ETHERS To make Williamson SN2 o Works well for Me 1 sometimes well for 2 o Hindered Substrate NO WILLIAMSON When 2 3 or hindered substrate E2 predominates WHY RO alkoxide ion is STRONG BASE STRONG NU o Given ether cut in half so that nucleophile strong base LG SCHEME occurs in two steps 1 Strong base OH H etc attacks partially positive H bonded to O DE protonates alcohol to form alkoxide ion R O o Is this step in ANTI FASHNION IF SO HOW It s SN2 so it doesn t matter It is a direct SN2 reaction RO is nucleophile adding directly to substrate 2 partially positive C in introduced alkyl halide gets attacked by alkoxide nucleophile via backside attack SN2 fashion to form ether o 1 ALKYL HALIDES OPTIMAL because least steric hindrance Condensation of an Alcohol 1 O from alcohol attacks partially positive H from nonprotonation of alcohol water leaves another R OH nucleophilic OR nucelophilic ACID protonation of the alcohol to attacks positive carbocation base removes H ether form a GOOD LEAVING GROUP WATER formed 2 Water leaves Stable carbocation is formed 3 Another R OH group in solution attacks positive charge O from new alcohol is positively charged unstable 4 Base removes the third Hydrogen ether product formed THREE products can form if you have unsymmetrical ether i e isopropyl t butyl ether WATER IS ALSO PRODUCT In ether synthesis 2 alcohols i e isopropanol and t butanol can both be reactants but in ACIDIC solution Must be acidic solution in order to protonate either alcohol and form stable carbocation Therefore only 2 3 alcohols involved in condensation Entropy is net 0 2 moles to 2 moles 2 moles of reagents to 1 mole water 1 mole ether o At LOWER TEMPERATURES non nucleophillic acid H2SO4 FSO3H condensation o At HIGHER TEMPERATURES non nucleophillic acid H2SO4 FSO3H elimination NAMING ethers Longest C chain is parent chain is priority Shorter C chain attached to O oxy substituent If you have OH group OH GETS PRIORITY ETHERS undergo Combustion ether CO2 H2O o Ether O2 SPARK CO2 H2O H X reactions X Br I Cl ether alcohol o O in ether attacks partial positive H kicks out halide LG O now has positive charge and is unstable o Halide now good nucleophile attacks partially positive C O o X bonds to R leaves behind R OH o O attacks H X again and reaction repeats OXACYCLOALKANES cyclic ethers NAMING O gets priority OH vs O either gets priority Grignard reactions Are good for 3 4 membered rings ONLY o In a 3 or 4 membered ring Negative C from RMgX attacks partial positive C attached to O and cause ring to open BECAUSE When you open you relieve all of the angle strain ring is already so unstable Does not work for 5 or 6 membered because they are more stable 3 and 4 membered rings have HIGH RING STRAIN sp3 C s want 109 bond angles but in 3 and 4 membered rings the angles are 60 and 90 respectively o Therefore they are very REACTIVE and C O bond can be broken o When C O bond breaks the negative carbon grabs electrons from C of C O and that C s electrons go to oxygen O has negative charge o C from C O bond is now C R group from RMgX To make OXACYCLOPROPANE o OH R X alcohol with a halide attached BASE oxide attacking C X LG and forming ring SN2 o SYNTHESIS Two ways to form OH R X HALOHYDRIN MCPBA REGIOSELECTIVITY Nu attacks several possible regions of a molecule preferentially Nu always adds at more substituted C Nu will attack most hindered C most substituted Because O in a ring does not pull on both C s equally the more substituted C is more electrons donating therefore more partial positive and therefore WEAKER So easier for Nu to break Regiochemical control When you form a carbocation the most stable rearrangement will happen Thus NO regiochemical control To have regiochemical control no carbocation formation Markovnikov s rule is to have regiochemical control o Because you are controlling where the H will end up Office hours Stereospecific If you have cis starting material and if you start with trans starting material you get a different set of products o You get unique set of stereoisomers for each