1 (532)

Prévia do material em texto

Solutions to Problems 469 As you can see, the gulose Fischer synthesized was the L-enantiomer (hydroxy group on C5 is on the left). 0 C 63. First, D-glucuronic acid y-lactone is HO H 0 H H HO H CHO L-gulonic acid y-lactone has already been illustrated [answer to (f) of Problem 62]. (a) NaBH₄ (b) CH₂OH (c) OH H OH CH₂OH HO H OH H OH 5 CH₂OH (d) 1.2 2. KMnO₄ to oxidize unprotected primary alcohol to carboxylic acid; (e) H₂O, (hydrolyze acetals); (f) (-H₂O) 64. Follow along with the actual structure of D-lactose in Section 24-11. 1. Mild acid cleaves acetal linkages. You know what kind of functional linkage connects the two monosaccharides, but you know nothing about which carbon atoms bear the oxygen atoms of the acetal group. 2. Test to see if the "unknown" is a reducing sugar. It is. proving that one of the two component monosaccharides retains a hemiacetal function. (Otherwise, it would be like sucrose. in which both anomeric carbons are connected by an acetal oxygen.) 3. Complete methylation [with of all free -OH groups followed by mild acid hydrolysis is revealing. The galactose portion adds four methyl groups. but the glucose adds only three. Therefore it is the galactose whose anomeric carbon is part of the acetal group linking the two. 4. and 5. Comparison of the tri-O-methylglucose product of the methylation-hydrolysis sequence with known compounds would be one way to ascertain that the hydroxy groups at C4 and C5 were not methylated and. therefore. one must be responsible for the disaccharide (acetal) linkage. and the other is part of the cyclic but we don't know which is which. We need to use other chemistry to open the hemiacetal ring and identify which hydroxy group is part of it. We have such a reaction: In Sec- tion 24-4 we learned that aldoses (which exist as cyclic hemiacetals) are oxidized at CI by bromine in water to give (acyclic) aldonic acids. Starting with we therefore would have