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UB CHE 201 Ch 10 A

Alcohol Synthesis(New Reactions). Given Reactant and Reagent Give Product

QuestionAnswer
R-X + Mg ---ether--> R-Mg-X, grignard reagent
R-X + 2Li ---> R-Li(organolithium reagent) + Li+-X
Formaldehyde + R-MgX --1)ether solvent, 2) H3O+ ---> 1* alcohol
Aldehyde + R-MgX --1)ether solvent, 2) H3O+ ---> 2* alcohol
Ketone + R-MgX --1)ether solvent, 2) H3O+ ---> 3* alcohol, 1 group added
acid chloride + 2R-MgX --1)ether solvent, 2) H3O+ ---> 3* alcohol, 2 groups added
ester + 2R-MgX --1)ether solvent, 2) H3O+ ---> 3* alcohol, 2 groups added
Ethylene Oxide + R-MgX --1)ether solvent, 2) H3O+ ---> 1* alcohol, 2 carbons added
R-MgX + compound containing O-H, N-H, S-H, or terminal alkyne protonated reagent + alkane
R-MgX + compound containing C=O, C=N, nitrile, S=O, N=O will be attacked by reagent
1* alcohol + Na(s) alkoxide
2* or 3* alcohol + K(s) alkoxide
difficult alcohol + NaH in THF alkoxide
Phenol + NaOH (aq) or KOH(aq) phenoxide
Why doesnt a phenol need to be treated with Na or K metal to form a phenoxide? Ion formation is favored due to resonance stabilization
aldehyde + 1)NaBH4, 2)H3O+ 1* alcohol
ketone + 1)NaBH4, 2)H3O+ 2* alcohol
carboxylic acid + 1)NaBH4, 2)H3O+ no reaction, NaBH4 is selective
ester + 1)NaBH4, 2)H3O+ no reaction, NaBH4 is selective
aldehyde + 1)LiAlH4, 2)H3O+ 1* alcohol
ketone + 1)LiAlH4, 2)H3O+ 2* alcohol
carboxylic acid + 1)LiAlH4, 2)H3O+ 1* alcohol
ester + 1)LiAlH4, 2)H3O+ 1* alcohol
alkene + 1)LiAlH4 no reaction
alkene + NaBH4 no reaction
aldehyde or ketone with double bonds + H2 --Raney Nickel--> alcohol without double bonds
Na+-S-H + R-X R-SH (thiol)
thiol + KMnO4 or HNO3 sulfonic acid (has 2 other resonance forms)
R2CuLi(gilman reagent) +R'-X R'-R + R-Cu + LiX
2R-Li + CuI R2CuI(gilman reagent formation) + LiI
Created by: HugsAndKisses on 2010-12-15



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