(a) Show the product that results when fructose is treated with an excess of methyl iodide and silver oxide.

(b) Show what happens when the product of part (a) is hydrolyzed using dilute acid.

(c) What do the results of parts (a) and (b) imply about the hemiacetal structure of fructose?

Fructose is a ketohexose which contain a ketone group. It is used as a sweetening agent in many confectionaries and also as a substitute of cane sugar.

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.

Furanose is a five-membered cyclic hemiacetal whose name is derived from the five-membered cyclic ether furan. Five membered ring of fructose is known as fructofuranose.

1. 
   Fructose on treatment with excess methyl iodide and silver oxide gives methyl 2,3,4,6, -tetra-O-methyl-α-D-fructofuranoside as the product as the product
2. 
   Treatment with dilute acid only hydrolyzes the acetal methyl group.
3. The results show that the acetals are easily hydrolyzed by dilute acid whereas ethers are stable under these conditions.