Rxn rates

[Pamplemousse123]

---

Members do not see this ad.

As I was studying the section on rates of rxns, I thought about a question:
many references state that sn1 rates do not change with the addition of extra nucleophile. but, doesn't the addition of more nucleophile increase the volume of the reaction, so the rate decreases? is this volume negligible that we can ignore it? i remember in orgo chem lab, we added around 2 mL of nucleophile solution so that was a relatively large amount. maybe the extra nu changes the arrhenius constant, and thus k, which makes up for the difference. but wouldn't the arrhenius constant also decrease because of the decreased collision frequency?

any ideas?

 

[sekistudent]

I think you have to look at the mechanism. SN1 is a two step process. The first step is done by the "electrophile" alone. In this step the leaving group has to leave on its own, and thus this step is much much slower than the following step in which the nucleophile attacks the carbocation(electrophile). Yes, theoretically, the speed of this second step increases with the addition of more Nu, but the speed of the whole reaction wont change much as it wont affect the 1st step.

Its kinda like riding a rollercoaster at 6 flags. Slow step would be waiting 2 hours in line. The fast step would be 10 mins riding it. Shortening the track so that your ride takes 8 mins wont change the total time much.

As far as your comments regarding Volume and collision freq, I would think that adding more Nu would increase the (2nd step) reaction rate and also increase the collision freq, because there is a higher conc. of Nu. More Nu means more chance that it will collide with the carbocat.

 

[Pamplemousse123]

I think you have to look at the mechanism. SN1 is a two step process. The first step is done by the "electrophile" alone. In this step the leaving group has to leave on its own, and thus this step is much much slower than the following step in which the nucleophile attacks the carbocation(electrophile). Yes, theoretically, the speed of this second step increases with the addition of more Nu, but the speed of the whole reaction wont change much as it wont affect the 1st step.

Its kinda like riding a rollercoaster at 6 flags. Slow step would be waiting 2 hours in line. The fast step would be 10 mins riding it. Shortening the track so that your ride takes 8 mins wont change the total time much.

As far as your comments regarding Volume and collision freq, I would think that adding more Nu would increase the (2nd step) reaction rate and also increase the collision freq, because there is a higher conc. of Nu. More Nu means more chance that it will collide with the carbocat.

I see your point about considering the reaction in 2 different steps and I love the analogy!

But I am thinking about the decrease in the rate of the first slow step itself, the formation of carbocation.

We have: Rate = k[alkylhalide] for Sn1. if we add more nucleophile, aren't we diluting the solution and thus [alkylhalide] would decrease because moles of alkyl halide are constant?

 

[Rabolisk]

Yes, I can see why adding a nucleophile, with the effect of diluting the solution, could slow down the reaction. Of course, this is only true if the change in volume is large enough to change the concentration of the substrate noticeably.

 

[orgohacks]

I see your point about considering the reaction in 2 different steps and I love the analogy!

But I am thinking about the decrease in the rate of the first slow step itself, the formation of carbocation.

We have: Rate = k[alkylhalide] for Sn1. if we add more nucleophile, aren't we diluting the solution and thus [alkylhalide] would decrease because moles of alkyl halide are constant?

The concentration of alkyl halide will still be very small relative to the concentration of nucleophile, even if you double, quadruple, or even multiply the concentration by 10.

For instance the concentration of H2O in 1L of water is 55 M, which is way higher than the concentrations of alkyl halide you will ever see. This is why we ignore the concentration of water.

The reaction still behaves as though it was a first-order reaction - this is sometimes called "pseudo-first order" conditions.