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