Question | Answer |
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 |