Question

Discussion Questions: 1. The open-chain form ofglucose constitutes only a small fraction of any aqueous glucose sample (-0.02%). Yet when Cu2 is used to oxidize the open-chain form, nearly all of the glucose in the sample reacts. Use Le Chatelier's Principle to explain this observation

Answer 1

Answer:

Ring structure of glucose <--------------> open chain form

The equilibrium lies far left with the open chain form only 0.02%

Bu when Cu+2 (Benedicts reagent) is added to it the open chain form starts reacting it to give the red precipitae of Cu2O and the clucose is oxidised to acid.

open chain form + Cu+2 -----------> Cu2O + gluconic acid

As the Cu+2 is reacting with open chain form and removing it, the concentration of productsin the equilibrium reaction decrease.

Thus according to Le-Chatlier principle as the product is removed ,the equilibrum shifts to right side to produce more products.

Again the Cu+2 reacts with all the product formed.

Then the equilibrium produces more of open chain form which further goes onreacting.

The process continues till all the glucose is converted to open chain form and is consumed by reaction with Cu+2.

Thus the removal product by reacting with Cu+2 shifts the equilibrium to product side according to Le-chatlier principle and makes the reaction possible.