Question

Draw the expected product of the reaction of the following sugars with excess methyl iodide and silver oxide.

1. α-D-fructofuranose
2. β-D-galactopyranose

Short answer

1. methyl 2,3,4,6,-tetra-O-methyl-α-D-fructofuranoside
2. methyl 2,3,4,6,-tetra-O-methyl-β-D-fructofuranoside

Step-by-step solution

Formation of methyl ether

Hydroxy groups of sugar can be converted to methyl ethers by treating with methyl iodide (CH₃I) and silver oxide (Ag₂O). CH₃-I group can be polarised by silver oxide (Ag₂O), which makes the methyl carbon more strongly electrophilic. Attack by the -OH group of carbohydrate is then followed by deprotonation which gives the ether.

Pyranose and furanose

Pyranose is a six-membered cyclic hemiacetal while a furanose is a five-membered cyclic hemiacetal.

Formation of product

1. α-D-fructofuranose on treatment with excess methyl iodide and silver oxide gives methyl 2,3,4,6, -tetra-O-methyl-α-D-fructofuranoside as the product.
   
2. β-D-galactopyranose on treatment with excess methyl iodide and silver oxide gives methyl 2,3,4,6,-tetra-O-methyl-β-D-galactopyranoside as the product.