Reaction guide by James Ashenhurst [1]

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Free Radical Chlorination of Alkanes Description: Alkanes treated with chlorine gas (Cl2) and light (hv) or heat will be converted into alkyl chlorides.

Notes: One equivalent of HCl is formed for every C–Cl bond that is formed. Also, the reaction is not particularly selective, which limits its usefulness (see below) Examples:

Notes: If this occurs at a stereocenter, a mixture of alkyl chlorides will be obtained. Also note that one equivalent of HCl is generated for every C–Cl bond that is formed. Chlorine is not particularly selective, so the mixture of products will reflect statistics. For example in the third example (propane) there are six C–H bonds on the methyl groups, and two C–H bonds on the CH2 group, so the product mixture will be 3:1 favoring 1chloropropane over 2-chloropropane. Here is a helpful table that lets you compare the selectivity of chlorine and bromine radicals on primary, secondary, and tertiary carbons:

Mechanism: When treated with light, Cl2 fragments to chlorine radicals (Step 1, arrow A). At any given time only a small amount of these radicals are present. There’s lots of leftover Cl2! Remember that, as it comes up in Step 3. Chlorine radical then abstracts a hydrogen from the alkane, leaving behind the alkyl radical (Step 2, arrows B and C). The alkyl radical then reacts with Cl2, giving the alkyl chloride (Step 3, arrows D and E) and a chlorine radical, which can be recycled to perform Step 2 with another equivalent of the alkane.

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Notes: A common mistake is to show the carbon combining with Cl• in the second step! That’s not a propagation step, that’s termination!

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

Notes: Light (hν) is the initiator for this reaction. Alternatively, sometimes peroxides (RO-OR) can serve as radical initiators along with heat. 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. In the fourth example, ROOR represents peroxides, with a weak O-O bond which is ruptured homolytically by heat (Δ). CH2Cl2 is the solvent here. 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)

The key takeaway here is that bromine radicals are almost 20,000 times more likely to remove a hydrogen from a tertiary C-H bond than from a primary C-H bond. 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).

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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). Additional examples:

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Allylic bromination of alkanes using NBS Description: When treated with N-bromosuccinimide (NBS) and light (hν) alkyl groups adjacent to alkenes will be converted into alkyl bromides.

Notes: The position next to a double bond is called the “allylic” position. See also: Benzylic bromination Examples:

Notes: Note that in example 2, peroxides (that’s “ROOR”) and heat has the same effect as “light”. Mechanism: NBS provides a constant, low-level concentration of Br2, which is present when HBr reacts with NBS.

When treated with light, Br2 fragments homolytically to bromine radicals (Step 1, arrow A). At any given time only a small concentration of these radicals are present.

Bromine radical then removes a hydrogen from the allylic carbon, leaving behind the resonance-stabilized allylic radical (Step 2, arrows B and C). The allylic radical then reacts with Br2, giving the allylic bromide (Step 3, arrows D and E).

Notes: The KEY to this whole process is that the allylic C-H bond is weak and allylic radicals are greatly stabilized due to resonance. This reaction does not work on ordinary alkanes unless they are tertiary (see also).

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Addition of HCl to Alkenes to Give Alkyl Chlorides 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:

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Notes: When a secondary carbocation is formed adjacent to a tertiary or quaternary carbon, rearrangements are possible. The second example results in a 1:1 mixture of enantiomers (a “racemic mixture”). Note the third example – where Markovnikoff’s rule gives no clear preference a mixture will be obtained. In this case 3-chloropropane is formed along with 2-chloropropane (as a 1:1 mixture of enantiomers). For rearrangement examples, see: Addition with 1,2-hydride shift

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Addition with 1,2-alkyl shift 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).

Notes:

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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:

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Notes: The first two examples show a simple addition of HBr to give the Markovnikov product. In the third example Markovnikov’s rule gives no clear preference, so a mixture is obtained. The fifth example gives a mixture of diastereomers, since addition of HBr across the alkene will not affect the initial (R) stereochemistry. The final (sixth) example will give a dominant product, despite identical substitution, because formation of the resonancestabilized carbocation will be favored! When a secondary carbocation is formed adjacent to a tertiary or quaternary carbon, rearrangements are possible. See these pages for more examples: Addition with 1,2- hydride shift

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Addition with 1,2- alkyl shift 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)

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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.

Rearrangements: When secondary (or primary) carbocations are formed adjacent to a more substituted carbon, adjacent hydrogen atoms or alkyl groups can shift, resulting in formation of a more stable carbocation. The first step of the reaction below is formation of a carbocation. Here, the formation of a carbocation via attack of the alkene upon H-Br 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 result 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.

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.

Rearrangements can also occur. See Additions To Alkenes Accompanied By 1,2Hydride Shifts

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Addition of aqueous acid to alkenes to give alcohols 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 HSO 4(-) 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.

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Addition of Hydroiodic Acid to Alkenes to Give Alkyl Iodides 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).

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Chlorination of alkenes with Cl2 to give vicinal dichlorides Description: Treatment of alkenes with chlorine (Cl2) in an inert solvent like CCl4, alkanes, or CH2Cl2 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. If there is a pre-existing chiral center on the molecule, chlorination will not affect it (see example 6). Therefore a mixture of diastereomers will be obtained. If a solvent is not indicated, assume that it is inert (like CCl4, CH2Cl2, hexanes, etc.) and won’t affect the reaction . If water or alcohols are the solvent, halohydrins will be obtained. See Formation of chlorohydrins from Cl2 and water 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).

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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.

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Bromination of alkenes with Br2 to give dibromides

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). Sometimes “dark” is mentioned to distinguish this reaction from cases where Br2 can promote bromination through a radical pathway. 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. Additional examples:

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These examples show what happens when a chiral center is already present on the molecule; a mixture of diastereomers is obtained. The bottom example looks complicated, but actually isn’t – we’re still just breaking C-C, and forming two C-Br bonds, just as with any other alkene. Try not to get freaked out by all the additional atoms in the molecule! (There are chiral centers in the starting material, so again we’re forming a mixture of diastereomers).

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Iodination of alkenes to give vicinal diiodides (1,2-diiodides) 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”). See also: Bromohydrin formation, Alkyne halogenation 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.

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Formation of chlorohydrins from alkenes using water and Cl2 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. The fourth example is an example of the formation of diastereomers in chlorohydrin formation. 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 chloronium ion undergoes backside attack by the water.

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Formation of Bromohydrins from alkenes using water and Br2 Description: Alkenes treated with bromine (Br2) or N-bromosuccinimide (NBS) 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. See also: Bromination of alkenes, Formation of epoxides from bromohydrins Examples:

Notes: Note how in these examples the oxygen is attached to the most substituted carbon (Markovnikoff). The transstereochemistry is evident in example 2. The third example this is an example of an intramolecular reaction – this often gives students trouble, so watch out for it on exams! If a pre-existing chiral center is present (such as in example 4) look out for the formation of diastereomers; note that using ethanol as a solvent here results in a haloether instead of a halohydrin (also true of example 6). The resulting halohydrins are often converted into epoxides with base. See Formation of epoxides from bromohydrins 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.

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Addition Of Alcohols To Alkenes With Acid Description: 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 Cladding 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.

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Oxymercuration: Alcohols from alkenes using Hg(OAc)2 and Water Description: Alkenes treated with mercuric acetate [ Hg(OAc)2 ] and water will be hydrated to form alcohols, after addition of NaBH4.

See also: Ether formation using oxymercuration 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. Hg(OTFA)2 is similar to Hg(OAc)2 except the more electron-withdrawing trifluoroacetate group is used. (For our purposes this reagent is equivalent to Hg(OAc) 2, although in practice Hg(OTFA)2 is slightly more reactive). Importantly, note example #4. If an alcohol were attempted to be formed using H 3O+, a rearrangement would occur resulting in ring expansion. Oxymercuration is thus a milder way of forming an alcohol from an alkene which will not result in rearrangement since it does not go through a free carbocation. The last example has a pre-existing chiral center on the ring which is not affected by oxymercuration. A mixture of diastereomers will result. 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 NaBH 4. The mechanism of this step is generally not considered to be important for Org 1 / Org 2. However it is shown separately below.

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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.

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Oxymercuration of Alkenes to form Ethers using Hg(OAc)2 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 NaBH 4.

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.

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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.

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Hydroboration of Alkenes 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 BH 3 have been replaced by carbon atoms. They will do the exact same reaction as BH 3. Examples:

Notes: Example 1 just shows a simple anti-Markovnikov addition of BH3 to an alkene. Examples 2 and 3 show the hydroboration of a cyclic alkene. Note the syn addition; the C-H bond and C-OH bond are formed on the same side of the ring (this results in a mixture of enantiomers in this case). Example 4 has an existing chiral center (the methyl group) which is unaffected by hydroboration, so the result will be a mixture of diastereomers. 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).

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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.

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Formation of epoxides from alkenes using m-CPBA 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 is stereospecific – that is, trans alkenes give trans products, and cis alkenes give cis products. The trans alkene above gives a 1:1 mixture of enantiomers. The cis alkene in example 4 gives a meso product. Also note there is no reaction with alkynes. The last example is similar to m-CPBA except it’s with perbenzoic acid. Slightly less reactive than m-CPBA but pretty much identical for our purposes. 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).

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Dihydroxylation of Alkenes to give 1,2-diols (vicinal diols) 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. Sometimes H2S is used in place of KHSO3. The only purpose of this last step is to break up the “osmate ester” that is formed during the procedure (see below). Also note that the reaction does not occur with alkynes. Occasionally you might also see “NMO” used along with OsO4. “NMO” is N-methylmorpholine N-oxide, an oxidant that regenerates OsO4 from the Os(VI) species formed after dihydroxylation. NMO allows expensive (and toxic) osmium to be used in catalytic amounts. See references 4 and 5. 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.

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Notes: There is also a variation where OsO4 is used in catalytic amounts and a stoichiometric oxidant such as Nmethylmorpholine N-oxide (NMO) is used to regenerate OsO4.

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Dihydroxylation of alkenes with cold, dilute KMnO4 to give vicinal diols Description: Treatment of alkenes with cold, dilute basic KMnO4 leads to 1,2-diols (vicinal diols).

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Notes: The reaction proceeds with “syn” stereochemistry of the alkene, meaning that the two alcohols end up on the same side of the alkene

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Also note that osmium tetroxide (OsO4) does exactly the same reaction.

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The purpose of the NaOH is to assist in breaking up the intermediate manganate ester that forms after dihydroxylation. If this manganate ester is allowed to sit around, oxidative cleavage of the diol may occur. This is also why the temperature is kept cold. 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 permanganate 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 NaHSO 3 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. Reference: For those curious about the mechanism I strongly suggest you read this paper: “Permanganate Peroxidation of Cyclohexene: Hydroxide Ion And Salt Effect Studies. Taylor, J. E. ; Green, R. Can. J. Chem. 63, 2777 (1985). “Good yields of cis-l,2-cyclohexanediol were formed by the reaction of cyclohexene and aqueous potassium permanganate under turbulent stirring conditions only in the presence of low concentrations of sodium hydroxide. Larger amounts had no further benefit or were deleterious.” Also, apparently under acidic or neutral conditions, more highly oxidized products such as alpha-hydroxy ketones can be formed “without going through the glycols”. See J. Am. Chem. Soc. 1981, 103, 938.

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Formation of cyclopropanes from alkenes using methylene carbene (:CH2) 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 (CH 2N2) 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? The first three examples use CH2I2 (di-iodomethane) and zinc-copper couple (Zn-Cu). This forms a “carbenoid” which performs the cyclopropanation. The byproduct is zinc iodide. The second three examples use an alternative source of carbene, diazomethane (CH 2N2). Irradiation of diazomethane with ultraviolet light leads to loss of nitrogen gas (N2) and formation of the carbene (:CH2) which performs the cyclopropanation. 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.

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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”

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Addition of Dichlorocarbene to alkenes to give dichlorocyclopropanes Description: Alkenes treated with chloroform (CHCl3) and strong base will form dichlorocarbenes, which add to alkenes to form cyclopropanes.

Notes: A strong base such as NaOH is required for this reaction. Examples:

Notes:Note that the syn product is always formed in these reactions (see example 2). Mechanism: Deprotonation of chloroform by NaOH (Step 1, arrows A and B) leads to an anion, which then loses Cl in an alpha-elimination reaction (Step 2, arrow C). This forms the dichlorocarbene, which is extremely reactive. Reaction with the alkene (Step 3, arrows D and E) results in the cyclopropane. Note that the cyclopropanation step occurs through a concerted transition state (box) and thus arrows D and E occur at the same time.

Notes: Step 2 (alpha elimination) results in loss of chloride ion to give an empty orbital on carbon. The resulting species (divalent carbon, with one lone pair and one empty orbital) is called a carbene.

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Ozonolysis of alkenes to ketones and aldehydes (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 an acyclic compound. In example 4, reduction with Me2S (DMS) is in no way different from reduction with Zn. Example 5 is the ozonolysis of a bridged bicyclic alkene, which results in opening up of the ring and formation of two ketones. 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).

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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).

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Oxidative cleavage of alkenes to give ketones/carboxylic acids using ozone (O3) – (“oxidative workup”) Description: 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).

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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

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Oxidative cleavage of alkenes to ketones/carboxylic acids using KMnO4 Description: Alkenes treated with KMnO4 are cleaved into carbonyl compounds. The type of carbonyl compound depends on the substitution pattern of the alkene.

Notes: The reaction is generally run with aqueous acid and heat. Examples:

Notes: Note that all C–H bonds attached to alkenes are oxidized to C–OH bonds, but alkene carbons with only carbons attached become ketones. Note that the last example is a cyclic alkene that goes to a linear product. Mechanism: The detailed mechanism for this reaction is generally not considered important for the purposes of Org 1 / Org 2, although the first step is dihydroxylation of the alkene.

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Hydrogenation of Alkenes to give Alkanes Description: 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). Deuterium (D2) the heavy isotope of hydrogen can also be used (example 3). It works exactly the same way! Example 4 shows a somewhat complex example with a steroid carbon skeleton. The only thing that is affected here is the double bond. A pre-existing chiral center (example 5) will not be affected by hydrogenation, so here a mixture of cis and trans diastereomers will be produced. 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.

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Additions to alkenes accompanied by 1,2-hydride shifts Description: 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.

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Addition to alkenes accompanied by 1,2-alkyl shift Description: When secondary (or primary) carbocations are formed adjacent to a quaternary carbon, 1,2-shifts of alkyl group may occur.

Notes: In addition to 1,2-shifts of simple alkyl groups, 1,2-shifts of ring carbons can lead to changes in ring size (e.g. ring expansion reactions) Examples:

Notes: The first two examples show migration of a methyl group. The third example shows preferential migration of a ring carbon, which releases ring strain from the cyclobutane. The fourth example shows that hydrogen migration is generally preferred over alkyl migration. 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 quaternary carbon, shift of an alkyl group to C2 results in a more stable tertiary carbocation. Breakage of the C3-C4 or C3-C6 bonds is preferred since both of these result in a release of ring strain to give a five-membered ring (Step 2, arrow C). Finally trapping of the carbocation with bromine ion results in the alkyl halide (Step 3, arrow D).

Notes: Note in this example that migration of a ring carbon is preferred over migration of the methyl group because the latter does not allow for a release of ring strain. Note that breaking C3-C4 is equivalent to breaking C3-C6 (either could be shown here)

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Free Radical Addition of HBr To Alkenes

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, “ROOR”. 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 (hν) 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

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Hydroboration of alkynes using BH3 to give aldehydes Description: When alkynes are treated with BH3 and subsequently treated with H2O2, they are converted into aldehydes.

Notes: Note that only terminal alkynes (i.e. alkynes with an H on one end) are converted into aldehydes. The reaction is anti-Markovnikoff selective. The addition occurs syn but the stereochemistry is lost at the end. See Also: Oxymercuration of alkynes Examples:

Notes: A common set of conditions to see is BH3 followed by NaOH/H2O2. This is OK for our purposes, but in reality, antiMarkovnikov selectivity isn’t great, so the bulky boranes disiamyl borane (example 4) or 9-BBN (example 5) are often preferred.* Many instructors will just put in “R2BH” where R is a relatively bulky alkyl group, which will accomplish the same thing. With internal alkynes a ketone will be produced. There is no anti-Markovnikov selectivity here, so unless the alkyne is symmetrical (example 2) mixtures of ketones will be obtained (example 3) since neither carbon is favored. *double hydroboration can also occur with BH3, but that’s not something you will typically have to worry about. Mechanism: This is a loooong mechanism. The first step is hydroboration, where boron adds to the less substituted carbon (anti-Markovnikoff) (Step 1, arrows A and B). When hydroboration is complete, addition of NaOH and H2O2 begins the oxidation step, which starts with deprotonation of hydrogen peroxide (Step 2, arrows C and D). The deprotonated hydrogen peroxide is a better nucleophile than HOOH, and it now attacks the boron to give a charged boron species (Step 3, arrow E). Then comes the weird step! The carbon-boron bond breaks, and the electrons from this bond migrate to the oxygen, which breaks the weak oxygen-oxygen bond (Step 4, arrows F and G). Then, cleavage of O–BH2 (the “boronic ester”) with NaOH (Step 5, arrows H and I) gives the free oxygen (an “enolate”) which is protonated (Step 6, arrows J and K) to give the enol. Finally (second weird step!) the enol rearranges into its more stable constitutional isomer, the aldehyde. This is called “tautomerization” (Step 7, arrows L, M, N, O. P).

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Notes: It’s probably not necessary to actually show Step 2 if you just show HOO(-) attacking boron in Step 3. It’s also reasonable to switch the order of the protonation and tautomerization (i.e. do step 7 before step 6) or to show an acid other than water in Step 6.

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Oxymercuration of Alkynes Description: Alkynes treated with mercury (usually HgSO4) and water will be hydrated to give ketones, via an enol intermediate.

Notes: This reaction follows Markovnkov’s rule. Only a catalytic amount of mercury is required in this reaction. Acid is also important. Often mercuric sulfate (HgSO4) is used in conjunction with sulfuric acid (H2SO4), although this exact combination is not crucial. Examples:

Notes: Note for that the third example a mixture of ketones is obtained because no clear preference can be established using Markovnikoff’s rule (i.e. both carbons are equally substituted) Mechanism: For the purposes of Org 1/Org 2 this mechanism is generally not considered “important” due to the weird “demercuration” step. Mercury adds to the alkyne (Step 1, arrow A) giving the mercuronium ion. This is then attacked by water at the more substituted carbon (“Markovnikoff selectivity”) (Step 2, arrows B and C) to give the mercurated “enol” after deprotonation (Step 3, arrows D and E). Removal of the mercury happens in two steps: first, protonation of the enol at carbon (Step 4, arrows F, G, H) gives the protonated carbonyl; attack at mercury by solvent (or other nucleophile) regenerates the enol (Step 5, arrows I, J K). Then, the enol is converted to the more stable ketone through tautomerization (Step 6, arrows L and M).

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• •

Notes: The ketone is formed because it is more stable than the enol. This process is called keto-enol tautomerism. Note that there’s actually quite a few ways this mechanism could be written and this is just one suggestion. The key steps to really be concerned with are steps 1, 2, and 6.

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Addition of HCl once to alkynes to give alkenyl chlorides Description: Addition of one equivalent of hydrogen chloride (HCl) to alkynes results in vinyl chlorides (also known as alkenyl chlorides)

Notes: This reaction proceeds with Markovnkoff selectivity (the chlorine attaches itself to the more substituted carbon) Examples:

Notes: Note how in all cases the chloride adds to the more substituted carbon (Markovnikoff selectivity). However if both ends of the acetylene are attached to carbon, mixtures of E and Z isomers will be obtained. Mechanism: Attack of the alkyne pi bond upon H–Cl leads to the vinyl carbocation being formed at the more substituted position (Markovnikoff selectivity) (Step 1, arrows A and B). Attack of chloride ion at the carbocation (Step 2, arrow C) provides the vinyl chloride.

Notes An alternative mechanism involving three molecules of H-Cl is sometimes invoked which avoids the vinyl carbocation.

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Addition of HCl to alkynes twice to give geminal dichlorides Description: Addition of two equivalents of hydrogen chloride (HCl) to alkynes results in geminal dichlorides (1,1-dichlorides)

Notes: The reaction proceeds with Markovnikoff selectivity. “Geminal” means that the two chlorides are attached to the same carbon. Examples:

Notes: Note how in examples 1 and 4 the chlorides add to the more substituted carbon (Markovnikoff selectivity). In example 2 both carbons are equally substituted, but the reaction still produces one product since the alkyne is symmetrical. In example 3 both carbons are equally substituted, but two different products are obtained since the alkyne is not symmetrical. The last example shows that use of DCl instead of HCl gives a deuterated product (deuterium, for our purposes, behaves identically to hydrogen in chemical reactions). Mechanism: Attack of the alkyne π bond on HCl leads to the vinyl carbocation at the more substituted position (Markovnikoff selectivity) (Step 1, arrows A and B) followed by attack of the chloride on the carbocation (Step 2, arrows C) to give the vinyl chloride. When a second equivalent of HCl is present, the process then repeats itself. Protonation of the alkene (Step 3, arrows D and E) gives a carbocation, which is then attacked by chloride (Step 4, arrow F).

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Hydrogenation of Alkynes to Alkanes using Pd/C Description: Palladium on carbon (Pd/C) catalyzes the addition of hydrogen to C–C multiple bonds. For alkynes, this reaction will occur twice to give alkanes.

Notes: Platinum (Pt) is also a good catalyst for this reaction. Note that reaction can be halted at the alkene stage if one uses Lindlar’s catalyst or Na/NH3. Delivery of the hydrogens occurs syn to the alkene, but since the final product is a linear alkane, this is of no consequence. Examples:

Notes: Note that platinum (Pt) does the exact same reaction as Pd/C (example 4). Any other alkenes present will also be hydrogenated (example 3), so this procedure is not selective. In place of hydrogen gas (H2), deuterium gas (D2) may be used, which behaves essentially identically to hydrogen. This is a way of incorporating deuterium into the molecule. Mechanism: The detailed mechanism of this transformation is generally not considered “important” for the purposes of Org 1 / Org 2 although it is important to know that the reaction occurs on the surface of palladium metal. Both the alkyne and hydrogen are adsorbed onto the surface of the metal, and the hydrogen atoms are delivered to the same face of the π bond (syn addition).

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Note: For hydrogenation of alkenes, the general trend of reactivity follows this pattern. Increasing substitution on the alkene leads to slower reaction, due to greater inaccessibility of the alkene to the surface of the Pd catalyst (steric hindrance, in other words). Thank you to Daniel in the comments for correcting an earlier error of mine .

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Oxidative Cleavage of Alkynes with Ozone (O3) Description: Alkynes treated with ozone (O3) will be cleaved at the triple bond to form carboxylic acids. Terminal alkynes will give carbon dioxide (CO2)

Notes: The exact same reaction can be performed by KMnO 4 See also: Oxidative cleavage of alkynes with KMnO4 Examples:

Notes: For terminal alkynes (examples 1 and 4) the byproduct is carbon dioxide (CO 2) Mechanism: For the purposes of Org 1 / Org 2 the detailed mechanism is generally not considered to be “important”. However the mechanistic pathway is similar to that for the ozonolysis of alkenes.

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Oxidative Cleavage of Alkynes with KMnO4 Description: 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.

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Formation of alkynes through double elimination of vicinal dibromides Description: Sodium amide will convert 1,2-dihalides (“vicinal dihalides”) into alkynes through two consecutive elimination reactions.

Notes: The reaction is greatly assisted by heat. Also note that vicinal dihalides are formed through bromination of alkenes. So this is an important step on the pathway from converting an alkene into an alkyne. Bromides, chlorides, and iodides all work fine. Examples:

Notes: The reaction works well for iodides, bromides, and chlorides. Note that the byproducts of this reaction are NH3 and the sodium salt of the respective halide. Example 4 shows the formation of an internal alkyne. Example 5 shows how this reaction can be applied to form alkynes from alkenes in two steps, a common exam problem. KOH is sometimes used instead of NaNH2 ; one difference is that it typically require more heat since it’s a weaker base. Mechanism: Strong base (NaNH2) removes a proton from carbon, leading to formation of a C–C double bond and loss of halide (Step 1, arrows A, B and C). [This goes through an E2 mechanism]. A second equivalent of NaNH2 removes a second proton from the alkene carbon, leading to formation of the triple bond (Step 2, arrows D, E and F).

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Although not shown, technically, three equivalents of NaNH2 are required for the elimination to give terminal alkynes. That’s because as soon as the alkyne is formed, the strong base (NaNH2) will react with the alkyne C-H (pKa of 25, a much stronger acid than NH3) to give the acetylide. In order for the reaction to proceed to completion, then, an additional equivalent of NaNH2 is required. The reaction is quenched with a weak acid workup. Notes: It’s also reasonable to show removal of proton from C-2 first and then from C-1 (i.e. to reverse the order of the eliminations).

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Halogenation of Alkynes Description: Treatment of alkynes with one equivalent of a halogen (Cl2, Br2, I2) results in an alkenyl dihalide. The alkenyl dihalide can be converted to a tetra-halide through use of a second equivalent of the halogen.

Notes: The halogen (X2) here can be Cl2, Br2, or I2. The solvent is generally an inert halogenated solvent like CH2Cl2 or CCl4. Note that the stereochemistry of addition is anti. Addition of a second equivalent of halogen leads to formation of the tetra-halide product. Examples:

Notes: In the first example, Cl2 adds to the alkyne to give the trans product. The solvent is CH2Cl2 (dichloromethane). The second example shows halogenation by bromine (Br2) to give the trans dibromide. The third example shows halogenation by iodine to give the trans product. The fourth example shows addition of “excess” Br 2 (more than 2 equivalents) which is sufficient to produce the tetrahalogenated product. If one is careful with the number of equivalents, one can even perform two successive additions, as is done in the fifth example (chlorination, then bromination). Mechanism: The mechanism is essentially the same as halogenation of alkenes. In the first step, a bromonium ion is formed from the alkene (Step 1, arrows A, B, and C). In the second step, this bromonium ion is attacked by bromide, resulting in formation of the trans dibromide.

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The alkenyl dihalide can be halogenated if a second equivalent of halogen is added. For example, a second equivalent of Br2 leads to the formation of the tetrabromo product shown below.

Notes: Alkynes are less reactive than alkenes towards halogenation because the resulting intermediates are quite strained. Additionally they would be expected to have some “anti aromatic” character, a concept that doesn’t generally come up until second-semester organic chemistry.

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SN2 reaction of alkyl halides with hydroxide ions to give alcohols Description: Alkyl halides (or tosylates) will react with hydroxide ions to give alcohols. This is an S N2 reaction.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs) The counter-ion for the hydroxide ion is generally an alkali metal such as lithium, sodium (shown) or potassium. In order to avoid competition with the E2 reaction, a polar aprotic solvent like DMSO or acetone can be used. This is particularly important for secondary alkyl halides/tosylates. Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon (note the 3rd example). Mechanism: In the SN2 reaction here the nucleophile (HO–) attacks the carbon with the good leaving group, forming a C–OH bond and breaking the C–Br bond (Step 1, arrows A and B).

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SN2 reaction of alkyl halides with hydroxide ions to give alcohols Description: Alkyl halides (or tosylates) will react with hydroxide ions to give alcohols. This is an S N2 reaction.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs) The counter-ion for the hydroxide ion is generally an alkali metal such as lithium, sodium (shown) or potassium. In order to avoid competition with the E2 reaction, a polar aprotic solvent like DMSO or acetone can be used. This is particularly important for secondary alkyl halides/tosylates. Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon (note the 3rd example). Mechanism: In the SN2 reaction here the nucleophile (HO–) attacks the carbon with the good leaving group, forming a C–OH bond and breaking the C–Br bond (Step 1, arrows A and B).

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SN2 reaction of water with alkyl halides to give alcohols Description: Primary or methyl halides (or tosylates) will react with water to give alcohols. This is an S N2 reaction.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs) Examples:

Notes: Water is a poor nucleophile and generally isn’t strong enough to displace a leaving group from a secondary alkyl halide. Mechanism: In this reaction water attacks the carbon in an SN2 reaction displacing the leaving group (bromide in this case) (Step 1, arrows A and B). The bromide ion then removes a proton from the oxygen, providing the neutral alcohol and HBr.

Notes: It’s also reasonable to use water as the base here in the second step instead of Br(-). In fact it’s probably more reasonable to do so since water is a stronger base than Br(-).

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SN2 of Cyanide with Alkyl Halides to give Nitriles Description: Alkyl halides (or tosylates) will react with cyanide ion to give alkyl cyanides (nitriles) in an S N2 reaction.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs, OTf). Any alkali metal can be used (Li, Na, K…) as the counter-ion for the CN. For our purposes the exact identity is unimportant.

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon (note the last three examples). Mechanism: In the SN2 reaction the nucleophile CN(-) attacks the carbon with the good leaving group (C-1 in this case), displacing chloride ion int this example (Step 1, arrows A and B). The reaction is concerted, meaning both steps occur at the same time.

Notes: As with all SN2 reactions, the reaction is facilitated by a polar aprotic solvent such as DMSO, acetone, or acetonitrile. (Advanced) References and Further Reading 1. BENZYL CYANIDE Roger Adams and A. F. Thal Org. Synth. 1922, 2, 9 DOI: 10.15227/orgsyn.002.0009

A classic SN2 reaction of cyanide ion.

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SN2 reaction of hydrosulfide ion with alkyl halides to give thiols Description: Alkyl halides (or tosylates) will react with the hydrosulfide ion (HS–) to give thiols. Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs). Although Na is shown here as the counter-ion for HS(-), any alkali metal can be used (e.g. Li, Na, K) Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon (note examples 3 and 4). Example 5 is a little bit tricky! It involves two successive SN2 reactions to eventually give a cyclic thioether product. Mechanism: In the SN2 reaction the nucleophile (HS-) attacks the carbon with the good leaving group, forming a C–S bond and breaking the C–Br bond (Step 1, arrows A and B).

Notes: For more information on the SN2 see the summary sheet on this reaction. It isn’t crucial to show the sodium (Na) here.

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SN2 reaction of alkoxide ions with alkyl halides to give ethers (Williamson synthesis) Description: Alkyl halides (or tosylates) will react with alkoxy ions to form ethers. This reaction is called the Williamson ether synthesis.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs). The counter-ion on the alkoxy ion can be any alkali metal (e.g. Li, Na, K) Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon (note the last two examples). The best choice of solvent is usually the conjugate acid of the alkoxide. Note that in the second example that this ether would be difficult to make the opposite way (CH3O- attacking a tertiary alkyl bromide) since SN2 reactions don’t work on tertiary centers. Example 4 shows a deprotonation with NaH to give the alkoxide followed by addition of the alkyl halide. The sixth example is an intramolecular Williamson ether synthesis! Watch out for these types of examples as they are common exam questions. Mechanism: In the SN2 reaction the nucleophile (RO-) attacks the carbon with the good leaving group, forming a C–O bond and breaking the C–Br bond (Step 1, arrows A and B).

Notes: Again, the Na(+) is not crucial here, it’s just a spectator ion. >

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SN2 reaction of thiolates with alkyl halides to give thioethers (sulfides) Description: Alkyl halides (or tosylates) will react with thiolate ions to form sulfides [also known as thioethers]

•

Notes: Sulfides are the sulfur-containing equivalents of ethers. This reaction is analogous to the Williamson synthesis of ethers.

•

X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs)

•

The counter-ion for RS(-) can be any alkali ion, such as Li, Na (as shown) or K. Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon. The fourth example shows deprotonation of a thiol with strong base, followed by addition of the alkyl chloride. Note that the identity of the counter ion (Na+ or K+ ) doesn’t matter for our purposes. Mechanism: In the SN2 reaction here the nucleophile (RS-) attacks the carbon with the good leaving group, forming a C– S bond and breaking the C–Br bond (Step 1, arrows A and B).

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SN2 reaction between azide ion and alkyl halides to give alkyl azides Description: Alkyl halides (or tosylates) will react with azide ions (such as NaN3 or KN3) in an SN2 reaction to give alkyl azides.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs) Although Na is often shown as the counter-ion any alkali metal can also be used (e.g. Li, Na, K) Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon. The second example is still (-)N3, it just shows the azide ion drawn out. Mechanisms: In the SN2 reaction the nucleophile (N3-) attacks the carbon with the good leaving group, forming a C–N bond and breaking the C–I bond (Step 1, arrows A and B).

Notes: For more information on the SN2 see the summary sheet on this reaction. For our purposes using Na or K is immaterial.

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SN2 reaction of carboxylate ions with alkyl halides to give esters Description: Alkyl halides (or tosylates) will react with carboxylate ions to form esters. This is an S N2 reaction.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs) The counter-ion on the carboxylate can be any alkali ion, such as Li, Na (pictured) or K. Examples:

Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuation at the carbon (note examples 3 and 4). The second-to-last example is an intramolecular example showing the formation of a new seven-membered ring, a cyclic ester called a lactone. The last example is actually a really simple example of a primary alkyl halide; note that despite the complex structure, only one bond is being formed and broken here! Mechanism: In the SN2 reaction the nucleophile (RCOO-) attacks the carbon with the good leaving group, forming a C–O bond and breaking the C–Br bond (Step 1, arrows A and B).

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SN2 Reaction of Acetylide Ions with Alkyl Halides Description: Alkyl halides treated with acetylide ions (the conjugate bases of acetylenes) will undergo S N2 reactions to give alkynes.

Notes: This works best for primary or methyl alkyl halides. Sodium (Na) is not crucial, it’s just a spectator ion. See also: Deprotonation of alkynes with strong base to give acetylide ions. Examples:

Notes: Be prepared to see the alkyl halide either as the reactant (example 1) or above the arrow (example 2). It is also common to see the alkyne as starting material with a two-step deprotonation/SN2 process (example 4). Since this is an SN2 reaction, everything you’ve previously learned about S N2 reactions still applies. Example 5 with one equivalent of nucleophile will react preferentially to displace the better leaving group Br(-). Finally this reaction works best for primary alkyl halides. Since the acetylide is such a strong base, secondary and tertiary alkyl halides tend to get deprotonated instead (example 6). Mechanism: Acetylide ions are great nucleophiles. They will perform S N2 reactions on alkyl halides (Step 1, arrows A and B).

Notes: Really important! There aren’t many ways of making C–C bonds in Org 1, but this is one of them. This is one of the most USEFUL ways of making C–C bonds. Furthermore, alkynes are really versatile and can be turned into all kinds of other functional groups. See: Alkynes Are A Blank Canvas

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SN2 reaction of organocuprates (Gilman reagents) with alkyl halides to give alkanes Description: Alkyl halides (or tosylates) will react with organocuprates (Gilman reagents) to form alkanes in an S N2 reaction

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs) Also note that this can be a very useful way of forming carbon-carbon bonds. Examples:

Notes: The byproduct of all of these reactions is the lithium salt (e.g. LiBr) and the organocopper species (e.g. CH3Cu). Note the bottom example. Unlike Grignards, no reaction occurs between Gilman reagents and aldehydes/ketones. Mechanism: This reaction is actually a fairly straightforward SN2 reaction, with breakage of the C–Cu bond accompanying attack at the alkyl halide (Step 1, arrows A and B).

Notes: Again note that this forms a carbon-carbon bond. This is important as there aren’t too many reactions that do so, through an SN2 (another example is the reaction of alkyl halides with acetylide ions).

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Acidic cleavage of ethers (SN2) Description: When ethers are treated with strong acid in the presence of a nucleophile, they can be cleaved to give alcohols and alkyl halides. If the ether is on a primary carbon this may occur through an S N2 pathway.

Notes: Common acids for this purpose are HI and other hydrogen halides, as well as H 2SO4 in the presence of H2O. In the case where the ether being cleaved is secondary and has a stereocenter, there will be inversion of configuration. Examples:

Notes: The third example could also be written “H3O+” . Note that excess HI will convert primary alcohols to alkyl halides via SN2, but not phenol (C6H5OH) since sp2 hybridized carbons do not undergo SN1 or SN2 reactions. Mechanism: Strong acid (HI) protonates the ether oxygen, which turns it into a better leaving group (Step 1, arrows A and B). Next, the iodide ion attacks the carbon in an SN2 reaction (Step 2, arrows C and D) to give the alcohol and methyl iodide.

Notes: In cases where the ether being cleaved is secondary and has a stereocenter, there will be inversion of configuration. Another example: Opening of tetrahydrofuran (THF) with aqueous acid:

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SN2 reaction of amines with alkyl chlorides to give ammonium salts Description: Amines will react with alkyl halides to give ammonium salts. Importantly, this occurs even if only one equivalent of the alkyl halide and amine are used.

Notes: alkylation of amines with alkyl halides is not generally considered to be a useful reaction, because the reaction cannot be stopped after just one addition of alkyl halide. It will continue until the ammonium salt is obtained, and these have limited utility. Note that X here is halide (Cl, Br, I) or sulfonate (OTs, OMs) Examples:

Notes: It’s generally not possible to “stop” the reaction after reaction with one equivalent of alkyl halide. Again, this is why it’s not generally considered useful since there aren’t many synthetic applications of ammonium salts. Note that since this is an SN2 reaction and proceeds through backside attack, there will be inversion of configuration at the carbon. Example 3 is an intramolecular reaction; DMSO (dimethyl sulfoxide) is a solvent. The last example is kind of a trick question; it’s just protonation of the amine with HBr. Mechanism: The amine performs an SN2 on the alkyl halide (Step 1, arrows A and B) forming a positively charged nitrogen (“ammonium”). This is deprotonated by a second equivalent of the amine (Step 2, arrows C and D) giving a neutral secondary amine which can now perform a second S N2 on the alkyl halide (Step 3, arrows E and F). This results in a new ammonium salt, which is then deprotonated by a second equivalent of base (Step 4, arrows G and H) giving a tertiary amine, which then performs a third nucleophilic substitution (Step 5, arrows I and J).

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Notes: It’s not possible to get this reaction to stop after one alkylation because the tertiary amine (formed after step 4) is a better nucleophile than the secondary amine (formed after step 2) which is a better nucleophile than the primary amine (in step 1).

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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.

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Sharpless Epoxidation Description: The Sharpless Epoxidation is an enantioselective epoxidation of allylic alcohols.

Notes: The Sharpless epoxidation only works for alkenes adjacent to an alcohol (CH 2OH). 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

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epoxidation of the top face, whereas (+)-DET will lead to epoxidation of the bottom face.

Notes: If you’re really interested in learning more about how this reaction works, along with many excellent examples, this handout by the Myers Group at Harvard University is phenomenal (as are their other handouts)

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