Foellinger Auditorium Hour Exam III Wednesday, Apr. 23 7:00 – 8:15 pm Foellinger Auditorium Conflict Exam 5:30-6:45 161 NL Help session: Mon. 7:00 – 9:00 100 MSEB

Hour Exam I Wednesday, Apr. 23 7:00 – 8:15 pm 141 Wohlers Ford (CQF, G, J ,L) Livingston (CQP) 100 Noyes Lab Gorski (CQC, H) Carberry (CQI, K) Tumuluru (CQD, E) 217 Noyes Lab Mohan (CQA, B) Help session: Mon. 7:00 – 9:00 100 MSEB Conflict Exam 5:30-6:45 161 NL

Carbohydrates CnH2nOn C (H2O) 1 degree of unsaturation I II II * * * * * I II II optically active? yes I optically active? yes how many possible stereoisomers? 23 = 8 how many possible stereoisomers? 4 how many actual stereoisomers? 8 how many actual stereoisomers? 4 Is this D- or L-? Is this D- or L-? Is this (+) or (-)? Is this (+) or (-)? ? ?

I II Carbohydrates can be: simple monosaccharides can’t be hydrolyzed complex disaccharides hydrolyzed to 2 monosaccharides polysaccharides hydrolyzed to 3 - 1000’s

Classifying monosaccharides II Classifying monosaccharides 1. carbonyl group C=O “aldo” or “keto” 2. number of carbon atoms “tri”, “tetr”, “pent”, “hex” 3. suffix “ose” I aldo tetr ose II keto hex ose 22 isomers 23 isomers

Physical properties Classify: aldo pentose 2-deoxyribose DNA * 2-deoxyribose * DNA What are IMF ? H-bond donors and acceptors high b.p. solids at room T soluble in H2O chiral 2-deoxyribose has ______ isomers 4 Is this L- or D- isomer

natural monosaccharides = D ______stereoisomers 24=16 CHO CHO CH2OH H HO OH * H OH * HO H * H OH * H OH CH2OH aldo hexose L-glucose ____stereocenters 4 natural monosaccharides = D ______stereoisomers 24=16 D- glucose

monosaccharides to know: D-glucose D-galactose D-fructose keto hexose aldohexose diastereomers

hemi-acetal formation OH .. H+ .. .. H+ + R’-OH H .. OR’ + H+ aldehyde alcohol cyclic hemi-acetal

Fischer Haworth _ _ _ = = = aldehyde + alcohol left = up right = down 6 1 HO C CH2OH OH C 2 O = C _ H 1 4 2 OH C OH C OH 3 5 3 HO 4 OH 6 HOH2C OH 5 H+ 5 OH OH C CH2OH .. 6 + CH2OH O = C _ H H+ OH C + OH C OH C aldehyde + alcohol left = up right = down

_ created a new C* at C1 25 isomers = differ only at C1 OH up =  6 5 * 25 isomers OH HO CH2OH O = C _ H 4 * * 1 1 * * differ only at C1 * 2 2 3 * 3 OH up =  * 4 -D-glucose 5 * OH down =  6 racemic mixture anomers C1 = anomeric C hemi-acetal unstable  -D-glucose

different physical properties -D-glucose 3 OH 1 2 4 5 6 O CH2OH H more stable 6 CH2OH 4 5 OH HO OH 2 b.p. = 150o 1 3 OH anomers diastereomers different physical properties -D-glucose 3 OH 1 2 4 5 6 O CH2OH H b.p. = 146o OH HO CH2OH OH -D-glucose

_ = 64% -D-glucose 36% -D-glucose 0.01% unstable hemi-acetals OH HO CH2OH O = C _ H OH HO CH2OH OH HO CH2OH 64% -D-glucose 36% -D-glucose 0.01% unstable hemi-acetals alcohol + aldehyde hemi-acetal [O] aldehyde + Cu2+ Ag+ carboxylic acid + Cu+ Ag Fehling’s reagent - reducing sugars hemi-acetals

cyclic hemi-ketals C2 is anomeric C - D-fructose -D-fructose 1 2 3 4 5 6 CH2OH 5 HOH2C 4 O OH OH CH2OH OH HO 2 3 OH CH2OH C2 is anomeric C - D-fructose -D-fructose 5 sided ring

oxidation of ketoses all monosaccharides are reducing sugars [O] ketone no reaction [O] ketose carboxylic acid ketose “enol” aldose all monosaccharides are reducing sugars