Alkenes as Lewis bases Dihalogenation Halonium ions

Transcription

1 Alkenes as Lewis bases Dihalogenation alonium ions

2 Polarizable Dihalogens 2 37!+! C 4 38 When cyclohexene is mixed with elemental bromine, in carbon tetrachloride as a solvent, the product is trans-1,2- dibromocyclohexane (38) isolated in 57% yield. Diatomic halogens such as chlorine ( ), bromine ( ) and iodine (I-I) are highly polarizable. In the absence of another molecule, there is no difference in electronegativity between the atoms in - and -, or I I. When the halogen is in close proximity to an electron rich atom or group, the halogen atom closest to the electron source becomes δ+ while the other becomes δ. This means that the atoms are polarized via an induced dipole.

3 Fluorine 3 For many years, elemental fluorine was thought to be too reactive and too dangerous for reaction with alkenes. To avoid such problems, fluorine is typically mixed with an inert gas such as nitrogen or argon. Diluted in this manner, fluorine does react with alkenes, but the yields are often poor, and in some cases solvents for the alkene, such as methanol, participate in the reaction. 1-Phenylethene (PhC=C 2 ), for example reacted with fluorine in methanol to give 51% of the corresponding difluoride, along with 49% of 2-fluoro-1-methoxy-1-phenylpropane. ecause of problems associated with fluorine, this chapter will report alkene reactions only for chlorine, bromine or iodine and not fluorine.

4 alonium Ions 4 induced dipole!+!" C 4!+!!+!!! transition state Cation 40 is the experimentally detected cation intermediate mentioned above. In the transition state for this reaction (39), the second carbon of the alkene π-bond develops positive character as the π-electrons are transferred to bromine. The bromine donates two electrons to that positive carbon (called back donation) to form a second C- bond in 40 and a threemembered ring with a formal charge of +1, with a bromide ion as the counterion. The product is not the carbocation (with the positive charge on carbon) but rather the three-membered ring cation 40 called a bromonium ion (one type of halonium ion), with the positive charge on bromine. 38

5 alonium Ions 5 induced dipole!+!"!!+!!! C 4! transition state In reactions with alkenes, bromine gives a bromonium ion, chlorine gives a chloronium ion, iodine gives an iodonium ion, and generically a halogen gives a halonium ion. It is important to understand that 40 Is generated in this nonpolar solvent because it is more stable than the carbocation intermediate.

6 trans-dihalides 6 In 40, the large bromine atom must reside on one side of the cyclohexane ring, or the other, due to the nature of the threemembered ring and the inability to undergo C C bond rotation. When bromide ion attacks an electrophilic carbon atom of the three-membered ring, it must do so from the sterically less hindered side opposite the first bromine atom (as in 40C) leading to formation of a new C bond, with cleavage of the three-membered ring to form the trans dibromide. Note that the other diastereomer (the cis-dibromide) is not formed in this reaction. 40A 40 40C

7 trans-dihalides 7 Only the trans diastereomer formed, so this reaction is diastereospecific. If cyclohexene is viewed from the "side", as in 41, it is clear that the initial reaction with bromine must deliver the of the bromonium ion to one on side of the ring or the other. The may be on either the "top" or the "bottom" since there is no facial bias in the C=C unit of cyclohexene. If bromine is arbitrarily drawn on the "bottom," as in 42 (compare this structure with 40C with the bromine on the top ), nucleophilic attack by the bromide ion will occur from the opposite "top" face since that is less sterically hindered. This backside attack leads to the trans stereochemistry in 43. Note that attack from the top or bottom can occur with equal facility, but a transdibromide is produced in both cases. Attack from the top face leads to one enantiomer and attack from the bottom face leads to the other enantiomer, and attack from either face occurs with equal facility. The reaction must produce a racemic mixture

8 Dihalogenation is Stepwise 8 2 C 4 2 C 4 + +

9 The Reaction with Acyclic Alkenes is Diastereospecific In cis-2-butene, the two methyl groups are "locked" on one side of the C=C unit since there is no rotation around those carbon atoms. The key to the stereoselectivty is the fact that the stereochemical relationship of the groups on the C=C unit is retained in the transition state that leads to the bromonium ion, in the bromonium ion and in the final product. reacts with 46 via backside (anti) attack at carbon, on the face opposite the bromine atom in 46, which fixes the stereochemistry of the two bromine atoms as anti. 9 (Z) - (S) (R) 46 (S) (S) rotate 60 44A (S) (S) rotate 60 (S) (S)

10 Diastereospecific but Racemic Attack of the bromonium ion from the face opposite the bromine atom will generate the (2R,3R) diastereomer (i.e. 44), which is the enantiomer of 44A generated from bromonium ion 46 in which bromine was on the "right". The bromine may add to either face of the alkene with equal facility so there is no facial selectivity in the addition of bromine to the alkene since approach from either face is equally likely. A racemic mixture is formed when there is no facial selectivity, but the reaction is still diastereospecific since only one diastereomer is formed. (Z) - (S) (R) (Z) 46 (R) (S) 47 (S) (S) rotate 60 44A (S) (S) rotate 60 (R) (R) 44 (S) (S) 10

11 Review of Diastereospecific Dihalogenation 2 C 4 11

12 Alkenes with hypohalous acids

13 Alkenes with ypohalous Acids It is known that dissolving chlorine in water leads to a solution that contains hypochlorous acid (O) and bromine dissolved in water contains hypobromous acid (O). 1-pentene is mixed with chlorine and water (O in aqueous media), and the major product is 1-chloro-2-pentanol (48), in 43% isolated yield. The polarization of O is O δ δ+, and chlorine is the electrophilic atom. A nucleophile would attack a halonium ion at the less substituted carbon and attack by the nucleophilic hydroxide ion at the less hindered carbon of 49 gives 2-chloro-1-pentanol. owever, the isolated product is 48 (1-chloro-2-pentanol), and formation of 48 must result from attack at the more substituted carbon atom of 49.!+! O 2 O O 48 43% + 13 O 2 O 2 O 49 O 50

14 Alkenes with ypohalous Acids 14 The major product is not consistent with nucleophilic attack with chloronium ion 49. In the reaction with 1-pentene, formation of 48 is only consistent with formation of a secondary carbocation such as 50 rather than 49. Carbocation 50 forms in water. Water not only generates O, but it also separates charge and stabilizes charge by solvation. Since O is generated in the presence of water, it is anticipated that 50 is more stable in the aqueous medium than 49, and attack by hydroxide at the positive carbon gives 48. Note that in a large excess of water (water is the solvent), 50 can also be attacked by water, and loss of a proton from the oxonium ion (hydroxide or water can function as the base in this reaction) also gives 48.!+! O 2 O O 48 43% + O 2 O 2 O 49 O 50

15 Alkenes as Lewis bases Reaction with borane ydroboration

16 ydroboration: Formation of orane 16 oron trifluoride (F 3 ) is a classic Lewis acid. orane, 3, is a highly reactive boron compound that also functions as a Lewis acid in the presence of a suitable electron donating species. Sodium borohydride (Na 4 ) is a reducing agent, and it reacts with boron trifluoride (F 3 ) to give a volatile product, borane. orane is a reactive species that is usually written as 3, but it is actually a dimeric species called diborane (51) that has hydrido bridges (bridging hydrogen atoms). There is an equilibrium between borane and 51, but for reactions presented in this section, the monomeric species ( 3 ) and the dimeric species 2 6 are used interchangeably. Na 4 + F 3 51

17 ydroboration: orane is a Lewis Acid 17 1-exene reacts with borane to give a new product known as an alkylborane, which is listed as unknown for the moment. The alkylborane product is treated with NaO and 2 O 2 in a second chemical reaction and the major product obtained after this two-reaction process (two-step process) is 1-hexanol (52), isolated in 81% yield. The second reaction with hydroxide and peroxide reacts with the new borane product of the first reaction to give the alcohol. In effect, O replaces the boron unit. The remainder of the 100% is alcohol 53 as a minor product, and this will be explained later. u Na 4 alkylborane NaO F 3 OEt 2 product 2 O 2 u O O + u 52 53

18 ydroboration: No Intermediate Experiments have shown that there is no intermediate, so there is no carbocation. 18 If there is no intermediate, the C- σ-bond must be formed almost simultaneously with the C- σ-bond. ased on the experimental evidence that there is no intermediate, the reaction of borane with an alkene is said to be a concerted reaction. Formally, it classified as concerted asynchronous rather than concerted synchronous. A reaction that has no intermediate is called a concerted reaction. If the bond-making and bond-breaking events do not occur simultaneously, the reaction is said to be asynchronous. An unknown alkylborane product is listed as a product, but there are two alcohol products after the second reaction, which means there must be two alkylborane products.

19 u ydroboration: Four-Center Transition State Na 4 alkylborane NaO F 3 OEt 2 product 2 O 2 Alcohols 52 and 53 are regioisomers, with the O unit attached to different carbon atoms. The alkylborane products contain C bonds, and it is reasonable to assume that the O unit in the alcohol product replaces boron in the alkylborane. orane product 56 must be the precursor to alcohol 52 and borane product 57 is the precursor to alcohol 53. Since alcohol 52 is the major product, the reaction shows a preference for the formation of 56. There is no intermediate, so the transition states for each alkylborane product (54 and 55,, respectively) must be examined to explain the preference for 56 over 57. u + 3 C 2 C 2 C 2 C ether 3 C 2 C 2 C 2 C u u O + O u u u u

20 ydroboration: Less indered Transition State 20 This reaction proceeds by a four-center transition state, represented by 54 or 55 for the two alkylborane products. The only way to rationalize 56 as the major product is for transition state 54 to be lower in energy than transition state 55. There is more steric hindrance in 55 than in 54, so 55 is higher in energy than 54. Since transition state 54 is lower in energy, it will lead to alkylborane 56, the observed major product, whereas transition state 55 leads to organoborane 57. in reactions with alkenes, borane will become attached to the less substituted carbon atom via a four-center transition state. The ether solvent catalyzes the reaction between borane and the alkene. u + 3 C 2 C 2 C 2 C ether 3 C 2 C 2 C 2 C u u u u

21 ydroboration: Trialkylborane Formation 21 Initial reaction of 1-hexene and borane gives monoalkylborane 56. One of the - units of 56 can react with a second molar equivalent of 1-hexene. In other words, the monoalkylborane R 2 also reacts as a Lewis acid to give a dialkylborane R 2, 58. Since 58 also has a - unit, reaction with a third molar equivalent of 1-hexene leads to the trialkylborane 59. For convenience, assume that the reaction of borane ( 3 ) and any alkene gives the trialkylborane. Using an excess of the alkene is a convenient way to ensure that the reactions proceeds to the trialkylborane. u 3 ether 3 u 56 ether 2 3 u 58 u ether

22 22 ydroboration: Specialized Alkylboranes When the alkene precursor is hindered (highly substituted), it may be possible to isolate monoalkylboranes or dialkylboranes. When 2-methyl-2-butene reacts with borane, the product is the dialkylborane 60, and its common name is disiamylborane (from the common name disecondary-isoamylborane). Similar reaction of borane with 2,3-dimethyl-2-butene leads to a monoalkylborane, 61. The common name of this product is thexylborane (from tertiary-hexylborane). The reaction of borane with 1,5-cyclooctadiene gives a product where two of the - units have reacted with the two C=C units, across the ring to give a bicyclic product 62= 9-borabicyclo[3.3.1]nonane, which is often abbreviated 9- N or drawn as the cartoon to the right

23 ydroboration: 9-N - More Selective An important reason for using is the stereochemical consequence of the borane-alkene, relative to the same reaction with 3. Since 9-N is so bulky, 64 is much less hindered and therefore lower in energy. 65 as the major product with <0.1% of the regioisomer. The same reaction with 3 gives about 85-90% of 65 and about 10% of the regioisomer. The larger and bulkier 9-N provides greater selectivity in the transition state due to increased steric hindrance, which means greater regioselectivity in formation of the product N C 3 2 C C 2 C 2 63 <0.1% formed ether C 3 2 C C 2 C

24 ydroboration. Oxidation to an Alcohol The reaction of 59 (the trialkylborane derived from 1- hexene) with a mixture of hydrogen peroxide ( 2 O 2 ) and aqueous sodium hydroxide (NaO) gave an alcohol (in this case 1-hexanol, 52) and boric acid [(O) 3 ]. Comparing the borane with alcohol product shows that the O has replaced the boron. 24 u ether 3 u 59 3 NaO 2 O 2 u 3 52 O

25 ydroboration. Oxidation to an Alcohol; chanism 25 The mechanism for this transformation of 59 to 52 involves several steps. Initial reaction is at boron rather than carbon, and the transformation involves a rearrangement from boron to oxygen. The sequence begins with the reaction of OO and NaO to give sodium hydroperoxide (Na + OO), which attacks the boron (a Lewis acid) to form "ate" complex 66 (the Lewis acid-lewis base complex). This product is not isolated because it undergoes a boron-to-carbon alkyl shift (a rearrangement), with loss of hydroxide (O ) to give 67; the alkyl group on boron migrates to oxygen. The final product is 68. If hydroxide attacks boron in 68, the RO group (an alkoxides) is displaced (three times in three successive steps) to give boric acid and three equivalents of the alcohol (RO). R R O-R O-R - OO R R O-O R R-O - O - R R R 66 O O-R 1. excess - O O + 3 RO 2 O O 2. 3 O +

26 ydroboration. Oxidation to an Alcohol 26 1st reaction 1. 3, ether reactant starting material 2. NaO, 2 O 2 reaction arrow 2nd reaction 70 O product

27 Reactions 1. 9-N, ether 2. 2 O 2, NaO O , ether 2. 2 O 2, NaO O