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UK CHE 232 - NOTES

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1 Cyanohydrin formation (nucleophilic addition of CN–) RC(R')HOKCN, HClH(R')RO-NCK+HCNH(R')ROHNC-KCN - Cyanohydrins are more easily formed with aldehydes than ketones; yet, another way to make C–C bond! - HCN is formed from acid-base reaction of –CN with HCl. - the reaction does not occur with HCN alone; cyanide is required. Thus, the reaction can be made to occur using HCN/catalytic base. Generally, use of HCN is avoided since it’s poisonous! Cyanohydrins can be converted back to the carbonyl compound by treatment with base: RC(R')HO- KCNH(R')RO-NCK+H(R')ROHNCKOH+ H2O - the cyano group of a cyanohydrin is readily hydrolyzed to a COOH group by heating with aqueous base or acid: H(R')ROHNCH2O(H+ or –OH)!H(R')ROHHOOC OHCNOHCOOHH3O+, ! Imines and Enamines Just as with alcohols, amines can add to ketones and aldehydes. Primary amines add to give imines while secondary amines give enamines (pronounced ene-amines). The reactions are generally catalyzed by a small amount of acid and need to be buffered to a pH of ~4.5. a) imines RC(R')HOH(R')CROH"RHNR"-NH2carbinolamineRC(R')HNR"imineH(R')CRO"RH2NH3O+H(R')CROH2"RHN-H2ORC(R')HNR"HOH2- H3O+-H2O2 i) The reactions involve nucleophilic addition to yield a carbinolamine (hemiaminal) followed by E1 elimination of H2O. Note that this reaction requires the presence of at least two hydrogens in the starting amine (H+ comes from N; O is protonated before it leaves) ii) Equilibrium favors imine formation when R” = hydroxy, alkoxy, or amino groups; the products called oximes, oxime ethers, or hydrazones, respectively. When R”= alkyl, equilibrium usually favors the carbonyl compound, but can be pushed toward imine by removal of H2O. iii) Fastest at near-neutral pH: acidic pH protonates starting material and slows first reaction, basic pH slows protonation of O in hemiaminal and prevents second reaction. b) Enamines In the case of secondary amines, we have a lack of protons that can easily be removed from the amine – the mechanism thus requires that the offending positive charge to be neutralized by removing a proton from the alkyl group. Thus, there must be at least one H atom bonded to the carbon alpha to the carbonyl RCR'H2COR2"NHRCR"2NCHR'enamineCH2R'CROR"2HNH3O+CH2R'CROH2R"2N-H2ORCR"2NCHR'HOH2CH2R'CROHR"2NcarbinolamineRCR"2NCH2R'-H3O+-H2O Equilibrium usually favors carbonyl, but can be pushed toward enamine by removal of H2O. Clearly, hydrolysis of enamines and imines are simply the reverse of these reactions (reverse mechanism) and hence form aldehydes and ketones. De-Oxygenation reactions. In Chapter 14, we saw two general reactions for the complete de-oxygenation of ketones and aldehydes. The general scheme for de-oxygenation is: The two methods are the Wolff-Kishner (runs under basic conditions) and the Clemmensen (under acidic conditions). For examples,3 The Witt ig Reaction is another deoxygenation reaction that’s useful for making mono-, di-, and trisubstituted alkenes, but not for tetrasubstituted alkenes. The reaction involves nucleophilic addition to the carbonyl compound followed by elimination. RCR'O+Ph3P CR"R*CR"R*RR'+ Ph3P=OPh3P CR"R*Either R' or R" Must be hydrogen i.e. H3CCHO+Ph3P CCH3CH3CCH3CH3H3CH+ Ph3P=OC O+Ph3P CCH3HCCH3H+ Ph3P=O Note that the alkene product must contain at least one hydrogen; another way to make C–C bond! In the most general scheme of this reaction, an alkyl halide (usually the bromide) and an aldehyde or ketone are taken to an alkene. In a more detailed picture of this reaction, the alkyl bromide is allowed to react with triphenyl phosphine to form an alkylphosphonium salt: Ph3P + BrCHR”R* → [Ph3P+–CHR”R*]Br– - this SN2 works best for CH3X and 1˚ alkyl halides; yields are lower for 2˚ alkyl hakides - alkylphosphonium salts are generally quite stable, and can be stored over long periods of time (i.e. many of these salts are even sold commercially). The phosphonium salt is then deprotonated (usually with sodium hydride, or4 butyllithium) to form the ylide. An ylide is simply a charge-separated species, as shown below: [Ph3P+–CHR”R*]Br– + BuLi → Ph3P+–C–R”R* <-> Ph3P=CR”R* This ylide is then allowed to react with a ketone or aldehyde, to form a betaine intermediate. This intermediate cyclizes to form another intermediate, an oxaphosphetane: OR R'Ph3PR"R*OPh3PRR'R*R"OPh3PRR'R*R"betaineoxaphosphetane The oxaphosphetane decomposes rapidly to form the alkene and triphenylphosphine oxide: OPh3PRR'R*R"OPRR'*RR"PhPhPh+ Thus, half of an alkene can come from an alkyl bromide, and the other half from a ketone or aldehyde. The main driving force for this reaction is the formation of the phosphorous-oxygen double bond; this is one of the strongest bonds known, and its formation drives the reaction to completion. A limitation of the Wittig reaction is that a mixture of stereoisomers sometimes forms (below): H3CH2CCHO+Ph3P CH(CH2)4CH3CH(CH2)4CH3H3CH2CH+CH(CH2)4CH3HH3CH2C59%41% This approach to alkene formation is better than addition of a Grignard to aldehyde or ketone followed by dehydration, because dehydration follows Saytzeff’s rule (thermodynamic ratio of products, more substituted favored) while Wittig reaction gives only one product: C OCH31. CH3Li2. H2OCOHCH3H3O+H2OCHHmajor+minor versus C O+Ph3P CHHCHH+ Ph3P=O5 Retrosynthetic analysis: RCH=CHR' ⇒ RCHO + BrCH2R' or RCH2Br + R'CHO. Remember, RBr ⇒ ROH; and we have seen that RCHO or R2CO ⇒ R”CH2OH or R”2CHOH (oxidation of aldehydes and ketones) Which starting materials would you use to prepare PhCH=C(CH3)2? PhCHO and (CH3)2CHBr versus PhCH2Br and (CH3)2CO? How would you prepare PhCH2Br from PhCOOMe? How would you prepare PhCHO from PhCOOMe? Below are a few examples of the Wittig reaction at work: A simple modification of the Wittig reaction leads to a dibromo-olefin, which is an excellent precursor to


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UK CHE 232 - NOTES

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