Textbook 7.9 [ENDORSED]

[005778617]

Hello everyone,

This was one of the suggested problems in Outline 6. I'm having trouble answering this question since I'm not sure what 'kinetic data' would mean in this case, besides maybe rate law. If someone understands, please explain! Thank you :)

The hydrolysis of sucrose (C12H22O11) produces fructose and glucose:
C12H22O11 (aq) + H2O(l) => C6H12O6 (glucose, aq) + C6H12O6 (fructose, aq). Two mechanisms are proposed for this reaction:

(i)
Step 1: C12H22O11 => C6H12O6 + C6H10O5 (slow)
Step 2: C6H10O5 + H2O => C6H12O6 (fast)
(ii)
C12H22O11 + H2O => C6H12O6 + C6H12O6 (slow)

Under what conditions can these two mechanisms be distinguished by using kinetic data ?

[Chem_Mod]

Mechanism (ii) has a slow step involving water. Therefore, under high sucrose concentrations the rate will depend on water concentration.

Whereas for mechanism (i) the rate of the reaction is independent of water.

One needs to study this reaction at high sucrose concentrations to determine if the rate depends on water (mechanism (ii)) or not (mechanism (i)).

Sometimes I cover this in class. If not, it makes for good discussion with TAs and UAs.