Dolbier HW Solutions 631

Dolbier HW Solutions 631 - shown in the f rst part of step...

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3. Both ( 1 )-glucose and ( 1 )-mannose are oxidized to optically active aldaric acids with nitric acid. Because both ( 1 )-glucose and ( 1 )-mannose yield optically active aldaric acids and both have the same con f guration at C-4, the hydroxyl group must lie at the right in the Fischer pro- jection at this carbon. The structures of the corresponding aldaric acids are Both are optically active. Had the C-4 hydroxyl group been to the left, one of the aldaric acids would have been a meso form. 4. There is another sugar, ( 1 )-gulose, that gives the same aldaric acid on oxidation as does ( 1 )-glucose. This is the last piece in the puzzle, the one that permits one of the Fischer projections
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Unformatted text preview: shown in the f rst part of step 3 to be assigned to ( 1 )-glucose and the other to ( 1 )-mannose. Consider f rst the structure CHO CH 2 OH HO H HO equivalent to HO H H OH H CH 2 OH CHO H H H H HO OH OH OH CO 2 H CO 2 H HO HO H H HO H H OH CO 2 H CO 2 H H HO H H HO OH H OH (This aldaric acid is optically inactive.) CO 2 H CO 2 H H H H H HO OH OH OH CO 2 H CO 2 H HO H H H HO H OH OH CHO CH 2 OH H H H H HO OH OH OH 1 2 3 4 5 6 CHO CH 2 OH HO H H H HO H OH OH 1 2 3 4 5 6 [One of these is ( 1 )-glucose, the other is ( 1 )-mannose.] CARBOHYDRATES 725...
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