Dolbier HW Solutions 625

Dolbier HW Solutions 625 - hydrolysis a In the case of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
a , a (1,1) a , b (1,1) b , b (1,1) a (1,2) b (1,2) a (1,3) b (1,3) a (1,4) (maltose) b (1,4) (cellobiose) a (1,6) b (1,6) ( b ) To be a reducing sugar, one of the anomeric positions must be a free hemiacetal. All except a , a (1,1), a , b (1,1), and b , b (1,1) are reducing sugars. 25.27 Because gentiobiose undergoes mutarotation, it must have a free hemiacetal group. Formation of two molecules of D -glucose indicates that it is a disaccharide and because that hydrolysis is catalyzed by emulsin, the glycosidic linkage is b . The methylation data, summarized in the following equation, require that the glucose units be present in pyranose forms and be joined by a b (1,6)-glycoside bond. 25.28 Like other glycosides, cyanogenic glycosides are cleaved to a carbohydrate and an alcohol on
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hydrolysis. ( a ) In the case of linamarin the alcohol is recognizable as the cyanohydrin of acetone. Once formed, this cyanohydrin dissociates to hydrogen cyanide and acetone. d-Glucose HO HOCH 2 OH HO O HO 1 H 2 O 1 Linamarin HO HOCH 2 OC(CH 3 ) 2 CN HO O HO H 1 or enzyme Acetone cyanohydrin HOCCH 3 CH 3 CN 1 Hydrogen cyanide HCN Acetone CH 3 CCH 3 O RO RO RO ROCH 2 CH 2 O O RO RO OR 1 O OR R 5 H: gentiobiose R 5 CH 3 : gentiobiose octamethyl ether H 3 O 1 CH 3 O CH 3 O OCH 3 CH 3 OCH 2 O OH 2,3,4,6-Tetra-O-methyl-d-glucose CH 3 O CH 3 O OCH 3 HOCH 2 O OH 2,3,4-Tri-O-methyl-d-glucose Site subject to mutarotation when R 5 H CARBOHYDRATES 719...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online