midterm 2 reagents Flashcards
HNO3 or H2SO4, H2O (alkene hydration)
-OH and H bonds to the alkene
-rearrangement can occur
-OH is attached to the MORE substituted carbon
- Hg(OAc)2, H2O
- NaBH4, H2O, NaOH (oxymercuration)
-OH and H bonds to the alkene
-NO rearrangement occurs
-anti addition where the OH is dash and H is wedge or vice versa
-OH is attached to the MORE substituted carbon
-if EtOH instead of just OH, OEt will go the MORE substituted carbon
- BH3,THF
- H2O2, NaOH (hydroboration)
-OH and H bonds to the alkene
-syn-addition where the OH is a dash and H is a dash or both are wedges
- PBr3 or Ph3PBr2
- polar aprotic solvent like DMF (reaction of alcohols)
-OH becomes a leaving group by attacking P and losing an H to one halide ion and then the other halide attacks the molecule
- SOCl2
- polar aprotic solvent like DMF (reactions of alcohols)
-OH gets protonated to become a leaving group and then Cl bind to the molecule
- TsCl or MsCl
- pyridine (reactions of alcohols)
-OH gets deprotonated by pyridine and the Cl leaves. the O now bind to the Ts or Ms group to give OTs or Ms and pyridine has an H now so the N carries a plus charge
dimethyl sulfate (SO4CH3) (reactions of alcohols)
-O gets attached to the alkane
-need the sulfate attached to an alkane and an alkoxide anion
-the sulfate becomes a leaving group and an alkoxide anion binds with the alkane
H3PO4 with heat (dehydration of alcohol) (reactions of alcohols)
-OPO3H2 leaves and OH gets an H. OH with the H now becomes a leaving group. it leaves behind a carbocation so the bond to H makes a double bond while the OPO3H2 takes the H back and gets regenerated
CrO3 or H2CrO4 w/ aq H2SO4, or K2Cr2O7 w/ H2O (oxidation of alcohols)
-bond with OH now becomes a double bonded O
-primary alcohol becomes an aldehyde (C with double bond O and bond to alkyl group and H)
-secondary alcohols give ketones (C with double bond O and bond to 2 alkyl groups)
-tertiary alcohols CAN’T BE OXIDIZED BECAUSE NO H’S on the alpha carbon
KMnO4 w/ aq. NaOH (oxidation of alcohols)
-KMnO4 used to oxidize PRIMARY alcohols to get carboxylic acid (C with double bond O and bonded to alkyl group and OH)
-KMnO4 is NOT USED TO OXIDIZE SECONDARY ALCOHOLS
PCC (dry w/ no H2O) (oxidation of alcohols)
-for primary alcohol, PCC doesn’t generate water and stop at the aldehyde stage (C with double bond O and bonded to alkyl group and H
Willimason Ether Synthesis (reagents is an –ROH with a strong base like NaH or NaOH with an alkyl electrophile)
-1st step: the strong base deprotonates the O
-2nd step: O attacks in SN2 while the leaving groups leaves
-SN2 is favored for methyl or primary (remember inversion for SN2)
-E2 is favored for secondary or tertiary
alkene with a peroxyacid aka MCPA which is RCO3H
-MCPBA is best for making epoxides
-make sure the stereochem is the same throughout
-if you have trans, product is anti addition
-if you have cis, product is syn addition
-make sure nucelophile and halogen are 180 degrees apart (OH is 2 carbons away from the halogen)
epoxide w/ ROH/NaOR
-if in basic condition (Nu^-), the nucleophile attacks the less substituted carbon
- ex. of strong bases: NaH, KOtBu
*in base, it’s protonated at the last step
-if in acidic condition (NuOH), the nucleophile attacks more substituted carbon
*in acid, the acid must be protonated FIRST
-ring opening is ANTI (halogen and nucleophile MUST be 180 degrees apart aka nucleophile is 2 carbons away from halogen)
epoxide+ acid (reagents: H2SO4/H2O, H2SO4/ROH, HX dil, aq, or H2O, and HX dry, anhydrous
-the OH leaves and the halide attacks
*if in acid, nucleophile attacks MORE substituted carbon
*if in base, nucleophile attacks LESS substituted carbon
