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Why am I getting higher conversion values of isopropanol dehydration over a zeolite catalyst than n-propanol dehydration?

Chemistry Asked by marietiara on December 10, 2021

The two process are run separetly but under the same condition, same activation of the catalyst, same reaction temperature, same flow.
The initial conversion of n-propanol is ~50% and isopropanol ~70%.

My expections was lower conversion values of isopropanol, because of it’s branched structure it has a more contrained diffusion to the acid sites of the zeolite.

2 Answers

You expect using zeolite might selectively allow n-propanol over isopropanol into the cage of zeolite. While it may be true, 0.38 nm window kind of zeolites might have activity on the external surface as well. For example, there are several research articles available where sugars like glucose and fructose, which are quite big compared to n-propanol or isopropanol, can be dehydrated over SAPO34, SAPO44 zeolites with chabazite structure (also 0.38 nm window).[1-2] And if you consider the reaction at external surface the previous answer by Andrew is spot on.

References for sugar dehydration over SAPO34

https://www.sciencedirect.com/science/article/pii/S1385894716312396 https://www.sciencedirect.com/science/article/pii/S0926669019302973 https://pubs.rsc.org/en/content/articlehtml/2013/ra/c3ra43197e

Answered by yayati naresh on December 10, 2021

In a typical acid-catalyzed dehydration, the reaction goes by an SN1 mechanism, and the rate-limiting step is

$$ce{R-OH2+ -> R+ + H2O}$$

The rate of this step is typically a function of the stability of the cation product. In your case, the secondary cation formed from isopropanol is more stable than the primary cation formed from n-propanol.

I'm using "typical" and "typically" here, because these are results from homogeneous reactions, and it sounds like you are doing heterogeneous catalysis with a solid zeolite. That can certainly change the reaction behavior, but in your case it appears that the zeolite is not selective for one molecule over the other, and the trend is the same as what is observed for homogeneous systems.

Answered by Andrew on December 10, 2021

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