# Hydrolysis of Sucrose over Sn1Ac or Sn2Ac

In "March's Advanced Organic Chemistry" I have read about the general mechanism of the acetal hydrolysis. It names the acid-catalyzed SN1 or SN2 as a possible mechanism for a Acetal-hydrolysis.

Now I am interested to know if SN1 or SN2 is mainly going on at the Hydrolysis of Sucrose to Glucose and Fructose. So I compared the different parameters which influence SN1/SN2.

• Polarity of the solvent: Water -> really polar -> SN1 preferred.

• Stabilization of a possible Carbokation-Intermediat: Primary (Anomeric C!), so not really stabilized by Hyperconjunction -> SN2 preferred. (Is there Hyperconjunction also in the ether-O?)

• Mesomeric stabilization There is also a mesomeric stabilization between the resulting carbocation and a Oxonium-Ion in the SN1-Mechanism.

• Steric Hindrance: In my opinion, through the planar region around the anomeric C (planar ether), there is no steric hindrance -> SN2 possible.

• Temperature: high temperature -> SN1 preferred.

So to conclude: Many different aspects. Because of the main points "steric hindrance" and the destabilized primary carbocation intermediate I would conclude that the hydrolysis of Sucrose goes mainly (maybe a little bit over SN1) over SN2.

Have I forgotten something important? Are my assumptions "No steric hindrance" and primary carbocation-intermediate correct?

I have attached my possible SN1-mechanism.

• Unstabilized primary cations never undergo $S_N1$ reactions so yes you are correct. – bon Apr 18 '16 at 12:45
• @bon So the mesomeric stabilization isn't enough stabilization? – Nilsfrank99 Apr 18 '16 at 13:25
• I would not call the intermediate a primary carbocation. Yes the primary carbocation is one of the resonance forms but it is an extremely minor one. The oxonium ion resonance form is a much larger contributor. In any case, both pathways are possible. The SN1 mechanism you have drawn is perfectly correct. In addition to the factors you considered, the SN1 pathway is particularly favoured for the β-anomer because of the anomeric effect (n -> σ* donation from ring oxygen lone pair into C-O σ*). – orthocresol Apr 18 '16 at 15:02
• @orthocresol Tanks for your answer! So the oxonium-resonance-structure stabilize the whole S<sub>N</sub>1 intermediat and so we can assume that also the S<sub>N</sub>1-mechanism happens also at a "big" amount? – Nilsfrank99 Apr 18 '16 at 15:17
• @orthocresol To the anomeric effekt: Sucrose has the Alpha-D-Glucose-monosaccharid, so the the anomeric OH-group makes the whole intermediat not more stable. But if the Glucose-rest is the leaving group, and Fructose works as the intermediat (is that the case?), there your spoken Anomeric effect works? – Nilsfrank99 Apr 18 '16 at 15:24