most substituted carbon).Examples:

Notes: Note that the addition is always Markovnikov; the proton adds to the least substituted carbon of the alkene, and the alcohol adds to the most substituted carbon. Example 1 shows H+ as acid; note how there is essentially no difference in changing to H2SO4 or p-TsOH.The reaction proceeds through a carbocation intermediate, so rearrangements can occur (example 4).Intramolecular reactions are possible (example 5).Mechanism: This reaction proceeds similarly to other “Markovnikov” additions to alkenes such as addition of hydrogen halides. In the first step the alkene is protonated by acid so as to produce the most stable carbocation (Step 1, arrows A and B). The carbocation then formed is attacked by the alcohol (Step 2, arrow C) leading to formation of the C-O bond. In the final step, the positively charged oxygen is deprotonated to give the neutral ether.

Notes: Note that although here -OSO3H is shown as doing the final deprotonation, it is just as correct (if not more correct) to show the alcohol solvent performing this deprotonation.

Notes: It’s probably more reasonable to show water as the base in step 3. Although not important for this molecule, the bromine and oxygen are in atrans orientation since water attacks the bromonium ion from the back.Oxymercuration ( Hg(OAc)2 / H2O )Description: Alkenes treated with mercuric acetate [ Hg(OAc)2 ] and water will be hydrated to form alcohols, after addition of NaBH4.

Notes: This reaction follows Markovnikov’s rule – the oxygen adds to the more substituted carbon. The purpose of the second step (NaBH4) is to remove the mercury.Examples:

Notes: See that the oxygen is attached to the more substituted carbon of the alkene. Also, there is no clear preference for syn or anti addition. In the third example notice that here Markovnikoff’s rule gives us no clear preference for which carbon to add the oxygen to, so we obtain a mixture.Mechanism: The alkene adds to Hg(OAc)2, displacing acetate and forming a mercuronium ion (Step 1, arrow A and B). Next, the more substituted carbon is attacked by water (Step 2, arrows C and D). The resulting protonated alcohol is then deprotonated by acetate (Step 3, arrows E and F). Removal of mercury is done with NaBH4. The mechanism of this step is generally not considered to be important for Org 1 / Org 2. However it is shown separately below.

Note that attack of alcohol (Step 2) occurs trans to the mercury, although that is not depicted here. Removal of mercury leads to the formation of a free radical (Step 6) so the reaction is not stereoselective.Here’s the mechanism of the reduction.

Oxymercuration ( Hg(OAc)2 / ROH )Description: When an alkene is treated with an alcohol in the presence of mercuric acetate [Hg(OAc)2] it will add to the more substituted position of the alkene (Markovnikoff) to give an ether. The mercury can then be removed using NaBH4.

Notes: The alcohol is generally used as the solvent here.Examples:

Notes: Note that in the second reaction, no rearrangement occurs (this would have happened for the acid-catalyzed reaction).The last example is an intramolecular reaction – often gives students a hard time.The byproducts in all cases are NaOAc, HOAc, BH3 and Hg(s)Mechanism: The alkene attacks Hg(OAc)2, displacing acetate and forming a mercuronium ion (Step 1, arrows A and B). Next, the more substituted carbon is attacked by the solvent alcohol (Step 2, arrows C and D). The resulting protonated ether is then deprotonated by acetate (Step 3, arrows E and F). Removal of mercury is done with NaBH4. The mechanism of this step is not generally considered to be “important” for Org 1/ Org 2. However, it is shown separately below.

Mechanism for removal of mercury:

Notes: Attack of alcohol (Step 2) occurs trans to the mercury, although that is not depicted here. Removal of mercury proceeds through formation of a free radical, so the reaction is not stereoselective.

Hydroboration-Oxidation (BH3 / H2O2)Description: Hydroboration-oxidation transforms alkenes into alcohols. It performs the net addition of water across an alkene.

Notes: Note that the oxygen is always attached at the less substituted carbon (anti-Markovnikoff). Furthermore the stereochemistry is always syn (H and OH add to same side of the alkene).The boron byproduct will depend on the # of equivalents of BH3 used reative to the alkene. Here their molar ratio is 1:1. One equivalent of BH3 can hydroborate up to 3 equivalents of alkene. BH3-THF is the same as BH3. Tetrahydrofuran (THF) is merely a solvent. Sometimes B2H6 is written, which is another form of BH3. It behaves in exactly the same way as BH3.You might also see 9-BBN or (Sia)2BH. These are hydroboration reagents in which two of the H atoms in BH3 have been replaced by carbon atoms. They will do the exact same reaction as BH3.Examples:

Notes: Note in example 2 that the hydrogen and oxygen add to the same side of the alkene (syn addition)Mechanism: The reaction begins with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon (Step 1, arrows A and B). In the second step, hydrogen peroxide and a base such as NaOH are added. the NaOH deprotonates the hydrogen peroxide (Step 2, arrows C and D) which makes the conjugate base of hydrogen peroxide (a better nucleophile than H2O2 itself). The resulting NaOOH then attacks the boron (Step 3, arrow E). This sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond (Step 4, arrows F and G). The OH expelled then comes back to form a bond on the boron (Step 5, arrows H and I) resulting in the deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species (Step 6, arrows J and K).

Notes: In step 1 the addition is syn and the reaction is concerted. If excess alkene is present the two remaining B-H bonds can do subsequent hydroborations. Note that only one enantiomer is shown here (but the product will be racemic) The migration step (Step 4, arrows F and G) occurs with retention of stereochemistry at the carbon.For the last step it’s reasonable to use water, HOOH or any other comparably acidic species as the acid.

Epoxides from alkenes (RCO3H e.g. mCPBA)Description: meta-chloroperoxybenzoic acid (m-CPBA) or other peroxyacids such as peracetic acids will convert alkenes into epoxides.

Notes: The stereochemistry of this reaction is always “syn” (both bonds to oxygen are formed on the same side of the alkene).Examples:

Notes: Note that example 2 just shows m-CPBA written out – it’s in no way different from the others. Examples 3 and 4 show that the reaction isstereospecific – that is, trans alkenes give trans products, and cis alkenes give cis products. Also note there is no reaction with alkynes.Mechanism:

This is a reaction which the arrow pushing formalism isn’t very good at describing. Instead, a transition state is shown here, which shows breaking of the C–C π bond at the same time the C–O bonds are forming. At the same time the O–O bond is breaking and the hydrogen from the OH is being picked up by the carbonyl oxygen.Notes: Note that the byproduct here is m-chlorobenzoic acid (in grey).Dihydroxylation (OsO4)Description: Treatment of alkenes with osmium tetroxide (OsO4) leads to the formation of 1,2-diols (“vicinal diols”).

Notes: The reaction proceeds with “syn” stereochemistry on the alkene, meaning that the two alcohols end up on the same side of the alkene. Sometimes a reducing agent like NaHSO3 or KHSO3 is added to remove the osmium at the end. Regardless, the key reaction is the same.Also note that cold, alkaline KMnO4 does exactly the same reaction.Examples:

Notes: The reaction is essentially the same regardless of whether KHSO3 is present. Also note that the reaction does not occur with alkynes.

Mechanism: Osmium tetroxide adds to one face of the alkene through a cycloaddition reaction (Step 1, arrows A, B, and C) to give a cyclic osmium compound (“osmate ester”). The osmate ester is then reduced with NaHSO3 or KHSO3 and hydrolyzed to the diol through a very long process that is, to be frank, excruciatingly tedious to write out and not generally bothered with in Org 1 and Org 2.

Notes: There is also a variation where OsO4 is used in catalytic amounts and a stoichiometric oxidant such as N-methylmorpholine N-oxide (NMO) is used to regenerate OsO4.Dihydroxylation of alkenes (KMnO4)Description: Treatment of alkenes with cold, dilute basic KMnO4 leads to 1,2-diols (vicinal diols).

Notes: The reaction proceeds with “syn” stereochemistry of the alkene, meaning that the two

alcohols end up on the same side of the alkene Also note that osmium tetroxide (OsO4) does exactly the same reaction.

Examples:

Notes: The reaction works well so long as it is kept cold. If higher temperatures are used, cleavage of the diol to give carbonyl compounds is observed. Note that the reaction does not occur with alkynes.Mechanisms: Potassium permanganage adds to one face of the alkene through a cycloaddition reaction (Step 1, arrows A, B and C) to give a cyclic manganese compound (“manganate ester”). The manganate ester is then reduced with NaHSO3 or KHSO3 and hydrolyzed to the diol through a very long process that is excruciatingly boring to write out and generally not bothered with in Org 1/ Org 2

Notes: The boring part goes something like this: water attacks Mn, transfer proton to O, break Mn–O bond, then add second equivalent of water to Mn, transfer proton to O, break Mn–O bond. This gives the free diol.Cyclopropanation (CH2N2 or CH2I2/Zn-Cu)Description: Alkenes treated with methylene carbene (:CH2) will form cyclopropanes.

Notes: In practice there are several ways to write this. Two common examples are with diazomethane (CH2N2) in the presence of heat (Δ) or light (hν), with diiodomethane (CH2I2) in the presence of zinc-copper couple (Zn-Cu). For our purposes the end result is exactly the same. The example with CH2I2 and Zn-Cu is called the Simmons-Smith reaction.Examples:

Notes: Note that the syn product is always formed. See how the hydrogens on the double bond have become dashes?Mechanism: Treatment of diazomethane with light (or heat) results in fragmentation of the C–N2 bond (Step 1, arrow A), liberating nitrogen gas. The resulting carbene then does the cyclopropanation of the alkene (Step 2, arrows B and C). Note that the reaction proceeds through a concerted transition state, so B and C happen simultaneously.

For the reaction with Zn–Cu the mechanism is essentially the same with a slight twist. First, Zn inserts itself into the C–I bond in a mechanism similar to that for Grignard formation but is not depicted here (Step 1). Then, breakage of the C–Zn bond (Step 2, arrow A) leads to a species that performs the cyclopropanation (Step 3, arrows B and C) to give the cyclopropane. It also proceeds through a concerted transition state.

Notes: This example with Zn–Cu is called the “Simmons Simith cyclopropanation”DichlorocyclopropanationDescription: Treatment of chloroform (CHCl3) with strong base (NaOH) leads to the formation of a dichlorocarbene, which then can form a cyclopropane when added to an alkene.

Notes: The byproducts of this reaction are NaCl and H2OExamples:

Notes:Mechanism: Deprotonation of chloroform (Step 1, arrows A and B) leads to formation of a short-lived anion. Loss of chloride ion from this ion (Step 2, arrow C) leads to formation of a carbene. When combined with an alkene, a cyclopropane will form (Step 3, arrows D and E).

Notes: This reaction proceeds through a concerted transition state (see bottom left).Ozonolysis of alkenes (reductive workup)Description: Ozone will cleave carbon-carbon double bonds to give carbonyl compounds such as aldehydes or ketones, after treatment with a reducing agent such as zinc or dimethyl sulfide (“reductive workup”)

Notes: Acid here could be written as “H3O+” , “H+”, “acid”, etc. It’s not crucial.Examples:

Notes: Note that any alkene carbon that has a C–H bond is converted into an aldehyde. Any alkene carbon with two C–C bonds is converted into a ketone. Example 3 is the cleavage of a

cyclic alkene to give a linear compound. In example 4, reduction with Me2S (DMS) is in no way different from reduction with Zn.Mechanism: The first step of the reaction is a cycloaddition of ozone with the alkene (Step 1, arrows A, B, and C). The second step is areverse cycloaddition, resulting in cleavage of the carbon-carbon single bond (Step 2, arrows D, E and F). The oxygen of the carbonyl oxide then performs a 1,2-addition on the other carbonyl (Step 3, arrows G and H) giving a negatively charged oxygen that performs a 1,2-addition on the carbonyl carbon of the carbonyl oxide to give the ozonide. (Step 4, arrows I and J).

For reductive workup the mechanism is generally not considered so important for the purposes of Org 1/ Org 2, but here it is! In the reduction step, zinc attacks one of the oxygens of the ozonide, breaking the weak O–O bond (Step 5, arrows K and L). The resulting anion then performs a 1,2-elimination (Step 6, arrows M and N) to give one of the fragments, followed by a second 1,2-elimination (Step 7, arrows O and P).

Ozonolysis (O3) – Oxidative WorkupDescription: Ozone will cleave carbon-carbon double bonds to give ketones/carboxylic acids after oxidative workup.

Notes: The initial product of this reaction is an ozonide. Treatment of the ozonide with acid and an oxidant such as hydrogen peroxide (H2O2) will convert any aldehydes to carboxylic acids. Note that any C–H bonds on the alkenes are converted to C–OH bonds, giving carboxylic acids.Examples:

Notes: Note how every C–H bond on the alkene is converted into a C–OH bond to give a carboxylic acid. Also note that example 3 shows cleavage of a cyclic alkene to give a linear compound. In example 4, cleavage of a terminal alkene results in CO2.Mechanism: The first step of the reaction is a cycloaddition of ozone with the alkene (Step 1, arrows A, B, and C). The second step is a reverse cycloaddition, resulting in cleavage of the carbon-carbon single bond (Step 2, arrows D, E, and F). The oxygen of the carbonyl oxide then performs a 1,2-addition on the other carbonyl (Step 3, arrows G and H) giving a negatively charged oxygen that performs a 1,2-addition on the carbonyl carbon of the carbonyl oxide to give the ozonide (Step 4, arrows I and J).

With warming, the ozonide breaks down to the aldehyde and a carbonyl oxide (Step 5, arrows K, L, and M). Addition of peroxide to the aldehyde then occurs (Step 6, arrows N and O). This is followed by proton transfer (Step 7, arrows P and Q) and then removal of a proton with base to give the carbonyl (C=O) (Step 8, arrows R, S, and T).

Notes: There are other reasonable ways to draw this mechanism, particularly other ways of drawing proton transfer in Step 7 and other species that could act as bases in Step 8.Oxidative Cleavage of Alkynes with KMnO4Description: Potassium permanganate (KMnO4) will cleave alkynes to give carboxylic acids.

Notes: The same reaction can be performed with ozone (O3)Examples:

Notes: Note where a terminal alkyne is being cleaved (examples 2 and 4) one of the products is carbon dioxide.Mechanism: For the purposes of Org 1 / Org 2 the mechanism of this reaction is not considered “important”. However the first step is almost certainly the cycloaddition of MnO4 to the triple bond, followed by oxidative cleavage.Hydrogenation of Alkenes to give AlkanesDescription: In the presence of a metal catalyst such as palladium, hydrogen will add to alkenes to give alkanes.

Notes: Many other metal catalysts can be used, such as platinum (Pt), nickel (Ni), or Rh (rhodium)Examples:

Notes: Note that the stereochemistry of addition to the alkene is syn (see the second example). Furthermore note that deuterium (the heavy isotope of hydrogen) can also be used instead of hydrogen. It works exactly the same way!Mechanism: For the purposes of Org 1/ Org 2 the exact mechanism of this reaction is generally not considered important. However it does follow the following sequence: hydrogen and the alkene are both adsorbed onto the surface of the metal, and the hydrogen-hydrogen bond is broken. Then, both hydrogen atoms are delivered syn to the alkene.

Additions to alkenes with 1,2-hydride shiftsDescription: When secondary (or primary) carbocations are formed, adjacent to a more substituted carbon, hydrogen atoms can shift. This leads to formation of a more stable carbocation.

Notes: Note how the blue hydrogen has shifted positions and the “X” [which could be Br, Cl, I, or OH] is attached to where the H used to be.Examples:

Notes: Note that the first two examples show rearrangement to give mroe stable tertiary carbocations, whereas the last example shows a rearrangement to give a resonance stabilized benzylic carbocation.Mechanism: The first step of the reaction is formation of a carbocation. Here, the formation of a carbocation from addition of HBr to an alkene is shown (Step 1, arrows A and B). Since we have a secondary carbocation adjacent to a tertiary carbon, shift of a hydrogen to the secondary carbocation will results in a (more stable) tertiary carbocation (Step 2, arrow C). The tertiary carbocation is then trapped, in this case, by Br(–) (Step 3, arrows D). The rearrangement step goes through a transition state such as the one pictured.

Free Radical Bromination of AlkanesDescription: When treated with bromine (Br2) and light (hν) alkanes are converted into alkyl bromides. In the absence of any double bonds, with Br2this is selective for tertiary carbons.

Notes: Light (hν) is the initiator for this reaction.

Examples:

Notes: Note that each of these reactions produces HBr as a byproduct. If this reaction occurs at a stereocenter, a mixture of alkyl bromides will be obtained.This table allows for the relative comparison of the selectivity of radical chlorination and bromination (reaction rates with typical primary, secondary, tertiary alkyl C-H bonds)

Mechanism: When treated with light, Br2 fragments to bromine radicals (Step 1, arrow A). At any given time only a small amount of these radicals is present. There’s lots of leftover Br2! Remember this as it comes up in step 3. Bromine radical then abstracts a hydrogen from the tertiary carbon, leaving behind the tertiary radical (Step 2, arrows B and C). The tertiary radical then reacts with Br2, giving the tertiary bromide (Step 3, arrows D and E).

Notes: Avoid the common mistake of having the radical react with Br• in the third step – that’s not a propagation reaction, that’s termination!Note that tertiary radicals are more stable than secondary or primary radicals and thus require the least energy to form (another way of saying this is that tertiary C-H bonds are weaker than secondary and primary C-H bonds). The higher selectivity of this reaction for tertiary C-H (when compared with chlorination) is a reflection of the fact that formation of the weaker H–Br bond provides less of a driving force for this reaction as compared to H–Cl (which is a stronger bond).Free Radical Addition of HBr To AlkenesBY JAMES LEAVE A COMMENT

Description: Treatment of an alkene with HBr in the presence of catalytic amounts of a radical initiator (such as peroxides) in addition to heat or light leads to addition of HBr to the alkene in anti-Markovnikov fashion (note that the Br adds to the less substituted side of the alkene and the H adds to the more substituted side).Notes: A common initiator for this purpose is t-butyl peroxide (t-BuO-Ot-Bu) but in general the peroxide will be written, “RO-OR”. Heat or light is required to break the weak O–O bond. Again, note that this reaction is in contrast to normal addition of HBr which proceeds in “Markovnikov” fashion.Again, the initiator is generally catalytic (less than 1 equiv required)

Examples:

Notes: Note that either light (hγ) or heat (Δ ) can be used to initiate the reaction, and the initiator can be written variably as “peroxides”, “RO-OR” or even given specifically, as in the case of t-butyl peroxide in the bottom example. The reaction is not stereospecific, so if new stereocenter(s) are formed, such as in the bottom example, a mixture of stereoisomers may be obtained.Note also that Br adds to the less substituted carbon of the alkene.Mechanism: The purpose of heat/light is to cause homolytic fragmentation of the weak O–O bond of the peroxide catalyst (Step 1 arrow A) . This leads to reversible formation of alkoxy radicals, which can then remove a hydrogen (homolytically) from H–Br to give the alcohol (R-OH) and bromine radical Br• (Step 2, arrows B and C) . The purpose of the peroxide is just to get the free radical chain reaction going (that is why only a catalytic amount is necessary). However, at least 1 equiv of H-Br is necessary to convert all the alkene to the addition product.

Once the bromine radical is formed, it can then add to the alkene. Note that since radical stability increases with increasing substitution (i.e. primary < secondary << tertiary) addition will occur such as to form the most substituted (i.e. most stable) radical which means that Br will add to the least substituted carbon of the alkene (step 3, arrows D and E). In this case a secondary radical is formed, which then reacts with H–Br to form the alkyl halide product and to regenerate bromine radical (Step 4, arrows F and G).

Although often not shown, termination occurs when the concentration of alkene or H-Br becomes low. Termination is the combination of two radicals to form a new bond. One possible termination step might be the combination of the alkyl radical with bromine radical, shown below:

Notes: This radical addition process only works for H-Br, not H-Cl or H-I.Since free-radicals can participate in reactions from either face, if a stereocenter is formed, a mixture of stereoisomers will be obtained. For example, in the alkene below, the first addition (of Br•) can occur from either face, as can the attack of the radical on H-Br. Four stereoisomers are obtained.

Sharpless Epoxidation

BY JAMES LEAVE A COMMENTDescription: The Sharpless Epoxidation is an enantioselective epoxidation of allylic alcohols.

Notes: The Sharpless epoxidation only works for alkenes adjacent to an alcohol (CH2OH). The oxidant is t-butyl hydroperoxide, sometimes written (CH3)3C–OOH or abbreviated TBHP. The catalyst is titanium tetraisopropoxide, written Ti[Oi-Pr]4 or Ti[OCH(CH3)2]4. The additive that imparts chirality is diethyl tartrate (DET). Choosing (+) or (–) diethyl tartrate [full names: L-(+)-diethyl tartrate and D(–)-diethyl tartrate – one can omit the L or D without penalty] allows one to choose the major enantiomer that is formed in this reaction [see “Mechanism” section for specifics].The major product can be predicted by use of a mnemonic (see below).Examples:

Notes: As far as the reagents are concerned, each of these examples just shows different ways of representing the key titanium catalyst and t-butyl hydroperoxide. The only difference is in the choice of diethyl tartrate (sometimes abbreviated DET) enantiomer.Example 1 shows an epoxidation of an alkene drawn in the “side view” which clearly shows that using (–)-diethyl tartrate leads to epoxidation from the top face when the allylic alcohol is drawn in this orientation.Example 2 shows that epoxidation only occurs on the alkene adjacent to CH2OH, the other alkene is untouched [BTW, this is because the OH coordinates to titanium]Examples 3 and 4 show the major products formed with each of these alkenes, predicted by using the mnemonic (below).Example 5 shows that no reaction occurs when an alkene without a neighbouring alcohol is used.Mechanism: Without going into specific details, what happens in the Sharpless epoxidation is that the titanium binds to the allylic alcohol, t-butylhydroperoxide, and the tartate in such a way as to provide a chiral environment whereby one face of the alkene is preferentially exposed to the oxidant. The outcome of a Sharpless epoxidation can be predicted by using the following mnemonic. Placing the CH2OH group in the upper right hand quadrant, using (–)-DET will lead to epoxidation of the top face, whereas (+)-DET will lead to epoxidation of the bottom face.