MeOH -----> H+
Glycoside formation. Turns all substituents into OMe but the CH2OH
CH3-I ------> Ag2O
Methylation. Turn all substituents into OMe including the CH2OH.
H2O-----> H+
Hydrolysis. Turns the anomeric substituent into an OH and leaves a floating H+ in solution.
HNO3
Reduces/Oxidizes. Turns the top and bottommost carbons into CO2H
An optically inactive molecule is ________
Symmetrical or has a plane of symmetry
An optically active molecule is ________
Not symmetrical. Does not have a line of symmetry.
NaBH4
Turns the top and bottom ends of the structure into OH
(1) HCN (2)Pd/H2 (3) H3O+
Adds an OH in both the D and L configuration. Results in 2 epimers. Kiliani Fischer Reaction
(1) H2NOH
(2) Ar2O
ArONa
(3) NaOMe
MeOH
Wohl's Degradation. An OH is removed from the structure.
LiAlH4
Gets rid of the C=O group in the structure, but leaves everything else the same
(1) NaBH4 or NaBH3CN or H2/Ni
Combines 2 structures at a Nitrogen and gets rid of an H2O (A =O and 2Hs)
NH3 and then NaBH4
Adds the NH3 to the alpha carbon to become NH4 and then H2O leaves.
Anytime you see a Br2 on top with another reagent at the bottom
Replace that alpha hydrogen with the Br. Do not have the Br be a leaving group.
Anytime you see an LDA with N(---)
It deprotonates the alpha hydrogen somewhere and makes an anion. Use when there is only 1 C=O because it is bulky
H3O+
Turns the important thing something into OH
When exposed to heat
A CO2 is lost, C=O with another OH. Crazy, I would have done so well on exams if I knew
A Br with NaOMe and OMeH
Replace the CH3s with Br and have is act as a LG
Claisen Condensation
Needs 2 alpha hydrogens to work. Then add that same molecule to itself at the alpha hydrogen position
NaOEt/EtOH
Use when there are 2 C=O present because its small
When it backside attacks, just combine the OH together
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