IV. Reactions of Organosamarium Compounds

1. The Samarium-Barbier Reaction

Because in the reaction shown in Scheme 4 the sulfone 4 and cyclo­hex­anone both are present in the reaction mixture from the outset, the reac­tion is described as a Barbier-type² or samar­ium-Barbier^3,34 reaction. The mech­anism pictured in Scheme 4 is a widely accepted one for this type of process­.^{1–9,34–37} The carbohydrate reactant frequently is a glycosyl sul­fone,^{28‑30,32,33,37‑43} but it also can be a glycosyl halide^{31,32,37,44-46} or phos­phate.⁴⁷ Possi­bil­ities for the carbonyl compound include ketones,^{2,28,32,39,42,43,46} alde­hydes,^{2,28–30,32,38–46,} and lac­tones.^{45, 48–50} Usually the carbonyl compound is a simple organic molecule, but some­times the carbonyl group is part of the more complex structure found in ­a carbo­hy­drate.^{38–40,42–44}

2. The Samarium-Grignard Reaction

The defining characteristic of the samarium-Barbier reaction is that all of the reactants are present in the reac­tion mix­ture at the outset. If an inter­mediate organosamarium com­pound is sufficiently stable, it can be formed prior to adding the carbonyl compound. When reaction takes place using such a procedure, it is described as a Grignard-type or samarium-Grignard reac­tion.^3,34 Many organo­sa­marium compounds are not stable enough to undergo reaction in this way; in particular, the reaction shown in Scheme 4 is only suc­cess­ful when run under samarium-Barbier conditions.³³

1. Internal Addition to a Carbon–Carbon Multiple Bond

A samarium ketyl that contains a properly positioned multiple bond readily forms a new ring system.^51–62 Examples indicating the range of reac­tivity of these ketyl intermediates are found in the reactions shown in equa­tions 4 and 5 and Scheme 5. In the reaction pictured in eq 4, an unsaturated aldehyde forms a samarium ketyl that cyclizes and then reacts with cyclo­hexanone.⁶⁰ Eq 5 describes the reaction of an unsat­ur­ated carbonyl compound that has a substituent on the carbon atom α to the carbonyl group. If the sub­stituent is a poor leaving group [e.g., (C₆H₅)₃CO], ring formation takes place, but when a better leaving group [e.g., (CH₃)₃CCO₂] is present, cyclization is replaced by elim­in­ation of the corresponding anion [e.g., (CH₃)₃CCO₂^-] followed by hydrogen-atom abstraction.⁶² The highly stereoselective cyclization shown in Scheme 5 is an internal addi­tion of a samarium ketyl to a triple bond.⁵¹ The resulting cyclic inter­mediate (7) either can react with another mole­cule of samarium(II) iodide or, since 7 is a highly reactive radical, abstract a hydrogen atom from the solvent (THF). Either reaction can be part of a two-step sequence leading to the final product (Scheme 5).

2. Internal Addition to a Carbon–Oxygen Double Bond (Pinacol Formation)

Reaction of samarium(II) iodide with a compound that has 1,4-,⁶³ 1,5‑,^64–68 or 1,6‑^69–77 related aldehydo or keto groups produces a samarium ketyl that then forms a cyclic pin­a­col. A typical example of such a reaction is shown in eq 6,⁶⁴ and a gen­eral mec­h­an­ism for pinacol formation is pro­posed in Scheme 6.⁷⁸ Based on this pro­posal, one would expect that the two hydroxyl groups in a pinacol should be found on the same side of the newly formed ring system because during reac­tion the oxygen atoms in these two groups interact simul­taneously with a single samarium ion. Further, one also would anticipate that reaction should place the hydroxyl groups stereo­se­lec­tively on the less-hindered face of the new ring. Both of these expec­ta­tions are realized not only in the reaction shown in eq 6 but in other, similar reac­tions, where the major products always are cis diols formed by minim­izing steric interactions during ring construc­tion.^63,65–77

3. Internal Addition to a Carbon–Nitrogen Double Bond

Reaction analogous to pinacol formation occurs when one of the car­bonyl groups in a reac­tant molecule is replaced by a group with a C–N double bond (eq 7⁷⁹).^79–83 A significant stereo­­chemical difference between this type of reaction and pinacol formation is that the hydroxyl and substi­tuted amino groups produced during cyclization are on opposite faces of the newly formed ring. This result indicates that complexation between the carbonyl groups and the samar­ium ion during pinacol formation has no anal­o­gous interaction in reactions of keto-oximes.

Sometimes the cyclization of a keto-oxime produces an amine rather than a substituted amine (eq 7).^79,84 This occurs when samarium(II) iodide, in excess of that needed for cyclization, transfers an electron to the N–O bond in the cyclic product leading to replacement of the amine substituent with a hydro­gen atom. This reaction is accelerated by addition of water to the reac­tion mixture.

4. Ring-Contraction Reactions

Scheme 7 describes a reaction in which a samarium ketyl is involved in ring contraction. This process begins with electron transfer from SmI₂ to the carbohydrate iodide 8 to gen­erate the radical 9.⁸⁵ Reaction of 9 with a second mole­cule of SmI₂ produces the organosamarium com­pound 10. Elimination of the elements of MeOSmI₂ from 10 causes the pyran­oid ring to open to give the unsaturated aldehyde 11, which reacts with SmI₂ to form a samarium ketyl that then cyclizes to give the substituted cyclo­pen­tanes 12 and 13. Similar ring con­trac­tions occur when 6-aldehydo hexo­pyran­osyl derivatives react with SmI₂.^86,87