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Summary Comprehensive final exam review: EVERYTHING you need to know from student who got 93% in Orgo 2223. Includes notes from all prep 101 sessions.$40.49
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,Carbohydrates
Monosaccharides
Glycoprotein: proteins + short, branched carbs
o Cell-surface receptors
Proteoglycan: proteins + long, linear carbs
o Connective tissue
Peptidoglycan: short oligopeptides cross-linking long, linear carbs
o Cell walls of bacteria
Lipopolysaccharides: fatty acids + carbs
o Outer membrane of gram-negative bacteria
Empirical formula: CH2O
Molecular formula: Cn(H2O)n
Aldose = aldehyde
o Anomeric C connected to 1H
Ketose = ketone
o Anomeric C not connected to an H (2C, 1OH, 1O)
“ulose” = ketone
Glyceraldehyde: D = +; L = -
D/L enantiomer: switch position of OH on every chiral C
Cyclization to hemiacetal = nucleophilic addition
Anomers:
o α = OH on anomeric C on opposite side as CH2OH
o β = OH on anomeric C on same side as CH2OH
At equilibrium, anomer with OH on anomeric C in more stable equatorial position
present at higher abundance
Hemiacetal = OH, OR will mutarotate
Acetal = OR, OR will not mutarotate
Acetal formation: hemiacetal + ROH (O-glycoside) or NR2H (N-glycoside)
o SN1 via C+ intermediate
o Makes both α and β but won’t mutarotate once formed
Fischer to Haworth:
i) Number carbons in Fischer
ii) Draw Haworth structure for aldose = pyranose (6-mem ring, including O) or ketose =
furanose (5-mem ring, including O with C1 branched off C2)
iii) Number carbons in Haworth ring
iv) Draw in C6: D = up, L = down
v) Draw in OH on anomeric C: α = opposite C6, β = same as C6
vi) Fill in remaining OH (right = down, left = up)
Oxidation:
1
, o Bromine water = oxidize just aldehyde to COOH
o HNO3 = oxidize aldehyde and primary alcohol on C6 to COOH
Reduction: reduce aldehyde to CH2OH or ketone to OH, H
o NaBH4
Glucose = epimer of mannose = const. isomer of fructose
o Deprot Cα, will get mix of all 3
o Epimerization: Deprot Cα, form enolate C-/C=C, reprotonate Cα on opposite face
o Isomerization: 2 tautomerizations
Deprot Cα, form C-, form O- (C=C), protonate O-, deprotonate other OH,
form C=O to form C-, protonate C-
Aldol Reaction:
i) Deprot Cα
ii) C- attacks C=O of another equivalent
iii) Protonate O-
iv) Forms β hydroxy ketone
Retro-aldol: cleave between α and β
Kiliani Fischer
Extends chain one carbon at a time
# rounds of synthesis = # carbons added just below C1 aldehyde
Makes new chiral center: 2 products
Step 1: nuc addition of CN- to carbonyl
Historic:
o Aqueous acid to hydrolyze CN to COOH
o Spontaneously form lactone (cyclic ester)
o NaOH to hydrolyze to carboxylate salt
o Recrystalize salts (diastereomers with diff physical properties)
o Protonate COO- (forms lactone = cyclic ester)
o Reduce ester to aldehyde with Na/Hg
Modern:
o Add H2 + Pd/BaSO4 catalyst to reduce nitrile to imine
o Add aqueous acid to hydrolyze imine to aldehyde
o Separate by HPLC
o Fewer steps / recrystallization / separations
o Fewer toxic reagents
2
, Disaccharides
Disaccharides: two monosaccharides (aldoses or ketoses) connected via an acetal linkage
= glycoside
o At least one sugar’s anomeric carbon is involved in the acetal, though both may
be involved
Acetals are cleaved by acid
Acetals are stable in neutral and basic conditions (unlike hemiacetals)
Acetals do not interconvert with their open chain forms
o Do not mutarotate
o Are not reducing
If one part of the disaccharide molecule is a hemiacetal (ie., one anomeric carbon is not
involved in the acetal) then the whole molecule will mutarotate and will be a reducing
sugar
Disaccharides of D-Glucose: Maltose and Cellobiose
Both have hemiacetals: mutarotate and are reducing
Maltose joined by α 1,4
Cellobiose joined by β 1, 4
Lactose
Lactose = D-galactose + D-glucose joined β(1,4)
Lactase aka. β-galactosidase cleaves lactose linkage
o Lactose intolerant individuals deficient in this enzyme
Sucrose
Sucrose = D-glucose + D-fructose
o Linkage = α(1,2) at glucose, β(2, 1) at fructose
No hemiacetals (both anomeric carbons involved in acetal) so does not mutarotate and is
not reducing
Sucrose (disaccharide) = “invert sugar” because when hydrolyzed to monosaccharides
(glucose and fructose), the direction of optical rotation changes / inverts
o Sucrose rotates PPL +
o Glucose + fructose mixture rotates PPL –
Glucose + fructose mixture is twice as sweet tasting than sucrose
3
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