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Hour Exam III Wednesday, Apr. 23 7:00 – 8:15 pm 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)

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Hour Exam III Wednesday, Apr. 23 7:00 – 8:15 pm Foellinger Auditorium

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  1. 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

  2. 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

  3. CnH2nOn C (H2O) Carbohydrates 1 degree of unsaturation * * * * * 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 (+) or (-)? Is this D- or L-? Is this (+) or (-)? ?

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

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

  6. Physical properties pentose Classify: aldo 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

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

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

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

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

  11. _ H O = C OH HO OH OH CH2OH created a new C* at C1 6 5 * 25 isomers 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

  12. 6 6 CH2OH CH2OH 5 5 O H O OH 4 1 4 1 OH OH 3 2 3 2 OH H OH OH OH OH CH2OH OH HO OH more stable 6 CH2OH 4 5 OH OH HO 2 b.p. = 150o 1 3 OH anomers diastereomers different physical properties -D-glucose b.p. = 146o OH -D-glucose

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

  14. 1 2 3 4 5 6 cyclic hemi-ketals 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

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

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