ochem mt2 rxns and reagents

switzer

glycoside formation

MeOH, H+ , adds an OMe to anomeric carbon

• mechanism

  • anomeric OH gets protonated and becomes a leaving group (H2O+)

  • double bond forms between anomeric carbon and oxygen

  • methanol/alcohol reacts with the anomeric position, and attaches itself to it

  • the hydrogen from the newly added alcohol leaves to stabilize the charge in MeO-

methylation

CH3I, Ag2O (turns all OH in sugar into OMe)

hydrolysis after methylation

methylation reagents (CH3I and Ag2O), followed by H2O and H+, reverts anomeric OMe to OH

• if methylating and hydrolyzing disaccharides, the bridge severs and the bridged O becomes an OH as well

redox of a sugar

HNO3, if symmetric = inactive, asymmetric = active

• rxn gets rid of the aldehyde and CH2OH and becomes carboxylic acid (CO2H) on both sides

kiliani fischer chain extension

HCN (CN-) , then Pd/H2, then H3O+ or H+, adds extra carbon to carbohydrate chain, should give two products with an OH on the new carbon facing left or right;

• mechanism

  • CN- attacks carbonyl (becomes OH) and attaches itself to carbohydrate

  • Pd/H2 turns R-CN into imine (R=NH)

  • H3O+ reverts imine to carbonyl and releases ammonia

wohl degradation

H2NOH, then Ac2O or AcONa, then NaOMe and MeOH, targets and removes carbonyl, so carbon underneath original carbonyl becomes the new one

reducing sugars

must have a free OH at the anomeric position to be considered reducing

alpha anomer

where the OH at the anomeric position is trans the alcohol at the 5 position

beta anomer

where the OH at the anomeric position is cis to the alcohol at the 6 position

forming alditols

NaBH4

• reacts aldose or ketose to reduce the carbonyl to OH, making sugar into alditol

head to head linkages

linkage between two or more sugars at the equatorial (up oriented) anomeric position (can be for only one or both sugars in a disaccharide)

there are three types

• beta-beta linkages (cis-cis)

• alpha-beta linkages (trans-cis)

• alpha-alpha linkages (trans-trans)

• potential fourth linkage if there are different sugars

head to tail linkages

linkage between two or more sugars, can be from one anomeric carbon, and an alcohol (R-OH2C) creating a 1,6 linkage

• can be either alpha or beta depending on the first sugar’s orientation

• head to tail linkage can vary in each position of each sugar 

  • 8 possible if same sugar

  • 16 possible if different sugar

enzyme needed to split specific disaccharide conformations

• for alpha-alpha linkage and alpha (1,6); alpha enzyme

• for beta-beta linkage and beta (1,6); beta enzyme

• for alpha-beta linkage; alpha and beta enzymes

amines

compounds containing nitrogen, bonded to R or H groups

• ex:

  • NH3, ammonia

  • RNH2, primary amine

  • R2NH, secondary amine

  • R3N, tertiary

  • R4N+, quaternary ammonium ion

rapid interconversion of amines

when N is attached to four different groups (one group is a lone pair), it becomes chiral

• amine doesn’t exist as one enantiomer, it interconverts from one enantiomeric form to another

• transition state of an NR3 group when interconverting is a trigonal planar, with N orbitals perpendicular to the planar bonds

  • at normal temp, you will get a racemic mix of amines (50-50)

special cases for amine interconversion

if amines are directly part of a ring (ex. pyridine), there is no interconversion, making its only visible configuration the only stable enantiomer (remains in current form)

• there will be no stable transition state, the amine cannot interconvert

basicity of amines

an amine is basic if pKa of conjugate acid is large

most basic to least basic amines

• most basic is an imine (R=NH)

• second most basic is a tertiary amine (NR3)

• third most basic is a secondary amine (R2NH)

• fourth most basic is a primary amine (RNH2)

• fourth LEAST basic is an imidizole (5 membered ring with an N that is one carbon apart from an NH)

• third LEAST basic is pyridine or pyridine-like nitrogens (6 membered ring with one nitrogen in it)

• second LEAST basic is aniline or aniline-like nitrogens (phenyl attached to NH2)

• LEAST basic is pyrrole or pyrrole-like nitrogens (5 membered ring with NH)

gabriel amine synthesis

OH-, then a primary alkylhalide (R-X), then H2O and H+

• mechanism

  • OH- deprotonates amine (in phthalic anhydride), and creates a lone pair in the attacked N

  • N attacks alkyl halide, kicking out the halogen and attaching alkyl chain to itself

  • if in ring, the other groups bonded to N get severed except for the new alkyl bond

  • reprotonate the N to make it into a primary amine (R-NH2), and protonate the rest of the severed structure (if it was a carbonyl attached to N originally, they become carboxylic acids)

    • rxn only works for primary and secondary amines, tertiary not used due to E2 reaction and formation of an alkene

alkyl halide + excess NH3

forms primary amine

reductive amination

requires aldehyde or ketone + primary (R-NH2) or secondary amines (R2NH), then NaBH4 or H2/Ni or NaBH3CN

• end product; react amine with aldehyde or ketone to form another specific amine

  • creates imine (R=NH) in first step, then becomes amine again

    • ketone/aldehyde + NH3 = primary amine

    • ketone/aldehyde + primary amine = secondary amine

    • ketone/aldehyde + secondary amine = tertiary amine

reductive amination retrosynthesis trick

• look at alpha carbons surrounding the amine (can be primary, secondary, or tertiary)

• then remove an alpha hydrogen from the alpha carbon to create a double bond between N and the alpha carbon

• insert O in between the double bond, and sever the N away

• protonate the N and it should give you an amine

hofmann rearrangement

used to synthesize primary amines from a primary amide (RC=O-NH2 to R-NH2), requires NaOH, Br2, and H3O+

• mechanism

  • deprotonate the amine (-OH), then react with bromine (Br-), then deprotonate amine again (-OH to R-NHBr)

  • localize charge to oxygen, and create a double bond for N and carbon

  • move the carbonyl attachment (not the O-) and stick onto N, and push out bromide, and then delocalize charge from O- to create an isocyanate (R-N=C=O)

  • react C with -OH and localize oxygen charge from the isocyanate

  • protonate with water or H+ (should protonate the -O, and protonates N)

  • deprotonate the newly formed carboxylic acid with -OH to kick out and create CO2 as byproduct

  • reprotonate the new amine and that should be a final product

reduction of azides

results in formation of primary amine; requires alkyl halide, NaN3 and LiAlH4

• mechanism

  • literally SN2 and then reducing it with LiAlH4 to become a primary amine

reduction of nitriles

requires NaCN, alkyl bromide, and LiAlH4

• mechanism;

  • SN2, and reducing C≡N to CH2-NH2

  • notice: you should be adding a new carbon to the original alkyl bromide structure

reduction of amides

requires all amines (1, 2, and 3) attached to carbonyl and LiAlH4, literally deletes (=O) part of carbonyl

hofmann elimination

literally E2 but wanting to form least sub product (Hofmann product), requires a primary amine, CH3I and NaHCO3, then Ag2O, H2O and heat

• mechanism:

  • CH3I and NaHCO3 turns R-NH2 to R-NMe3+

  • deprotonate least substituted (beta) carbon’s hydrogen to form double bond with C-NMe3+ and kick out the NMe3+

  • major product should be least substituted alkene