Anyone able to help with drawing the mechanisms for:

2-Bromopropane + Sodium Iodide in Acetone
2-Bromopropane + Silver Nitrate in Ethanol

For the first case, I know that: R-Br + NaI ---> R-I + NaBr
Pretty simple and straight forward. There should be an arrow going from the Iodine to the secondary carbon and an arrow from the single bond pushing the electrons onto the bromine. That should be it for the mechanism, correct?

Second case, I know that: R-Br --> R-OC2H5 + AgBr
I'm just not sure how to accurately draw this one. I'm assuming that this one is a two step?

Thanks

This post was edited by Micrographia on Jan 17 2015 03:04pm

if i remember correctly, sn1 is a rear attack... the br comes off and a + is left it in spot. the I comes in from the backside of the plus, pushes everything around and attaches, essentially mirroring the chirality at that location

For someone who has never taken chemistry your post looks almost like a rape story

Other things to remember about alkyl halides: In this case the 2-Bromopropane is symmetric. But for chiral molecules (2-Bromobutane, etc) the resulting molecule will have its chirality inversed.
The transition state has 5 members to it (typical of Sn2).
The resulting NaBr will be insoluble in acetone.
Speed of the reaction: primary > secondary > tertiary due to steric considerations.
This reactants must be absolutely dry (no protic nor polar residual solvent).
Acetone is aprotic. Acetone has a low dielectric polarization, so it won't participate in the reaction (i.e. avoids solvolysis).


Yes, two-step Sn1 reaction. The alkyl can form a stable carbocation where the nitrate is the counter ion -- definitely Sn1. Then the nucleophilic ethanol attacks the carbocation, leading to the formation of an ether. The leftover proton will acidify the ethanol solvent.
Although the 2-Bromopropane is not chiral, you should a racemic mixture if the reaction does involve a chiral alkyl halide (2-Bromobutane, etc).
The resulting silver bromine is insoluble in ethanol.

Thanks for the replies guys. So here's what I go thus far:

This one makes sense to me, actually. Like I had said earlier, it's pretty straight forward.

One question, however, is what happens to the acetone in this case? Why is it not part of a product (like in the reaction below).

This one, however, still perplexes me. I do understand that an ether will be formed along with silver bromide - but how do I go about showing this in a mechanism? Also, where is the nitrate going?


If you could draw a picture showing the electron flow and correct setup for this case that would be a huge help

This post was edited by Micrographia on Jan 18 2015 10:49am

First the drawings are not quite right. The first reaction has a transition state involved. In other words the 5-membered species occurs while the reaction proceeds, i.e. in transition. In this case the iodine nucleophile attacks from one side only where it displaces the bromine in one substitution. The acetone is very weak nucleophile, while the I- is a strong nucleophile. The iodine charge is not solvated in this case. Beside, the bromine precipitates so it's no longer available as a nucleophile to compete with I- .


In the second reaction, the carbocation is stabilized by the more polar solvent (ethanol is definitely more polar than acetone) and by the nitrate as a counterion. Ethanol is a better nucleophile than NO3- , hence why it reacts. Beside, the ethanol concentration is much higher than the (pinch) of nitrate.
Also in this case the carbocation is an Intermediate, i.e. it does exist as an actual species in solution, although it's quite unstable. Because it is a actual species in solution, the nucleophile ethanol can attack the carbocation from both sides, equally.


I'm a bit short on time, but from the drawings above, I'm sure you'll figure out where the electrons go.

Alright, I think that I might be getting somewhere now.
How does this look?

(Br- is included in the second one, it got cut off for some reason)

Better. The ethanol molecule attacks the electrophile as CH3CH2OH, not as the conjugated base. A proton is then released, thus acidifying the solution and slowing it down.

That was how I originally had it drawn, but I wasn't sure how to depict the remaining proton. So, just to clarify, the final products in that mechanism would be 4 total - right? There would be the 50:50 racemic, Br-, and a H+?

Also, just to make sure I'm understanding this, in drawing a mechanism I don't necessarily show the final products - right? For example, I would leave the Br- instead of showing it as AgBr.

Thanks for the help man

The final products should be stoichiometric with the reactants. Given a reaction that consumes one mol 2-bromopropane, one mol ethanol and one mol NaI, the products are one mol 2-iodopropane (half of each enantiomer), one mol NaBr(s) and one mol H+ . The subscript (s) means solid. I suppose the answer is yes to your question about 4 total, but it was a bit confusing to me.

List them all, either using split arrows or spelling them out on the same line, including the solid AgBr. The intermediate or transition state will be shown along with the reactants and the products. I should have listed them on my drawing.



This post was edited by Dmon_Hunter on Jan 18 2015 03:06pm