Lecture Outline 4. Hydroboration Oxidation

Name ______________________________________ Wow. That was an extremely long break. Let's start by cleaning out the cobwebs, as they say... Recall that we've learned two ways to do "Markovnikov" addition of H and OH to a double bond. Alcohols can also be made by hydroboration/oxidation. This is covered in section 11.7. The beginning of that section covers all sorts of things that you should know about but that aren't directly relevant to the regiochemistry and stereochemistry of the overall reaction. Note that, while there is experimental evidence for the asymmetrical transition states δ + and δ– charges, steric effects are generally believed to be the more important factor influencing the regiochemistry of the addition. 1 Write the reagents that you would use to hydroborate/oxidize isobutylene, and draw the alcohol produced. 2 How would you make the following alcohol by hydroboration/oxidation? Draw the alkene that you would use, and write the reagents. OH 3 The alcohol in #2 can also be made from a different alkene by a reaction that we learned before the break. (hint: it's either oxymercuration or acid-catalyzed hydration — think carefully about which one would work.) Draw that alkene and write the reagents. Lecture outline 4. Hydroboration / oxidation R2C CH2 + H BH3 borane BH2 H2O2 HO– , H2 O R2C CH2 OH H R2C CH2 — overall anti-Markovnikov add'n of H and OH, meaning OH ends up on the lesssubstituted d.b. C in simple cases, but of course understanding the reason it goes this way is a much more powerful predictive tool than blindly applying labels or rules — syn stereochemistry — H and OH go on the same face of the π-bond (It's actually a bit more complicated than shown — each B–H bond of BH 3 can react — so that thing that's formed after the first step is actually a "trialkylborane", or maybe even a dialkylborane in some cases. 3 H2C CH2 + 1 BH3 B (or "Et 3B") It doesn't matter exactly what it is since the rest of the reaction works the same way regardless of how many "R"s vs Hs are attached to boron. In the lab, one generally uses 2 - 3 equivalents of alkene for each equivalent of borane.) Addition of B–H to p-bond is concerted — bond-making and bond-breaking occur simultaneously, i.e., rxn goes in one step with no intermediate R2C H CH2 R2C H B CH2 transition state: B H or alkyl !+ R2C H CH B !– (this reaction does not involve a cation intermediate or any other intermediate) Steric effects control regiochemistry — boron adds to the less sterically hindered end of the double bond; H to the more hindered end (electronic effects, i.e. partial charges, etc, are less important) R R C C H B R H H R C B H H C H 2nd step — R2CH CH2 B H2O2 R2CH HO– , H2 O CH2 OH HO—H mech: R2CH HOO CH2 O R2CH CH2 B – HO O OH R2CH CH2 B R2CH O OH CH2 O B OH this is the key step — the rest is uninteresting there's no loss of stereochem in this step ... so (unlike the oxymerc/red) the 2-step hydroboration/oxidation sequence results in overall syn add'n of H and OH Recall that in the hydration and oxymercuration, we could replace water with an alcohol (ROH) and end up with an ether (via overall addition of –H and –OR) instead of an alcohol product (via overall addition of –H and –OH). Will that work here? Here are some real experimental data ... 1. BH3 OH 2. H2O2, aq. HO– + 94 % 6% OH (6 % is small enough that we can ignore the second compound and — in the interest of simplicity — just say the first is "the" product.) More examples on p 429 of the text. ... and steric effects also control the π-facial selectivity of the addition. CH3 H3C 1. BH3 H3C CH3 CH3 H 2. H2O2, aq. HO– CH3 OH H 89% !-pinene Q: I smell turpentine. Do you smell turpentine? A: That's the α-pinene. Turpentine consists of about 60% α-pinene and 30% β-pinene, the isomer that has the double bond to the exocyclic C. 5. Catalytic hydrogenation C H2 C Pd (or Pt or Ni) catalyst H C H C — rxn occurs by coordination of the alkene π-bond to the catalyst surface, bonding of two Hs to the catalyst surface, and then bonding of the Hs to the side of the alkene that is coordinated with the surface. The result is syn add'n of 2 Hs CH3 CH3 H2 Pd Here's an example that demonstrates the π-facial selectivity of the reaction. H3C CH3 CH3 H3C CH3 H H2 CH3 Pt/C EtOH H !-pinene H 87% 6. Hydroxylation — can be accomplished with osmium tetroxide, OsO4 (followed by aqueous sodium sulfite) or with dilute permanganate, MnO4–, in cold, basic solution. (permanganate is a relatively brutal oxidant — under acidic conditions it chews organic molecules to bits, spits them out, then stomps on them. Dilute permanganate in basic soln and cold is sort of like a partially tranquilized tiger on a sturdy leash in a cage.) C C 1. OsO4 2. aq. Na2SO3 — rxn goes via osmate ester intermediate, and the osmium gets lopped off in the second step. Because both new C–O bonds are formed at once, the result is syn add'n of 2 OH groups OsO4 HO C OH C O O C O Os O C Na2SO3 H2O The text uses "[OsO4]" and tBuOOH to imply an alternative method with catalytic OsO4 and tert-butyl hydroperoxide as the "stoichiometric oxidant". That's fine — that's the way most chemists do the reaction because OsO4 is expensive and highly toxic. However putting something in brackets does not imply that it's a catalyst — please don't use that notation. 7. Epoxidation — Epoxides can be made by reaction of alkenes with peroxycarboxylic acids, R–CO3H O C C + RCO3 H C C + RCO2 H — commonly used "R–CO3H"s — Cl CH3–CO3H CF3–CO3H CO3H m-chloroperoxybenzoic acid (mCPBA) mech: O C C O C H R O Rxn is concerted, so stereospecifically syn Ph CF3CO3H Ph CF3CO3H Ph Ph Recall that nucleophiles can open epoxide rings under acidic or basic conditions. Ring opening with aqueous acid or base provides a method for overall anti addition of two –OH groups. This is stereochemically complementary to syn hydroxylation with osmium tetroxide. OH O OH Note that's stereochemically complementary — they give different stereoisomers — as opposed to stereochemically complimentary, where one molecule might walk into a bar and say to another "What lovely stereochemistry you have... Your "R"s just seem to glisten in the polarized light."