Ozonolysis of Alkenes 1

When 2-methyl-2-pentene reacts with ozone, the initial 1,2,3-trioxolane product is 144, but thisrearranges to ozonide 145. If 145 is treated with hydrogen peroxide as above, one of the cleavage products is acetone, asexpected, but the other product is propanoic acid. One might expect that treatment of ozonide 145 with hydrogen peroxide should give acetoneand the aldehyde. However, aldehydes are susceptible to oxidation by many reagents, includingoxygen in the air (see chapter 17 for oxidations). if an ozonide is formed from an alkene that has a hydrogen on a C=C carbon, and thentreated with an oxidizing agent such as hydrogen peroxide, the final product is acarboxylic acid not an aldehyde. If 145 is treated with dimethyl sulfide, the products are acetone and propanal rather thanacetone and propanoic acid. Hydrogen peroxide is an oxidizing agent and dimethyl sulfide is a reducing agent when usedwith an ozonide. When there is a hydrogen atom on the C=C unit, ozonolysis followed by oxidation leadsto a carboxylic acid, but ozonolysis and then reduction leads to an aldehyde.

OOHO

H2O2

O

O

OH

OOO

HH

O3

CH3SCH3 OO

H

+144 145

+

1. O3 , 78C

1. O3 , 78C

1. O3 , 78C

2. CH3-S-CH3

2. H2O2

2. Me2S

1. O3 , 78C2. H2O2

OO H

OO OHH

O

O

H

O

O

HOH

Ozonolysis - Reactions

Formation of Radicals 3

A radical contains one electron in an orbital, and it cantheoretically be tetrahedral, planar, or 'in between'. It is generally conceded that carbon radicals without significantsteric encumbrance are planar, as represented by 147. In terms of its reactivity, radical 147 may be considered electronrich or electron poor. In most of its reactions, the electron-deficient characterization isuseful for predicting products. In other words, radicals such as this are not nucleophilic.

C RRR

147

Formation of Radicals 4

Radicals can be formed by dissociative homolytic cleavage(one electron is transferred to each adjacent atom from the bond)as a key step, as depicted by X-Y, leaving two radical products. Another major route to radical intermediates involves theequilibrium reaction of a radical (X) and a neutral molecule (X-Y),producing a new radical (Y) and a new neutral molecule (X-X).

X Y + YX

X X Y X X + Y

Formation of Radicals 5

There are a number of reagents that generate radicals when heated or exposed to light (thesymbol for a photon of light is h). When heated, many peroxides such as alkyl hydroperoxide 79 and dialkylperoxide 80 undergohomolytic cleavage to generate radicals RO + OH or two molar equivalents of RO, respectively. Heating of tert-butylhydroperoxide (148) gives the tert-butoxy radical and the hydroxyl radical. Heating dibenzoyl peroxide (149) gives two molar equivalents of the acyl radical 150. Peroxide 149 is classified as a diacyl peroxide, but acyl compounds of this type will bediscussed in chapter 16 (section 16.7) in the context of carboxylic acid derivatives. A useful reagent that will generate radicals is an azo compound called azobis-isobutyronitrile(AIBN, 151). When heated, AIBN easily decomposes to produce two molar equivalents of the radical 152,along with nitrogen gas, which escapes from the reaction.

OO

H

N NC!N

MeMe

N!CMe

Me

OO

O

O

OO

H

O

O

C!N

MeMe

2heat + N!N

151 152

148

149

heat

heat

2

150

Radicals, HBr and Alkenes 6

Hydrogen bromide (HBr) is added to undec-10-enoic acid (153) in a hydrocarbonsolvent, in the presence of benzoyl peroxide (149). When the product isolated, 11-bromoundecanoic acid (154) is obtained in 70% yield.The bromine is attached to the less substituted carbon. Since Markovnikov's rule places the hydrogen atom on the less substituted carbonatom of the C=C unit, and the bromine on the more substituted, formation of 154 isexactly the opposite result, an anti-Markovnikov addition.

(CH2)8CO2H

Ph O O PhO

O(CH2)8

CO2HBr

153

149

154hydrocarbon solvent HBr

If a carbocation were formed, the reaction must follow Markovnikov addition,consistent with formation of the more stable secondary carbocation.

Therefore, the reaction must follow a different mechanism, presumably with adifferent intermediate, but what?

Note that a peroxide is added, and this appears to be responsible forchanging the normal reaction of HBr and an alkene.

Radicals, HBr and Alkenes 7

Some chemical process must occur before HBr can react with the alkene. Logically, this event involves the new additive, the peroxide, which is known toundergo homolytic bond fragmentation to produce radicals. If radicals are involved, does the bromine go in first or second? If a secondary radicalis assumed to be more stable than a primary radical (as with carbocations which areelectron deficient), the most reasonable mechanism generates a bromine radical (Br),which reacts with the C=C unit of the alkene. The bromine must add to the C=C unit before the H in order to generate a secondaryradical, and this will place bromine on the less substituted carbon.

(CH2)8CO2H

Ph O O PhO

O(CH2)8

CO2HBr

153

149

154hydrocarbon solvent HBr

Radicals, HBr and Alkenes 8

The mechanism involves heating 149to give the radical, which reacts with HBrto form benzoic acid and the bromineradical, Br. The bromine radical reacts with thealkene unit in 153 to give a secondarycarbon radical rather than a less stableprimary radical. The secondary radicalis expected to be more stable than aprimary radical. In the radical process a covalentbond is formed, but there are twoelectrons in the new -bond. Oneelectron is donated by Br and theother by the C=C -bond, leavingbehind an unpaired electron onthe carbon, a carbon radical.

All that remains is for radical 155 toreact with the hydrogen atom from another molecule of HBr to give 154 and another equivalent of the bromineradical. This mechanistic sequence is described as a chain radical reaction, and it is important to note than when areaction generates a radical product, the process continues. When the reaction generates a neutral product, as Br + Br Br-Br, the radical reaction stops.

Br

Ph OO

(CH2)8CO2HBr

(CH2)8CO2H

Ph O O PhO

O

HBr

HBr

Ph OHO

(CH2)8CO2HBr

Ph OO

(CH2)8CO2HBr

153

149

154

2

155

+ Br

155

+ Br

Br Br Br-Br radical coupling

H

benzoic acid

This reaction work wells with HBr, but not with HCl or HI because the bromineradical reacts in a selective manner.

Carbocation versus Radical 9

HBr

HBr

HBrt-BuOOt-Bu

BrH

Br

Br

156 157

158 159 160Br

H HH

The reaction of 3-methyl-1-pentene and HBr in the presence of di-tert-butyl peroxide(Me3COOCMe3, another peroxide radical precursor) leads to 157 via a radical chainmechanism and radical intermediate 156.

Note that unlike carbocation intermediates, radical intermediates such as 156 donot rearrange.

By contrast, when 3-methyl-1-pentene reacts with HBr without the peroxide, reactiongives the expected secondary carbocation 158, which rearranges to the more stabletertiary carbocation 159.

Final coupling with bromide ion gives 160.

Reaction with HBr via the carbocation leads to 160 and reaction with HBr andperoxide via the radical leads to 157.

Chapter 11. Nucleophiles: LewisBase-Like Reactions At sp3

Carbon

10

Reaction of a nucleophile with an electrophilic carbon atom is not limited to carbocations. A nucleophile can also donate electrons to a polarized carbon atom (C+) such as the one in 3,where the presence of an electronegative atom X (such as Cl or Br) generates an induced dipole atcarbon. The nucleophilic iodide ion donates two electrons to the positive carbon in 3 to form a new CI -bond in 4. However, if a new bond is formed to a carbon that has four covalent bonds, one of those bondsmust break. The relatively weak CX bond breaks as the CI bond is formed, and the products are alkyl iodide4 and the X ion. The conversion of 3 to 4 constitutes a new type of reaction, a substitution at an sp3 hybridizedcarbon.

I C X!+ !

CI X

3 4

+

To begin, you should know: 11

Understand IUPAC nomenclature for alkyl halides, ethers, alcohols, and alkynes.(chapter 4, section 4.6, chapter 5, sections 5.2, 5.6) Understand the role of bond polarization and dipole to generate a +CX species.(chapter 3, section 3.7) Understand the basics of conformation applied to both acyclic molecules and cyclicmolecules of ring sizes of 3-6. (chapter 8, sections 8.1, 8.2, 8.4, 8.5) Define and recognize a nucleophile. (chapter 6, section 6.7) Understand that a nucleophile can donate two electrons to a positive carbon or to acarbon with a positive dipole, to form a new bond to that carbon. (chapter 6,, section 6.7) An alkene reacts with HCl, HBr or HI to form a carbocation. (chapter 10, section 10.2) Be able to determine the relative stability of carbocation intermediates. (chapter 7,section 7.4.A and chapter 10, section 10.2) Once formed, an intermediate carbocation reacts with chloride ion, bromide ion oriodide in to form a new C-X bond. (chapter 10, section 10.2) Be able to recognize a transition state. (chapter 7, section 7.3). Understand that a mechanism is a step by step description of a reaction that includesall intermediates and all chemical steps along the way to the final, isolated product.(chapter 7, section 7.8) Identify a stereogenic center, absolute configuration, and name stereoisomers. (chapter9, sections 9.1, 9.3) Identify and name diastereomers. (chapter 9, section 9.5)

When completed, you should know: 12

The C-X bond of alkyl halides and sulfonate esters is polarized such that the carbon has apositive dipole. Halides and sulfonate anions are good leaving groups. Nucleophilesattack primary and secondary alkyl halides, displacing the leaving group in what is knownas aliphatic, bimolecular nucleophilic substitution, the SN2 reaction. The SN2 reaction isfollows second order kinetics, has a transition state rather than an intermediate, andproceeds via backside attack of the nucleophile on the halide and inversion ofconfiguration. Due to steric hindrance in the penta-coordinate transition state, tertiary halides do notundergo the SN2 reaction and primary halides undergo the reaction fastest, with secondaryhalides reactive but less so than primary halides. SN2 Reactions are faster in aprotic solvents and slower in protic solvents. Water tendsto promote ionization. In protic media, particularly aqueous media, ionization of tertiaryhalides, and more slowly with secondary halides, occurs to give a carbocationintermediate. Carbocation intermediates can be trapped by nucleophiles in what is known at an SN1reaction. An SN1 reaction proceeds by ionization to a planar carbocation containing ansp2-hybridized carbon, follows first order kinetics, and proceeds with racemization of achiral center. Carbocations are subject to rearrangement o a more stable cation via 1,2-hydrogen or alkyl shifts. A variety of nucleophiles can be used in the substitution reactions, includes halides,alkoxides, amines, phosphines, azides, cyanide, acetylides, and enolate anions. Alkyl halides are prepared by the reaction of alcohols with mineral acids (HCl and HBr) orwith reagents such as thionyl chloride, thionyl bromide, PX3 and PX5. Alcohols are alsoprepared by the reaction of alkenes with HCl, HBr or HI, but alkyl sulfates, alkyl nitratesand alkyl perchlorates from those mineral acids tend to be unstable.

When completed, you should know: 13

Ethers are generally unreactive except with strong acids such as HI and HBr, which leadsto cleavage of the ether to an alcohol and an alkyl halide. Epoxides are particularlyreactive with nucleophiles, which open the three-membered ring at the less substitutedcarbon. Epoxides also react with an acid catalyst and weak nucleophiles such as water oralcohols, as well as with cyanide, azide. etc. Exposure to light or heating to 300C leads to homolytic cleavage of diatomic chlorineand bromine to give chlorine or bromine radicals. Chlorine and bromine radicals react withalkanes, removing hydrogen atoms via a radical process that leads to substitution andformation of alkyl chlorides and alkyl bromides. Both NBS and NCS can be used forcontrolled radical bromination or chlorination of allylic and benzylic systems. A molecule with a particular functional group can be prepared from molecules containingdifferent functional groups by a series of chemical steps (reactions). This process iscalled synthesis, such that the new molecule is synthesized from the old one (see chapter25). Spectroscopy can be used to determine the structure of a particular molecule, and candistinguish the structure and functionality of one molecule when compared with another.See chapter 14.

lc

Iodide Ion displaces a Bromide 14

An old and interesting experiment mixes 5 with sodium iodide (NaI) using acetone asa solvent. This mixture is heated at the boiling point of acetone (refluxed) and the isolatedproduct is 1-iodo-3-methylbutane (6), in 66% yield. Sodium bromide (NaBr) is formed during the reaction, as the sodium iodide isconsumed. In terms of the structural changes, iodide substitutes for the bromine, producingbromide ion (Br). Iodide is a nucleophile since it reacts at C+, breaking the CBr bond and transferringthe electrons in the CBr bond to bromine. This transformation constitutes a new type of reaction that is This transformation constitutes a new type of reaction that is calledcalled nucleophilic nucleophilicaliphatic substitutionaliphatic substitution

Br I+ Na Br+

acetone

5 6

!+

!"

I Na

Nucleophiles and Alkyl Halides 15

An early experiment by Hughes, Ingold [Edward D. Hughes, England (Wales), 1906-1962;Christopher K. Ingold, England, (1893-1970)] and a student named and Masterman reacted (+)-2S-bromooctane (7) with sodium ethoxide (NaOEt) in ethanol, and obtained the correspondingether, 2R-ethoxyoctane, 8. This experiment introduces alkoxide nucleophile (RO), which is the conjugate base of analcohol. The electron rich oxygen of the ethoxide ion (OCH2CH3) attacks the carbon bearing thebromine in 7, displacing the bromine. Ethanol is a solvent and does not appear in the final product. The interesting observation is than the product, ether 8, is also optically active but the absoluteconfiguration is R. The actual experiment reacted 7, with a measured specific rotation of 24.54, with sodiumethoxide to give 8, and the specific rotation was measured to be +17.1. The inversion of configuration is observed by monitoring the specific rotation of both thereactant (7) and the final product, (7). The change in sign indicated that inversion ofconfiguration occurred during the conversion of the S-bromide to the R-ether. The experimental data shows that complete inversion of configuration is observed. Apart from the substitution of a leaving group with a nucleophile, a mechanism must explainthis result.

Br HNa+ OEt

EtOH

H OCH2CH3

7 8+ NaBr

Nucleophilic strength:

I > Br > Cl > F (OPPOSITE BASICITY)

Charged >> neutral: HO >> HOH H2N >> H3N

Increases to the left in the Periodic Table F < HO < H2N >>> H4C

Nucleophilicity

Leaving Groups 17

The C-X bond (X = halogen) is polarized and it is relativelyweak. Moreover, when groups such as Cl, Br or I "leave" theyform a very stable halide ion. Since halogens are electronegative, they can easilyaccommodate the negative charge, and the larger ions dispersethe charge more effectively than smaller ions over a larger surfacearea, again leading to stabilization. Therefore, when a nucleophile attacks the sp3 carbon of thehalide, breaking the C-X bond generates a very stable ionicproduct, which facilitates the substitution. Most halides are good leaving groups, with iodide the best andfluoride the poorest leaving group in this series.