1 zonolysis of Alkenes 1 When 2-methyl-2-pentene reacts with ozone, the initial 1,2,3-trioxolane product is 144, but this rearranges to ozonide 145. If 145 is treated with hydrogen peroxide as above, one of the cleavage products is acetone, as expected, but the other product is propanoic acid. ne might expect that treatment of ozonide 145 with hydrogen peroxide should give acetone and the aldehyde. owever, aldehydes are susceptible to oxidation by many reagents, including oxygen 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 then treated with an oxidizing agent such as hydrogen peroxide, the final product is a carboxylic acid not an aldehyde. If 145 is treated with dimethyl sulfide, the products are acetone and propanal rather than acetone and propanoic acid. ydrogen peroxide is an oxidizing agent and dimethyl sulfide is a reducing agent when used with an ozonide. When there is a hydrogen atom on the C=C unit, ozonolysis followed by oxidation leads to a carboxylic acid, but ozonolysis and then reduction leads to an aldehyde C 3 SC
2 zonolysis - Reactions 1. 3, 78 C 2. C 3 -S-C , 78 C , 78 C 2. Me 2 S 1. 3, 78 C
3 Formation of Radicals A radical contains one electron in an orbital, and it can theoretically be tetrahedral, planar, or 'in between'. It is generally conceded that carbon radicals without significant steric encumbrance are planar, as represented by 147. In terms of its reactivity, radical 147 may be considered electron rich or electron poor. In most of its reactions, the electron-deficient characterization is useful for predicting products. In other words, radicals such as this are not nucleophilic. 3 R R C 147 R
4 Formation of Radicals 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 the equilibrium reaction of a radical (X) and a neutral molecule (X-Y), producing a new radical (Y) and a new neutral molecule (X-X). 4 X Y X + Y X X Y X X + Y
5 Formation of Radicals 5 There are a number of reagents that generate radicals when heated or exposed to light (the symbol for a photon of light is hν). When heated, many peroxides such as alkyl hydroperoxide 79 and dialkylperoxide 80 undergo homolytic cleavage to generate radicals R + or two molar equivalents of R, respectively. eating of tert-butylhydroperoxide (148) gives the tert-butoxy radical and the hydroxyl radical. eating 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 be discussed 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. 148 heat heat N!C Me Me C!N C!N heat 2 N N Me Me Me Me N!N
6 Radicals, Br and Alkenes 6 ydrogen bromide (Br) is added to undec-10-enoic acid (153) in a hydrocarbon solvent, 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 carbon atom of the C=C unit, and the bromine on the more substituted, formation of 154 is exactly the opposite result, an anti-markovnikov addition. (C 2 ) 8 C Ph Ph 149 hydrocarbon solvent Br Br (C 2 ) 8 C 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 a different intermediate, but what? Note that a peroxide is added, and this appears to be responsible for changing the normal reaction of Br and an alkene.
7 Radicals, Br and Alkenes 7 Some chemical process must occur before Br can react with the alkene. Logically, this event involves the new additive, the peroxide, which is known to undergo homolytic bond fragmentation to produce radicals. If radicals are involved, does the bromine go in first or second? If a secondary radical is assumed to be more stable than a primary radical (as with carbocations which are electron 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 in order to generate a secondary radical, and this will place bromine on the less substituted carbon. Ph Ph (C 2 ) 8 C hydrocarbon solvent Br Br (C 2 ) 8 C 2 154
8 The mechanism involves heating 149 to give the radical, which reacts with Br to form benzoic acid and the bromine radical, Br. The bromine radical reacts with the alkene unit in 153 to give a secondary carbon radical rather than a less stable primary radical. The secondary radical is expected to be more stable than a primary radical. In the radical process a covalent Br bond is formed, but there are two electrons in the new σ-bond. ne electron is donated by Br and the other by the C=C π-bond, leaving behind an unpaired electron on the carbon, a carbon radical. Radicals, Br and Alkenes Ph Br Ph Ph Br (C 2 ) 8 C (C 2 ) 8 C Br 2 Ph Br Ph Br C 2 (C 2 ) C 2 (C 2 ) radical coupling Br Br Br-Br benzoic acid 8 + Br + Br All that remains is for radical 155 to react with the hydrogen atom from another molecule of Br to give 154 and another equivalent of the bromine radical. This mechanistic sequence is described as a chain radical reaction, and it is important to note than when a reaction generates a radical product, the process continues. When the reaction generates a neutral product, as Br + Br Br-Br, the radical reaction stops. This reaction work wells with Br, but not with Cl or I because the bromine radical reacts in a selective manner.
9 Carbocation versus Radical 9 Br Br Br Br t-but-bu Br Br Br The reaction of 3-methyl-1-pentene and Br in the presence of di-tert-butyl peroxide (Me 3 CCMe 3, another peroxide radical precursor) leads to 157 via a radical chain mechanism and radical intermediate 156. Note that unlike carbocation intermediates, radical intermediates such as 156 do not rearrange. By contrast, when 3-methyl-1-pentene reacts with Br without the peroxide, reaction gives the expected secondary carbocation 158, which rearranges to the more stable tertiary carbocation 159. Final coupling with bromide ion gives 160. Reaction with Br via the carbocation leads to 160 and reaction with Br and peroxide via the radical leads to 157.
10 Chapter 11. Nucleophiles: Lewis Base-Like Reactions At sp 3 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 δ+ ) such as the one in 3, where the presence of an electronegative atom X (such as Cl or Br) generates an induced dipole at carbon. The nucleophilic iodide ion donates two electrons to the positive carbon in 3 to form a new C I σ- bond in 4. owever, if a new bond is formed to a carbon that has four covalent bonds, one of those bonds must break. The relatively weak C X bond breaks as the C I bond is formed, and the products are alkyl iodide 4 and the X ion. The conversion of 3 to 4 constitutes a new type of reaction, a substitution at an sp 3 hybridized carbon. I C X I!+! 3 4 C + X
11 To begin, you should know: 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 δ+ C X δ species. (chapter 3, section 3.7) Understand the basics of conformation applied to both acyclic molecules and cyclic molecules 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 a carbon with a positive dipole, to form a new bond to that carbon. (chapter 6,, section 6.7) An alkene reacts with Cl, Br or I 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) nce formed, an intermediate carbocation reacts with chloride ion, bromide ion or iodide 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 includes all 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. (chapter 9, sections 9.1, 9.3) Identify and name diastereomers. (chapter 9, section 9.5) 11
12 When completed, you should know: 12 The C-X bond of alkyl halides and sulfonate esters is polarized such that the carbon has a positive dipole. alides and sulfonate anions are good leaving groups. Nucleophiles attack primary and secondary alkyl halides, displacing the leaving group in what is known as aliphatic, bimolecular nucleophilic substitution, the S N 2 reaction. The S N 2 reaction is follows second order kinetics, has a transition state rather than an intermediate, and proceeds via backside attack of the nucleophile on the halide and inversion of configuration. Due to steric hindrance in the penta-coordinate transition state, tertiary halides do not undergo the S N 2 reaction and primary halides undergo the reaction fastest, with secondary halides reactive but less so than primary halides. S N 2 Reactions are faster in aprotic solvents and slower in protic solvents. Water tends to promote ionization. In protic media, particularly aqueous media, ionization of tertiary halides, and more slowly with secondary halides, occurs to give a carbocation intermediate. Carbocation intermediates can be trapped by nucleophiles in what is known at an S N 1 reaction. An S N 1 reaction proceeds by ionization to a planar carbocation containing an sp 2 -hybridized carbon, follows first order kinetics, and proceeds with racemization of a chiral 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 (Cl and Br) or with reagents such as thionyl chloride, thionyl bromide, PX 3 and PX 5. Alcohols are also prepared by the reaction of alkenes with Cl, Br or I, but alkyl sulfates, alkyl nitrates and alkyl perchlorates from those mineral acids tend to be unstable.
13 When completed, you should know: Ethers are generally unreactive except with strong acids such as I and Br, which leads to cleavage of the ether to an alcohol and an alkyl halide. Epoxides are particularly reactive with nucleophiles, which open the three-membered ring at the less substituted carbon. Epoxides also react with an acid catalyst and weak nucleophiles such as water or alcohols, as well as with cyanide, azide. etc. Exposure to light or heating to 300 C leads to homolytic cleavage of diatomic chlorine and bromine to give chlorine or bromine radicals. Chlorine and bromine radicals react l c with alkanes, removing hydrogen atoms via a radical process that leads to substitution and formation of alkyl chlorides and alkyl bromides. Both NBS and NCS can be used for controlled radical bromination or chlorination of allylic and benzylic systems. A molecule with a particular functional group can be prepared from molecules containing different functional groups by a series of chemical steps (reactions). This process is called synthesis, such that the new molecule is synthesized from the old one (see chapter 25). Spectroscopy can be used to determine the structure of a particular molecule, and can distinguish the structure and functionality of one molecule when compared with another. See chapter
14 Iodide Ion displaces a Bromide 14 An old and interesting experiment mixes 5 with sodium iodide (NaI) using acetone as a solvent. This mixture is heated at the boiling point of acetone (refluxed) and the isolated product is 1-iodo-3-methylbutane (6), in 66% yield. Sodium bromide (NaBr) is formed during the reaction, as the sodium iodide is consumed. In terms of the structural changes, iodide substitutes for the bromine, producing bromide ion (Br ). Iodide is a nucleophile since it reacts at C δ+, breaking the C Br bond and transferring the electrons in the C Br bond to bromine. This transformation constitutes a new type of reaction that is called nucleophilic aliphatic substitution!+ + I Na acetone Br!" 5 6 I + Na Br
15 Nucleophiles and Alkyl alides 15 An early experiment by ughes, Ingold [Edward D. ughes, England (Wales), ; Christopher K. Ingold, England, ( )] and a student named and Masterman reacted (+)- 2S-bromooctane (7) with sodium ethoxide (NaEt) in ethanol, and obtained the corresponding ether, 2R-ethoxyoctane, 8. This experiment introduces alkoxide nucleophile (R ), which is the conjugate base of an alcohol. The electron rich oxygen of the ethoxide ion ( C 2 C 3 ) attacks the carbon bearing the bromine 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 absolute configuration is R. The actual experiment reacted 7, with a measured specific rotation of 24.54, with sodium ethoxide to give 8, and the specific rotation was measured to be The inversion of configuration is observed by monitoring the specific rotation of both the reactant (7) and the final product, (7). The change in sign indicated that inversion of configuration 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 explain this result. Br 7 Na + Et Et 8 C 2 C 3 + NaBr
16 Nucleophilic strength: Nucleophilicity I > Br > Cl > F Charged >> neutral: (PPSITE BASICITY) >> 2 N >> 3 N Increases to the left in the Periodic Table F < < 2 N <<< 3 C note C 4 is not a nucleophile - no electrons to donate so, 2 N >>>>> 4 C
17 Leaving Groups The C-X bond (X = halogen) is polarized and it is relatively weak. Moreover, when groups such as Cl, Br or I "leave" they form a very stable halide ion. Since halogens are electronegative, they can easily accommodate the negative charge, and the larger ions disperse the charge more effectively than smaller ions over a larger surface area, again leading to stabilization. Therefore, when a nucleophile attacks the sp 3 carbon of the halide, breaking the C-X bond generates a very stable ionic product, which facilitates the substitution. Most halides are good leaving groups, with iodide the best and fluoride the poorest leaving group in this series. 17