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Foellinger Auditorium
Hour Exam III Wednesday, Apr. 23 7:00 – 8:15 pm Foellinger Auditorium Conflict Exam 5:30-6: NL Help session: Mon. 7:00 – 9:00 100 MSEB
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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: NL
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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 (-)? ? ?
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I II Carbohydrates can be: simple monosaccharides can’t be hydrolyzed complex disaccharides hydrolyzed to 2 monosaccharides polysaccharides hydrolyzed to ’s
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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
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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
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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
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monosaccharides to know:
D-glucose D-galactose D-fructose keto hexose aldohexose diastereomers
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hemi-acetal formation
OH .. H+ .. .. H+ + R’-OH H .. OR’ + H+ aldehyde alcohol cyclic hemi-acetal
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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
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_ 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
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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
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_ = 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
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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
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oxidation of ketoses all monosaccharides are reducing sugars [O]
ketone no reaction [O] ketose carboxylic acid ketose “enol” aldose all monosaccharides are reducing sugars
![oxidation of ketoses all monosaccharides are reducing sugars [O]](http://slideplayer.com./slide/16516422/96/images/15/oxidation+of+ketoses+all+monosaccharides+are+reducing+sugars+%5BO%5D.jpg "ketone. no reaction. [O] ketose. carboxylic acid. ketose. enol aldose. all monosaccharides are reducing sugars.")
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